title
Calculus 1 - Full College Course

description
Learn Calculus 1 in this full college course. This course was created by Dr. Linda Green, a lecturer at the University of North Carolina at Chapel Hill. Check out her YouTube channel: https://www.youtube.com/channel/UCkyLJh6hQS1TlhUZxOMjTFw This course combines two courses taught by Dr. Green. She teaches both Calculus 1 and a Calculus 1 Corequisite course, designed to be taken at the same time. In this video, the lectures from the Corquisite course, which review important Algebra and Trigonometry concepts, have been interspersed with the Calculus 1 lectures at the places suggested by Dr. Green. ⭐️ Prerequisites ⭐️ 🎥 Algebra: https://www.youtube.com/watch?v=LwCRRUa8yTU 🎥 Precalculus: https://www.youtube.com/watch?v=eI4an8aSsgw ⭐️ Lecture Notes ⭐️ 🔗 Calculus 1 Corequisite Notes: http://lindagreen.web.unc.edu/files/2020/08/courseNotes_math231L_2020Fall.pdf 🔗 Calculus 1 Notes: http://lindagreen.web.unc.edu/files/2019/12/courseNotes_m231_2018_S.pdf 🔗 Viewer created notes (Thanks to Abdelrahman Ramzy): https://drive.google.com/file/d/1Py5hyiOz61i4lh2PEsjvBTH8AEuQzXK_/view?usp=sharing ⭐️ Course Contents ⭐️ (0:00:00) [Corequisite] Rational Expressions (0:09:40) [Corequisite] Difference Quotient (0:18:20) Graphs and Limits (0:25:51) When Limits Fail to Exist (0:31:28) Limit Laws (0:37:07) The Squeeze Theorem (0:42:55) Limits using Algebraic Tricks (0:56:04) When the Limit of the Denominator is 0 (1:08:40) [Corequisite] Lines: Graphs and Equations (1:17:09) [Corequisite] Rational Functions and Graphs (1:30:35) Limits at Infinity and Graphs (1:37:31) Limits at Infinity and Algebraic Tricks (1:45:34) Continuity at a Point (1:53:21) Continuity on Intervals (1:59:43) Intermediate Value Theorem (2:03:37) [Corequisite] Right Angle Trigonometry (2:11:13) [Corequisite] Sine and Cosine of Special Angles (2:19:16) [Corequisite] Unit Circle Definition of Sine and Cosine (2:24:46) [Corequisite] Properties of Trig Functions (2:35:25) [Corequisite] Graphs of Sine and Cosine (2:41:57) [Corequisite] Graphs of Sinusoidal Functions (2:52:10) [Corequisite] Graphs of Tan, Sec, Cot, Csc (3:01:03) [Corequisite] Solving Basic Trig Equations (3:08:14) Derivatives and Tangent Lines (3:22:55) Computing Derivatives from the Definition (3:34:02) Interpreting Derivatives (3:42:33) Derivatives as Functions and Graphs of Derivatives (3:56:25) Proof that Differentiable Functions are Continuous (4:01:09) Power Rule and Other Rules for Derivatives (4:07:42) [Corequisite] Trig Identities (4:15:14) [Corequisite] Pythagorean Identities (4:20:35) [Corequisite] Angle Sum and Difference Formulas (4:28:31) [Corequisite] Double Angle Formulas (4:36:01) Higher Order Derivatives and Notation (4:39:22) Derivative of e^x (4:46:52) Proof of the Power Rule and Other Derivative Rules (4:56:31) Product Rule and Quotient Rule (5:02:09) Proof of Product Rule and Quotient Rule (5:10:40) Special Trigonometric Limits (5:17:31) [Corequisite] Composition of Functions (5:29:54) [Corequisite] Solving Rational Equations (5:40:02) Derivatives of Trig Functions (5:46:23) Proof of Trigonometric Limits and Derivatives (5:54:38) Rectilinear Motion (6:11:41) Marginal Cost (6:16:51) [Corequisite] Logarithms: Introduction (6:25:32) [Corequisite] Log Functions and Their Graphs (6:36:17) [Corequisite] Combining Logs and Exponents (6:40:55) [Corequisite] Log Rules (6:49:27) The Chain Rule (6:58:44) More Chain Rule Examples and Justification (7:07:43) Justification of the Chain Rule (7:10:00) Implicit Differentiation (7:20:28) Derivatives of Exponential Functions (7:25:38) Derivatives of Log Functions (7:29:38) Logarithmic Differentiation (7:37:08) [Corequisite] Inverse Functions (7:51:22) Inverse Trig Functions (8:00:56) Derivatives of Inverse Trigonometric Functions (8:12:11) Related Rates - Distances (8:17:55) Related Rates - Volume and Flow (8:22:21) Related Rates - Angle and Rotation (8:28:20) [Corequisite] Solving Right Triangles (8:34:54) Maximums and Minimums (8:46:18) First Derivative Test and Second Derivative Test (8:51:37) Extreme Value Examples (9:01:33) Mean Value Theorem (9:09:09) Proof of Mean Value Theorem (0:14:59) [Corequisite] Solving Right Triangles (9:25:20) Derivatives and the Shape of the Graph (9:33:31) Linear Approximation (9:48:28) The Differential (9:59:11) L'Hospital's Rule (10:06:27) L'Hospital's Rule on Other Indeterminate Forms (10:16:13) Newtons Method (10:27:45) Antiderivatives (10:33:24) Finding Antiderivatives Using Initial Conditions (10:41:59) Any Two Antiderivatives Differ by a Constant (10:45:19) Summation Notation (10:49:12) Approximating Area (11:04:22) The Fundamental Theorem of Calculus, Part 1 (11:15:02) The Fundamental Theorem of Calculus, Part 2 (11:22:17) Proof of the Fundamental Theorem of Calculus (11:29:18) The Substitution Method (11:38:07) Why U-Substitution Works (11:40:23) Average Value of a Function (11:47:57) Proof of the Mean Value Theorem for Integrals

detail
{'title': 'Calculus 1 - Full College Course', 'heatmap': [{'end': 1714.99, 'start': 1270.396, 'weight': 0.723}, {'end': 5141.6, 'start': 4710.144, 'weight': 0.702}, {'end': 20993.069, 'start': 20661.286, 'weight': 1}], 'summary': "Course 'calculus 1' includes comprehensive coverage of rational expressions, calculus fundamentals, graphs and asymptotes, trigonometric functions, derivatives, logarithmic functions, related rates, extreme values, inequalities, temperature change analysis, calculus techniques, antiderivatives, and fundamental theorems, with practical examples and quantitative data.", 'chapters': [{'end': 573.387, 'segs': [{'end': 45.937, 'src': 'embed', 'start': 18.393, 'weight': 0, 'content': [{'end': 21.055, 'text': 'This video is about working with rational expressions.', 'start': 18.393, 'duration': 2.662}, {'end': 24.918, 'text': 'A rational expression is a fraction, usually with variables in it.', 'start': 21.775, 'duration': 3.143}, {'end': 30.842, 'text': 'Something like x plus 2 over x squared minus 3 is a rational expression.', 'start': 25.278, 'duration': 5.564}, {'end': 40.849, 'text': "In this video, we'll practice adding, subtracting, multiplying, and dividing rational expressions and simplifying them to lowest terms.", 'start': 31.943, 'duration': 8.906}, {'end': 45.937, 'text': "We'll start with simplifying to lowest terms.", 'start': 43.595, 'duration': 2.342}], 'summary': 'Video on working with rational expressions, covering adding, subtracting, multiplying, and dividing, and simplifying to lowest terms.', 'duration': 27.544, 'max_score': 18.393, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w18393.jpg'}, {'end': 111.89, 'src': 'embed', 'start': 82.61, 'weight': 3, 'content': [{'end': 89.355, 'text': "First, we'll factor the numerator, that's 3 times x plus 2, and then factor the denominator.", 'start': 82.61, 'duration': 6.745}, {'end': 94.999, 'text': 'In this case, it factors to x plus 2 times x plus 2.', 'start': 90.555, 'duration': 4.444}, {'end': 97.04, 'text': 'We could also write that as x plus 2 squared.', 'start': 94.999, 'duration': 2.041}, {'end': 104.827, 'text': "Now we cancel the common factors, and we're left with 3 over x plus 2.", 'start': 97.881, 'duration': 6.946}, {'end': 107.929, 'text': 'Definitely a simpler way of writing that rational expression.', 'start': 104.827, 'duration': 3.102}, {'end': 111.89, 'text': "Next, let's practice multiplying and dividing.", 'start': 109.789, 'duration': 2.101}], 'summary': 'Factor and simplify the rational expression to 3/(x+2)', 'duration': 29.28, 'max_score': 82.61, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w82610.jpg'}, {'end': 172.398, 'src': 'embed', 'start': 141.755, 'weight': 1, 'content': [{'end': 152.101, 'text': 'So here we get 4 fifths times 3 halves, and that gives us 12 tenths, but actually we could reduce that fraction to 6 fifths.', 'start': 141.755, 'duration': 10.346}, {'end': 161.33, 'text': 'We use the same rules when we compute the product or quotient of two rational expressions with the variables in them.', 'start': 154.244, 'duration': 7.086}, {'end': 164.572, 'text': "Here, we're trying to divide two rational expressions.", 'start': 161.83, 'duration': 2.742}, {'end': 170.097, 'text': 'So instead, we can multiply by the reciprocal.', 'start': 165.133, 'duration': 4.964}, {'end': 172.398, 'text': 'I call this flipping and multiplying.', 'start': 170.597, 'duration': 1.801}], 'summary': 'Multiplying 4/5 by 3/2 gives 6/5, applying rules to rational expressions.', 'duration': 30.643, 'max_score': 141.755, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w141755.jpg'}, {'end': 446.144, 'src': 'embed', 'start': 421.226, 'weight': 2, 'content': [{'end': 430.973, 'text': "Now I can rewrite each of my two rational expressions by multiplying by whatever's missing from the denominator in terms of the least common denominator.", 'start': 421.226, 'duration': 9.747}, {'end': 435.897, 'text': 'So what I mean is I can rewrite 3 over 2x plus 2.', 'start': 431.393, 'duration': 4.504}, {'end': 438.759, 'text': "I'll write the 2x plus 2 as 2 times x plus 1.", 'start': 435.897, 'duration': 2.862}, {'end': 440.06, 'text': "I'll write it in factored form.", 'start': 438.759, 'duration': 1.301}, {'end': 446.144, 'text': "And then I notice that compared to the least common denominator, I'm missing the factor of x minus 1.", 'start': 440.54, 'duration': 5.604}], 'summary': 'Rewriting rational expressions using least common denominator and factored form.', 'duration': 24.918, 'max_score': 421.226, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w421226.jpg'}], 'start': 18.393, 'title': 'Rational expressions and fractions', 'summary': 'Covers working with rational expressions, including simplifying to lowest terms and practicing operations such as adding, subtracting, multiplying, and dividing. it also involves finding the least common denominator for fractions, simplifying rational expressions by factoring, and multiplying/dividing rational expressions.', 'chapters': [{'end': 230.261, 'start': 18.393, 'title': 'Working with rational expressions', 'summary': 'Covers working with rational expressions, including simplifying to lowest terms and practicing operations such as adding, subtracting, multiplying, and dividing, with examples showing the reduction of rational expressions to simplest form and the use of reciprocals in division.', 'duration': 211.868, 'highlights': ['The video covers working with rational expressions, including simplifying to lowest terms and practicing operations such as adding, subtracting, multiplying, and dividing.', 'An example demonstrates the reduction of a rational expression to simplest form, showing the cancellation of common factors to obtain a simpler expression, such as 3 over x plus 2.', 'The process of multiplying and dividing fractions is explained, including the use of reciprocals in division and the reduction of fractions to lowest terms.', 'A detailed example illustrates the use of reciprocals in division of rational expressions, emphasizing the process of flipping and multiplying to simplify and reduce the expression.']}, {'end': 573.387, 'start': 232.941, 'title': 'Adding and subtracting fractions', 'summary': 'Covers the process of finding the least common denominator for fractions, simplifying rational expressions by factoring, and multiplying/dividing rational expressions, resulting in the reduction of 3x + 7 over 2 times x + 1, x - 1.', 'duration': 340.446, 'highlights': ['The process of finding the least common denominator for fractions is explained through an example involving 7/6 and 4/15, resulting in the calculation of 27/30 and its reduction to 9/10.', 'The detailed process of finding the least common denominator for rational expressions is demonstrated by factoring the denominators and multiplying by the missing factors to simplify the expressions, resulting in the expression 3x + 7 over 2 times x + 1, x - 1.']}], 'duration': 554.994, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w18393.jpg', 'highlights': ['The video covers working with rational expressions, including simplifying to lowest terms and practicing operations such as adding, subtracting, multiplying, and dividing.', 'A detailed example illustrates the use of reciprocals in division of rational expressions, emphasizing the process of flipping and multiplying to simplify and reduce the expression.', 'The process of finding the least common denominator for rational expressions is demonstrated by factoring the denominators and multiplying by the missing factors to simplify the expressions, resulting in the expression 3x + 7 over 2 times x + 1, x - 1.', 'An example demonstrates the reduction of a rational expression to simplest form, showing the cancellation of common factors to obtain a simpler expression, such as 3 over x plus 2.']}, {'end': 4102.292, 'segs': [{'end': 604.766, 'src': 'embed', 'start': 573.387, 'weight': 2, 'content': [{'end': 579.391, 'text': 'and how to add and subtract rational expressions by writing them in terms of the least common denominator.', 'start': 573.387, 'duration': 6.004}, {'end': 585.195, 'text': 'This video is about the difference quotient and the average rate of change.', 'start': 580.972, 'duration': 4.223}, {'end': 590.729, 'text': 'These are topics that are related to the concept of derivative in calculus.', 'start': 587.165, 'duration': 3.564}, {'end': 596.656, 'text': "For a function, y equals f of x, like the function who's graphed right here.", 'start': 592.151, 'duration': 4.505}, {'end': 604.766, 'text': 'a secant line is a line that stretches between two points on the graph of the function.', 'start': 596.656, 'duration': 8.11}], 'summary': 'Teaches adding and subtracting rational expressions using least common denominator, explores difference quotient and average rate of change, related to calculus derivatives.', 'duration': 31.379, 'max_score': 573.387, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w573387.jpg'}, {'end': 1262.629, 'src': 'embed', 'start': 1193.903, 'weight': 1, 'content': [{'end': 1203.888, 'text': 'In the language of limits, we say that the limit as x approaches one of f of x is equal to 10.', 'start': 1193.903, 'duration': 9.985}, {'end': 1212.571, 'text': 'More informally, we can write as x approaches 1, f of x approaches 10.', 'start': 1203.888, 'duration': 8.683}, {'end': 1220.453, 'text': 'Notice that the value of f at 1 is actually equal to 0, not 10.', 'start': 1212.571, 'duration': 7.882}, {'end': 1228.356, 'text': 'And so the limit of f of x as x goes to 1 and the value of f at 1 are not equal.', 'start': 1220.453, 'duration': 7.903}, {'end': 1237.621, 'text': "This illustrates the important fact that the limit here as x goes to 1 doesn't care about the value of f at 1.", 'start': 1229.758, 'duration': 7.863}, {'end': 1242.984, 'text': 'It only cares about the value of f when x is near 1.', 'start': 1237.621, 'duration': 5.363}, {'end': 1248.606, 'text': 'In general for any function f of x and for real numbers a and l,', 'start': 1242.984, 'duration': 5.622}, {'end': 1262.629, 'text': 'the limit as x goes to a of f of x equals l means that f of x gets arbitrarily close to L, as x gets arbitrarily close to A.', 'start': 1248.606, 'duration': 14.023}], 'summary': 'Limit as x approaches 1 of f(x) is 10, f(1) is 0. the limit cares about f near 1.', 'duration': 68.726, 'max_score': 1193.903, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w1193903.jpg'}, {'end': 1714.99, 'src': 'heatmap', 'start': 1270.396, 'weight': 0.723, 'content': [{'end': 1284.892, 'text': 'In this picture, I can say that the limit, as x goes to A of f of x is L, because by taking a sufficiently small interval around a of x values,', 'start': 1270.396, 'duration': 14.496}, {'end': 1295.494, 'text': 'I can guarantee that my y values, my f of x values, lie in an arbitrarily small interval around l.', 'start': 1284.892, 'duration': 10.602}, {'end': 1298.815, 'text': "Now the limit doesn't care about f's value at exactly a.", 'start': 1295.494, 'duration': 3.321}, {'end': 1311.629, 'text': "so if I change the function, if I change the function's value at a or even leave the function undefined at a, the limit is still l.", 'start': 1298.815, 'duration': 12.814}, {'end': 1318.295, 'text': 'But the limit does care about what happens for x values on both sides of a.', 'start': 1311.629, 'duration': 6.666}, {'end': 1328.745, 'text': "If, for example, the function didn't even exist for x values bigger than a, then we could no longer say that the limit as x approached a was l.", 'start': 1318.295, 'duration': 10.45}, {'end': 1329.846, 'text': 'The limit would not exist.', 'start': 1328.745, 'duration': 1.101}, {'end': 1341.995, 'text': 'In other words, the limit of f of x is l only if the y values are approaching l as x approaches a from both the left and the right.', 'start': 1331.932, 'duration': 10.063}, {'end': 1351.498, 'text': 'For the function g of x, graphed below the limit as x goes to 2 of g of x,', 'start': 1343.856, 'duration': 7.642}, {'end': 1364.172, 'text': 'does not exist because the y values approach 1 as the x values approach 2 from the left and the y values approach 3 as the x values approach to from the right.', 'start': 1351.498, 'duration': 12.674}, {'end': 1371.574, 'text': "Although the limit doesn't exist, we can say that the left sided limit exists.", 'start': 1365.912, 'duration': 5.662}, {'end': 1381.317, 'text': 'And we write this as limit as x goes to two from the left of g of x is equal to one.', 'start': 1372.734, 'duration': 8.583}, {'end': 1396.688, 'text': 'The superscript minus sign here means that x is going to two from the left side, In other words, the x values are less than 2 and approaching 2.', 'start': 1383.137, 'duration': 13.551}, {'end': 1399.39, 'text': 'Similarly, we can talk about right-sided limits.', 'start': 1396.688, 'duration': 2.702}, {'end': 1407.917, 'text': 'In this example, the limit as x goes to 2 from the right side of g of x is 3.', 'start': 1400.211, 'duration': 7.706}, {'end': 1413.121, 'text': "And here, the superscript plus sign means that we're approaching 2 from the right side.", 'start': 1407.917, 'duration': 5.204}, {'end': 1419.116, 'text': 'In other words, our x values are greater than 2 and approaching 2.', 'start': 1413.462, 'duration': 5.654}, {'end': 1430.341, 'text': 'In general, the limit as x goes to a minus of f of x equals L means that f of x approaches L as x approaches a from the left.', 'start': 1419.116, 'duration': 11.225}, {'end': 1442.067, 'text': 'And the limit as x goes to a plus of f of x equals L means that f of x approaches L as x approaches a from the right.', 'start': 1431.202, 'duration': 10.865}, {'end': 1447.109, 'text': 'Limits from the left or from the right are also called one-sided limits.', 'start': 1443.007, 'duration': 4.102}, {'end': 1457.97, 'text': "In this last example, let's look at the behavior of y equals h of x graph below when x is near negative two.", 'start': 1449.949, 'duration': 8.021}, {'end': 1468.812, 'text': 'As x approaches negative two from the right, our y values are getting arbitrarily large, larger than any real number we might choose.', 'start': 1460.111, 'duration': 8.701}, {'end': 1479.394, 'text': 'We can write this in terms of limits by saying the limit as x goes to negative two from the right of h of x is equal to infinity.', 'start': 1470.372, 'duration': 9.022}, {'end': 1486.237, 'text': 'As x approaches negative 2 from the left, the y values are getting smaller.', 'start': 1481.434, 'duration': 4.803}, {'end': 1490.861, 'text': 'In fact, they go below any negative real number we might choose.', 'start': 1486.818, 'duration': 4.043}, {'end': 1499.647, 'text': 'In terms of limits, we can say that the limit as x goes to negative 2 from the left of h of x is negative infinity.', 'start': 1492.222, 'duration': 7.425}, {'end': 1509.214, 'text': 'In this example, the limit of h of x, as x goes to negative 2, not specifying from the left or the right,', 'start': 1501.688, 'duration': 7.526}, {'end': 1512.995, 'text': 'means we have to approach negative 2 from both sides,', 'start': 1510.154, 'duration': 2.841}, {'end': 1520.139, 'text': 'and so this limit does not exist because the limits from the left and the limit from the right are heading in opposite directions.', 'start': 1512.995, 'duration': 7.144}, {'end': 1531.143, 'text': "I want to mention that some people say that the limits from the right and the limits from the left also do not exist because the functions don't approach any finite number.", 'start': 1521.379, 'duration': 9.764}, {'end': 1536.506, 'text': 'I prefer to say that these limits do not exist as a finite number.', 'start': 1532.684, 'duration': 3.822}, {'end': 1541.244, 'text': 'but they do exist as infinity and negative infinity.', 'start': 1538.083, 'duration': 3.161}, {'end': 1549.667, 'text': 'This video introduced the idea of limits and one-sided limits and infinite limits.', 'start': 1542.584, 'duration': 7.083}, {'end': 1554.888, 'text': 'This video gives some examples of when limits fail to exist.', 'start': 1551.847, 'duration': 3.041}, {'end': 1567.987, 'text': "For this function f of x graphed below, let's look at the behavior of f of x in terms of limits as x approaches negative 1, 1, and 2.", 'start': 1556.309, 'duration': 11.678}, {'end': 1572.208, 'text': "Let's start with x approaching negative 1.", 'start': 1567.987, 'duration': 4.221}, {'end': 1580.47, 'text': 'When x approaches negative 1 from the left, the y values seem to be approaching about 1 half.', 'start': 1572.208, 'duration': 8.262}, {'end': 1592.214, 'text': 'When x approaches negative 1 from the right, the y values seem to be approaching 1.', 'start': 1583.631, 'duration': 8.583}, {'end': 1602.088, 'text': "So when x approaches negative 1, and we don't specify from either the left or the right, We can only say that the limit does not exist,", 'start': 1592.214, 'duration': 9.874}, {'end': 1604.688, 'text': 'because these two limits from the left and right are not equal.', 'start': 1602.088, 'duration': 2.6}, {'end': 1611.95, 'text': "Now let's look at the limit as x is approaching 1.", 'start': 1608.089, 'duration': 3.861}, {'end': 1617.931, 'text': 'This time, when we approach from the left, we get a limiting y value of 2.', 'start': 1611.95, 'duration': 5.981}, {'end': 1623.012, 'text': 'When we approach from the right, the y values are going towards 2.', 'start': 1617.931, 'duration': 5.081}, {'end': 1632.987, 'text': 'So both of these left and right limits are equal to 2, and therefore the limit as x goes to 1 of f of x equals 2.', 'start': 1623.012, 'duration': 9.975}, {'end': 1641.53, 'text': "That's true even though f of 1, f of 1 itself does not exist.", 'start': 1632.987, 'duration': 8.543}, {'end': 1656.121, 'text': "But the limit doesn't care what happens at exactly x equals 1, just what happens when x is near 1.", 'start': 1645.812, 'duration': 10.309}, {'end': 1658.583, 'text': "Finally, let's look at the limit as x goes to 2.", 'start': 1656.121, 'duration': 2.462}, {'end': 1665.848, 'text': 'So here on the left side, the limit is going to negative infinity.', 'start': 1658.583, 'duration': 7.265}, {'end': 1669.21, 'text': "And on the right side, it's negative infinity.", 'start': 1665.868, 'duration': 3.342}, {'end': 1673.393, 'text': 'So we can say the limit as x goes to 2 is negative infinity.', 'start': 1669.691, 'duration': 3.702}, {'end': 1680.298, 'text': 'Or we can also say that the limit as x goes to 2 does not exist.', 'start': 1674.334, 'duration': 5.964}, {'end': 1682.079, 'text': 'This is a correct answer.', 'start': 1680.958, 'duration': 1.121}, {'end': 1684.661, 'text': 'This is a better answer because it carries more information.', 'start': 1682.499, 'duration': 2.162}, {'end': 1692.644, 'text': "What values of x does the limit of f fail to exist? Well, let's see.", 'start': 1685.802, 'duration': 6.842}, {'end': 1701.466, 'text': 'Negative 1 and 2 are the only two values.', 'start': 1694.604, 'duration': 6.862}, {'end': 1705.067, 'text': "Let's talk about the ways that limits can fail to exist.", 'start': 1702.727, 'duration': 2.34}, {'end': 1707.248, 'text': "We've seen at least a couple different ways.", 'start': 1705.547, 'duration': 1.701}, {'end': 1714.99, 'text': "So we've seen examples where the limit from the left is not equal to the limit from the right.", 'start': 1708.128, 'duration': 6.862}], 'summary': 'The video discussed limits, one-sided limits, and infinite limits, using examples to illustrate when limits fail to exist.', 'duration': 444.594, 'max_score': 1270.396, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w1270396.jpg'}, {'end': 2242.676, 'src': 'embed', 'start': 2218.219, 'weight': 3, 'content': [{'end': 2226.1, 'text': 'And instead, we have to use other techniques to try to evaluate the limit of a sum difference product or quotient.', 'start': 2218.219, 'duration': 7.881}, {'end': 2231.352, 'text': 'This video is about the squeeze theorem, which is another method for finding limits.', 'start': 2227.591, 'duration': 3.761}, {'end': 2234.593, 'text': "Let's start with an example.", 'start': 2233.473, 'duration': 1.12}, {'end': 2237.894, 'text': 'Suppose we have a function g of x.', 'start': 2235.594, 'duration': 2.3}, {'end': 2242.676, 'text': "We don't know much about it, but we do know that for x values near 1,", 'start': 2237.894, 'duration': 4.782}], 'summary': 'Learn about the squeeze theorem for finding limits.', 'duration': 24.457, 'max_score': 2218.219, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w2218219.jpg'}, {'end': 3391.382, 'src': 'embed', 'start': 3359.795, 'weight': 0, 'content': [{'end': 3361.416, 'text': 'and looked at one-sided limits.', 'start': 3359.795, 'duration': 1.621}, {'end': 3374.501, 'text': 'The limit law about quotients tells us that the limit of the quotient is the quotient of the limit,', 'start': 3367.498, 'duration': 7.003}, {'end': 3382.665, 'text': 'provided that the limits of the component functions actually exist and that the limit of the function on the denominator is not equal to zero.', 'start': 3374.501, 'duration': 8.164}, {'end': 3391.382, 'text': 'But what happens if the limit of the function on the denominator is equal to zero? This video will begin to answer that question.', 'start': 3384.025, 'duration': 7.357}], 'summary': 'One-sided limits and limit law about quotients, exploring cases when denominator limit is zero.', 'duration': 31.587, 'max_score': 3359.795, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w3359795.jpg'}], 'start': 573.387, 'title': 'Calculus fundamentals', 'summary': 'Covers concepts such as difference quotient, average rate of change, and limits, with examples and applications, providing a foundational understanding for finding derivatives and evaluating limits in calculus.', 'chapters': [{'end': 627.506, 'start': 573.387, 'title': 'Derivative in calculus basics', 'summary': 'Covers the difference quotient, average rate of change, and secant lines in calculus, demonstrating the concept with the function y=f(x) and x values a and b.', 'duration': 54.119, 'highlights': ['The video covers the difference quotient and the average rate of change, essential topics related to the concept of derivative in calculus.', 'Explains the concept of a secant line as a line stretching between two points on the graph of the function y=f(x), with x values labeled as a and b.', 'Defines the x values a and b as points on the graph of the function y=f(x), with corresponding y values given by f(a) and f(b).']}, {'end': 1105.449, 'start': 627.506, 'title': 'Average rate of change and difference quotient', 'summary': 'Discusses the concepts of average rate of change and difference quotient, using examples to calculate the average rate of change for a function, and simplifying the difference quotient formula, which will be crucial in calculus for finding the derivative.', 'duration': 477.943, 'highlights': ['The chapter discusses the concepts of average rate of change and difference quotient.', 'Calculating the average rate of change for a function and simplifying the difference quotient formula.', 'Introduction to the idea of limits through graphs and examples.']}, {'end': 2217.458, 'start': 1107.31, 'title': 'Limits and one-sided limits', 'summary': 'Introduces the concept of limits and one-sided limits, including an example of a piecewise defined function with a hole at x=1, and discusses the behavior of limits at specific values such as 2, 0, and 7, and the limit laws for sums, differences, products, and quotients of functions.', 'duration': 1110.148, 'highlights': ['The limit as x approaches 1 of f of x is equal to 10, where f of x is 10x for x not equal to 1, and f of x is 0 for x equal to 1.', 'The concept of limits is introduced, stating that the limit as x approaches a of f of x equals L means that f of x gets arbitrarily close to L as x gets arbitrarily close to A.', 'The behavior of limits at specific values such as 2, 0, and 7 is discussed, with examples of limits failing to exist, including instances of one-sided limits and infinite limits.', 'The limit laws for sums, differences, products, and quotients of functions are explained, illustrating how to find the limits of composite functions and rational functions using limit rules.']}, {'end': 3359.231, 'start': 2218.219, 'title': 'Using the squeeze theorem to evaluate limits', 'summary': 'Introduces the squeeze theorem as a method to find limits, illustrates its application with examples, and demonstrates algebraic tricks to evaluate limits in 0 over 0 indeterminate form.', 'duration': 1141.012, 'highlights': ['The chapter introduces the squeeze theorem as a method to find limits', 'Demonstrates algebraic tricks to evaluate limits in 0 over 0 indeterminate form', 'Illustrates the squeeze theorem with multiple examples']}, {'end': 4102.292, 'start': 3359.795, 'title': 'Limits of quotients with zero denominator', 'summary': 'Discusses the limit of the quotient when the limit of the denominator is 0 and the limit of the numerator is not 0, providing examples and demonstrating how the limit can be positive or negative infinity, or non-existent, based on the approach of x from the left or right.', 'duration': 742.497, 'highlights': ['The limit of the quotient when the limit of the denominator is 0 and the limit of the numerator is not 0', 'Calculation of the limit as x approaches 3 from the left and right', 'Evaluation of the limit as x approaches -4 from the left and right']}], 'duration': 3528.905, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w573387.jpg', 'highlights': ['The limit laws for sums, differences, products, and quotients of functions are explained, illustrating how to find the limits of composite functions and rational functions using limit rules.', 'The concept of limits is introduced, stating that the limit as x approaches a of f of x equals L means that f of x gets arbitrarily close to L as x gets arbitrarily close to A.', 'The video covers the difference quotient and the average rate of change, essential topics related to the concept of derivative in calculus.', 'The chapter introduces the squeeze theorem as a method to find limits.', 'The limit as x approaches 1 of f of x is equal to 10, where f of x is 10x for x not equal to 1, and f of x is 0 for x equal to 1.']}, {'end': 5533.887, 'segs': [{'end': 4214.044, 'src': 'embed', 'start': 4190.024, 'weight': 1, 'content': [{'end': 4198.672, 'text': "The coordinates of the first point are 1, 2, and the next point, this is, let's see, 5, negative 1.", 'start': 4190.024, 'duration': 8.648}, {'end': 4202.295, 'text': 'Now I can find the slope by looking at the rise over the run.', 'start': 4198.672, 'duration': 3.623}, {'end': 4209.441, 'text': 'So as I go through a run of this distance, I go through a rise of that distance.', 'start': 4203.016, 'duration': 6.425}, {'end': 4214.044, 'text': "it's actually going to be a negative rise or a fall, because my line is pointing down.", 'start': 4209.441, 'duration': 4.603}], 'summary': 'Finding slope between two points: (1,2) and (5,-1), showing negative slope.', 'duration': 24.02, 'max_score': 4190.024, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w4190024.jpg'}, {'end': 4603.343, 'src': 'embed', 'start': 4576.127, 'weight': 2, 'content': [{'end': 4581.21, 'text': 'In this video, we saw that you can find the equation for a line, if you know the slope.', 'start': 4576.127, 'duration': 5.083}, {'end': 4586.513, 'text': 'And you know one point.', 'start': 4583.251, 'duration': 3.262}, {'end': 4590.916, 'text': 'you can also find the equation for the line if you know two points,', 'start': 4586.513, 'duration': 4.403}, {'end': 4598.8, 'text': 'because you can use the two points to get the slope and then plug in one of those points to figure out the rest of the equation.', 'start': 4590.916, 'duration': 7.884}, {'end': 4603.343, 'text': 'We saw two standard forms for the equation of a line.', 'start': 4600.801, 'duration': 2.542}], 'summary': 'Finding equations for a line using slope and points.', 'duration': 27.216, 'max_score': 4576.127, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w4576127.jpg'}, {'end': 5146.542, 'src': 'heatmap', 'start': 4688.582, 'weight': 0, 'content': [{'end': 4691.584, 'text': 'But this rational function has a different type of n behavior.', 'start': 4688.582, 'duration': 3.002}, {'end': 4698.109, 'text': 'Notice as x gets really big, the y values are leveling off at about a y value of 3.', 'start': 4692.145, 'duration': 5.964}, {'end': 4705.894, 'text': 'And similarly, as x values get really negative, our graph is leveling off near the line y equals 3.', 'start': 4698.109, 'duration': 7.785}, {'end': 4709.783, 'text': "I'll draw that line, y equals 3, on my graph.", 'start': 4705.894, 'duration': 3.889}, {'end': 4712.209, 'text': 'That line is called a horizontal asymptote.', 'start': 4710.144, 'duration': 2.065}, {'end': 4724.183, 'text': 'A horizontal asymptote is a horizontal line that our graph gets closer and closer to as x goes to infinity or as x goes to negative infinity, or both.', 'start': 4714.218, 'duration': 9.965}, {'end': 4728.445, 'text': "There's something else that's different about this graph from a polynomial graph.", 'start': 4725.403, 'duration': 3.042}, {'end': 4733.107, 'text': 'Look at what happens as x gets close to negative 5.', 'start': 4729.305, 'duration': 3.802}, {'end': 4739.13, 'text': 'As we approach negative 5 with x values on the right, our y values are going down towards negative infinity.', 'start': 4733.107, 'duration': 6.023}, {'end': 4745.717, 'text': 'And as we approach the x value of negative 5 from the left, our y values are going up towards positive infinity.', 'start': 4739.89, 'duration': 5.827}, {'end': 4753.847, 'text': 'We say that this graph has a vertical asymptote at x equals negative 5.', 'start': 4746.719, 'duration': 7.128}, {'end': 4758.193, 'text': 'A vertical asymptote is a vertical line that the graph gets closer and closer to.', 'start': 4753.847, 'duration': 4.346}, {'end': 4763.499, 'text': "Finally, there's something really weird going on at x equals 2.", 'start': 4759.537, 'duration': 3.962}, {'end': 4768.221, 'text': "There's a little open circle there, like the value at x equals 2 is dug out.", 'start': 4763.499, 'duration': 4.722}, {'end': 4769.942, 'text': "That's called a hole.", 'start': 4768.881, 'duration': 1.061}, {'end': 4776.425, 'text': "A hole is a place along the curve of the graph where the function doesn't exist.", 'start': 4770.922, 'duration': 5.503}, {'end': 4787.51, 'text': "Now that we've identified some of the features of our rational functions graph I want to look back at the equation and see how we could have predicted those features just by looking at the equation.", 'start': 4777.825, 'duration': 9.685}, {'end': 4796.629, 'text': 'To find horizontal asymptotes, we need to look at what our function is doing when x goes through really big positive or really big negative numbers.', 'start': 4788.723, 'duration': 7.906}, {'end': 4806.677, 'text': 'Looking at our equation for our function, the numerator is going to be dominated by the 3x squared term, when x is really big right?', 'start': 4797.73, 'duration': 8.947}, {'end': 4813.802, 'text': 'Because 3 times x squared is going to be absolutely enormous compared to this negative 12 if x is a big positive or negative number.', 'start': 4806.737, 'duration': 7.065}, {'end': 4816.445, 'text': 'in the denominator.', 'start': 4815.003, 'duration': 1.442}, {'end': 4819.428, 'text': 'the denominator will be dominated by the x squared term.', 'start': 4816.445, 'duration': 2.983}, {'end': 4825.834, 'text': 'again, if x is a really big positive or negative number, like a million, a million squared will be much,', 'start': 4819.428, 'duration': 6.406}, {'end': 4829.558, 'text': 'much bigger than three times a million or negative 10..', 'start': 4825.834, 'duration': 3.724}, {'end': 4835.744, 'text': 'For that reason, to find the end behavior or the horizontal asymptote for our function,', 'start': 4829.558, 'duration': 6.186}, {'end': 4843.149, 'text': 'We just need to look at the term on the numerator and the term on the denominator that have the highest exponent.', 'start': 4836.545, 'duration': 6.604}, {'end': 4846.13, 'text': 'Those are the ones that dominate the expression in size.', 'start': 4843.249, 'duration': 2.881}, {'end': 4857.016, 'text': "So as x gets really big, our function's y values are going to be approximately 3x squared over x squared, which is 3.", 'start': 4846.891, 'duration': 10.125}, {'end': 4861.877, 'text': "That's why we have a horizontal asymptote at y equals 3.", 'start': 4857.016, 'duration': 4.861}, {'end': 4868.844, 'text': 'Now our vertical asymptotes, those tend to occur where our denominator of our function is zero.', 'start': 4861.877, 'duration': 6.967}, {'end': 4873.929, 'text': "That's because the function doesn't exist when our denominator is zero.", 'start': 4869.785, 'duration': 4.144}, {'end': 4880.195, 'text': "And when we get close to that place where our denominator is zero, we're going to be dividing by tiny, tiny numbers,", 'start': 4874.59, 'duration': 5.605}, {'end': 4882.878, 'text': 'which will make our y values really big in magnitude.', 'start': 4880.195, 'duration': 2.683}, {'end': 4887.758, 'text': "So to check where our denominator is 0, let's factor our function.", 'start': 4884.113, 'duration': 3.645}, {'end': 4891.703, 'text': "In fact, I'm going to go ahead and factor the numerator and the denominator.", 'start': 4888.619, 'duration': 3.084}, {'end': 4897.591, 'text': "So the numerator factors, let's see, pull out the 3, I get x squared minus 4.", 'start': 4892.104, 'duration': 5.487}, {'end': 4904.515, 'text': 'Factoring the denominator, That factors into x plus 5 times x minus 2.', 'start': 4897.591, 'duration': 6.924}, {'end': 4906.936, 'text': 'I can factor the numerator a little further.', 'start': 4904.515, 'duration': 2.421}, {'end': 4915.999, 'text': "That's 3 times x minus 2 times x plus 2 over x plus 5 x minus 2.", 'start': 4907.276, 'duration': 8.723}, {'end': 4924.322, 'text': 'Now, when x is equal to negative 5, my denominator will be 0, but my numerator will not be 0.', 'start': 4915.999, 'duration': 8.323}, {'end': 4929.89, 'text': "That's what gives me the vertical asymptote at x equals negative 5.", 'start': 4924.322, 'duration': 5.568}, {'end': 4935.933, 'text': 'Notice that when x equals two, the denominator is zero, but the numerator is also zero.', 'start': 4929.89, 'duration': 6.043}, {'end': 4942.417, 'text': 'In fact, if I canceled the x minus two factor from the numerator and denominator,', 'start': 4937.214, 'duration': 5.203}, {'end': 4950.861, 'text': 'I get a simplified form for my function that agrees with my original function, as long as x is not equal to two.', 'start': 4942.417, 'duration': 8.444}, {'end': 4958.767, 'text': "That's because when x equals 2, this simplified function exists, but the original function does not.", 'start': 4952.659, 'duration': 6.108}, {'end': 4960.168, 'text': "It's 0 over 0.", 'start': 4958.847, 'duration': 1.321}, {'end': 4960.929, 'text': "It's undefined.", 'start': 4960.168, 'duration': 0.761}, {'end': 4967.665, 'text': 'But for every other x value, including x values near x equals 2,.', 'start': 4961.91, 'duration': 5.755}, {'end': 4974.151, 'text': "Our original function is just the same as this function, and that's why our function only has a vertical asymptote.", 'start': 4967.665, 'duration': 6.486}, {'end': 4981.999, 'text': 'at x equals negative 5, not one at x equals 2, because the x minus 2 factor is no longer in the function after simplifying.', 'start': 4974.151, 'duration': 7.848}, {'end': 4989.106, 'text': 'It does have a hole at x equals 2 because the original function is not defined there, even though the simplified version is.', 'start': 4982.68, 'duration': 6.426}, {'end': 4997.848, 'text': 'If we want to find the y value of our whole, we can just plug in x equals 2 into our simplified version of our function.', 'start': 4990.563, 'duration': 7.285}, {'end': 5008.716, 'text': 'That gives a y value of 3 times 2 plus 2 over 2 plus 7, or 12 ninths, which simplifies to 4 thirds.', 'start': 4998.689, 'duration': 10.027}, {'end': 5013.62, 'text': 'So our whole is at 2 4 thirds.', 'start': 5010.177, 'duration': 3.443}, {'end': 5019.044, 'text': "Now that we've been through one example in detail, let's summarize our findings.", 'start': 5015.361, 'duration': 3.683}, {'end': 5025.936, 'text': 'we find the vertical asymptotes and the holes by looking where the denominator is zero.', 'start': 5020.455, 'duration': 5.481}, {'end': 5032.737, 'text': 'The holes happen where the denominator and numerator are both zero and those factors cancel out.', 'start': 5026.956, 'duration': 5.781}, {'end': 5038.458, 'text': 'The vertical asymptotes are all other x values where the denominator is zero.', 'start': 5033.797, 'duration': 4.661}, {'end': 5045.919, 'text': 'We find the horizontal asymptotes by considering the highest power term on the numerator and the denominator.', 'start': 5039.558, 'duration': 6.361}, {'end': 5049.82, 'text': "I'll explain this process in more detail in three examples.", 'start': 5046.759, 'duration': 3.061}, {'end': 5061.868, 'text': 'In the first example, if we circle the highest power terms, that simplifies to 5x over 3x squared, which is 5 over 3x.', 'start': 5051.499, 'duration': 10.369}, {'end': 5067.152, 'text': 'As x gets really big, the denominator is going to be huge.', 'start': 5063.069, 'duration': 4.083}, {'end': 5070.595, 'text': "So I'm going to be dividing 5 by a huger and huger number.", 'start': 5067.372, 'duration': 3.223}, {'end': 5073.958, 'text': "That's going to be going very close to 0.", 'start': 5071.175, 'duration': 2.783}, {'end': 5079.153, 'text': 'And therefore, we have a horizontal asymptote at y equals 0.', 'start': 5073.958, 'duration': 5.195}, {'end': 5087.544, 'text': 'In the second example, the highest power terms, 2x cubed over 3x cubed, simplify to 2 thirds.', 'start': 5079.153, 'duration': 8.391}, {'end': 5094.954, 'text': "So as x gets really big, we're going to be heading towards 2 thirds, and we have a horizontal asymptote at y equals 2 thirds.", 'start': 5088.165, 'duration': 6.789}, {'end': 5106.039, 'text': 'In the third example, the highest power terms, x squared over 2x, simplifies to x over 2.', 'start': 5096.716, 'duration': 9.323}, {'end': 5109.96, 'text': 'As x gets really big, x over 2 is getting really big.', 'start': 5106.039, 'duration': 3.921}, {'end': 5113.781, 'text': "And therefore, we don't have a horizontal asymptote at all.", 'start': 5110.68, 'duration': 3.101}, {'end': 5120.443, 'text': 'This is going to infinity when x goes through big positive numbers.', 'start': 5114.461, 'duration': 5.982}, {'end': 5127.43, 'text': "and it's going to negative infinity when x goes through big negative numbers.", 'start': 5121.363, 'duration': 6.067}, {'end': 5134.499, 'text': "So in this case, the end behavior is kind of like that of a polynomial, and there's no horizontal asymptote.", 'start': 5128.352, 'duration': 6.147}, {'end': 5141.6, 'text': 'In general, when the degree of the numerator is smaller than the degree of the denominator.', 'start': 5136.317, 'duration': 5.283}, {'end': 5146.542, 'text': "we're in this first case where the denominator gets really big compared to the numerator and we go to zero.", 'start': 5141.6, 'duration': 4.942}], 'summary': 'Rational function behavior: horizontal asymptote at y=3, vertical at x=-5, hole at x=2.', 'duration': 457.96, 'max_score': 4688.582, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w4688582.jpg'}, {'end': 4813.802, 'src': 'embed', 'start': 4788.723, 'weight': 3, 'content': [{'end': 4796.629, 'text': 'To find horizontal asymptotes, we need to look at what our function is doing when x goes through really big positive or really big negative numbers.', 'start': 4788.723, 'duration': 7.906}, {'end': 4806.677, 'text': 'Looking at our equation for our function, the numerator is going to be dominated by the 3x squared term, when x is really big right?', 'start': 4797.73, 'duration': 8.947}, {'end': 4813.802, 'text': 'Because 3 times x squared is going to be absolutely enormous compared to this negative 12 if x is a big positive or negative number.', 'start': 4806.737, 'duration': 7.065}], 'summary': "To find horizontal asymptotes, look at function's behavior for large x values. the 3x squared term dominates the numerator, making it enormous compared to the negative 12 for large x values.", 'duration': 25.079, 'max_score': 4788.723, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w4788723.jpg'}], 'start': 4102.292, 'title': 'Graphs, equations, and asymptotes', 'summary': 'Covers finding equations of lines, rational functions, and explores the concept of asymptotes, including horizontal and vertical asymptotes, with examples and quantifiable data on slope, intercepts, and end behavior.', 'chapters': [{'end': 4327.57, 'start': 4102.292, 'title': 'Graphs and equations of lines', 'summary': 'Discusses finding the equation of a line using the standard format y=mx+b, with the slope as -3/4 and the y-intercept as 11/4, by utilizing points with integer coordinates and applying the formula y2-y1/x2-x1.', 'duration': 225.278, 'highlights': ['The slope of the line is -3/4, calculated by finding the rise over the run from two convenient points with integer coordinates (1, 2) and (5, -1).', 'The y-intercept of the line is 11/4, determined by substituting the x and y coordinates of a point into the equation y=-3/4x+b and solving for b.', 'The standard format for the equation of a line is y=mx+b, where m represents the slope and b represents the y-intercept.']}, {'end': 4806.677, 'start': 4328.431, 'title': 'Equations for lines and rational functions', 'summary': "Covers finding equations for lines using slope-intercept and point-slope forms, showcasing two methods and also explains the features of rational functions' graph such as horizontal asymptotes, vertical asymptotes, and holes.", 'duration': 478.246, 'highlights': ['The chapter covers finding equations for lines using slope-intercept and point-slope forms, showcasing two methods. For example, finding the equation of a line using the slope-intercept form y equals mx plus b, and the point-slope form y minus y-naught equals m times x minus x-naught.', "It explains the features of rational functions' graph such as horizontal asymptotes, vertical asymptotes, and holes. For instance, the horizontal asymptote is a horizontal line that the graph gets closer and closer to as x goes to infinity or as x goes to negative infinity, and a vertical asymptote is a vertical line that the graph gets closer and closer to.", "The end behavior of a rational function graph differs from a polynomial graph, as the rational function's graph levels off at certain y values for large positive or negative x values, unlike the polynomial graph.", 'The chapter demonstrates finding the equation for a line using the slope-intercept form and the point-slope form, providing a step-by-step example with calculations and final equations.']}, {'end': 5134.499, 'start': 4806.737, 'title': 'Asymptotes and exponential dominance', 'summary': 'Discusses the concept of horizontal and vertical asymptotes, finding them by examining the highest power terms in the numerator and denominator, and provides examples of end behavior and asymptote determination with quantifiable data.', 'duration': 327.762, 'highlights': ['The chapter discusses the concept of horizontal and vertical asymptotes, finding them by examining the highest power terms in the numerator and denominator.', 'Examples of end behavior and asymptote determination with quantifiable data are provided.', "The function's y values are approximately 3x squared over x squared, resulting in a horizontal asymptote at y equals 3.", 'The chapter explains that when x gets really big, the denominator is going to be huge, resulting in a horizontal asymptote at y equals 0.', 'The highest power terms in the third example simplify to x over 2, leading to no horizontal asymptote, and the function tends to infinity for big positive and negative numbers.']}, {'end': 5533.887, 'start': 5136.317, 'title': 'Asymptotes and limits in rational functions', 'summary': 'Discusses the behavior of rational functions in relation to horizontal and vertical asymptotes, as well as the concept of limits at infinity, using an example to demonstrate the application of these concepts.', 'duration': 397.57, 'highlights': ['The highest power term in the numerator and denominator determines the horizontal asymptote, with a ratio of their leading coefficients.', 'Identification of vertical asymptotes and holes by factoring the function, with common factors indicating the presence of holes.', 'Demonstration of finding horizontal asymptotes, vertical asymptotes, and holes for a rational function through a step-by-step example.', 'Explanation of the concept of limits at infinity and their relevance in understanding the behavior of functions as x approaches arbitrarily large positive and negative values.']}], 'duration': 1431.595, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w4102292.jpg', 'highlights': ["The function's y values are approximately 3x squared over x squared, resulting in a horizontal asymptote at y equals 3.", 'The slope of the line is -3/4, calculated by finding the rise over the run from two convenient points with integer coordinates (1, 2) and (5, -1).', 'The chapter covers finding equations for lines using slope-intercept and point-slope forms, showcasing two methods.', 'The highest power term in the numerator and denominator determines the horizontal asymptote, with a ratio of their leading coefficients.', 'The chapter discusses the concept of horizontal and vertical asymptotes, finding them by examining the highest power terms in the numerator and denominator.']}, {'end': 6714.547, 'segs': [{'end': 5562.483, 'src': 'embed', 'start': 5535.248, 'weight': 6, 'content': [{'end': 5542.411, 'text': 'An infinite limit means that the y values or the f of x values go to infinity or negative infinity.', 'start': 5535.248, 'duration': 7.163}, {'end': 5553.576, 'text': 'Limits at infinity correspond to horizontal asymptotes, as drawn above, while infinite limits correspond to vertical asymptotes.', 'start': 5545.33, 'duration': 8.246}, {'end': 5562.483, 'text': "The only exception to this, when we don't have a horizontal or vertical asymptote, is when x and f are both going infinite at the same time.", 'start': 5554.797, 'duration': 7.686}], 'summary': 'Infinite limits lead to horizontal or vertical asymptotes.', 'duration': 27.235, 'max_score': 5535.248, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w5535248.jpg'}, {'end': 5875.861, 'src': 'embed', 'start': 5851.893, 'weight': 5, 'content': [{'end': 5858.239, 'text': 'This video gives some algebraic techniques for computing the limits at infinity of rational functions.', 'start': 5851.893, 'duration': 6.346}, {'end': 5863.986, 'text': "Let's find the limit as x goes to infinity of this rational function.", 'start': 5860.342, 'duration': 3.644}, {'end': 5872.54, 'text': 'The numerator and the denominator of this rational function are each getting arbitrarily large as x goes to infinity.', 'start': 5865.578, 'duration': 6.962}, {'end': 5875.861, 'text': 'One way to see this is by estimating the graphs.', 'start': 5873.22, 'duration': 2.641}], 'summary': 'Algebraic techniques for computing limits of rational functions at infinity.', 'duration': 23.968, 'max_score': 5851.893, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w5851893.jpg'}, {'end': 6214.927, 'src': 'embed', 'start': 6189.516, 'weight': 7, 'content': [{'end': 6201.267, 'text': 'then we can rewrite our limit as the limit of five x squared over two x cubed, which is the same as the limit, as x goes to infinity,', 'start': 6189.516, 'duration': 11.751}, {'end': 6203.93, 'text': "of five over two x, just by canceling x's.", 'start': 6201.267, 'duration': 2.663}, {'end': 6206.741, 'text': 'which is 0 as x goes to infinity.', 'start': 6204.619, 'duration': 2.122}, {'end': 6214.927, 'text': 'Similarly, if we just focus on the highest power terms in the numerator and denominator in the second example,', 'start': 6207.801, 'duration': 7.126}], 'summary': 'Limit of 5x^2/2x^3 equals 0 as x goes to infinity.', 'duration': 25.411, 'max_score': 6189.516, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w6189516.jpg'}, {'end': 6391.948, 'src': 'embed', 'start': 6364.178, 'weight': 0, 'content': [{'end': 6371.02, 'text': 'Please pause this video and try to draw graphs of at least two different functions that fail to be continuous in different ways.', 'start': 6364.178, 'duration': 6.842}, {'end': 6376.482, 'text': 'One common kind of discontinuity is called a jump discontinuity.', 'start': 6373.001, 'duration': 3.481}, {'end': 6383.625, 'text': 'A function has a jump discontinuity if its graph separates into two pieces with a jump in between them.', 'start': 6378.043, 'duration': 5.582}, {'end': 6391.948, 'text': 'This particular function can be described as a piecewise defined function with two linear equations.', 'start': 6385.986, 'duration': 5.962}], 'summary': 'Video instructs to draw graphs of at least two functions with different types of discontinuities, such as jump discontinuity.', 'duration': 27.77, 'max_score': 6364.178, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w6364178.jpg'}], 'start': 5535.248, 'title': 'Understanding limits and continuity', 'summary': 'Covers limits at infinity, infinite limits, and rational functions, discussing computation of limits, continuity conditions, and types of discontinuities, with examples showcasing behavior as x approaches infinity or negative infinity and possibilities of limits being 0, any other number, infinity, or not existing.', 'chapters': [{'end': 5816.513, 'start': 5535.248, 'title': 'Limits at infinity', 'summary': 'Discusses limits at infinity and infinite limits, including examples of functions and their limits at infinity, highlighting specific values and the behavior of functions as x approaches infinity or negative infinity.', 'duration': 281.265, 'highlights': ['The limit as x goes to infinity of g of x is 0, while the limit as x goes to minus infinity of g of x is infinity, demonstrating the behavior of the function as x approaches positive and negative infinity.', 'The limit as x goes to infinity of h of x is negative infinity, while the limit as x goes to minus infinity of h of x does not exist, showcasing the behavior of the function as x approaches positive and negative infinity.', 'The limit as x goes to infinity of 1 over x, 1 over x squared, and 1 over the square root of x are all equal to 0, demonstrating the behavior of these functions as x approaches infinity.']}, {'end': 6156.556, 'start': 5816.513, 'title': 'Limits at infinity of rational functions', 'summary': 'Discusses the computation of limits at infinity of rational functions, demonstrating how to rewrite expressions in a different form to evaluate them and applying limit laws to find the limits, with examples showcasing the possibilities of limits being 0, any other number, infinity, negative infinity, or not existing.', 'duration': 340.043, 'highlights': ['The limit as x goes to negative infinity of a rational expression with the highest power of x in the numerator as x to the fourth and in the denominator as x squared results in a final limit of negative infinity, due to the magnitudes getting arbitrarily large.', 'The limit as x goes to infinity of a rational expression with the highest power of x in the numerator as x cubed and in the denominator as x cubed yields a final limit of 3 halves, as all the parts cancel out and go to 0.', 'The chapter explains how to rewrite expressions in a different form to make it easier to evaluate, such as factoring out the highest power of x from the numerator and the denominator, and applying limit laws to find the limits, showcasing the possibilities of limits being 0, any other number, infinity, negative infinity, or not existing.']}, {'end': 6445.544, 'start': 6157.877, 'title': 'Computing limits of rational functions', 'summary': 'Explains how to compute limits of rational functions by focusing on the highest power terms, providing examples and shortcut rules, and discussing the definition of continuity based on limits, including common types of discontinuities like jump and removable discontinuities.', 'duration': 287.667, 'highlights': ['The chapter emphasizes the importance of focusing on the highest power terms when computing limits of rational functions, providing examples and shortcut rules, such as predicting limits of infinity and negative infinity based on the degrees of the numerator and denominator.', 'The video introduces two methods for computing limits of rational functions: formal method of factoring out highest power terms and simplifying, and informal method of looking at the degree of the numerator and the denominator and focusing on the highest power terms.', 'The chapter discusses the definition of continuity based on limits and provides examples of common types of discontinuities, including jump discontinuity and removable discontinuity, with specific functions and their graphical representations.']}, {'end': 6714.547, 'start': 6445.544, 'title': 'Understanding function continuity', 'summary': 'Explains the concept of function continuity and its conditions, including how to avoid jump discontinuity, removable discontinuity, infinite discontinuity, and wild discontinuity, and highlights the places where the function is not continuous.', 'duration': 269.003, 'highlights': ["The function's not continuous at x equals 3 because the limit doesn't exist, and at x equals 1 due to a jump discontinuity.", "The function is not continuous at negative 3 because the function's simply not defined there, and at x equals negative 2 because the limit doesn't exist from the left.", 'The function is continuous at x equals negative 2 from the left, but not from the right, as the value and the limit from the right are not equal.']}], 'duration': 1179.299, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w5535248.jpg', 'highlights': ['The limit as x goes to infinity of 1 over x, 1 over x squared, and 1 over the square root of x are all equal to 0, demonstrating the behavior of these functions as x approaches infinity.', 'The limit as x goes to infinity of g of x is 0, while the limit as x goes to minus infinity of g of x is infinity, demonstrating the behavior of the function as x approaches positive and negative infinity.', 'The limit as x goes to infinity of h of x is negative infinity, while the limit as x goes to minus infinity of h of x does not exist, showcasing the behavior of the function as x approaches positive and negative infinity.', 'The limit as x goes to negative infinity of a rational expression with the highest power of x in the numerator as x to the fourth and in the denominator as x squared results in a final limit of negative infinity, due to the magnitudes getting arbitrarily large.', 'The limit as x goes to infinity of a rational expression with the highest power of x in the numerator as x cubed and in the denominator as x cubed yields a final limit of 3 halves, as all the parts cancel out and go to 0.', 'The chapter emphasizes the importance of focusing on the highest power terms when computing limits of rational functions, providing examples and shortcut rules, such as predicting limits of infinity and negative infinity based on the degrees of the numerator and denominator.', 'The chapter discusses the definition of continuity based on limits and provides examples of common types of discontinuities, including jump discontinuity and removable discontinuity, with specific functions and their graphical representations.', "The function's not continuous at x equals 3 because the limit doesn't exist, and at x equals 1 due to a jump discontinuity."]}, {'end': 9481.713, 'segs': [{'end': 6753.707, 'src': 'embed', 'start': 6716.394, 'weight': 0, 'content': [{'end': 6721.857, 'text': 'By the same reasoning at x equals 1, the function is not continuous.', 'start': 6716.394, 'duration': 5.463}, {'end': 6727.78, 'text': 'but it is continuous from the right because the limit from the right is equal to the value of the function.', 'start': 6721.857, 'duration': 5.923}, {'end': 6740.876, 'text': 'Notice that f is not continuous from the left here, because the limit from the left is about 1 and 1 half, while the value of the function is 1.', 'start': 6730.782, 'duration': 10.094}, {'end': 6753.707, 'text': 'In general, we say that a function f is continuous from the left at x equals a if the limit as x goes to a from the left of f of x is equal to f of a.', 'start': 6740.876, 'duration': 12.831}], 'summary': 'The function is not continuous at x=1 from the left, limit is 1.5, value is 1.', 'duration': 37.313, 'max_score': 6716.394, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w6716394.jpg'}, {'end': 9199.873, 'src': 'embed', 'start': 9162.059, 'weight': 2, 'content': [{'end': 9171.816, 'text': 'One way to find cosine of t is to use the fact that cosine squared t plus sine squared t is equal to one.', 'start': 9162.059, 'duration': 9.757}, {'end': 9181.282, 'text': 'That is cosine of t squared plus negative two seventh squared is equal to one.', 'start': 9173.457, 'duration': 7.825}, {'end': 9199.873, 'text': 'I can write this as cosine of t squared plus four 49th is equal to one and so cosine of t squared is equal to one minus four 49th, which is 45 49.', 'start': 9183.004, 'duration': 16.869}], 'summary': 'Cosine of t squared equals 45/49 using trigonometric identity.', 'duration': 37.814, 'max_score': 9162.059, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w9162059.jpg'}, {'end': 9402.517, 'src': 'embed', 'start': 9372.609, 'weight': 4, 'content': [{'end': 9375.331, 'text': 'Or I could just use the fact that the cosine values repeat.', 'start': 9372.609, 'duration': 2.722}, {'end': 9383.715, 'text': "If I add or subtract two pi to the my angle t, I'll be at the same place on the unit circle.", 'start': 9376.371, 'duration': 7.344}, {'end': 9385.917, 'text': 'So my cosine will be exactly the same.', 'start': 9384.116, 'duration': 1.801}, {'end': 9393.173, 'text': 'Therefore, my values of cosine, which are represented by my y values on this graph, repeat themselves.', 'start': 9386.991, 'duration': 6.182}, {'end': 9402.517, 'text': 'For example, when my t value is 2 pi plus pi over 6, about like here, its cosine is the same as the cosine of just pi over 6.', 'start': 9394.254, 'duration': 8.263}], 'summary': 'Cosine values repeat every 2pi, leading to repeated y values on the graph.', 'duration': 29.908, 'max_score': 9372.609, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w9372609.jpg'}], 'start': 6716.394, 'title': 'Properties of trig functions', 'summary': 'Discusses continuity in functions, intermediate value theorem, trigonometric functions, special angles, unit circle trigonometry, and properties of trig functions, covering concepts, computations, and applications.', 'chapters': [{'end': 6981.35, 'start': 6716.394, 'title': 'Continuity in functions', 'summary': 'Discusses the concept of continuity in functions, including the definition of continuity at a point, continuity on intervals, and examples of functions that are continuous on specific intervals and everywhere.', 'duration': 264.956, 'highlights': ["Functions are continuous at a point if the limit as x goes to a of the function is equal to the function's value at a.", 'A function f is continuous on the open interval BC if f is continuous at every point in that interval.', 'Functions like polynomials, sine x, cosine x, and the absolute value of x are examples of functions that are continuous everywhere.', 'Rational functions and trig functions are examples of functions that are continuous on their domains.']}, {'end': 7547.127, 'start': 6981.35, 'title': 'Continuous functions and intermediate value theorem', 'summary': 'Explains the properties of continuous functions, including their compositions, and the application of the intermediate value theorem to find roots of equations and other scenarios.', 'duration': 565.777, 'highlights': ['The sums, differences, products, and quotients of continuous functions are continuous on their domains.', 'The compositions of continuous functions are continuous on their domains.', 'The intermediate value theorem can be applied to prove the existence of roots of equations by selecting an interval where the function changes sign.']}, {'end': 7996.911, 'start': 7547.127, 'title': 'Trigonometric functions and applications', 'summary': 'Explains the definitions and applications of trigonometric functions, including the computation of sine and cosine for specific angles and an application involving the calculation of the height of a kite flying at a 75-degree angle with a 100-meter string.', 'duration': 449.784, 'highlights': ['The sine of 45 degrees is the square root of 2 over 2.', 'The exact values of all six trig functions for the angle theta in the given triangle are calculated.', 'The height of the kite at a 75-degree angle with a 100-meter string is calculated to be 96.59 meters.']}, {'end': 8450.458, 'start': 8001.212, 'title': 'Trigonometric computations and special angles', 'summary': 'Covers the computation of sine and cosine for special angles 30, 45, and 60 degrees, emphasizing the connection between right triangles and the unit circle, and recommends memorizing key trigonometric values and their corresponding angles.', 'duration': 449.246, 'highlights': ['The sine and cosine of 45 degrees are both equal to the square root of 2 over 2, which is rationalized to 5 square root of 2 over 2 when the hypotenuse of the triangle is 5, demonstrating the consistent ratios of sides in similar triangles.', 'The computation of the sine and cosine of 30 degrees is shown using a 30-60-90 right triangle, yielding the values of 1/2 and square root of 3 over 2, and the relationship of these angles with radians is emphasized.', 'The recommendation to memorize key trigonometric values such as 1/2, square root of 2 over 2, and square root of 3 over 2, and their association with specific angles, enabling the reconstruction of triangles and identification of their angles.']}, {'end': 8725.635, 'start': 8450.478, 'title': 'Unit circle trigonometry', 'summary': 'Explains the concept of the unit circle, defining sine, cosine, and tangent in terms of the x and y coordinates of a point on the unit circle, and provides a method for calculating these trigonometric functions in terms of the unit circle.', 'duration': 275.157, 'highlights': ['The video explains the concept of the unit circle, defining the trigonometric functions sine, cosine, and tangent in terms of the x and y coordinates of a point on the unit circle.', 'The properties of the trig functions sine and cosine are deduced from the unit circle definition, including the periodic property stating that the values of cosine and sine are periodic with period two pi.', 'The calculation of sine, cosine, and tangent in terms of the unit circle is demonstrated with an example, providing the specific value of tangent of phi as -0.3639.']}, {'end': 9481.713, 'start': 8726.035, 'title': 'Properties of trig functions', 'summary': 'Covers periodic property, even odd property, and pythagorean property of trig functions, with examples and graphs, showing how cosine and sine values repeat and how to compute values using the pythagorean theorem and unit circle.', 'duration': 755.678, 'highlights': ['The periodic property states that cosine of an angle plus two pi is equal to cosine of the angle, and the similar statement for sine is that sine of an angle plus two pi is equal to sine of the original angle, both measured in radians.', 'The even odd property explains that cosine is an even function, while sine is an odd function, and the tangent of negative theta is the negative of the tangent of theta, indicating symmetry and direction of angles.', 'The Pythagorean property states that cosine of theta squared plus sine of theta squared is equal to one, derived from the Pythagorean theorem, and is used to compute values of cosine given values of sine and vice versa.']}], 'duration': 2765.319, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w6716394.jpg', 'highlights': ['The height of the kite at a 75-degree angle with a 100-meter string is calculated to be 96.59 meters.', 'The computation of the sine and cosine of 30 degrees is shown using a 30-60-90 right triangle, yielding the values of 1/2 and square root of 3 over 2, and the relationship of these angles with radians is emphasized.', 'The sums, differences, products, and quotients of continuous functions are continuous on their domains.', 'The compositions of continuous functions are continuous on their domains.', 'The intermediate value theorem can be applied to prove the existence of roots of equations by selecting an interval where the function changes sign.']}, {'end': 12975.634, 'segs': [{'end': 9535.414, 'src': 'embed', 'start': 9511.694, 'weight': 14, 'content': [{'end': 9518.842, 'text': 'since adding pi over two on the inside moves the graph horizontally to the left by pi over two.', 'start': 9511.694, 'duration': 7.148}, {'end': 9529.089, 'text': 'Or we can think of the graph of sine as being constructed from the graph of cosine by shifting the cosine graph right by pi over two.', 'start': 9520.183, 'duration': 8.906}, {'end': 9535.414, 'text': 'That means we can write sine of x as equal to cosine of x minus pi over two,', 'start': 9529.63, 'duration': 5.784}], 'summary': 'Shifting sine graph left by pi/2, or cosine graph right by pi/2.', 'duration': 23.72, 'max_score': 9511.694, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w9511694.jpg'}, {'end': 9581.802, 'src': 'embed', 'start': 9555.885, 'weight': 9, 'content': [{'end': 9563.33, 'text': 'That makes sense, because sine and cosine come from the unit circle, the input values for the domain come from angles.', 'start': 9555.885, 'duration': 7.445}, {'end': 9569.694, 'text': 'And you can use any numbers and angle positive or negative as big as you want, just by wrapping a lot of times around the circle.', 'start': 9563.91, 'duration': 5.784}, {'end': 9577.219, 'text': 'The output values for the range, that is the actual values of sine and cosine, come from the coordinates on the unit circle.', 'start': 9570.475, 'duration': 6.744}, {'end': 9581.802, 'text': "And those coordinates can't be any bigger than 1 or any smaller than negative 1.", 'start': 9577.639, 'duration': 4.163}], 'summary': 'Sine and cosine values come from unit circle, limited to -1 to 1.', 'duration': 25.917, 'max_score': 9555.885, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w9555885.jpg'}, {'end': 11762.979, 'src': 'embed', 'start': 11629.073, 'weight': 0, 'content': [{'end': 11635.517, 'text': 'Now this process of picking points closer and closer to our original point from the left and from the right should remind you of limits.', 'start': 11629.073, 'duration': 6.444}, {'end': 11647.886, 'text': 'And indeed, the slope of the tangent line is the limit as x goes to 1.5 of the slope of my secant lines, which are given by this expression.', 'start': 11636.018, 'duration': 11.868}, {'end': 11651.968, 'text': "This quantity is so important that it's given its own name.", 'start': 11649.307, 'duration': 2.661}, {'end': 11656.171, 'text': "It's called the derivative of f of x at x equals 1.5.", 'start': 11652.329, 'duration': 3.842}, {'end': 11673.986, 'text': 'So in other words, the derivative, which is written as f prime at 1.5 is the limit as x goes to 1.5 of f of x minus f of 1.5 over x minus 1.5.', 'start': 11656.171, 'duration': 17.815}, {'end': 11682.493, 'text': 'Now, based on our numerical tables, for example, here we can see that that limit seems to be heading towards 3,', 'start': 11673.986, 'duration': 8.507}, {'end': 11685.855, 'text': 'whether x approaches 1.5 from the right or from the left.', 'start': 11682.493, 'duration': 3.362}, {'end': 11691.498, 'text': "So I'll write down the answer of 3.", 'start': 11686.895, 'duration': 4.603}, {'end': 11694.059, 'text': 'You know, numeric 11 is pretty strong.', 'start': 11691.498, 'duration': 2.561}, {'end': 11700.262, 'text': "If we wanted to have a really precise argument, we'd actually need to use algebra to compute this limit.", 'start': 11694.399, 'duration': 5.863}, {'end': 11706.725, 'text': 'exactly using the formula for the function itself, f of x equals x squared.', 'start': 11700.262, 'duration': 6.463}, {'end': 11709.487, 'text': "And we'll do examples like that in a future video.", 'start': 11707.126, 'duration': 2.361}, {'end': 11719.36, 'text': 'But for now, the main point is just that the slope of the tangent line is the limit of the slope of the secant lines, which is given by this formula.', 'start': 11710.848, 'duration': 8.512}, {'end': 11727.929, 'text': "For now, let's look at an animation that shows how the slope of our secant lines approach the slope of our tangent line.", 'start': 11722.824, 'duration': 5.105}, {'end': 11732.233, 'text': 'So this black curve here is the function y equals x squared.', 'start': 11728.729, 'duration': 3.504}, {'end': 11737.858, 'text': 'The red line is the tangent line through the point where x equals 1.5.', 'start': 11732.813, 'duration': 5.045}, {'end': 11745.865, 'text': 'And the blue line is a secant line that goes through the point with x coordinate 1.5 and a second point with x coordinate 2.5.', 'start': 11737.858, 'duration': 8.007}, {'end': 11747.226, 'text': 'The points are shown here on the right.', 'start': 11745.865, 'duration': 1.361}, {'end': 11752.933, 'text': "So I'm going to use this slider here and drag my second point closer to my first point.", 'start': 11748.39, 'duration': 4.543}, {'end': 11762.979, 'text': 'So notice how as the x-coordinate of my second point gets closer and closer to 1.5, my secant line is getting closer and closer to my tangent line.', 'start': 11754.154, 'duration': 8.825}], 'summary': 'Derivative at x=1.5 is the limit of secant slopes, approaching 3 for f(x)=x^2.', 'duration': 133.906, 'max_score': 11629.073, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w11629073.jpg'}, {'end': 11827.855, 'src': 'embed', 'start': 11789.213, 'weight': 8, 'content': [{'end': 11795.295, 'text': 'We saw in our example that the slope of our tangent line, or the derivative at 1.5 was given by the limit,', 'start': 11789.213, 'duration': 6.082}, {'end': 11808.651, 'text': 'as x goes to 1.5 of f of x minus f of 1.5, divided by x minus 1.5..', 'start': 11795.295, 'duration': 13.356}, {'end': 11812.692, 'text': 'Well, in general, the derivative of a function y equals f of x.', 'start': 11808.651, 'duration': 4.041}, {'end': 11813.832, 'text': 'at an x value.', 'start': 11812.692, 'duration': 1.14}, {'end': 11816.472, 'text': 'a is given by f.', 'start': 11813.832, 'duration': 2.64}, {'end': 11827.855, 'text': 'prime of a equals the limit, as x goes to a, of f of x minus f of a over x minus a.', 'start': 11816.472, 'duration': 11.383}], 'summary': "Derivative of a function at x=a is f'(a) = lim(x->a) [f(x)-f(a)] / (x-a)", 'duration': 38.642, 'max_score': 11789.213, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w11789213.jpg'}, {'end': 12082.025, 'src': 'embed', 'start': 12048.222, 'weight': 10, 'content': [{'end': 12053.065, 'text': 'And we need the numerator here to look like f of x minus f of a.', 'start': 12048.222, 'duration': 4.843}, {'end': 12057.638, 'text': "Well, let's just try the simplest thing we can.", 'start': 12054.537, 'duration': 3.101}, {'end': 12060.799, 'text': "Let's try f of x equals x plus 5 squared.", 'start': 12057.898, 'duration': 2.901}, {'end': 12075.183, 'text': 'Then x plus 5 squared is our f of x, and f of a is our f of negative 1, is going to be negative 1 plus 5 squared, which is 16.', 'start': 12062.099, 'duration': 13.084}, {'end': 12076.384, 'text': 'So that matches up perfectly.', 'start': 12075.183, 'duration': 1.201}, {'end': 12082.025, 'text': "We've got our f of x here, our f of a here, and our x minus a at the bottom.", 'start': 12076.664, 'duration': 5.361}], 'summary': 'Using f(x) = (x+5)^2, f(-1) = 16 matches f(x) - f(a)', 'duration': 33.803, 'max_score': 12048.222, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w12048222.jpg'}, {'end': 12519.94, 'src': 'embed', 'start': 12486.263, 'weight': 12, 'content': [{'end': 12486.763, 'text': 'All right.', 'start': 12486.263, 'duration': 0.5}, {'end': 12489.383, 'text': "So here we've got.", 'start': 12487.223, 'duration': 2.16}, {'end': 12503.127, 'text': "so as h goes to 0, I'm just going to get 1 over 2 times the square root of 4 times 2, plus the square root of 4, which equals 1 16th.", 'start': 12489.383, 'duration': 13.744}, {'end': 12509.353, 'text': 'So by now, you may have forgotten what the original problem was.', 'start': 12506.331, 'duration': 3.022}, {'end': 12510.914, 'text': 'I think I have.', 'start': 12509.373, 'duration': 1.541}, {'end': 12512.575, 'text': "But let's go back up here.", 'start': 12511.515, 'duration': 1.06}, {'end': 12519.94, 'text': 'We were looking for the derivative of f of x, which was 1 over a square root of 3 minus x at x equals negative 1.', 'start': 12513.136, 'duration': 6.804}], 'summary': 'Finding the derivative of f(x) at x = -1 yields 1/16th.', 'duration': 33.677, 'max_score': 12486.263, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w12486263.jpg'}], 'start': 9483.135, 'title': 'Trigonometry and derivatives', 'summary': 'Covers properties, graphing, and trigonometric functions, including sine and cosine graphs, transformations, graphing trig functions, solving trig equations, and interpreting derivatives with specific examples and calculations, emphasizing similarities, ranges, symmetries, and practical interpretations.', 'chapters': [{'end': 9554.484, 'start': 9483.135, 'title': 'Properties of sine and cosine graphs', 'summary': 'Discusses the properties of sine and cosine graphs, highlighting their similarities and the horizontal shifts by pi over two, the domain and range of sine and cosine functions, and their range being from negative one to one.', 'duration': 71.349, 'highlights': ['The range of sine and cosine functions is from negative one to one.', 'The domain of sine and cosine functions is all real numbers, expressed as negative infinity to infinity.', 'The graph of cosine can be thought of as the graph of sine shifted to the left by pi over two, and the graph of sine can be constructed from the graph of cosine by shifting it right by pi over two.']}, {'end': 10267.913, 'start': 9555.885, 'title': 'Graphing sine and cosine functions', 'summary': 'Explains the properties and transformations of sine and cosine functions, including their ranges, symmetries, midline, amplitude, and period, and how to graph and analyze transformed functions. it also demonstrates the process of graphing the function y = 3 sin(2x-pi/4)-1 and explains the effects of different transformations on the midline, amplitude, period, and phase shift.', 'duration': 712.028, 'highlights': ['The output values for the range, that is the actual values of sine and cosine, come from the coordinates on the unit circle.', 'The absolute maximum values of these two functions is one and the absolute minimum value is negative one.', 'The midline is the horizontal line halfway in between the maximum and minimum points.', 'The amplitude is the vertical distance between a maximum point and the midline.', 'For y equals cosine of x, the period is two pi.']}, {'end': 11427.38, 'start': 10268.313, 'title': 'Graphing trig functions & solving basic trig equations', 'summary': 'Covers graphing trig functions, including the graphs of tangent, secant, cotangent, and cosecant, and solving basic trig equations by finding principal solutions and using the unit circle for tan x and secant x.', 'duration': 1159.067, 'highlights': ['The graphs of tangent, secant, cotangent, and cosecant are covered.', 'Understanding the graph of y equals tangent of x.', 'Graphing of y equals tan x and understanding its periodicity.', 'Explanation of the graph of y equals secant x and its properties.', 'Solving basic trig equations using the unit circle and finding principal solutions.', 'Introducing the idea of the derivative using graphs, secant lines, and tangent lines.']}, {'end': 12023.969, 'start': 11428.908, 'title': 'Slope of tangent line approximation', 'summary': 'Discusses the process of obtaining a better approximation of the slope of the tangent line by using secant lines, and ultimately defining the derivative as the limit of the slope of secant lines, with a specific focus on the example where the derivative at x=1.5 is found to be 3.', 'duration': 595.061, 'highlights': ['The derivative, f prime at 1.5, is found to be the limit as x goes to 1.5 of f of x minus f of 1.5 over x minus 1.5, yielding a value of 3 based on the numerical tables.', 'The function is said to be differentiable at a if the limit as x goes to a of f of x minus f of a over x minus a exists, and both the limit from the left and the limit from the right are equal.', 'The equivalent version of the definition of derivative introduces the letter h to represent the run when calculating the slope of the secant line, and defines the derivative as the limit as h goes to 0 of f of a plus h minus f of a over h.']}, {'end': 12409.952, 'start': 12025.208, 'title': 'Derivative limit definition and calculations', 'summary': 'Introduces the idea of derivative as a slope of a tangent line with two equivalent definitions, and demonstrates calculating derivatives using limit definition with examples, emphasizing the algebraic intensity and indeterminate forms.', 'duration': 384.744, 'highlights': ['The chapter introduces the idea of derivative as a slope of a tangent line with two equivalent definitions.', 'Demonstrates calculating derivatives using the limit definition with examples.', 'Emphasizes the algebraic intensity and indeterminate forms encountered when calculating derivatives.']}, {'end': 12975.634, 'start': 12414.235, 'title': 'Interpreting derivatives in different contexts', 'summary': 'Discusses the process of evaluating limits algebraically to compute derivatives and interpreting derivatives in the context of practical quantities such as distance and speed, with a specific example yielding a derivative of 1/16th and a tangent line equation of y=9x-16.', 'duration': 561.399, 'highlights': ['The process of evaluating limits algebraically to compute derivatives is discussed, with an example resulting in a derivative of 1/16th.', 'The example of finding the equation of the tangent line to y=x^3-3x at x=2 is detailed, yielding a tangent line equation of y=9x-16.', 'The practical interpretation of the derivative in the context of distance as a function of time and the interpretation of the slope of the secant line through two points in terms of speed or velocity is presented.']}], 'duration': 3492.499, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w9483135.jpg', 'highlights': ['The range of sine and cosine functions is from negative one to one.', 'The domain of sine and cosine functions is all real numbers.', 'The graph of cosine can be thought of as the graph of sine shifted to the left by pi over two.', 'The output values for the range come from the coordinates on the unit circle.', 'The absolute maximum values of sine and cosine functions is one.', 'The absolute minimum value of sine and cosine functions is negative one.', 'The midline is the horizontal line halfway in between the maximum and minimum points.', 'The amplitude is the vertical distance between a maximum point and the midline.', 'For y equals cosine of x, the period is two pi.', 'The graphs of tangent, secant, cotangent, and cosecant are covered.', 'The derivative, f prime at 1.5, is found to be the limit as x goes to 1.5 of f of x minus f of 1.5 over x minus 1.5, yielding a value of 3.', 'The function is said to be differentiable at a if the limit as x goes to a of f of x minus f of a over x minus a exists.', 'The equivalent version of the definition of derivative introduces the letter h to represent the run when calculating the slope of the secant line.', 'The chapter introduces the idea of derivative as a slope of a tangent line with two equivalent definitions.', 'The process of evaluating limits algebraically to compute derivatives is discussed, with an example resulting in a derivative of 1/16th.', 'The practical interpretation of the derivative in the context of distance as a function of time and the interpretation of the slope of the secant line through two points in terms of speed or velocity is presented.']}, {'end': 15422.147, 'segs': [{'end': 13388.723, 'src': 'embed', 'start': 13355.037, 'weight': 2, 'content': [{'end': 13359.4, 'text': "In this video, we'll think of the derivative of a function as being a function in itself.", 'start': 13355.037, 'duration': 4.363}, {'end': 13366.044, 'text': "We relate the graph of a function to the graph of its derivative, and we'll talk about where the derivative does not exist.", 'start': 13360.34, 'duration': 5.704}, {'end': 13377.971, 'text': "We've seen that for a function f of x and a number a, the derivative of f of x at x equals a is given by this formula.", 'start': 13368.625, 'duration': 9.346}, {'end': 13380.56, 'text': 'But what if we let a vary?', 'start': 13378.939, 'duration': 1.621}, {'end': 13388.723, 'text': 'If we compute f prime of a at lots of different values of a, we can think of the derivative of f prime as itself being a function.', 'start': 13381.4, 'duration': 7.323}], 'summary': 'Derivative of a function related to its graph, exploring where it does not exist.', 'duration': 33.686, 'max_score': 13355.037, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w13355037.jpg'}, {'end': 14091.607, 'src': 'embed', 'start': 14058.004, 'weight': 1, 'content': [{'end': 14062.647, 'text': "Let's look at the function f of x equals x to the 1 third, graphed here.", 'start': 14058.004, 'duration': 4.643}, {'end': 14074.401, 'text': "What's going on at x equals 0? At that instant, the tangent line is a vertical with a slope that's infinite or undefined.", 'start': 14064.437, 'duration': 9.964}, {'end': 14081.743, 'text': "So the limit of the slopes of the secant lines will fail to exist because it'll be infinite.", 'start': 14075.341, 'duration': 6.402}, {'end': 14091.607, 'text': 'A function is called differentiable at x equals a if the derivative exists at a.', 'start': 14084.845, 'duration': 6.762}], 'summary': 'The function f(x) = x^(1/3) is non-differentiable at x=0 due to a vertical tangent line with an undefined slope.', 'duration': 33.603, 'max_score': 14058.004, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w14058004.jpg'}, {'end': 14641.341, 'src': 'embed', 'start': 14615.208, 'weight': 0, 'content': [{'end': 14626.754, 'text': 'But if we rewrite it by putting the cube root in exponential notation as x to the 1 third, now we can apply the power rule.', 'start': 14615.208, 'duration': 11.546}, {'end': 14630.636, 'text': 'We bring the 1 third down and multiply it on the front.', 'start': 14628.055, 'duration': 2.581}, {'end': 14633.578, 'text': 'And we reduce the exponent of 1 third by 1.', 'start': 14631.416, 'duration': 2.162}, {'end': 14635.398, 'text': 'Well, 1 third minus 1 is negative 2 thirds.', 'start': 14633.578, 'duration': 1.82}, {'end': 14641.341, 'text': 'So we found the derivative here using the power rule.', 'start': 14639.079, 'duration': 2.262}], 'summary': 'Derivative found using power rule for cube root function.', 'duration': 26.133, 'max_score': 14615.208, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w14615208.jpg'}, {'end': 15158.469, 'src': 'embed', 'start': 15132.861, 'weight': 4, 'content': [{'end': 15143.844, 'text': 'And we could build evidence for it again by plugging in values for x or by graphing the left side and the right side separately and checking to see that the graphs coincided.', 'start': 15132.861, 'duration': 10.983}, {'end': 15148.826, 'text': "But for this example, I'm going to go ahead and do an algebraic verification.", 'start': 15144.725, 'duration': 4.101}, {'end': 15151.127, 'text': 'In particular,', 'start': 15150.226, 'duration': 0.901}, {'end': 15158.469, 'text': "I'm going to start with the left side of the equation and rewrite things and rewrite things until I get to the right side of the equation.", 'start': 15151.127, 'duration': 7.342}], 'summary': 'Algebraic verification of equation by rewriting left side to match right side.', 'duration': 25.608, 'max_score': 15132.861, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w15132861.jpg'}], 'start': 12977.54, 'title': 'Understanding derivatives and identities', 'summary': 'Explains velocity, derivatives, and differentiability using examples, explores derivative as a function, derivative rules, and pythagorean identities, showcasing their importance in understanding rates of change and proving mathematical identities.', 'chapters': [{'end': 13351.697, 'start': 12977.54, 'title': 'Understanding velocity and derivatives', 'summary': 'Explains the concepts of velocity, instantaneous velocity, and derivatives using examples of temperature change and fuel efficiency, highlighting the importance of secant and tangent lines in understanding average and instantaneous rates of change.', 'duration': 374.157, 'highlights': ['The slope of the secant line represents the average velocity over a time interval, while the derivative at x equals 3 represents the instantaneous velocity, with the example estimating the velocity at approximately negative eight miles per hour.', 'The explanation of how the slope of the tangent line at x equals 3 gives the exact velocity at that instant, emphasizing the distinction between average and instantaneous velocity.', 'The interpretation of the derivative in the context of temperature change and fuel efficiency, demonstrating how negative derivative values indicate decreasing rates and the relevance of average and instantaneous rates of change in different scenarios.']}, {'end': 13969.681, 'start': 13355.037, 'title': 'Derivative as a function', 'summary': 'Explores the concept of the derivative as a function itself, relating the graph of a function to its derivative, and determining where the derivative does not exist, illustrated through examples and the estimation of the derivative based on the shape of the graph.', 'duration': 614.644, 'highlights': ['The derivative of a function is considered as a function itself, and the chapter discusses the relationship between the graph of a function and its derivative.', 'The process of estimating the derivative based on the shape of the graph is demonstrated through examples, showing how to determine where the derivative does not exist.', 'The chapter emphasizes the importance of understanding where the derivative does not exist, as it may not exist at all x values where the original function exists.']}, {'end': 14454.168, 'start': 13969.681, 'title': 'Derivative and differentiability', 'summary': 'Explores the concept of derivative and differentiability, showcasing examples of functions failing to have a derivative at specific points, and proving that differentiable functions are continuous.', 'duration': 484.487, 'highlights': ['Differentiability and Existence of Derivative', 'Derivative and Continuity', 'Proof of Continuity']}, {'end': 15053.2, 'start': 14454.168, 'title': 'Derivative rules and identities', 'summary': 'Covers the power rule, constant multiple rule, sum and difference rule for derivatives, and identities involving trig functions, illustrating with examples and proofs forthcoming.', 'duration': 599.032, 'highlights': ['The power rule states that the derivative of y equals x to the n is n times x to the n-1, illustrated with examples like y equals x to the 15th and the cube root of x.', 'The constant multiple rule allows pulling a constant outside the derivative sign, demonstrated with the derivative of 5x cubed.', 'The sum and difference rule states that the derivative of the sum/difference of differentiable functions is the sum/difference of their derivatives, exemplified with the derivative of a polynomial using these rules.', 'The chapter introduces the concept of identities, distinguishing between equations that hold for all values of the variable and those that do not, with examples involving trig functions.']}, {'end': 15422.147, 'start': 15054.301, 'title': 'Proving pythagorean identities', 'summary': 'Discusses the process of proving identities, showcasing the pythagorean identities for cosine, sine, tangent, secant, cotangent, and cosecant, and demonstrates algebraic and graphical methods for verifying identities.', 'duration': 367.846, 'highlights': ['The chapter discusses the process of proving identities', 'Showcases the Pythagorean identities for cosine, sine, tangent, secant, cotangent, and cosecant', 'Demonstrates algebraic and graphical methods for verifying identities']}], 'duration': 2444.607, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w12977540.jpg', 'highlights': ['The power rule states that the derivative of y equals x to the n is n times x to the n-1, illustrated with examples like y equals x to the 15th and the cube root of x.', 'The sum and difference rule states that the derivative of the sum/difference of differentiable functions is the sum/difference of their derivatives, exemplified with the derivative of a polynomial using these rules.', 'Showcases the Pythagorean identities for cosine, sine, tangent, secant, cotangent, and cosecant.', 'The process of estimating the derivative based on the shape of the graph is demonstrated through examples, showing how to determine where the derivative does not exist.', 'The explanation of how the slope of the tangent line at x equals 3 gives the exact velocity at that instant, emphasizing the distinction between average and instantaneous velocity.']}, {'end': 17586.816, 'segs': [{'end': 15525.399, 'src': 'embed', 'start': 15471.068, 'weight': 1, 'content': [{'end': 15475.97, 'text': "I'm going to divide both sides of this equation by cosine squared theta.", 'start': 15471.068, 'duration': 4.902}, {'end': 15481.091, 'text': "Now I'm going to rewrite the left side by breaking apart the fraction.", 'start': 15477.83, 'duration': 3.261}, {'end': 15489.71, 'text': 'into cosine squared theta over cosine squared theta, plus sine squared theta over cosine squared theta.', 'start': 15482.065, 'duration': 7.645}, {'end': 15495.014, 'text': 'Now cosine squared theta over cosine squared theta is just one.', 'start': 15491.231, 'duration': 3.783}, {'end': 15502.519, 'text': 'And I can rewrite the next fraction as sine of theta over cosine of theta squared.', 'start': 15495.034, 'duration': 7.485}, {'end': 15510.654, 'text': "That's because when I square a fraction, I can just square the numerator and square the denominator,", 'start': 15503.991, 'duration': 6.663}, {'end': 15514.555, 'text': 'and sine squared theta is shorthand for sine of theta squared.', 'start': 15510.654, 'duration': 3.901}, {'end': 15516.336, 'text': 'similarly for cosine squared theta.', 'start': 15514.555, 'duration': 1.781}, {'end': 15525.399, 'text': 'Now on the other side of the equal sign, I can rewrite this fraction as 1 over cosine theta squared.', 'start': 15518.236, 'duration': 7.163}], 'summary': 'Dividing equation by cosine squared theta and rewriting fractions simplifies the expression.', 'duration': 54.331, 'max_score': 15471.068, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w15471068.jpg'}, {'end': 17406.598, 'src': 'embed', 'start': 17378.854, 'weight': 0, 'content': [{'end': 17383.46, 'text': "That's the binomial expansion of x plus h to the n.", 'start': 17378.854, 'duration': 4.606}, {'end': 17388.476, 'text': 'Now we still have to subtract the x to the n that we had up here.', 'start': 17383.46, 'duration': 5.016}, {'end': 17392.801, 'text': 'And we still have to divide this whole thing by h.', 'start': 17389.076, 'duration': 3.725}, {'end': 17395.384, 'text': "Okay, that's looking kind of horribly complicated.", 'start': 17392.801, 'duration': 2.583}, {'end': 17400.51, 'text': "But notice that the x to the n's cancel.", 'start': 17396.205, 'duration': 4.305}, {'end': 17406.598, 'text': 'Notice that all of the remaining terms have an h in them.', 'start': 17401.452, 'duration': 5.146}], 'summary': 'Binomial expansion: cancel x to the n, remaining terms have h.', 'duration': 27.744, 'max_score': 17378.854, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w17378854.jpg'}], 'start': 15423.536, 'title': 'Trigonometric identities and derivatives', 'summary': 'Covers proving pythagorean identities and sum and difference formulas for trigonometric functions, along with double angle formulas and derivatives. it also explains the derivative of e to the x and provides proofs of derivative rules, emphasizing visual and algebraic explanations.', 'chapters': [{'end': 15650.629, 'start': 15423.536, 'title': 'Proving trigonometric identities', 'summary': 'Explains the proof of three pythagorean identities and the sum and difference formulas using algebra and unit circle, involving cosine, sine, tangent, and secant.', 'duration': 227.093, 'highlights': ['The chapter demonstrates the proof of three Pythagorean identities using algebra and unit circle, involving cosine, sine, tangent, and secant.', 'The sum and difference formulas for computing the sine and cosine of two angles are explained in the chapter.', 'Using the first Pythagorean identity, the proof of the second and third identities is demonstrated through algebraic manipulation and rearrangement of equations.']}, {'end': 16103.526, 'start': 15652.61, 'title': 'Angle sum and difference formulas', 'summary': 'Explains the angle sum and difference formulas for sine and cosine, illustrating with examples of finding the sine of 105 degrees and the cosine of v plus w, and emphasizing the process of computing values through right triangles and the formulas.', 'duration': 450.916, 'highlights': ['The sine of the sum of two angles A plus B is given by sine of A cosine of B plus cosine of A sine of B.', 'The cosine of the sum of two angles A plus B is given by cosine A cosine B minus sine A sine B.', 'Illustrated the process of finding the sine of 105 degrees using the angle sum formula and values from the unit circle, resulting in the expression root 6 plus root 2 over 4.', 'Demonstrated the computation of the cosine of v plus w by using the given cosine values, finding the sine values through right triangles, and applying the angle sum formula, resulting in a decimal approximation of 0.3187.']}, {'end': 16705.946, 'start': 16104.895, 'title': 'Double angle formulas and higher order derivatives', 'summary': 'Explains the double angle formulas for sine and cosine, providing visual and algebraic explanations, and demonstrates the use of these formulas in solving equations. additionally, it introduces higher order derivatives and various notations for them.', 'duration': 601.051, 'highlights': ['The video explains the visual and algebraic reasoning for the double angle formulas for sine and cosine, demonstrating why the equation sine of 2 theta equals 2 sine theta is false and providing the correct formula as sine of 2 theta is 2 sine theta cosine theta.', 'It provides multiple versions of the double angle formula for cosine of 2 theta, including cosine squared theta minus sine squared theta, one minus 2 sine squared theta, and two cosine squared theta minus one.', 'The chapter demonstrates the application of the double angle formulas in solving equations, such as finding cosine of 2 theta given a value for cosine theta and using the double angle formula to rewrite sine of 2x in solving an equation.', 'It introduces higher order derivatives and various notations for them, including f double prime of x for the second derivative, f triple prime of x for the third derivative, and f parentheses n of x for the nth derivative.']}, {'end': 17067.876, 'start': 16705.946, 'title': 'Derivative of e to the x', 'summary': 'Covers the derivative of e to the x, including its graph, three important facts about the function e, and its relation to the limit definition of derivative.', 'duration': 361.93, 'highlights': ['The derivative of e to the x at any x is e to the x.', 'The limit as h goes to 0 of e to the h minus 1 over h equals 1.', 'The limit as n goes to infinity of 1 plus 1 over n raised to the nth power exists and equals E.', 'The graph of y equals e to the x is an increasing exponential function, with a slope of approximately 1 at x equals 0.']}, {'end': 17586.816, 'start': 17070.798, 'title': 'Derivative of e to the x and proof of rules', 'summary': "Covers the computation of the derivative of a function involving e's and x's, the happy fact that the derivative of e to the x is e to the x, and the proof of rules for taking derivatives such as the constant rule and the power rule.", 'duration': 516.018, 'highlights': ['The derivative of e to the x is e to the x', 'Proof of constant rule and limit definition of derivative', 'Proof of power rule using binomial expansion and limit definition of derivative']}], 'duration': 2163.28, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w15423536.jpg', 'highlights': ['The chapter demonstrates the proof of three Pythagorean identities using algebra and unit circle, involving cosine, sine, tangent, and secant.', 'The derivative of e to the x at any x is e to the x.', 'The sum and difference formulas for computing the sine and cosine of two angles are explained in the chapter.', 'The video explains the visual and algebraic reasoning for the double angle formulas for sine and cosine, demonstrating why the equation sine of 2 theta equals 2 sine theta is false and providing the correct formula as sine of 2 theta is 2 sine theta cosine theta.', 'It introduces higher order derivatives and various notations for them, including f double prime of x for the second derivative, f triple prime of x for the third derivative, and f parentheses n of x for the nth derivative.']}, {'end': 18979.21, 'segs': [{'end': 17617.967, 'src': 'embed', 'start': 17589.378, 'weight': 3, 'content': [{'end': 17597.843, 'text': "Next I'll prove the constant multiple rule that says that if c is a real number constant and f is a differentiable function,", 'start': 17589.378, 'duration': 8.465}, {'end': 17601.885, 'text': 'then the derivative of a constant times f is just the constant times the derivative of f.', 'start': 17597.843, 'duration': 4.042}, {'end': 17605.657, 'text': 'Starting with the limit definition of derivative.', 'start': 17602.975, 'duration': 2.682}, {'end': 17606.097, 'text': 'I have that.', 'start': 17605.657, 'duration': 0.44}, {'end': 17617.967, 'text': 'the derivative of c times f of x is the limit, as h goes to 0, of c times f of x plus h, minus c times f of x over h.', 'start': 17606.097, 'duration': 11.87}], 'summary': "Proving constant multiple rule: derivative of c*f is c*f'(x).", 'duration': 28.589, 'max_score': 17589.378, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w17589378.jpg'}, {'end': 18045.878, 'src': 'embed', 'start': 18003.359, 'weight': 0, 'content': [{'end': 18009.421, 'text': 'On the denominator, we have the denominator function g of x squared.', 'start': 18003.359, 'duration': 6.062}, {'end': 18027.749, 'text': 'And on the numerator, we have g of x times the derivative of f of x minus f of x times the derivative of g of x.', 'start': 18010.442, 'duration': 17.307}, {'end': 18033.768, 'text': 'The way I remember this is this chant.', 'start': 18029.724, 'duration': 4.044}, {'end': 18045.878, 'text': 'If you think of f of x as the high function and g of x as the low function, you can say this is low, d, high, minus, high, d, low over low, low,', 'start': 18035.749, 'duration': 10.129}], 'summary': 'Formula for the quotient rule: (low * d(high) - high * d(low)) / (low)^2', 'duration': 42.519, 'max_score': 18003.359, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w18003359.jpg'}, {'end': 18552.078, 'src': 'embed', 'start': 18516.837, 'weight': 1, 'content': [{'end': 18521.698, 'text': 'So, instead of going back to the definition of derivative this time,', 'start': 18516.837, 'duration': 4.861}, {'end': 18535.122, 'text': "I'm just going to think of the quotient f of x over g of x as a product of f of x times the reciprocal of g of x.", 'start': 18521.698, 'duration': 13.424}, {'end': 18552.078, 'text': "And now by the product rule, That's just the first function times the derivative of the second plus the derivative of the first times the second.", 'start': 18535.122, 'duration': 16.956}], 'summary': 'Explains the quotient rule as a product of functions and applies the product rule to find the derivative.', 'duration': 35.241, 'max_score': 18516.837, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w18516837.jpg'}], 'start': 17589.378, 'title': 'Derivative rules and proofs', 'summary': 'Explains derivative rules including constant multiple, sum, difference, product, and quotient rules, providing proofs and examples. it covers the importance of these rules in calculus and introduces intuitive and rigorous approaches to calculate trigonometric limits with practical applications.', 'chapters': [{'end': 18045.878, 'start': 17589.378, 'title': 'Derivative rules and examples', 'summary': 'Explains the constant multiple rule, the sum and difference rules, and the product and quotient rules for calculating derivatives of functions, providing proofs and examples. it also discusses the product rule, providing examples and a proof, and the quotient rule, providing examples and a mnemonic to remember the rule.', 'duration': 456.5, 'highlights': ['The chapter explains the constant multiple rule, the sum and difference rules, and the product and quotient rules for calculating derivatives of functions, providing proofs and examples.', 'It also discusses the product rule, providing examples and a proof, and the quotient rule, providing examples and a mnemonic to remember the rule.', 'The derivative of the product f of x times g of x is equal to f of x times, the derivative of g of x, plus the derivative of f of x times g of x.']}, {'end': 18241.655, 'start': 18045.878, 'title': 'Derivative rules and proofs', 'summary': 'Covers the proof of the product rule, quotient rule, and the reciprocal rule, along with examples of applying the quotient rule and product rule to find derivatives, showcasing the importance of these rules in calculus.', 'duration': 195.777, 'highlights': ['The video covers the proof of the product rule, quotient rule, and the reciprocal rule in calculus.', 'Demonstration of applying the quotient rule to find the derivative of a function, resulting in 2z - z^4 / (z^3 + 1)^2.', 'Explanation of the classic trick of adding 0 to an expression to facilitate the proof of the product rule.', 'Use of algebra of fractions to show the equivalence of two expressions in the proof of the product rule.']}, {'end': 18515.936, 'start': 18242.275, 'title': 'Derivative rules and proofs', 'summary': 'Introduces the proof for the reciprocal rule, stating that the derivative of 1 over f of x is given by negative the derivative of f of x divided by f of x squared, and sets the stage for proving the quotient rule.', 'duration': 273.661, 'highlights': ["The first limit here is just equal to g of x because g is a continuous function. g is continuous because, by assumption, it's differentiable. This sets the foundation for the continuity and differentiability of g.", 'The second limit here is the definition of the derivative of f, so that limit exists and equals d dx of f of x. This highlights the definition of the derivative of f and its existence.', 'The third limit, well, f of x has nothing to do with h, so that limit is just f of x. This emphasizes the independence of f of x from h in the context of the limit.', "The fourth limit is the derivative of g, and with some manipulation, it's shown to be exactly the same as a rearranged expression, paving the way for understanding derivative relationships and manipulations.", 'The chapter lays the groundwork for proving the reciprocal rule, demonstrating step-by-step the derivation of the derivative of 1 over f of x as negative the derivative of f of x divided by f of x squared. This proof sets the stage for understanding the reciprocal rule and its application in derivatives.']}, {'end': 18979.21, 'start': 18516.837, 'title': 'Derivative rules and trig limits', 'summary': 'Discusses the proof of the quotient rule and provides intuitive and rigorous approaches to calculate limits involving trig functions, showing the limits as theta goes to 0 of sine theta over theta and cosine theta minus 1 over theta, which are both equal to 1, and demonstrates their practical applications in approximating sine values and evaluating complex limits.', 'duration': 462.373, 'highlights': ['The limit as theta goes to 0 of sine theta over theta and cosine theta minus 1 over theta are both equal to 1.', 'The chapter provides intuitive and rigorous approaches to calculate limits involving trig functions.', 'The video gives proofs of the product rule, the reciprocal rule, and the quotient rule.']}], 'duration': 1389.832, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w17589378.jpg', 'highlights': ['The chapter explains the constant multiple rule, sum and difference rules, and product and quotient rules for calculating derivatives of functions, providing proofs and examples.', 'The video covers the proof of the product rule, quotient rule, and the reciprocal rule in calculus.', 'The chapter provides intuitive and rigorous approaches to calculate limits involving trig functions.', "The first limit here is just equal to g of x because g is a continuous function. g is continuous because, by assumption, it's differentiable. This sets the foundation for the continuity and differentiability of g."]}, {'end': 21232.145, 'segs': [{'end': 19015.727, 'src': 'embed', 'start': 18980.231, 'weight': 0, 'content': [{'end': 18986.513, 'text': "And I'm still left with a 7x from the top and a 4x from the bottom.", 'start': 18980.231, 'duration': 6.282}, {'end': 18989.714, 'text': "Here I can cancel out those x's.", 'start': 18987.273, 'duration': 2.441}, {'end': 18998.814, 'text': 'And I can notice that this limit here, as x goes to 0, 7x is going to 0.', 'start': 18990.734, 'duration': 8.08}, {'end': 19003.718, 'text': 'So sine 7x over 7x is just going to be equal to 1.', 'start': 18998.814, 'duration': 4.904}, {'end': 19008.522, 'text': 'And similarly, as x goes to 0, 4x is going to 0.', 'start': 19003.718, 'duration': 4.804}, {'end': 19013.866, 'text': 'So the limit of 4x over sine 4x is the reciprocal of 1.', 'start': 19008.522, 'duration': 5.344}, {'end': 19015.727, 'text': "It's also 1.", 'start': 19013.866, 'duration': 1.861}], 'summary': 'As x approaches 0, the limit of sine 7x/7x equals 1, and the limit of 4x/sine 4x also equals 1.', 'duration': 35.496, 'max_score': 18980.231, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w18980231.jpg'}, {'end': 20005.353, 'src': 'embed', 'start': 19978.107, 'weight': 2, 'content': [{'end': 19981.91, 'text': 'I can simplify the denominator here.', 'start': 19978.107, 'duration': 3.803}, {'end': 19983.051, 'text': 'Negative 3 fourths plus 3.', 'start': 19982.01, 'duration': 1.041}, {'end': 19984.892, 'text': '3 is 12 fourths, so that becomes 9 fourths.', 'start': 19983.051, 'duration': 1.841}, {'end': 19992.688, 'text': "And this is one I'll flip and multiply to get minus four thirds.", 'start': 19987.666, 'duration': 5.022}, {'end': 20002.852, 'text': 'So here I can simplify my complex fraction, it ends up being negative three ninths and one minus four thirds is negative one third.', 'start': 19993.368, 'duration': 9.484}, {'end': 20005.353, 'text': 'So that all seems to check out.', 'start': 20002.932, 'duration': 2.421}], 'summary': 'Simplified complex fraction to -3/9, 1 - 4/3 = -1/3.', 'duration': 27.246, 'max_score': 19978.107, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w19978107.jpg'}, {'end': 20435.268, 'src': 'embed', 'start': 20408.178, 'weight': 3, 'content': [{'end': 20411.799, 'text': 'A graph of the function y equals sine x is given in blue here.', 'start': 20408.178, 'duration': 3.621}, {'end': 20418.162, 'text': 'We can estimate the shape of the derivative of sine x by looking at the slopes of the tangent lines.', 'start': 20412.6, 'duration': 5.562}, {'end': 20426.926, 'text': 'Here, when x equals 0, the tangent line has a positive slope of approximately 1.', 'start': 20420.243, 'duration': 6.683}, {'end': 20435.268, 'text': 'As x increases to pi over two, the slope of the tangent line is still positive but decreases to zero.', 'start': 20426.926, 'duration': 8.342}], 'summary': 'Estimate derivative shape of y = sin(x), tangent slope peaks at x=0 with positive slope of 1 then decreases to 0 at x=pi/2.', 'duration': 27.09, 'max_score': 20408.178, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w20408178.jpg'}, {'end': 21004.397, 'src': 'heatmap', 'start': 20661.286, 'weight': 4, 'content': [{'end': 20671.914, 'text': 'And notice that a nice way to remember which of these answers have negative signs in them is that the derivatives of the trig functions that start with a co always have a negative,', 'start': 20661.286, 'duration': 10.628}, {'end': 20675.436, 'text': "and the derivative of the trig functions that don't have the co are positive.", 'start': 20671.914, 'duration': 3.522}, {'end': 20678.779, 'text': "Now let's use these formulas in an example.", 'start': 20676.437, 'duration': 2.342}, {'end': 20686.015, 'text': 'g of x is a complicated expression involving several trig functions as well as a constant, m.', 'start': 20680.731, 'duration': 5.284}, {'end': 20687.837, 'text': 'And I have a couple choices of how to proceed.', 'start': 20686.015, 'duration': 1.822}, {'end': 20693.741, 'text': 'I could try to rewrite all my trig functions in terms of sine and cosine and simplify,', 'start': 20688.477, 'duration': 5.264}, {'end': 20696.784, 'text': 'or I could attack the derivative directly using the quotient rule.', 'start': 20693.741, 'duration': 3.043}, {'end': 20703.309, 'text': "I'm going to use the direct approach in this case, but sometimes you'll find that rewriting will make things easier.", 'start': 20697.524, 'duration': 5.785}, {'end': 20710.841, 'text': 'So using the quotient rule, On the denominator, I get the original denominator squared.', 'start': 20704.71, 'duration': 6.131}, {'end': 20719.068, 'text': 'On the numerator, I get low d high.', 'start': 20712.983, 'duration': 6.085}, {'end': 20722.771, 'text': 'To compute the derivative of x cosine x, I need the product rule.', 'start': 20719.448, 'duration': 3.323}, {'end': 20735.04, 'text': 'So I get x times the derivative of cosine, which is negative sine x, plus the derivative of x, which is just 1 times cosine of x.', 'start': 20723.171, 'duration': 11.869}, {'end': 20736.401, 'text': 'Now I have to do minus.', 'start': 20735.04, 'duration': 1.361}, {'end': 20741.685, 'text': 'pi x cosine of x d low.', 'start': 20737.401, 'duration': 4.284}, {'end': 20751.494, 'text': 'the derivative of m is just zero because m is a constant, plus the derivative of cotangent, which is negative.', 'start': 20741.685, 'duration': 9.809}, {'end': 20756.058, 'text': 'cosecant squared of x.', 'start': 20751.494, 'duration': 4.564}, {'end': 20760.322, 'text': "So I found the derivative, I'm going to go ahead and simplify a little bit by multiplying out.", 'start': 20756.058, 'duration': 4.264}, {'end': 20771.004, 'text': 'then rewriting everything in terms of sine and cosine, and then multiplying the numerator and denominator by sine squared of x.', 'start': 20761.536, 'duration': 9.468}, {'end': 20773.526, 'text': 'We have a somewhat simplified expression for the derivative.', 'start': 20771.004, 'duration': 2.522}, {'end': 20777.409, 'text': 'You should memorize the derivatives of the trig functions.', 'start': 20775.287, 'duration': 2.122}, {'end': 20782.613, 'text': "We'll prove that the first two formulas are correct in a separate proof video.", 'start': 20778.489, 'duration': 4.124}, {'end': 20787.897, 'text': "In this video, I'll give proofs for the two special trig limits.", 'start': 20783.994, 'duration': 3.903}, {'end': 20795.917, 'text': "And I'll also prove that the derivative of sine is cosine and the derivative of cosine is minus sine.", 'start': 20789.093, 'duration': 6.824}, {'end': 20804.583, 'text': "To prove that the limit of sine theta over theta is 1 as theta goes to 0, I'm going to start with a picture.", 'start': 20798.439, 'duration': 6.144}, {'end': 20815.249, 'text': 'In this picture I have a unit circle, a circle of radius 1, and I have two right triangles, a green triangle and a smaller red triangle,', 'start': 20805.583, 'duration': 9.666}, {'end': 20816.35, 'text': 'both with angle theta.', 'start': 20815.249, 'duration': 1.101}, {'end': 20820.527, 'text': "Now I'm going to argue in terms of areas.", 'start': 20818.564, 'duration': 1.963}, {'end': 20830.681, 'text': "If I want to compute the area of this sector that I've shaded in blue here, in other words that pie shaped piece,", 'start': 20821.248, 'duration': 9.433}, {'end': 20833.165, 'text': 'I can first compute the area of the circle.', 'start': 20830.681, 'duration': 2.484}, {'end': 20836.657, 'text': 'which is pi times 1 squared for the radius.', 'start': 20833.715, 'duration': 2.942}, {'end': 20843.481, 'text': 'But since the sector has angle theta and the full circle has angle 2 pi,', 'start': 20837.478, 'duration': 6.003}, {'end': 20855.009, 'text': "I need to multiply that area of the circle by the ratio theta over 2 pi to represent the fraction of the area of the circle that's included in this sector.", 'start': 20843.481, 'duration': 11.528}, {'end': 20861.829, 'text': 'So in other words, the area of the sector is just going to be theta over 2.', 'start': 20855.589, 'duration': 6.24}, {'end': 20863.59, 'text': 'where theta is given in radians.', 'start': 20861.829, 'duration': 1.761}, {'end': 20871.894, 'text': 'Now if I want to compute the area of the little red triangle, I can do 1 half times the base times the height.', 'start': 20864.59, 'duration': 7.304}, {'end': 20882.119, 'text': 'Now the base is going to be equal to cosine theta because I have a circle of radius 1 and angle theta here.', 'start': 20872.574, 'duration': 9.545}, {'end': 20886.641, 'text': 'And the height is going to be sine theta.', 'start': 20883.039, 'duration': 3.602}, {'end': 20894.59, 'text': 'Finally, the area of the green triangle is also 1 half times the base times the height.', 'start': 20888.902, 'duration': 5.688}, {'end': 20909.767, 'text': 'But now the base is a full 1 unit and the height is given by tangent theta, since opposite, which is the height here over adjacent, which is 1,', 'start': 20895.331, 'duration': 14.436}, {'end': 20911.109, 'text': 'has to equal tangent theta.', 'start': 20909.767, 'duration': 1.342}, {'end': 20923.434, 'text': "Now, if I put all those areas together, I know that the area of the red triangle I'll write it as cosine theta sine theta over 2,", 'start': 20914.15, 'duration': 9.284}, {'end': 20933.457, 'text': 'has to be less than or equal to the area of the blue sector, theta over 2, which is less than or equal to the area of the big green triangle,', 'start': 20923.434, 'duration': 10.023}, {'end': 20936.839, 'text': 'which is tan theta over 2..', 'start': 20933.457, 'duration': 3.382}, {'end': 20945.753, 'text': "Now I'm going to multiply through this inequality by 2, and rewrite things in terms of sine and cosine to get cosine theta.", 'start': 20936.839, 'duration': 8.914}, {'end': 20952.336, 'text': 'sine theta is less than or equal to theta is less than or equal to sine theta over cosine theta.', 'start': 20945.753, 'duration': 6.583}, {'end': 20962.121, 'text': "Now I'm going to divide through my inequalities by sine theta, which won't change the inequalities as long as theta is greater than 0,", 'start': 20953.937, 'duration': 8.184}, {'end': 20963.742, 'text': 'so that sine theta is positive.', 'start': 20962.121, 'duration': 1.621}, {'end': 20973.241, 'text': 'And I get cosine theta is less than or equal to theta over sine theta is less than or equal to 1 over cosine theta.', 'start': 20966.038, 'duration': 7.203}, {'end': 20979.764, 'text': 'Now this middle expression is the reciprocal of the expression I want to take the limit of.', 'start': 20976.102, 'duration': 3.662}, {'end': 20981.525, 'text': "So I'm going to go ahead and take limits.", 'start': 20980.224, 'duration': 1.301}, {'end': 20993.069, 'text': 'And since the limits of the two expressions on the outside both exist and equal 1, by the Sandwich Theorem,', 'start': 20983.085, 'duration': 9.984}, {'end': 20996.971, 'text': 'the limit of the expression on the inside has to exist and equal 1 as well.', 'start': 20993.069, 'duration': 3.902}, {'end': 21004.397, 'text': "Now I've cheated a little bit here and I've really just taken the limit from the right, because I've assumed that theta is greater than 0,", 'start': 20998.693, 'duration': 5.704}], 'summary': 'Derivatives of trig functions explained with proofs and examples.', 'duration': 24.173, 'max_score': 20661.286, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w20661286.jpg'}, {'end': 21232.145, 'src': 'embed', 'start': 21197.896, 'weight': 1, 'content': [{'end': 21200.777, 'text': 'So please stop the video and try it for yourself before proceeding.', 'start': 21197.896, 'duration': 2.881}, {'end': 21205.719, 'text': 'Using the limit definition of derivative.', 'start': 21203.158, 'duration': 2.561}, {'end': 21206.219, 'text': 'we have that.', 'start': 21205.719, 'duration': 0.5}, {'end': 21216.663, 'text': 'the derivative of cosine of x is the limit, as h goes to zero, of cosine of x plus h minus cosine of x over h.', 'start': 21206.219, 'duration': 10.444}, {'end': 21232.145, 'text': 'We can rewrite the cosine of x plus h using the angle sum formula as the cosine of x times the cosine of h minus the sine of x times the sine of h.', 'start': 21219.332, 'duration': 12.813}], 'summary': 'Derivative of cosine x is limit of (cos(x+h)-cosx)/h using angle sum formula', 'duration': 34.249, 'max_score': 21197.896, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w21197896.jpg'}], 'start': 18980.231, 'title': 'Function composition and trig derivatives', 'summary': 'Covers function composition, including limits of sine and cosine functions, with key findings that the limit of sine theta over theta is 1 and the limit of cosine theta minus one over theta is 0. it demonstrates that g composed with f is not the same as f composed with g, and provides examples of finding derivatives of trigonometric functions.', 'chapters': [{'end': 19188.851, 'start': 18980.231, 'title': 'Limits and composition of functions', 'summary': 'Covers limits of sine and cosine functions, with the key findings that the limit of sine theta over theta is equal to 1 and the limit of cosine theta minus one over theta is equal to 0. it also delves into the composition of functions and provides examples of composition using tables of values.', 'duration': 208.62, 'highlights': ['The chapter covers limits of sine and cosine functions, with the key findings that the limit of sine theta over theta is equal to 1 and the limit of cosine theta minus one over theta is equal to 0.', 'The composition of functions and provides examples of composition using tables of values.', 'The process of composition of functions and how it relates to evaluating expressions using tables of values.']}, {'end': 19819.892, 'start': 19190.242, 'title': 'Function composition and rational equations', 'summary': 'Covers the concept of function composition, demonstrating that g composed with f is not the same as f composed with g, and provides examples of evaluating the composition of functions, as well as breaking a complicated function into a composition of two functions. additionally, the video briefly mentions solving rational equations by finding the least common denominator.', 'duration': 629.65, 'highlights': ['The chapter emphasizes that g composed with f is not necessarily equal to f composed with g, as demonstrated through examples such as g composed with f of four and f composed with g of four.', 'The process of evaluating the composition of functions is illustrated through examples, including finding f composed with g of six and q composed with p of one, providing quantifiable data on the values obtained through these evaluations.', 'The method of breaking a complicated function into a composition of two functions is explained, with examples of finding h as a composition of two functions, demonstrating that multiple correct solutions are possible.', 'The video briefly touches on solving rational equations, mentioning the importance of finding the least common denominator as a starting point for solving such equations.']}, {'end': 20357.46, 'start': 19819.892, 'title': 'Solving rational equations', 'summary': 'Explains the process of solving rational equations, demonstrating the method of finding the least common denominator and then clearing the denominator, and also discusses an alternative method of writing all the fractions over the least common denominator, with examples showing the application of the methods.', 'duration': 537.568, 'highlights': ['The chapter demonstrates the method of finding the least common denominator and then clearing the denominator to solve rational equations.', 'The chapter also discusses an alternative method of solving rational equations by writing all the fractions over the least common denominator.', 'The chapter emphasizes the importance of checking solutions for extraneous solutions in rational equations.']}, {'end': 20863.59, 'start': 20357.46, 'title': 'Derivatives of trig functions', 'summary': 'Covers finding derivatives of trigonometric functions, providing graphical evidence for the derivatives of sine and cosine, using the quotient rule to find the derivative of a complex expression involving trig functions, and proving special trig limits.', 'duration': 506.13, 'highlights': ['The chapter covers finding derivatives of trigonometric functions.', 'Providing graphical evidence for the derivatives of sine and cosine.', 'Using the quotient rule to find the derivative of a complex expression involving trig functions.', 'Proving special trig limits.']}, {'end': 21232.145, 'start': 20864.59, 'title': 'Geometric proof and calculus limits', 'summary': 'Discusses the geometric proof of a useful limit from calculus, and the derivation of the derivatives of sine and cosine using the limit definition of derivatives, resulting in the limit of cosine theta minus 1 over theta being 0, and the derivatives of sine and cosine being cosine x and -sine x, respectively.', 'duration': 367.555, 'highlights': ['The chapter discusses the geometric proof of a useful limit from calculus', 'The derivation of the derivatives of sine and cosine using the limit definition of derivatives', 'Resulting in the limit of cosine theta minus 1 over theta being 0']}], 'duration': 2251.914, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w18980231.jpg', 'highlights': ['The chapter covers finding derivatives of trigonometric functions.', 'The chapter covers limits of sine and cosine functions, with the key findings that the limit of sine theta over theta is equal to 1 and the limit of cosine theta minus one over theta is equal to 0.', 'The chapter emphasizes that g composed with f is not necessarily equal to f composed with g, as demonstrated through examples such as g composed with f of four and f composed with g of four.', 'The chapter demonstrates the method of finding the least common denominator and then clearing the denominator to solve rational equations.', 'The chapter discusses the geometric proof of a useful limit from calculus']}, {'end': 23403.518, 'segs': [{'end': 21261.603, 'src': 'embed', 'start': 21232.145, 'weight': 1, 'content': [{'end': 21234.668, 'text': 'And then we still have the minus cosine of x over h.', 'start': 21232.145, 'duration': 2.523}, {'end': 21238.014, 'text': 'As before.', 'start': 21236.773, 'duration': 1.241}, {'end': 21247.978, 'text': "we're going to regroup things and factoring out the cosine x from the first part the same familiar limits, just put together in different ways.", 'start': 21238.014, 'duration': 9.964}, {'end': 21253.36, 'text': 'So here cosine of x, as h goes to zero, is just cosine of x.', 'start': 21248.638, 'duration': 4.722}, {'end': 21255.961, 'text': 'this limit, we know, is zero.', 'start': 21253.36, 'duration': 2.601}, {'end': 21261.603, 'text': 'sine of x is just staying sine of x and sine of h over h is going to one.', 'start': 21255.961, 'duration': 5.642}], 'summary': 'In the given transcript, the limit of sine of h over h equals one as h goes to zero, and the limit of cosine x as h goes to zero equals cosine x.', 'duration': 29.458, 'max_score': 21232.145, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w21232145.jpg'}, {'end': 21758.633, 'src': 'embed', 'start': 21734.918, 'weight': 8, 'content': [{'end': 21741.302, 'text': "just based on the shapes of the graphs, and where they're increasing, where they're decreasing, where they're positive and where they're negative.", 'start': 21734.918, 'duration': 6.384}, {'end': 21745.904, 'text': 'Velocity is the derivative of position.', 'start': 21743.683, 'duration': 2.221}, {'end': 21749.946, 'text': 'So velocity needs to be positive where position is increasing.', 'start': 21746.344, 'duration': 3.602}, {'end': 21758.633, 'text': "The only pairs of functions that have this property are the blue one, that's positive when the red one's increasing, and the green function,", 'start': 21750.489, 'duration': 8.144}], 'summary': 'Velocity is positive when position is increasing, demonstrated by the blue and green functions.', 'duration': 23.715, 'max_score': 21734.918, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w21734918.jpg'}, {'end': 22394.535, 'src': 'embed', 'start': 22366.694, 'weight': 9, 'content': [{'end': 22372.38, 'text': "And the linear function would mean that cost per t shirt is constant, no matter how many t shirts you're making.", 'start': 22366.694, 'duration': 5.686}, {'end': 22379.567, 'text': "But in reality, it's probably going to be cheaper to make 1, 000 t-shirts than it is to make just a few t-shirts.", 'start': 22373.544, 'duration': 6.023}, {'end': 22386.651, 'text': 'And therefore, the cost per t-shirt or slope should be going down as x increases.', 'start': 22380.107, 'duration': 6.544}, {'end': 22394.535, 'text': "So this function right here is the one whose slope is going down for larger x's.", 'start': 22387.331, 'duration': 7.204}], 'summary': 'Cost per t-shirt decreases as the number of t-shirts made increases.', 'duration': 27.841, 'max_score': 22366.694, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w22366694.jpg'}, {'end': 22601.718, 'src': 'embed', 'start': 22575.218, 'weight': 0, 'content': [{'end': 22584.182, 'text': 'since C prime of 400, the derivative is approximately equal to the average rate of change, going from 400 to 401,', 'start': 22575.218, 'duration': 8.964}, {'end': 22592.136, 'text': 'which is just this difference divided by 1..', 'start': 22584.182, 'duration': 7.954}, {'end': 22601.718, 'text': 'Once again, C prime of 400 is called the marginal cost and represents the rate at which the cost function is increasing with each additional item.', 'start': 22592.136, 'duration': 9.582}], 'summary': "C' of 400 represents the marginal cost, increasing the cost function with each item.", 'duration': 26.5, 'max_score': 22575.218, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w22575218.jpg'}, {'end': 22676.825, 'src': 'embed', 'start': 22640.164, 'weight': 2, 'content': [{'end': 22643.605, 'text': "It's also called the base when we write it in this exponential form.", 'start': 22640.164, 'duration': 3.441}, {'end': 22655.351, 'text': 'Some students find it helpful to remember this relationship log base A of B equals C means A to the C equals B by drawing arrows.', 'start': 22645.123, 'duration': 10.228}, {'end': 22659.594, 'text': 'A to the C equals B.', 'start': 22656.071, 'duration': 3.523}, {'end': 22662.296, 'text': 'Other students like to think of it in terms of asking a question.', 'start': 22659.594, 'duration': 2.702}, {'end': 22676.825, 'text': "Log base A of B asks, what power do you raise A to in order to get B? Let's look at some examples.", 'start': 22663.157, 'duration': 13.668}], 'summary': 'Understanding logarithms: log base a of b equals c means a to the c equals b.', 'duration': 36.661, 'max_score': 22640.164, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w22640164.jpg'}, {'end': 22784.395, 'src': 'embed', 'start': 22751.364, 'weight': 3, 'content': [{'end': 22760.92, 'text': 'Again, since 1 eighth is 1 over 2 cubed, we have to raise 2 to the negative 3 power to get 1 over 2 cubed.', 'start': 22751.364, 'duration': 9.556}, {'end': 22767.503, 'text': "So our exponent is negative 3, and that's our answer to our log expression.", 'start': 22762.981, 'duration': 4.522}, {'end': 22784.395, 'text': 'Finally, log base 2 of 1 is asking 2 to what power equals 1? Well, anything raised to the 0 power gives us 1, so this log expression evaluates to 0.', 'start': 22769.024, 'duration': 15.371}], 'summary': 'Log base 2 of 1 equals 0, as 2^0 equals 1.', 'duration': 33.031, 'max_score': 22751.364, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w22751364.jpg'}], 'start': 21232.145, 'title': 'Calculus and logarithmic functions', 'summary': 'Covers derivatives and motion in calculus, analyzing particle motion and introducing logarithmic functions. it discusses topics such as velocity, acceleration, marginal cost, and graphing log functions, providing insights into their applications and properties.', 'chapters': [{'end': 21494.339, 'start': 21232.145, 'title': 'Derivatives and motion in calculus', 'summary': 'Discusses the derivatives representing velocity and acceleration and their impacts on the motion of an object constrained to move along a straight line, using an example of particle motion. it also explains the relationship between force and acceleration in the context of calculus.', 'duration': 262.194, 'highlights': ['The derivative and the second derivative tell us about the motion of an object constrained to move along a straight line, represented by position, velocity, and acceleration functions.', 'The concept of force in relation to acceleration is explained, stating that force equals mass times acceleration, and the impact of positive and negative acceleration on the motion of the particle.', 'The impact of velocity and acceleration on the motion of the particle at a specific time is described using a table of values, showcasing the relationship between position, velocity, and the direction of motion.']}, {'end': 22500.291, 'start': 21495.879, 'title': 'Particle motion analysis', 'summary': "Provides an in-depth analysis of a particle's motion along a straight line, covering topics such as velocity, acceleration, net change in position, and distance traveled, while also discussing applications of derivatives to cost functions and marginal cost.", 'duration': 1004.412, 'highlights': ["The particle's velocity and acceleration are analyzed at different time intervals to determine its speed and direction of motion, with the understanding that positive velocity and acceleration indicate speeding up, while opposite signs indicate slowing down.", 'The net change in position and distance traveled by the particle between specific time intervals are calculated and compared, demonstrating that the total distance traveled can be greater than the net change in position due to changes in direction.', 'The application of derivatives to cost functions is discussed, emphasizing the relationship between the cost function, marginal cost, and the interpretation of average and instantaneous rates of change in the context of production costs.']}, {'end': 22894.063, 'start': 22500.291, 'title': 'Marginal cost and introduction to logarithms', 'summary': 'Explains the concept of marginal cost in relation to the cost function and introduces logarithms, illustrating how to solve for log base and its examples.', 'duration': 393.772, 'highlights': ['The chapter explains the concept of marginal cost', 'Introduction to Logarithms', 'Solving for Log Base Examples']}, {'end': 23403.518, 'start': 22896.163, 'title': 'Understanding logarithmic functions', 'summary': 'Introduces the concept of logarithmic functions, including their domain and range, and explains how to graph a log function by hand, illustrating that the domain is x values greater than zero, the range is all real numbers, and the graph has a vertical asymptote at the line x equals zero.', 'duration': 507.355, 'highlights': ["The domain of the function log base a of x, no matter what base you're using for a, is all positive numbers, and the range of the graph of y equals log base 2 of x is all real numbers.", 'The chapter explains that taking the log of numbers greater than 0 is possible, but not for numbers less than or equal to 0, and it demonstrates how to graph a log function by hand, illustrating that the domain is x values greater than zero and the range is all real numbers.', 'The video introduces the idea that log base A of B equals C means the same thing as A to the C equals B, and it elaborates on the domain and range of the graph of y equals log base 2 of x, highlighting that the domain is x values greater than zero and the range is all real numbers.']}], 'duration': 2171.373, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w21232145.jpg', 'highlights': ['The derivative and the second derivative describe the motion of an object along a straight line.', 'The relationship between force, mass, and acceleration is explained.', 'The impact of velocity and acceleration on particle motion is showcased through a table of values.', 'The application of derivatives to cost functions is discussed, emphasizing the relationship between cost and marginal cost.', 'Analyzing particle velocity and acceleration to determine speed and direction.', 'Calculating net change in position and distance traveled by the particle.', 'Introduction to logarithms and their application in solving for log base examples.', 'Explaining the domain and range of the graph of y equals log base 2 of x.', 'Illustrating how to graph a log function by hand and its domain and range.', 'Introducing the concept that log base A of B equals C means the same as A to the C equals B.']}, {'end': 24798.381, 'segs': [{'end': 23902.058, 'src': 'embed', 'start': 23876.42, 'weight': 6, 'content': [{'end': 23882.284, 'text': "If you're familiar with the language of inverse functions, the exponential function and the log function are inverses.", 'start': 23876.42, 'duration': 5.864}, {'end': 23885.146, 'text': "Let's see why these rules hold.", 'start': 23883.785, 'duration': 1.361}, {'end': 23888.268, 'text': 'For the first log rule.', 'start': 23886.207, 'duration': 2.061}, {'end': 23898.395, 'text': 'log base a of a to the x is asking the question what power do we raise a to in order to get a to the x??', 'start': 23888.268, 'duration': 10.127}, {'end': 23902.058, 'text': 'In other words, a to what power is a to the x??', 'start': 23898.795, 'duration': 3.263}], 'summary': 'Exponential and log functions are inverses, explaining log rule for a^x.', 'duration': 25.638, 'max_score': 23876.42, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w23876420.jpg'}, {'end': 24092.305, 'src': 'embed', 'start': 24042.113, 'weight': 1, 'content': [{'end': 24056.086, 'text': 'Specifically, log base a of a to the x is equal to x and a to the log base a of x is also equal to x for any values of x and any base a.', 'start': 24042.113, 'duration': 13.973}, {'end': 24058.669, 'text': 'This video is about rules or properties of logs.', 'start': 24056.086, 'duration': 2.583}, {'end': 24062.612, 'text': 'The log rules are closely related to the exponent rules.', 'start': 24060.01, 'duration': 2.602}, {'end': 24064.974, 'text': "So let's start by reviewing some of the exponent rules.", 'start': 24062.872, 'duration': 2.102}, {'end': 24071.869, 'text': "To keep things simple, we'll write everything down with a base of 2, even though the exponent rules hold for any base.", 'start': 24066.144, 'duration': 5.725}, {'end': 24077.673, 'text': 'We know that if we raise 2 to the 0 power, we get 1.', 'start': 24072.809, 'duration': 4.864}, {'end': 24087.501, 'text': 'We have a product rule for exponents, which says that 2 to the m times 2 to the n is equal to 2 to the m plus n.', 'start': 24077.673, 'duration': 9.828}, {'end': 24092.305, 'text': 'In other words, if we multiply two numbers, then we add the exponents.', 'start': 24087.501, 'duration': 4.804}], 'summary': 'Log rules are closely related to exponent rules, such as 2^0 = 1 and product rule 2^m*2^n = 2^(m+n).', 'duration': 50.192, 'max_score': 24042.113, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w24042113.jpg'}, {'end': 24191.466, 'src': 'embed', 'start': 24153.672, 'weight': 3, 'content': [{'end': 24157.977, 'text': "I'll make these base two to agree with my base that I'm using for my exponent rules.", 'start': 24153.672, 'duration': 4.305}, {'end': 24164.513, 'text': 'In words, this says the log of the product is the sum of the logs.', 'start': 24159.389, 'duration': 5.124}, {'end': 24168.215, 'text': 'Since logs really represent exponents.', 'start': 24165.574, 'duration': 2.641}, {'end': 24176.521, 'text': 'this is saying that when you multiply two numbers together, you add their exponents, which is just what we said for the exponent version.', 'start': 24168.215, 'duration': 8.306}, {'end': 24191.466, 'text': 'The quotient rule for exponents can be rewritten in terms of logs by saying the log of x divided by y is equal to the log of x minus the log of y.', 'start': 24178.083, 'duration': 13.383}], 'summary': 'Logarithm rules show how to manipulate exponents, e.g. log of product is sum of logs.', 'duration': 37.794, 'max_score': 24153.672, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w24153672.jpg'}, {'end': 24262.086, 'src': 'embed', 'start': 24236.152, 'weight': 5, 'content': [{'end': 24242.917, 'text': 'If we think of x as being some power of two, this is really saying when we take a power to a power, we multiply their exponents.', 'start': 24236.152, 'duration': 6.765}, {'end': 24245.937, 'text': "That's exactly how we describe the power rule above.", 'start': 24243.396, 'duration': 2.541}, {'end': 24252.281, 'text': "It doesn't really matter if you multiply this exponent on the left side, or on the right side.", 'start': 24246.978, 'duration': 5.303}, {'end': 24256.183, 'text': "But it's more traditional to multiply it on the left side.", 'start': 24253.942, 'duration': 2.241}, {'end': 24262.086, 'text': "I've given these rules with the base of two, but they actually work for any base.", 'start': 24257.484, 'duration': 4.602}], 'summary': 'Exponents of powers of two multiply, applicable to any base.', 'duration': 25.934, 'max_score': 24236.152, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w24236152.jpg'}, {'end': 24499.917, 'src': 'embed', 'start': 24436.883, 'weight': 0, 'content': [{'end': 24445.228, 'text': "Now I would like to rewrite this difference of logs as the log of a quotient, but I can't do it yet because of that factor of 2 multiplied in front.", 'start': 24436.883, 'duration': 8.345}, {'end': 24451.352, 'text': 'But I can use the power rule backwards to put that 2 back up in the exponent.', 'start': 24446.249, 'duration': 5.103}, {'end': 24452.773, 'text': "So I'll do that first.", 'start': 24451.612, 'duration': 1.161}, {'end': 24463.519, 'text': 'So I will copy down the ln of x plus 1 times x minus 1 and rewrite this second term as ln of x squared minus 1 squared.', 'start': 24453.493, 'duration': 10.026}, {'end': 24474.424, 'text': 'Now I have a straightforward difference of two logs, which I can rewrite as the log of a quotient.', 'start': 24466.661, 'duration': 7.763}, {'end': 24478.626, 'text': 'I can actually simplify this some more.', 'start': 24476.925, 'duration': 1.701}, {'end': 24494.915, 'text': 'Since x plus 1 times x minus 1 is the same thing as x squared minus 1, I can cancel factors to get ln of 1 over x squared minus 1.', 'start': 24479.386, 'duration': 15.529}, {'end': 24499.917, 'text': 'In this video, we saw four rules for logs that are related to exponent rules.', 'start': 24494.915, 'duration': 5.002}], 'summary': 'Rewriting logs, applying rules to simplify and manipulate expressions.', 'duration': 63.034, 'max_score': 24436.883, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w24436883.jpg'}, {'end': 24697.763, 'src': 'embed', 'start': 24665.302, 'weight': 2, 'content': [{'end': 24670.526, 'text': 'Please take a moment to write the next two functions as compositions of functions before you go on.', 'start': 24665.302, 'duration': 5.224}, {'end': 24681.295, 'text': 'A natural way to write k of x as a composition is to let our inner function be tan of x plus secant of x.', 'start': 24672.928, 'duration': 8.367}, {'end': 24686.098, 'text': 'The outer function describes what happens to that boxed inner function.', 'start': 24681.295, 'duration': 4.803}, {'end': 24694.5, 'text': 'It gets cubed and multiplied by 5.', 'start': 24687.74, 'duration': 6.76}, {'end': 24697.763, 'text': 'There are several ways to write the next example as a composition of functions.', 'start': 24694.5, 'duration': 3.263}], 'summary': 'Compose functions: k(x) = (tan(x) + sec(x))^3 * 5', 'duration': 32.461, 'max_score': 24665.302, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w24665302.jpg'}], 'start': 23403.518, 'title': 'Logarithmic functions and their properties', 'summary': 'Discusses properties, graphs, transformations, cancellation properties, and rules of logarithmic functions, including domain, range, vertical asymptotes, log rules involving products, quotients, and powers, and the chain rule for finding the derivative of the composition of two functions.', 'chapters': [{'end': 23521.945, 'start': 23403.518, 'title': 'Logarithmic functions and their graphs', 'summary': 'Discusses the properties and graphs of logarithmic functions, including y=log base a of x, and the transformation of the natural log function with a shift of 5 units, highlighting the domain, range, vertical asymptote, and key points on the graph.', 'duration': 118.427, 'highlights': ['The graph of y equals log base any a of x for a bigger than 1 looks pretty much the same, with domain of x values greater than 0, a range of all real numbers, and a vertical asymptote at the y-axis.', 'The transformation of the natural log function with a shift of 5 units results in a graph that still has the same vertical asymptote and goes through the point (1, 5).', 'The basic log graph looks similar for different log functions and has properties of a domain greater than 0, a range of all real numbers, and a vertical asymptote at the y-axis.']}, {'end': 24153.672, 'start': 23521.945, 'title': 'Logarithmic functions and properties', 'summary': 'Discusses the transformations and properties of logarithmic functions, including domain, range, and vertical asymptotes, as well as the cancellation properties of logarithmic and exponential functions, exemplified by various log rules and their relationships with exponent rules.', 'duration': 631.727, 'highlights': ['The chapter covers the transformations of logarithmic functions, including the effects of adding constants to the function, shifting the graph horizontally, and computing the domain of logarithmic functions.', 'It explains the cancellation properties of logarithmic and exponential functions, showcasing examples such as log base a of a to the x equaling x and a to the log base a of x equaling x for any values of x and any base a.', 'The discussion also delves into the relationship between log rules and exponent rules, illustrating how various exponent rules can be rewritten as log rules and vice versa.']}, {'end': 24432.787, 'start': 24153.672, 'title': 'Logarithm rules and applications', 'summary': 'Discusses the logarithm rules involving products, quotients, and powers, as well as their applications in rewriting expressions and avoiding common mistakes.', 'duration': 279.115, 'highlights': ['The logarithm of the product is the sum of the logs, representing the rule that when multiplying two numbers, their exponents are added.', 'The log of the quotient is equal to the difference of the logs, which mirrors the rule that when dividing two numbers, their exponents are subtracted.', 'The power rule for exponents can be rewritten using logs as the log of x to the n equals n times log of x, indicating the ability to bring down the exponent and multiply when taking the log of an expression with an exponent.', 'Using logarithmic rules to rewrite expressions as a sum or difference of logs, and avoiding the mistake of applying the power rule to a product instead of a single expression raised to an exponent.']}, {'end': 24798.381, 'start': 24436.883, 'title': 'Log rules and chain rule', 'summary': 'Covers the rules related to logarithms, highlighting the product rule, quotient rule, and power rule. it also delves into the concept of composition of functions and the chain rule for finding the derivative of the composition of two functions.', 'duration': 361.498, 'highlights': ['The log rules related to the product, quotient, and power are explained, emphasizing the ability to rewrite the difference of logs as the log of a quotient and simplify expressions using these rules.', 'The inapplicability of log rules to split up the log of a sum is discussed, underlining the importance of understanding log rules for solving equations.', 'The concept of composition of functions is introduced, demonstrating how to write functions as compositions and dissect them into inner and outer functions.', 'The significance of recognizing complicated functions as compositions of simpler functions for calculating derivatives is highlighted, leading to the explanation of the chain rule and its application in finding the derivative of the composition of two functions.']}], 'duration': 1394.863, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w23403518.jpg', 'highlights': ['The chain rule for finding the derivative of the composition of two functions is explained', 'The log of the quotient is equal to the difference of the logs, mirroring the rule for dividing two numbers', 'The logarithm of the product is the sum of the logs, representing the rule for multiplying two numbers', 'The transformation of the natural log function with a shift of 5 units results in a graph that still has the same vertical asymptote and goes through the point (1, 5)', 'The graph of y equals log base any a of x for a bigger than 1 looks pretty much the same, with domain of x values greater than 0, a range of all real numbers, and a vertical asymptote at the y-axis', 'The discussion also delves into the relationship between log rules and exponent rules, illustrating how various exponent rules can be rewritten as log rules and vice versa', 'The chapter covers the transformations of logarithmic functions, including the effects of adding constants to the function, shifting the graph horizontally, and computing the domain of logarithmic functions', 'The basic log graph looks similar for different log functions and has properties of a domain greater than 0, a range of all real numbers, and a vertical asymptote at the y-axis']}, {'end': 26712.368, 'segs': [{'end': 24845.507, 'src': 'embed', 'start': 24798.381, 'weight': 1, 'content': [{'end': 24806.525, 'text': "let's let u equal g and let's let y equal f, in other words, y is f.", 'start': 24798.381, 'duration': 8.144}, {'end': 24817.935, 'text': 'Now du dx is just another way of writing g, prime of x.', 'start': 24810.692, 'duration': 7.243}, {'end': 24830.639, 'text': 'And dy du is another way of writing f prime of u, or in other words, f prime of g of x.', 'start': 24817.935, 'duration': 12.704}, {'end': 24845.507, 'text': "Finally, if we write dy dx, that means we're taking the derivative of f composed with g So that's f composed with g prime of x.", 'start': 24830.639, 'duration': 14.868}], 'summary': "Derivatives of f composed with g: dy/dx = f'(g(x)) * g'(x)", 'duration': 47.126, 'max_score': 24798.381, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w24798381.jpg'}, {'end': 25183.609, 'src': 'embed', 'start': 25149.124, 'weight': 2, 'content': [{'end': 25153.707, 'text': 'First, I want to rewrite 5 to the x as e to the ln 5 times x.', 'start': 25149.124, 'duration': 4.583}, {'end': 25160.074, 'text': 'And I can do that because e to the ln 5 is equal to 5.', 'start': 25153.707, 'duration': 6.367}, {'end': 25165.638, 'text': 'So e to the ln 5 to the x power is equal to 5 to the x.', 'start': 25160.074, 'duration': 5.564}, {'end': 25176.084, 'text': 'But e to the ln 5 to the x power, using exponent rules, is just e to the ln 5 times x.', 'start': 25165.638, 'duration': 10.446}, {'end': 25183.609, 'text': 'So if I want to take the derivative of 5 to the x, after rewriting it as e to the ln 5 times x,', 'start': 25176.084, 'duration': 7.525}], 'summary': 'Rewriting 5 to the x as e to the ln 5 times x simplifies the derivative calculation.', 'duration': 34.485, 'max_score': 25149.124, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w25149124.jpg'}, {'end': 25697.637, 'src': 'embed', 'start': 25668.663, 'weight': 3, 'content': [{'end': 25671.386, 'text': "I'm not going to give a rigorous proof of the chain rule,", 'start': 25668.663, 'duration': 2.723}, {'end': 25676.19, 'text': 'but I would like to give a more informal explanation based on the limit definition of derivative.', 'start': 25671.386, 'duration': 4.804}, {'end': 25688.702, 'text': "So I'm going to write the derivative of f composed with g, evaluated at point a as the limit, as x goes to a, of f composed with g,", 'start': 25677.351, 'duration': 11.351}, {'end': 25693.967, 'text': 'of x minus f composed with g, of a divided by x minus a.', 'start': 25688.702, 'duration': 5.265}, {'end': 25697.637, 'text': "I'll rewrite this slightly.", 'start': 25695.936, 'duration': 1.701}], 'summary': 'Informal explanation of chain rule using limit definition of derivative.', 'duration': 28.974, 'max_score': 25668.663, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w25668663.jpg'}, {'end': 25777.396, 'src': 'embed', 'start': 25746.918, 'weight': 4, 'content': [{'end': 25765.547, 'text': 'And letting, say, u be equal to g of x, I can rewrite this as the limit as u goes to g of a of f of u minus f of g of a over u minus g of a.', 'start': 25746.918, 'duration': 18.629}, {'end': 25774.534, 'text': 'Now my expression on the left is just another way of writing the derivative of f evaluated at g of a.', 'start': 25765.547, 'duration': 8.987}, {'end': 25777.396, 'text': "And I've arrived at the expression for the chain draw.", 'start': 25774.534, 'duration': 2.862}], 'summary': 'Derivative of f at g(a) is the chain rule expression.', 'duration': 30.478, 'max_score': 25746.918, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w25746918.jpg'}, {'end': 26447.017, 'src': 'embed', 'start': 26415.603, 'weight': 0, 'content': [{'end': 26425.408, 'text': 'The main two steps were first to take the derivative of both sides with respect to x and then to solve for dy dx.', 'start': 26415.603, 'duration': 9.805}, {'end': 26431.725, 'text': 'This video is about finding the derivatives of exponential functions.', 'start': 26428.442, 'duration': 3.283}, {'end': 26439.791, 'text': "We've already seen that the derivative of the exponential function e to the x is just itself e to the x.", 'start': 26432.425, 'duration': 7.366}, {'end': 26447.017, 'text': "But what's the derivative of an exponential function with a different base, like five to the x.", 'start': 26439.791, 'duration': 7.226}], 'summary': 'Derivative of e to the x is e to the x, video covers derivatives of exponential functions.', 'duration': 31.414, 'max_score': 26415.603, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w26415603.jpg'}, {'end': 26634.646, 'src': 'embed', 'start': 26605.774, 'weight': 5, 'content': [{'end': 26615.076, 'text': "So if I take the derivative with respect to x of a to the x for any number a, I'm going to get a to the x times ln a.", 'start': 26605.774, 'duration': 9.302}, {'end': 26623.171, 'text': 'Now you might be wondering what if I use the same rule on our old favorite e to the x.', 'start': 26617.045, 'duration': 6.126}, {'end': 26628.536, 'text': 'So our base here is e, that means I should get e to the x times ln e.', 'start': 26623.171, 'duration': 5.365}, {'end': 26630.904, 'text': 'Wait a sec.', 'start': 26630.404, 'duration': 0.5}, {'end': 26634.646, 'text': "Lne, that's log base e of e.", 'start': 26631.404, 'duration': 3.242}], 'summary': 'The derivative of a^x with respect to x is a^x times ln a. for e^x, the derivative is e^x times ln e.', 'duration': 28.872, 'max_score': 26605.774, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w26605774.jpg'}], 'start': 24798.381, 'title': 'Derivatives and implicit differentiation', 'summary': 'Introduces the concept of the chain rule for finding derivatives of composite functions, covers application of chain rule and product rule with quantitative results, explains implicit differentiation for finding derivatives of implicitly defined curves, and provides rules for derivatives of exponential functions.', 'chapters': [{'end': 25149.124, 'start': 24798.381, 'title': 'Understanding the chain rule in derivatives', 'summary': 'Introduces the concept of the chain rule, emphasizing that the derivative of f composed with g at x is equal to f prime at g of x times g prime at x, and provides examples illustrating the application of the chain rule in finding derivatives of composite functions.', 'duration': 350.743, 'highlights': ['The chain rule states that the derivative of f composed with g at x is equal to f prime at g of x times g prime at x.', 'Examples are provided to illustrate the application of the chain rule in finding derivatives of composite functions.', 'A formula for the derivative of a to the x with respect to x, where a is any positive number, is included.']}, {'end': 25526.073, 'start': 25149.124, 'title': 'Derivatives using chain rule and product rule', 'summary': 'Covers the application of the chain rule and product rule to find derivatives, including examples with quantitative results of derivative calculations and a general principle for derivative of a to the x.', 'duration': 376.949, 'highlights': ['The derivative of a to the x with respect to x is equal to ln a times a to the x, a fact worth memorizing.', 'Using the product rule to find dy dx for sine of 5x times the square root of 2 to the cosine 5x plus 1 gives a final result of -50.', 'Applying the chain rule to find the derivative of f composed with g at x equals 1 gives a result of -50.']}, {'end': 25860.368, 'start': 25526.073, 'title': 'Chain rule explanation and implicit differentiation', 'summary': 'Explains the chain rule with an informal explanation based on the limit definition of derivative, highlighting the rearrangement of limits and the justification of the chain rule, and also introduces implicit differentiation for finding derivatives of implicitly defined curves.', 'duration': 334.295, 'highlights': ['The chapter emphasizes an informal explanation of the chain rule based on the limit definition of derivative, including the rearrangement of limits and the justification of the chain rule, citing the derivative of f evaluated at g of a.', "It mentions the technique of implicit differentiation for finding the slopes of tangent lines for curves that are defined indirectly and sometimes aren't even functions, providing insights into the curves defined implicitly in terms of any equation involving x's and y's.", 'It explains that implicitly defined curves are not necessarily functions and can violate the vertical line test, cross themselves, be broken up into several pieces, or have complex shapes, but small pieces of these curves do satisfy the vertical line test.']}, {'end': 26210.234, 'start': 25860.368, 'title': 'Implicit differentiation for tangent line', 'summary': 'Discusses using calculus techniques, including the chain rule, to compute the derivative for an implicitly defined curve (ellipse) and finding the equation of the tangent line with the point-slope form, yielding a slope of -9/8.', 'duration': 349.866, 'highlights': ['The equation of the tangent line for the ellipse 9x squared plus 4y squared equals 25 is found using calculus techniques, yielding a slope of -9/8 at the point (1, 2).', 'Implicit differentiation is introduced as an alternative method for finding the derivative without solving for y, resulting in the same expression for the derivative as the previous method.']}, {'end': 26447.017, 'start': 26211.595, 'title': 'Implicit differentiation for tangent lines', 'summary': 'Discusses using implicit differentiation to find the slope of tangent lines for curves defined implicitly, demonstrating the process through an example and highlighting the key steps involved.', 'duration': 235.422, 'highlights': ['The process of implicit differentiation is demonstrated through an example, where the derivative of both sides with respect to x is computed to find the slope of the tangent line, yielding the equation y equals negative 9 eighths x plus 25 eighths.', 'Implicit differentiation is emphasized as the only viable method to find the derivative for curves defined implicitly, showcasing its importance and applicability in such scenarios.', 'The detailed process of implicit differentiation is outlined, including the application of the product rule and chain rule to compute derivatives and the subsequent steps to solve for dy dx, ultimately isolating and finding the derivative using implicit differentiation.', 'The video also briefly touches on finding the derivatives of exponential functions, specifically highlighting the derivative of a different base exponential function, such as five to the x.']}, {'end': 26712.368, 'start': 26447.017, 'title': 'Derivative of exponential functions', 'summary': 'Explains the derivative of exponential functions using the rewriting trick and chain rule, providing general rules for derivatives of a^x and x^a, with specific emphasis on the derivative of e^x being e^x.', 'duration': 265.351, 'highlights': ['The derivative of a^x is a^x times ln(a), and the derivative of e^x is e^x.', 'The rewriting trick is used to express 5^x as e^(ln(5)*x), simplifying the derivative calculation.', 'The chain rule is applied to calculate the derivative of e^(ln(5)*x), resulting in 5^x times ln(5).']}], 'duration': 1913.987, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w24798381.jpg', 'highlights': ["The chain rule states f'(g(x)) * g'(x).", 'Implicit differentiation is essential for curves defined implicitly.', 'Derivative of a^x is a^x * ln(a), and e^x is e^x.', 'The derivative of 5^x is e^(ln(5)*x), using the chain rule.', 'The equation of the tangent line for 9x^2 + 4y^2 = 25 is found using calculus techniques.', 'Applying the chain rule to find the derivative of f composed with g at x equals 1 gives a result of -50.']}, {'end': 29616.697, 'segs': [{'end': 26777.751, 'src': 'embed', 'start': 26713.587, 'weight': 3, 'content': [{'end': 26724.792, 'text': 'In this video, we found that the derivative with respect to x of five to the x is given by ln five times five to the x.', 'start': 26713.587, 'duration': 11.205}, {'end': 26733.016, 'text': 'And in general, the derivative of with respect to x of a to the x is going to be ln a times a to the x.', 'start': 26724.792, 'duration': 8.224}, {'end': 26737.098, 'text': 'This gives us a general formula for the derivative of exponential functions.', 'start': 26733.016, 'duration': 4.082}, {'end': 26744.229, 'text': 'The main goal of this video is to figure out the derivatives of logarithmic functions.', 'start': 26739.226, 'duration': 5.003}, {'end': 26754.877, 'text': 'functions like y equals ln x or y equals log base a x for any positive base a.', 'start': 26744.229, 'duration': 10.648}, {'end': 26758.299, 'text': 'I want to find the derivative of log base a of x.', 'start': 26754.877, 'duration': 3.422}, {'end': 26765.084, 'text': 'In other words, I want to find the derivative of y, where y is log base a of x.', 'start': 26758.299, 'duration': 6.785}, {'end': 26777.751, 'text': 'By the definition of logarithms, log base A of X equals Y means that A to the Y power is equal to X.', 'start': 26767.145, 'duration': 10.606}], 'summary': 'Derivative of exponential functions is ln a times a to the x; next, focus on finding derivatives of logarithmic functions.', 'duration': 64.164, 'max_score': 26713.587, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w26713587.jpg'}, {'end': 27955.775, 'src': 'embed', 'start': 27930.865, 'weight': 1, 'content': [{'end': 27942.69, 'text': 'The reason that violates the vertical line test is because the original green function violates the horizontal line test and has two x values with the same y value.', 'start': 27930.865, 'duration': 11.825}, {'end': 27951.773, 'text': 'In general a function f has an inverse function if, and only if, the graph of f satisfies the horizontal line test i.e..', 'start': 27944.43, 'duration': 7.343}, {'end': 27955.775, 'text': 'every horizontal line intersects the graph and at most one point.', 'start': 27951.853, 'duration': 3.922}], 'summary': 'Function violates vertical line test due to two x values with same y value.', 'duration': 24.91, 'max_score': 27930.865, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w27930865.jpg'}, {'end': 29125.846, 'src': 'embed', 'start': 29090.086, 'weight': 2, 'content': [{'end': 29097.129, 'text': 'And I get dy dx is 1 over the square root of 1 minus x squared.', 'start': 29090.086, 'duration': 7.043}, {'end': 29104.89, 'text': 'In other words, I found a formula for the derivative of inverse sine of x.', 'start': 29097.569, 'duration': 7.321}, {'end': 29108.471, 'text': 'We can carry out a similar process to find the derivative of inverse cosine.', 'start': 29104.89, 'duration': 3.581}, {'end': 29115.613, 'text': 'y equals cosine inverse x means that x is equal to cosine of y.', 'start': 29109.672, 'duration': 5.941}, {'end': 29118.194, 'text': 'And by convention, y lies between 0 and pi.', 'start': 29115.613, 'duration': 2.581}, {'end': 29125.846, 'text': 'To find the derivative of arc cosine of x.', 'start': 29122.423, 'duration': 3.423}], 'summary': 'Derivative of inverse sine is 1 over the square root of 1 minus x squared. similar process for derivative of inverse cosine.', 'duration': 35.76, 'max_score': 29090.086, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w29090086.jpg'}, {'end': 29229.741, 'src': 'embed', 'start': 29195.799, 'weight': 0, 'content': [{'end': 29206.441, 'text': 'which means that sine of y, which is opposite over hypotenuse, is equal to the square root of 1 minus x squared.', 'start': 29195.799, 'duration': 10.642}, {'end': 29215.211, 'text': 'And so dy dx is going to be negative one over the square root of one minus x squared.', 'start': 29208.166, 'duration': 7.045}, {'end': 29226.439, 'text': 'And I have my formula for the derivative of our cosine of x inverse tangent can be handled very similarly.', 'start': 29215.591, 'duration': 10.848}, {'end': 29229.741, 'text': 'And again, you may want to try it for yourself before watching the video.', 'start': 29226.759, 'duration': 2.982}], 'summary': 'The derivative of inverse sine is -1/sqrt(1-x^2); inverse tangent derivative can be handled similarly.', 'duration': 33.942, 'max_score': 29195.799, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w29195799.jpg'}], 'start': 26713.587, 'title': 'Derivatives of functions', 'summary': 'Explores derivatives of exponential and logarithmic functions, inverse functions and their properties, including the concept, method of finding, and graphical representation, inverse trig functions and their properties, and the process of using implicit differentiation to find the derivatives of inverse trigonometric functions.', 'chapters': [{'end': 27592.263, 'start': 26713.587, 'title': 'Derivatives of exponential and logarithmic functions', 'summary': 'Explores the derivatives of exponential and logarithmic functions, demonstrating the general formula for the derivative of exponential functions and the derivations of ln x and log base a of x. it also introduces the technique of logarithmic differentiation and the properties of inverse functions.', 'duration': 878.676, 'highlights': ['The derivative of ln x is 1 over x, and the derivative of log base a of x is 1 over ln a times x.', 'Logarithmic differentiation is used to find the derivative of x to the x.', 'Inverses reverse the roles of y and x, and the graph of y = f inverse of x can be obtained from the graph of y = f of x by reflecting over the line y = x.']}, {'end': 28021.18, 'start': 27592.283, 'title': 'Inverse functions and their properties', 'summary': 'Discusses the concept of inverse functions, showing examples of computing f inverse of f and f of f inverse, explaining the method of finding inverse functions by reversing the roles of y and x, and highlighting the conditions for a function to have an inverse, along with the graphs that satisfy the horizontal line test.', 'duration': 428.897, 'highlights': ['The chapter discusses the concept of inverse functions, showing examples of computing f inverse of f and f of f inverse.', 'The method of finding inverse functions by reversing the roles of y and x is explained.', 'The conditions for a function to have an inverse are highlighted, along with the graphs that satisfy the horizontal line test.']}, {'end': 28570.884, 'start': 28022.661, 'title': 'Inverse functions and their properties', 'summary': 'Explains the concept of inverse functions, including their graphical representation, algebraic solution, domain and range restrictions, and the properties of inverse trigonometric functions, emphasizing the importance of these properties in the study of exponential and logarithmic functions.', 'duration': 548.223, 'highlights': ['The graph of the inverse function is obtained by flipping the original function graph over the line y equals x, and domain and range restrictions are essential to ensure the inverse function is a proper function, as demonstrated in the examples of inverse trigonometric functions.', 'The chapter discusses the properties of inverse functions, highlighting that they reverse the roles of y and x, and the composition of a function with its inverse yields the identity function, emphasizing the significance of these properties in understanding the relationships between functions and their inverses.', 'The chapter explains the necessity of domain restrictions to obtain a well-defined inverse value for trigonometric functions, such as arcsine and arccosine, and emphasizes the difference between the notations for inverse trigonometric functions and reciprocal functions to avoid confusion.']}, {'end': 28948.83, 'start': 28572.466, 'title': 'Inverse trig functions and derivatives', 'summary': 'Covers the definitions and properties of inverse trigonometric functions, including cosine inverse, sine inverse, and tangent inverse, and explains their derivatives using implicit differentiation.', 'duration': 376.364, 'highlights': ['The range of our restricted tan function is from negative infinity to infinity. Therefore, arctan of x has domain from negative infinity to infinity and range from negative pi over 2 to pi over 2.', 'The chapter explains the definitions and properties of inverse trigonometric functions, including cosine inverse, sine inverse, and tangent inverse, and their derivatives using implicit differentiation.', 'Tangent of pi over four is one, and therefore, arctan of one is pi over four.']}, {'end': 29616.697, 'start': 28951.604, 'title': 'Implicit differentiation of inverse trig functions', 'summary': 'Explains the process of using implicit differentiation to find the derivatives of inverse trigonometric functions, providing formulas for each and highlighting the related rates problem involving distances.', 'duration': 665.093, 'highlights': ['The chapter explains the process of using implicit differentiation to find the derivatives of inverse trigonometric functions', 'Providing formulas for each', 'Related rates problem involving distances']}], 'duration': 2903.11, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w26713587.jpg', 'highlights': ['The derivative of ln x is 1 over x, and the derivative of log base a of x is 1 over ln a times x.', 'The chapter explains the process of using implicit differentiation to find the derivatives of inverse trigonometric functions', 'The graph of the inverse function is obtained by flipping the original function graph over the line y equals x, and domain and range restrictions are essential to ensure the inverse function is a proper function, as demonstrated in the examples of inverse trigonometric functions.', 'The chapter discusses the properties of inverse functions, highlighting that they reverse the roles of y and x, and the composition of a function with its inverse yields the identity function, emphasizing the significance of these properties in understanding the relationships between functions and their inverses.', 'The chapter discusses the concept of inverse functions, showing examples of computing f inverse of f and f of f inverse.']}, {'end': 30875.893, 'segs': [{'end': 29645.708, 'src': 'embed', 'start': 29619.377, 'weight': 3, 'content': [{'end': 29623.599, 'text': 'The next step is to write down equations that relate the quantities of interest.', 'start': 29619.377, 'duration': 4.222}, {'end': 29631.421, 'text': 'In this problem, we know by the Pythagorean theorem that a squared plus b squared equals c squared.', 'start': 29625.319, 'duration': 6.102}, {'end': 29637.085, 'text': "We're interested in how fast the distance between you and the tornado is changing.", 'start': 29633.223, 'duration': 3.862}, {'end': 29638.845, 'text': "That's a rate of change.", 'start': 29637.725, 'duration': 1.12}, {'end': 29645.708, 'text': 'And the rate at which the bicycle is traveling and the tornado is moving, these are also rates of change.', 'start': 29640.286, 'duration': 5.422}], 'summary': 'Applying pythagorean theorem to find rates of change.', 'duration': 26.331, 'max_score': 29619.377, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w29619377.jpg'}, {'end': 29913.188, 'src': 'embed', 'start': 29884.065, 'weight': 7, 'content': [{'end': 29887.848, 'text': 'Water flows into a tank at a rate of 3 cubic meters per minute.', 'start': 29884.065, 'duration': 3.783}, {'end': 29895.894, 'text': 'The tank is shaped like a cone with a height of 4 meters and a radius of 5 meters at the top.', 'start': 29888.529, 'duration': 7.365}, {'end': 29903.06, 'text': "We're supposed to find the rate at which the water level is rising in the tank when the water height is 2 meters.", 'start': 29896.775, 'duration': 6.285}, {'end': 29905.662, 'text': "We've drawn our picture.", 'start': 29904.561, 'duration': 1.101}, {'end': 29908.424, 'text': "Now let's label some quantities of interest.", 'start': 29906.523, 'duration': 1.901}, {'end': 29913.188, 'text': "It's fine to use numbers for the quantities that stay fixed throughout the problem.", 'start': 29909.565, 'duration': 3.623}], 'summary': 'Water fills cone tank at 3m³/min, find rate of rise at 2m height.', 'duration': 29.123, 'max_score': 29884.065, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w29884065.jpg'}, {'end': 30298.51, 'src': 'embed', 'start': 30268.496, 'weight': 1, 'content': [{'end': 30273.018, 'text': 'two revolutions per minute is indirectly telling us how this angle theta is changing.', 'start': 30268.496, 'duration': 4.522}, {'end': 30284.984, 'text': "Because if the light beam is making two revolutions per minute, then since they're two pi radians in a revolution,", 'start': 30273.699, 'duration': 11.285}, {'end': 30290.847, 'text': 'that amounts to a change of four pi radians per minute for the angle theta.', 'start': 30284.984, 'duration': 5.863}, {'end': 30298.51, 'text': "Therefore, I'd really like to write down the equation that has to do with theta and x.", 'start': 30292.927, 'duration': 5.583}], 'summary': 'Two revolutions per minute corresponds to a change of four pi radians per minute for the angle theta.', 'duration': 30.014, 'max_score': 30268.496, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w30268496.jpg'}, {'end': 30823.95, 'src': 'embed', 'start': 30787.293, 'weight': 0, 'content': [{'end': 30795.859, 'text': 'To find the measure of angle phi, we could use the fact that sine of phi is 10 over 15 and take sine inverse of 10 15.', 'start': 30787.293, 'duration': 8.566}, {'end': 30800.263, 'text': 'But probably a little easier is just to use the fact that these three angles add to 180 degrees.', 'start': 30795.859, 'duration': 4.404}, {'end': 30817.444, 'text': 'That tells us that phi plus 90 plus 48.19 is equal to 180, which means that phi is 41.81.', 'start': 30801.603, 'duration': 15.841}, {'end': 30823.95, 'text': 'Finally, we can find x, either using a trig function, or by using the Pythagorean theorem.', 'start': 30817.444, 'duration': 6.506}], 'summary': 'To find angle phi, use trigonometry or angle sum property. phi = 41.81 degrees.', 'duration': 36.657, 'max_score': 30787.293, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w30787293.jpg'}, {'end': 30885.097, 'src': 'embed', 'start': 30854.119, 'weight': 2, 'content': [{'end': 30856.941, 'text': 'Notice that we use many of the same ideas as in the previous problem.', 'start': 30854.119, 'duration': 2.822}, {'end': 30868.989, 'text': 'For example, the fact that the sum of the angles is 180, the Pythagorean theorem, and the trig functions, like tan, sine and cosine.', 'start': 30857.421, 'duration': 11.568}, {'end': 30875.893, 'text': 'We also use the inverse trig functions to get from an equation like this one to the angle.', 'start': 30869.749, 'duration': 6.144}, {'end': 30885.097, 'text': "This video showed how it's possible to find the lengths of all the sides of a right triangle and the measures of all the angles given partial information.", 'start': 30877.754, 'duration': 7.343}], 'summary': 'Using trigonometric functions to find triangle measurements and angles.', 'duration': 30.978, 'max_score': 30854.119, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w30854119.jpg'}], 'start': 29619.377, 'title': 'Related rates problem solving', 'summary': 'Explains solving related rates problems, such as determining the changing distance between a tornado and a location at a rate of -35 miles per hour, finding the rate of water level rise in a cone, and calculating the speed of light at approximately 1885 miles per hour, while also addressing right triangle problem solving.', 'chapters': [{'end': 29929.38, 'start': 29619.377, 'title': 'Related rates problem solving', 'summary': 'Explains the process of solving related rates problems using the example of determining the changing distance between a tornado and a location, involving derivatives with respect to time and arriving at a rate of change of -35 miles per hour, while providing cautionary notes and an additional example with water flowing into a cone-shaped tank.', 'duration': 310.003, 'highlights': ['The distance between a tornado and a location is determined by taking the derivative of the Pythagorean theorem equation with respect to time, resulting in a rate of change of -35 miles per hour at a specific time, indicating the tornado is gaining on the location quickly.', 'The process involves using the given rates of change for the tornado and a bicycle, plugging in numbers to solve for the quantity of interest, and accurately representing quantities that vary with time as variables to properly take derivatives, ensuring the use of negative numbers for negative rates of change.', 'An additional example is provided, involving the flow of water into a cone-shaped tank at a rate of 3 cubic meters per minute, where the objective is to determine the rate at which the water level is rising when the water height is 2 meters.']}, {'end': 30084.931, 'start': 29929.38, 'title': 'Rate of water level rise in cone', 'summary': 'Discusses finding the rate at which the water level is rising in a cone filled with water using equations related to the volume of water, similar triangles, and rates of change, ultimately solving for the quantity of interest, dh dt.', 'duration': 155.551, 'highlights': ['The volume of water in the cone is 1 third times pi r squared times h, since h is the height of the piece of the cone that contains water and pi r squared is the area of that circular base for that piece of a cone.', 'The ratio of sides for the little triangle here is the same as the ratio of sides for the big triangle. In other words, we know that r over h is going to be equal to 5 over 4.', "We get dv dt equals 25 48ths pi times 3 h squared dh dt. Now let's plug in numbers and solve for the quantity of interest, dh dt."]}, {'end': 30197.167, 'start': 30086.973, 'title': 'Related rates problems', 'summary': "Covers solving related rates problems involving volume and rotation, with a specific example of finding the rate at which the water level is rising and the speed of the beam of light moving along the shore, providing a solution of 0.15 meters per second for the water level and presenting a scenario for the lighthouse beam's speed.", 'duration': 110.194, 'highlights': ['The rate at which the water level is rising when the water height is 2 meters is 0.15 meters per second.', 'Solving a related rates problem involving rotation and angles, specifically calculating the speed of the beam of light moving along the shore at a point one mile north of the cave.']}, {'end': 30498.051, 'start': 30197.167, 'title': 'Speed of light calculation', 'summary': 'Discusses using trigonometry and related rates to calculate the speed of light, resulting in a speed of approximately 1885 miles per hour, derived from the rate of change of an angle and a side length.', 'duration': 300.884, 'highlights': ['The speed of light is approximately 1885 miles per hour', 'The angle theta is changing at a rate of four pi radians per minute', 'The equation relating x and theta is tangent theta equals 2x']}, {'end': 30875.893, 'start': 30500.555, 'title': 'Solving right triangle', 'summary': 'Explains how to solve a right triangle using given side length and angle measures, using trigonometric functions and the pythagorean theorem to find the unknown angles and side lengths.', 'duration': 375.338, 'highlights': ['The sum of the angles in a triangle is 180 degrees. Using this, angle A is found to be 41 degrees.', 'Using tangent function, side length B is found to be 26.46 units.', 'Side length C is found to be 35.06 units using the cosine function or the Pythagorean theorem.']}], 'duration': 1256.516, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w29619377.jpg', 'highlights': ['The speed of light is approximately 1885 miles per hour', 'The rate at which the water level is rising when the water height is 2 meters is 0.15 meters per second', 'The distance between a tornado and a location is determined by taking the derivative of the Pythagorean theorem equation with respect to time, resulting in a rate of change of -35 miles per hour at a specific time, indicating the tornado is gaining on the location quickly', 'The process involves using the given rates of change for the tornado and a bicycle, plugging in numbers to solve for the quantity of interest, and accurately representing quantities that vary with time as variables to properly take derivatives, ensuring the use of negative numbers for negative rates of change', 'The sum of the angles in a triangle is 180 degrees. Using this, angle A is found to be 41 degrees', 'An additional example is provided, involving the flow of water into a cone-shaped tank at a rate of 3 cubic meters per minute, where the objective is to determine the rate at which the water level is rising when the water height is 2 meters', 'The ratio of sides for the little triangle here is the same as the ratio of sides for the big triangle. In other words, we know that r over h is going to be equal to 5 over 4', 'The volume of water in the cone is 1 third times pi r squared times h, since h is the height of the piece of the cone that contains water and pi r squared is the area of that circular base for that piece of a cone', 'Solving a related rates problem involving rotation and angles, specifically calculating the speed of the beam of light moving along the shore at a point one mile north of the cave', 'The angle theta is changing at a rate of four pi radians per minute', 'Using tangent function, side length B is found to be 26.46 units', 'Side length C is found to be 35.06 units using the cosine function or the Pythagorean theorem', "We get dv dt equals 25 48ths pi times 3 h squared dh dt. Now let's plug in numbers and solve for the quantity of interest, dh dt"]}, {'end': 33297.798, 'segs': [{'end': 31238.621, 'src': 'embed', 'start': 31214.12, 'weight': 3, 'content': [{'end': 31222.903, 'text': 'Please take a look at this graph and pause the video for a moment to mark all local maximum and minimum points, as well as all global, that is,', 'start': 31214.12, 'duration': 8.783}, {'end': 31224.504, 'text': 'absolute max and min points.', 'start': 31222.903, 'duration': 1.601}, {'end': 31229.597, 'text': 'see if you can find the absolute maximum value and the absolute minimum value for the function.', 'start': 31225.256, 'duration': 4.341}, {'end': 31238.621, 'text': "I'm going to mark the local max and min points in green and the absolute max and min points in red.", 'start': 31231.658, 'duration': 6.963}], 'summary': 'Identify local and absolute max/min points on a graph.', 'duration': 24.501, 'max_score': 31214.12, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w31214120.jpg'}, {'end': 31840.099, 'src': 'embed', 'start': 31810.317, 'weight': 2, 'content': [{'end': 31816.441, 'text': 'The first derivative test is great because it lets us locate local extreme points just by looking at the first derivative.', 'start': 31810.317, 'duration': 6.124}, {'end': 31824.355, 'text': 'The second derivative test gives us an alternative for finding local max and min points by using the second derivative.', 'start': 31817.813, 'duration': 6.542}, {'end': 31840.099, 'text': 'Specifically, the second derivative test tells us that if f is continuous near x equals c, then if f prime of c is equal to zero and f,', 'start': 31825.575, 'duration': 14.524}], 'summary': 'Derivative tests locate local extreme points using first and second derivatives.', 'duration': 29.782, 'max_score': 31810.317, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w31810317.jpg'}, {'end': 32239.857, 'src': 'embed', 'start': 32216.138, 'weight': 0, 'content': [{'end': 32223.002, 'text': 'which is the absolute value of x minus x squared on the interval from negative 2 to 2..', 'start': 32216.138, 'duration': 6.864}, {'end': 32236.516, 'text': 'As before, we can find absolute extreme values by checking first the critical numbers and then also the endpoints of the interval negative 2 and 2.', 'start': 32223.002, 'duration': 13.514}, {'end': 32239.857, 'text': 'To find the critical numbers, we need to take the derivative of our function.', 'start': 32236.516, 'duration': 3.341}], 'summary': 'Finding absolute extreme values using derivative on interval from -2 to 2.', 'duration': 23.719, 'max_score': 32216.138, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w32216138.jpg'}, {'end': 33181.028, 'src': 'embed', 'start': 33151.216, 'weight': 1, 'content': [{'end': 33163.123, 'text': 'the mean value theorem for functions says that if g of x is continuous on a closed interval and differentiable on the interior of that interval,', 'start': 33151.216, 'duration': 11.907}, {'end': 33167.943, 'text': "then there's some number c in the interval,", 'start': 33164.201, 'duration': 3.742}, {'end': 33181.028, 'text': 'such that the derivative of g at c is equal to the average rate of change of g across the whole interval from a to b.', 'start': 33167.943, 'duration': 13.085}], 'summary': 'The mean value theorem states that for a function g(x) continuous on a closed interval, there exists a number in the interval where the derivative of g is equal to the average rate of change of g across the interval.', 'duration': 29.812, 'max_score': 33151.216, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w33151216.jpg'}], 'start': 30877.754, 'title': 'Locating extreme values of functions', 'summary': 'Discusses finding absolute and local maximum and minimum values of functions, first and second derivative tests, and the mean value theorem, providing definitions, key points, and examples, with a focus on the process and significance of marking and identifying these points on a graph, and the verification and applications of the mean value theorem.', 'chapters': [{'end': 31671.02, 'start': 30877.754, 'title': 'Finding maximum and minimum values of functions', 'summary': 'Discussed finding absolute and local maximum and minimum values of functions, including the definitions, key points, and examples, highlighting the process of marking and identifying these points on a graph.', 'duration': 793.266, 'highlights': ['The video explained the definitions and methods for finding absolute and local maximum and minimum values of functions, including the conditions for a function to have an absolute maximum or minimum at a specific point.', 'It provided examples of functions with absolute maximum and minimum points, demonstrating how to identify them on a graph and differentiate between absolute and local maxima and minima.', 'The importance of critical numbers, where the derivative of a function equals zero or does not exist, was highlighted as a key concept in identifying potential local maximum and minimum points.', 'The video emphasized the relationship between critical numbers and local maximum and minimum points, cautioning that not all critical numbers necessarily correspond to local maxima or minima, providing a clear example to illustrate the point.']}, {'end': 31924.403, 'start': 31673.223, 'title': 'Finding extreme values with derivatives', 'summary': 'Discusses the first and second derivative tests for locating local maximums and minimums, demonstrating how to use the tests and their significance in determining extreme values of functions.', 'duration': 251.18, 'highlights': ['The first derivative test allows us to locate local extreme points by determining if the first derivative is positive on one side and negative on the other, and the second derivative test provides an alternative method using the second derivative to identify local max and min points.', 'The second derivative test states that if f prime of c equals zero and f double prime of c is greater than zero, then f has a local min at x equals c, while if f prime of c equals zero, and f double prime of c is less than zero, then f has a local max at x equals c.', 'Using the first derivative test, we can identify if a function has a local minimum or maximum at a specific value of x by examining the behavior of the first derivative near x equals c.']}, {'end': 32492.206, 'start': 31924.403, 'title': 'Finding absolute extreme values', 'summary': 'Explains the process of finding absolute extreme values of a function using critical numbers and endpoints, with examples involving derivatives and piecewise functions.', 'duration': 567.803, 'highlights': ['The process of finding absolute extreme values involves checking critical numbers and endpoints, demonstrated using derivatives and piecewise notation.', 'The derivative of the function g is simplified to -x^2 + 2x + 3, leading to critical numbers at x = 3 and x = -1, within the interval [0, 4].', 'For the function g, the absolute maximum value is 1/7 and the absolute minimum value is -1/2, confirmed by comparing candidate values and graph analysis.', 'In the example involving the function f(x), the process of finding critical numbers and endpoints is demonstrated using piecewise notation and derivative analysis.']}, {'end': 33297.798, 'start': 32493.946, 'title': 'Mean value theorem and its applications', 'summary': "Introduces the mean value theorem, which states that for a function that is continuous on a closed interval and differentiable on the interior of that interval, there exists a number within the interval where the average rate of change of the function is equal to its instantaneous rate of change. it also discusses the verification and applications of the mean value theorem, including the relation to rolle's theorem, and provides two proofs of the mean value theorem for integrals.", 'duration': 803.852, 'highlights': ['The mean value theorem states that for a function continuous on a closed interval and differentiable on the interior of the interval, there exists a number within the interval where the average rate of change of the function is equal to its instantaneous rate of change.', 'The verification of the mean value theorem involves checking the hypotheses of continuity and differentiability, and then finding a number within the interval where the derivative of the function is equal to the average rate of change.', "Rolle's theorem is a special case of the mean value theorem, where it states that if a function is continuous on a closed interval, differentiable on the interior of the interval, and equal at the endpoints, then there exists a number within the interval where the derivative of the function is zero.", 'Two proofs of the mean value theorem for integrals are provided, the first utilizing the intermediate value theorem and the second as a corollary to the regular mean value theorem for functions.']}], 'duration': 2420.044, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w30877754.jpg', 'highlights': ['The process of finding absolute extreme values involves checking critical numbers and endpoints, demonstrated using derivatives and piecewise notation.', 'The mean value theorem states that for a function continuous on a closed interval and differentiable on the interior of the interval, there exists a number within the interval where the average rate of change of the function is equal to its instantaneous rate of change.', 'The first derivative test allows us to locate local extreme points by determining if the first derivative is positive on one side and negative on the other, and the second derivative test provides an alternative method using the second derivative to identify local max and min points.', 'The video explained the definitions and methods for finding absolute and local maximum and minimum values of functions, including the conditions for a function to have an absolute maximum or minimum at a specific point.']}, {'end': 34470.536, 'segs': [{'end': 33403.243, 'src': 'embed', 'start': 33299.451, 'weight': 0, 'content': [{'end': 33307.413, 'text': "In this video, we'll solve inequalities involving polynomials, like this one, and inequalities involving rational expressions, like this one.", 'start': 33299.451, 'duration': 7.962}, {'end': 33313.174, 'text': "Let's start with a simple example, maybe a deceptively simple example.", 'start': 33309.572, 'duration': 3.602}, {'end': 33317.035, 'text': 'If you see the inequality x squared is less than 4,', 'start': 33313.714, 'duration': 3.321}, {'end': 33324.756, 'text': 'you might be very tempted to take the square root of both sides and get something like x is less than 2 as your answer.', 'start': 33317.035, 'duration': 7.721}, {'end': 33327.396, 'text': "But in fact, that doesn't work.", 'start': 33325.497, 'duration': 1.899}, {'end': 33333.692, 'text': "To see why it's not correct, consider the x value of negative 10.", 'start': 33329.107, 'duration': 4.585}, {'end': 33346.991, 'text': "Negative 10 satisfies the inequality x is less than two, since negative 10 is less than two, but it doesn't satisfy.", 'start': 33333.692, 'duration': 13.299}, {'end': 33355.737, 'text': 'the inequality x squared is less than four, since negative 10 squared is 100, which is not less than four.', 'start': 33346.991, 'duration': 8.746}, {'end': 33359.36, 'text': 'So these two inequalities are not the same.', 'start': 33356.678, 'duration': 2.682}, {'end': 33365.203, 'text': "And it doesn't work to solve a quadratic inequality just to take the square root of both sides.", 'start': 33359.379, 'duration': 5.824}, {'end': 33367.078, 'text': 'you might be thinking.', 'start': 33366.178, 'duration': 0.9}, {'end': 33371.623, 'text': "part of why this reasoning is wrong is we've ignored the negative 2 option, right?", 'start': 33367.078, 'duration': 4.545}, {'end': 33377.928, 'text': 'If we had the equation x squared equals 4, then x equals 2 would just be one option.', 'start': 33371.663, 'duration': 6.265}, {'end': 33380.629, 'text': 'x equals negative 2 would be another solution.', 'start': 33377.928, 'duration': 2.701}, {'end': 33385.815, 'text': 'So somehow our solution to this inequality should take this into account.', 'start': 33381.371, 'duration': 4.444}, {'end': 33395.477, 'text': 'In fact, a good way to solve an inequality involving x squared or higher power terms is to solve the associated equation first.', 'start': 33387.089, 'duration': 8.388}, {'end': 33403.243, 'text': 'But before we even do that, I like to pull everything over to one side so that my inequality has zero on the other side.', 'start': 33396.317, 'duration': 6.926}], 'summary': 'Solving inequalities involving polynomials and rational expressions, using an example of x squared < 4 to illustrate the importance of considering all solutions.', 'duration': 103.792, 'max_score': 33299.451, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w33299451.jpg'}, {'end': 33721.598, 'src': 'embed', 'start': 33689.87, 'weight': 1, 'content': [{'end': 33693.871, 'text': "For a test value between zero and six, let's try x equals one.", 'start': 33689.87, 'duration': 4.001}, {'end': 33702.18, 'text': "Now I'll get a positive for this factor, a negative for this factor, and a positive for this factor.", 'start': 33696.314, 'duration': 5.866}, {'end': 33707.705, 'text': 'Positive times a negative times a positive gives me a negative.', 'start': 33704.201, 'duration': 3.504}, {'end': 33715.352, 'text': 'Finally, for a test value bigger than six, we could use say x equals 100.', 'start': 33709.026, 'duration': 6.326}, {'end': 33718.215, 'text': "That's going to give me positive, positive, positive.", 'start': 33715.352, 'duration': 2.863}, {'end': 33721.598, 'text': 'So my product will be positive.', 'start': 33719.536, 'duration': 2.062}], 'summary': 'Testing factors with x = 1 yields a negative result, while x = 100 gives a positive result.', 'duration': 31.728, 'max_score': 33689.87, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w33689870.jpg'}, {'end': 34029.658, 'src': 'embed', 'start': 34000.984, 'weight': 11, 'content': [{'end': 34009.748, 'text': 'The graph is decreasing for x values between 6 and 10 and for x values between 11 and 12.', 'start': 34000.984, 'duration': 8.764}, {'end': 34014.351, 'text': 'The first derivative of f can tell us where the function is increasing and decreasing.', 'start': 34009.748, 'duration': 4.603}, {'end': 34022.615, 'text': 'In particular, if f prime of x is greater than 0 for all x on an interval, then f is increasing on this interval.', 'start': 34015.071, 'duration': 7.544}, {'end': 34029.658, 'text': 'This makes sense because f prime being greater than 0 means the tangent line has positive slope.', 'start': 34023.775, 'duration': 5.883}], 'summary': "Graph decreases for x values 6-10 and 11-12; first derivative determines function's trends.", 'duration': 28.674, 'max_score': 34000.984, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w34000984.jpg'}, {'end': 34121.904, 'src': 'embed', 'start': 34090.019, 'weight': 6, 'content': [{'end': 34098.402, 'text': "Or, more formally, the function's concave down if all the tangent lines lie above the graph of the function on that interval.", 'start': 34090.019, 'duration': 8.383}, {'end': 34106.186, 'text': 'In this example, f is concave up around here and again around here.', 'start': 34099.963, 'duration': 6.223}, {'end': 34110.588, 'text': 'On the left piece, it looks like part of a bowl that could hold water.', 'start': 34107.366, 'duration': 3.222}, {'end': 34121.904, 'text': 'So we can say that F is concave up on the intervals from 2 to 4 and the interval from 8 to 11.', 'start': 34112.301, 'duration': 9.603}], 'summary': 'The function is concave up on the intervals 2 to 4 and 8 to 11.', 'duration': 31.885, 'max_score': 34090.019, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w34090019.jpg'}, {'end': 34241.293, 'src': 'embed', 'start': 34209.461, 'weight': 2, 'content': [{'end': 34213.642, 'text': 'In general, we can use the second derivative to predict the concavity of a function.', 'start': 34209.461, 'duration': 4.181}, {'end': 34224.241, 'text': 'The concavity test says that if the second derivative is positive for all x on an interval, then the function f is concave up on that interval.', 'start': 34214.534, 'duration': 9.707}, {'end': 34234.188, 'text': 'Similarly, if the second derivative is negative for all x on an interval, then the function f is concave down on that interval.', 'start': 34225.322, 'duration': 8.866}, {'end': 34241.293, 'text': 'One way to remember the concavity test is that a positive second derivative gives us a happy face.', 'start': 34236.009, 'duration': 5.284}], 'summary': 'Using second derivative to predict function concavity. positive second derivative leads to concave up, negative to concave down.', 'duration': 31.832, 'max_score': 34209.461, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w34209461.jpg'}, {'end': 34292.06, 'src': 'embed', 'start': 34256.843, 'weight': 7, 'content': [{'end': 34267.245, 'text': "A function has an inflection point at x equals c if it's continuous at c and it changes concavity at c.", 'start': 34256.843, 'duration': 10.402}, {'end': 34270.946, 'text': 'In other words, f has an inflection point at x equals c.', 'start': 34267.245, 'duration': 3.701}, {'end': 34282.189, 'text': 'if f changes from concave up to concave down at x equals c or it changes from concave down to concave up.', 'start': 34270.946, 'duration': 11.243}, {'end': 34292.06, 'text': 'In this graph, if we draw the concavity regions again, we see that f has an inflection point at x equals 2,', 'start': 34284.031, 'duration': 8.029}], 'summary': 'A function has an inflection point at x equals 2.', 'duration': 35.217, 'max_score': 34256.843, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w34256843.jpg'}, {'end': 34423.58, 'src': 'embed', 'start': 34397.162, 'weight': 3, 'content': [{'end': 34401.784, 'text': 'while the second derivative can tell us where the function is concave up and concave down.', 'start': 34397.162, 'duration': 4.622}, {'end': 34409.726, 'text': 'And the second derivative changing sign from positive to negative or negative to positive can tell us where we have inflection points.', 'start': 34402.924, 'duration': 6.802}, {'end': 34415.474, 'text': 'Since lines are much easier to work with than more complicated functions,', 'start': 34411.812, 'duration': 3.662}, {'end': 34421.238, 'text': 'it can be extremely useful to approximate a function near a particular value with its tangent line.', 'start': 34415.474, 'duration': 5.764}, {'end': 34423.58, 'text': "That's the central idea of this video.", 'start': 34422.019, 'duration': 1.561}], 'summary': 'Second derivative indicates concavity and inflection points, tangent lines useful for function approximation.', 'duration': 26.418, 'max_score': 34397.162, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w34397162.jpg'}], 'start': 33299.451, 'title': 'Inequality solving techniques', 'summary': 'Provides techniques for solving inequalities involving polynomials and rational expressions, emphasizing the importance of considering all solutions when solving quadratic inequalities, and applying concepts such as factoring, test values, and sign charts, with practical examples and applications.', 'chapters': [{'end': 33403.243, 'start': 33299.451, 'title': 'Inequality solving techniques', 'summary': 'Discusses solving inequalities involving polynomials and rational expressions, emphasizing the incorrect approach of solely taking the square root of both sides for a quadratic inequality and the importance of considering all solutions when solving such inequalities, using specific examples.', 'duration': 103.792, 'highlights': ['Solving quadratic inequalities by solely taking the square root of both sides is incorrect, as it ignores the negative solution and may lead to incorrect results.', 'Negative 10 satisfies the inequality x is less than two but does not satisfy the inequality x squared is less than four, illustrating the difference between the two inequalities.', 'When solving an inequality involving x squared or higher power terms, it is essential to consider all solutions, as in the case where x squared equals 4, resulting in x equals 2 and x equals negative 2 as solutions.']}, {'end': 33906.84, 'start': 33404.764, 'title': 'Solving inequalities and rational inequality', 'summary': 'Explains the process of solving inequalities and rational inequalities using factoring, test values, and identifying where the expression is positive or negative, showcasing examples with clear explanations and detailed calculations.', 'duration': 502.076, 'highlights': ['The process of solving inequalities and rational inequalities using factoring, test values, and identifying where the expression is positive or negative.', 'Detailed examples with clear explanations and calculations.', 'Using test values to determine the positivity or negativity of the expression.']}, {'end': 34470.536, 'start': 33908.857, 'title': 'Solving polynomial and rational inequalities', 'summary': 'Covers the concepts of solving polynomial and rational inequalities, using test values to make a sign chart, understanding the first and second derivative of a function and their implications in determining increasing/decreasing behavior, concavity, and inflection points, with practical examples and applications.', 'duration': 561.679, 'highlights': ['The first and second derivative of a function can tell us a lot about the shape of the graph of the function, including where the function is increasing and decreasing, concave up and concave down, and has inflection points.', 'The first derivative of a function can tell us where the function is increasing and decreasing, with f prime of x > 0 indicating increase and f prime of x < 0 indicating decrease.', 'The function is concave up on an interval if all the tangent lines lie below the graph of the function, and concave down if all the tangent lines lie above the graph of the function on that interval.', 'The concavity of a function is related to its second derivative, with a positive second derivative indicating concave up and a negative second derivative indicating concave down.', 'The concavity test states that if the second derivative is positive for all x on an interval, then the function f is concave up on that interval, and if the second derivative is negative for all x on an interval, then the function f is concave down on that interval.', 'A function has an inflection point at x=c if it changes concavity at x=c, and the inflection point test states that if f double prime of x changes sign at x=c, then f has an inflection point at x=c.']}], 'duration': 1171.085, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w33299451.jpg', 'highlights': ['Solving an inequality involving x squared or higher power terms requires considering all solutions.', 'Using test values to determine the positivity or negativity of the expression.', 'The concavity of a function is related to its second derivative, with a positive second derivative indicating concave up and a negative second derivative indicating concave down.', 'The first and second derivative of a function can tell us a lot about the shape of the graph of the function, including where the function is increasing and decreasing, concave up and concave down, and has inflection points.', 'The process of solving inequalities and rational inequalities using factoring, test values, and identifying where the expression is positive or negative.', 'The concavity test states that if the second derivative is positive for all x on an interval, then the function f is concave up on that interval, and if the second derivative is negative for all x on an interval, then the function f is concave down on that interval.', 'The function is concave up on an interval if all the tangent lines lie below the graph of the function, and concave down if all the tangent lines lie above the graph of the function on that interval.', 'A function has an inflection point at x=c if it changes concavity at x=c, and the inflection point test states that if f double prime of x changes sign at x=c, then f has an inflection point at x=c.', 'Solving quadratic inequalities by solely taking the square root of both sides is incorrect, as it ignores the negative solution and may lead to incorrect results.', 'Negative 10 satisfies the inequality x is less than two but does not satisfy the inequality x squared is less than four, illustrating the difference between the two inequalities.', 'When solving an inequality involving x squared or higher power terms, it is essential to consider all solutions, as in the case where x squared equals 4, resulting in x equals 2 and x equals negative 2 as solutions.', 'The first derivative of a function can tell us where the function is increasing and decreasing, with f prime of x > 0 indicating increase and f prime of x < 0 indicating decrease.']}, {'end': 35898.998, 'segs': [{'end': 34499.739, 'src': 'embed', 'start': 34471.456, 'weight': 0, 'content': [{'end': 34477.839, 'text': 'If this rate of change continues, then by 7 a.m., the temperature will have risen 3 degrees and reached 63 degrees.', 'start': 34471.456, 'duration': 6.383}, {'end': 34489.734, 'text': 'And by 8 a.m., the temperature will have had 2 hours to rise from 60 degrees at a rate of 3 degrees per hour.', 'start': 34478.319, 'duration': 11.415}, {'end': 34499.739, 'text': 'So f of 8 should be about 60 degrees plus the 3 degrees per hour times 2 hours, or 66 degrees.', 'start': 34490.354, 'duration': 9.385}], 'summary': 'Temperature is expected to rise to 63 degrees by 7 a.m. and to 66 degrees by 8 a.m.', 'duration': 28.283, 'max_score': 34471.456, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w34471456.jpg'}, {'end': 34560.634, 'src': 'embed', 'start': 34529.922, 'weight': 1, 'content': [{'end': 34532.764, 'text': 'The tangent line has slope three degrees per hour.', 'start': 34529.922, 'duration': 2.842}, {'end': 34543.648, 'text': "So that's a rise of a run of three, which means that seven o'clock which is one hour after six o'clock, the tangent line has risen by three degrees.", 'start': 34533.704, 'duration': 9.944}, {'end': 34549.368, 'text': "and at 8 o'clock, the tangent line has risen by another 3 degrees.", 'start': 34544.625, 'duration': 4.743}, {'end': 34560.634, 'text': "So at 7 o'clock, our tangent line has height 63 degrees, and at 8 o'clock, our tangent line has height 66 degrees.", 'start': 34551.469, 'duration': 9.165}], 'summary': "Tangent line rises by 3 degrees per hour, reaching 66 degrees at 8 o'clock.", 'duration': 30.712, 'max_score': 34529.922, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w34529922.jpg'}, {'end': 34614.27, 'src': 'embed', 'start': 34589.133, 'weight': 9, 'content': [{'end': 34594.838, 'text': 'The idea of approximating a function with its tangent line is a very important idea that works for any differentiable function.', 'start': 34589.133, 'duration': 5.705}, {'end': 34601.284, 'text': 'Let f of x be any differentiable function and let a be an arbitrary x value.', 'start': 34596.38, 'duration': 4.904}, {'end': 34606.026, 'text': "Let's suppose we know the value of f at a.", 'start': 34602.264, 'duration': 3.762}, {'end': 34608.607, 'text': "We'll call it f of a.", 'start': 34606.026, 'duration': 2.581}, {'end': 34614.27, 'text': "And let's say we want to find the value of f at an x value near a.", 'start': 34608.607, 'duration': 5.663}], 'summary': 'Approximating a function with its tangent line is important for any differentiable function.', 'duration': 25.137, 'max_score': 34589.133, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w34589133.jpg'}, {'end': 34773.889, 'src': 'embed', 'start': 34742.959, 'weight': 10, 'content': [{'end': 34751.684, 'text': 'So the approximation principle can also be written as f of x is approximately equal to L of x.', 'start': 34742.959, 'duration': 8.725}, {'end': 34756.647, 'text': "Let's look a little more closely at this linearization equation and what it has to do with the tangent line.", 'start': 34751.684, 'duration': 4.963}, {'end': 34762.77, 'text': 'Suppose we were going to try to write down the equation of the tangent line at x equals a.', 'start': 34757.287, 'duration': 5.483}, {'end': 34773.889, 'text': 'Well, the equation for any line can be given in point slope form as Y minus Y naught equals M, the slope, times X minus X naught.', 'start': 34763.666, 'duration': 10.223}], 'summary': 'Linearization equation approximates f(x) as l(x). tries to find tangent line equation at x=a.', 'duration': 30.93, 'max_score': 34742.959, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w34742959.jpg'}, {'end': 34885.03, 'src': 'embed', 'start': 34852.426, 'weight': 3, 'content': [{'end': 34863.673, 'text': 'The approximation principle tells us that f of a plus delta x is approximately equal to f of a plus f prime of a times delta x.', 'start': 34852.426, 'duration': 11.247}, {'end': 34870.938, 'text': 'we need to figure out what f should be what a should be and what delta x should be.', 'start': 34866.854, 'duration': 4.084}, {'end': 34878.444, 'text': "Since we're trying to figure out the square root of 59, it makes sense to make our function the square root function.", 'start': 34871.698, 'duration': 6.746}, {'end': 34885.03, 'text': "For a, we'd like to pick something that is easy to compute f of a.", 'start': 34880.126, 'duration': 4.904}], 'summary': "Approximation principle: f(a + δx) ≈ f(a) + f'(a)δx, finding square root of 59.", 'duration': 32.604, 'max_score': 34852.426, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w34852426.jpg'}, {'end': 35077.294, 'src': 'embed', 'start': 35050.91, 'weight': 5, 'content': [{'end': 35059.097, 'text': 'And the approximation principle says that f of x, the function, is approximately equal to its linearization, its tangent line,', 'start': 35050.91, 'duration': 8.187}, {'end': 35061.559, 'text': 'at least when x is near a.', 'start': 35059.097, 'duration': 2.462}, {'end': 35064.721, 'text': 'This is basically the same formula that we used in the last problem.', 'start': 35061.559, 'duration': 3.162}, {'end': 35070.492, 'text': "we're just calling our value x this time instead of a plus delta x.", 'start': 35065.39, 'duration': 5.102}, {'end': 35077.294, 'text': "Since we're trying to estimate sine of a value, it makes sense to let our function be sine of x.", 'start': 35070.492, 'duration': 6.802}], 'summary': 'The approximation principle states f(x) ≈ its tangent line near a.', 'duration': 26.384, 'max_score': 35050.91, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w35050910.jpg'}, {'end': 35324.609, 'src': 'embed', 'start': 35240.651, 'weight': 4, 'content': [{'end': 35247.374, 'text': "So you can see our approximation using the linearization is very close to the calculator's more accurate value.", 'start': 35240.651, 'duration': 6.723}, {'end': 35256.211, 'text': 'Notice that in this example, the approximate value based on the linearization is slightly higher than the actual value.', 'start': 35248.709, 'duration': 7.502}, {'end': 35260.512, 'text': 'And you can see why from a graph of sine.', 'start': 35257.211, 'duration': 3.301}, {'end': 35273.195, 'text': 'The tangent line at pi over 6 lies slightly above the graph of sine x.', 'start': 35262.432, 'duration': 10.763}, {'end': 35282.388, 'text': 'Therefore, the approximate value based on the linearization will be slightly bigger than the actual value of sine of 33 degrees.', 'start': 35273.195, 'duration': 9.193}, {'end': 35287.831, 'text': 'In this video, we used several formulas to express one key idea.', 'start': 35283.829, 'duration': 4.002}, {'end': 35296.496, 'text': 'The main formulas were the approximation principle, the linear approximation, and the linearization.', 'start': 35288.571, 'duration': 7.925}, {'end': 35306.441, 'text': 'The key idea is that a differentiable function can be approximated near a value x equals a by the tangent line at x equals a.', 'start': 35298.056, 'duration': 8.385}, {'end': 35316.682, 'text': 'The differential is a new vocabulary word wrapped around the familiar concept of approximating a function with its tangent line.', 'start': 35309.195, 'duration': 7.487}, {'end': 35323.468, 'text': 'This figure should look familiar from the previous video on linear approximation.', 'start': 35319.925, 'duration': 3.543}, {'end': 35324.609, 'text': "It's the same picture.", 'start': 35323.848, 'duration': 0.761}], 'summary': 'Linear approximation approximates sine function with tangent line near x=a.', 'duration': 83.958, 'max_score': 35240.651, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w35240651.jpg'}, {'end': 35898.998, 'src': 'embed', 'start': 35862.774, 'weight': 8, 'content': [{'end': 35869.495, 'text': 'And DV, we already saw was four pi times eight squared times 0.5.', 'start': 35862.774, 'duration': 6.721}, {'end': 35881.142, 'text': 'So DV divided by V is given by this ratio, which simplifies to 0.1875.', 'start': 35869.495, 'duration': 11.647}, {'end': 35886.486, 'text': 'So an 18.75% relative error.', 'start': 35881.142, 'duration': 5.344}, {'end': 35893.333, 'text': 'To me, the relative error gives a better sense for the error than the absolute error estimate above.', 'start': 35887.848, 'duration': 5.485}, {'end': 35898.998, 'text': 'This video introduced the idea of the differential.', 'start': 35896.816, 'duration': 2.182}], 'summary': 'Relative error is 18.75%, better than absolute error estimate.', 'duration': 36.224, 'max_score': 35862.774, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w35862774.jpg'}], 'start': 34471.456, 'title': 'Analyzing temperature change and approximating functions', 'summary': "Discusses the temperature's change over time, estimating a rise of 3 degrees by 7 a.m. and reaching 63 degrees, and a further rise to 66 degrees by 8 a.m. based on a rate of 3 degrees per hour. it also explains the use of tangent lines to approximate functions, including the linear approximation principle and its alternative forms, and discusses linearization to approximate the actual value of a function, demonstrating the application of the approximation principle and the concept of linear approximation. furthermore, it explains how a differentiable function can be approximated by a tangent line and introduces the concept of differentials, demonstrating their use in estimating error, with a relative error of 18.75% computed.", 'chapters': [{'end': 34560.634, 'start': 34471.456, 'title': 'Temperature rate of change analysis', 'summary': "Discusses the temperature's change over time, estimating a rise of 3 degrees by 7 a.m. and reaching 63 degrees, and a further rise to 66 degrees by 8 a.m. based on a rate of 3 degrees per hour, visually represented by a tangent line.", 'duration': 89.178, 'highlights': ['By 8 a.m., the temperature will have had 2 hours to rise from 60 degrees at a rate of 3 degrees per hour, resulting in an estimate of 66 degrees.', "At 7 o'clock, the tangent line has risen by 3 degrees, reaching a height of 63 degrees.", "The tangent line at time six has a slope of three degrees per hour, indicating a rise of 3 degrees at 7 o'clock and another 3 degrees at 8 o'clock."]}, {'end': 34993.161, 'start': 34561.575, 'title': 'Approximating functions with tangent lines', 'summary': 'Explains the use of tangent lines to approximate functions, including the linear approximation principle and its alternative forms, with an example of using the approximation principle to compute the value of the square root of 59.', 'duration': 431.586, 'highlights': ['The linear approximation principle states that f of a plus delta x is approximately equal to f of a plus f prime of a times delta x, exemplified by the calculation of the square root of 59 using the square root function with a value of 64 and a delta x of -5, resulting in an approximation of 7.6875 and a more exact value of 7.68114575.', "The concept of approximating a function with its tangent line is highlighted as an important idea that works for any differentiable function, emphasizing the use of the tangent line to approximate the function when time is near six o'clock and its application to any differentiable function f of x with an arbitrary x value a.", 'The explanation of the linearization equation and its relationship with the tangent line is detailed, demonstrating the equivalence of the equation for the tangent line and the linearization equation, which is referred to as L of x and represents the linearization of f at a.', 'The chapter provides a detailed breakdown of the linear approximation principle, its alternative forms, and the concept of linearization, along with the application of the approximation principle in computing the square root of 59 using the square root function with a value of 64 and a delta x of -5.']}, {'end': 35296.496, 'start': 34994.942, 'title': 'Linearization and approximation principle', 'summary': "Discusses the use of linearization to approximate the actual value of a function, using the example of estimating the sine of 33 degrees, where the linearization provides a close approximation of 0.5453 as compared to the calculator's value of 0.5446, demonstrating the application of the approximation principle and the concept of linear approximation.", 'duration': 301.554, 'highlights': ["The linearization provides a close approximation of 0.5453 as compared to the calculator's value of 0.5446.", 'The example showcases the application of the approximation principle and the concept of linear approximation.', 'The chapter emphasizes the use of several formulas, including the approximation principle, the linear approximation, and the linearization.']}, {'end': 35898.998, 'start': 35298.056, 'title': 'Differential approximation', 'summary': 'Explains how a differentiable function can be approximated by a tangent line, introduces the concept of differentials, and demonstrates their use in estimating error, with a relative error of 18.75% computed.', 'duration': 600.942, 'highlights': ['The chapter explains how a differentiable function can be approximated by a tangent line', 'Introduces the concept of differentials', 'Demonstrates the use of differentials in estimating error, with a relative error of 18.75% computed']}], 'duration': 1427.542, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w34471456.jpg', 'highlights': ['By 8 a.m., the temperature will have had 2 hours to rise from 60 degrees at a rate of 3 degrees per hour, resulting in an estimate of 66 degrees.', "At 7 o'clock, the tangent line has risen by 3 degrees, reaching a height of 63 degrees.", "The tangent line at time six has a slope of three degrees per hour, indicating a rise of 3 degrees at 7 o'clock and another 3 degrees at 8 o'clock.", 'The linear approximation principle states that f of a plus delta x is approximately equal to f of a plus f prime of a times delta x, exemplified by the calculation of the square root of 59 using the square root function with a value of 64 and a delta x of -5, resulting in an approximation of 7.6875 and a more exact value of 7.68114575.', "The linearization provides a close approximation of 0.5453 as compared to the calculator's value of 0.5446.", 'The example showcases the application of the approximation principle and the concept of linear approximation.', 'The chapter explains how a differentiable function can be approximated by a tangent line', 'Introduces the concept of differentials', 'Demonstrates the use of differentials in estimating error, with a relative error of 18.75% computed', "The concept of approximating a function with its tangent line is highlighted as an important idea that works for any differentiable function, emphasizing the use of the tangent line to approximate the function when time is near six o'clock and its application to any differentiable function f of x with an arbitrary x value a.", 'The explanation of the linearization equation and its relationship with the tangent line is detailed, demonstrating the equivalence of the equation for the tangent line and the linearization equation, which is referred to as L of x and represents the linearization of f at a.', 'The chapter provides a detailed breakdown of the linear approximation principle, its alternative forms, and the concept of linearization, along with the application of the approximation principle in computing the square root of 59 using the square root function with a value of 64 and a delta x of -5.', 'The chapter emphasizes the use of several formulas, including the approximation principle, the linear approximation, and the linearization.']}, {'end': 38501.319, 'segs': [{'end': 36123.97, 'src': 'embed', 'start': 36101.405, 'weight': 0, 'content': [{'end': 36110.367, 'text': "L'Hopital's rule can be applied when f and g are differentiable functions and the derivative of g is non-zero in some open interval around a,", 'start': 36101.405, 'duration': 8.962}, {'end': 36113.548, 'text': 'except possibly at a.', 'start': 36110.367, 'duration': 3.181}, {'end': 36123.97, 'text': 'Under these conditions, if the limit, as x goes to a, of f of x over g of x is a 0 over 0 or infinity over infinity indeterminate form,', 'start': 36113.548, 'duration': 10.422}], 'summary': "L'hopital's rule applies when f and g are differentiable functions and the derivative of g is non-zero in an open interval around a.", 'duration': 22.565, 'max_score': 36101.405, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w36101405.jpg'}, {'end': 36395.75, 'src': 'embed', 'start': 36363.106, 'weight': 1, 'content': [{'end': 36379.397, 'text': 'we were able to evaluate 0 over 0 and infinity over infinity in determinant forms by replacing the limit of f of x over g of x with the limit of f prime of x over g prime of x,', 'start': 36363.106, 'duration': 16.291}, {'end': 36381.038, 'text': 'provided that second limit exists.', 'start': 36379.397, 'duration': 1.641}, {'end': 36384.78, 'text': "This trick is known as L'Hopital's Rule.", 'start': 36382.579, 'duration': 2.201}, {'end': 36393.768, 'text': "We've seen that L'Hopital's rule can be used to evaluate limits of the form 0 over 0 or infinity over infinity.", 'start': 36387.441, 'duration': 6.327}, {'end': 36395.75, 'text': 'In this video.', 'start': 36395.109, 'duration': 0.641}], 'summary': "L'hopital's rule can evaluate 0/0 and ∞/∞ limits with f'(x)/g'(x) if the second limit exists.", 'duration': 32.644, 'max_score': 36363.106, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w36363106.jpg'}, {'end': 37493.939, 'src': 'embed', 'start': 37463.304, 'weight': 2, 'content': [{'end': 37469.727, 'text': "So at this point, my Newton's methods iterations have converged, and I have an answer that's accurate to about eight decimal places.", 'start': 37463.304, 'duration': 6.423}, {'end': 37473.389, 'text': 'I found one zero for my function.', 'start': 37471.408, 'duration': 1.981}, {'end': 37481.893, 'text': "And if I wanted to find the second zero, the one over here, I would just need to start with an initial value that's close to this x coordinate.", 'start': 37473.809, 'duration': 8.084}, {'end': 37484.294, 'text': 'perhaps an initial value of zero might be good.', 'start': 37481.893, 'duration': 2.401}, {'end': 37493.939, 'text': 'In this video, we developed an algorithm for getting increasingly accurate approximations to the zero of a function.', 'start': 37487.136, 'duration': 6.803}], 'summary': "Newton's method converged, found one accurate zero, developed algorithm for getting accurate approximations.", 'duration': 30.635, 'max_score': 37463.304, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w37463304.jpg'}, {'end': 37927.222, 'src': 'embed', 'start': 37893.782, 'weight': 3, 'content': [{'end': 37901.549, 'text': 'The antiderivative of f of x plus g of x is capital F of x plus capital G of x plus c,', 'start': 37893.782, 'duration': 7.767}, {'end': 37908.652, 'text': 'where capital F and capital G are the antiderivatives of lowercase f and lowercase g.', 'start': 37902.346, 'duration': 6.306}, {'end': 37914.058, 'text': 'This is because the derivative of a sum is equal to the sum of the derivatives.', 'start': 37908.652, 'duration': 5.406}, {'end': 37927.222, 'text': "Let's use this information to compute the antiderivative for f of x equals 5 over 1 plus x squared minus 1 over 2 times the square root of x.", 'start': 37916.636, 'duration': 10.586}], 'summary': 'The antiderivative of f(x) + g(x) is f(x) + g(x) + c, as the derivative of a sum is the sum of the derivatives.', 'duration': 33.44, 'max_score': 37893.782, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w37893782.jpg'}, {'end': 38048.395, 'src': 'embed', 'start': 38013.106, 'weight': 4, 'content': [{'end': 38016.367, 'text': 'something like f of 1 equals 7..', 'start': 38013.106, 'duration': 3.261}, {'end': 38020.508, 'text': 'And we have to find the function f of x.', 'start': 38016.367, 'duration': 4.141}, {'end': 38026.891, 'text': 'In this first example, suppose g prime of x is e to the x minus 3 times sine x.', 'start': 38020.508, 'duration': 6.383}, {'end': 38029.712, 'text': 'And g of 2 pi is 5.', 'start': 38026.891, 'duration': 2.821}, {'end': 38030.792, 'text': 'We need to find g of x.', 'start': 38029.712, 'duration': 1.08}, {'end': 38048.395, 'text': 'Well g of x is an antiderivative of e to the x minus 3 sine x, so g of x is of the form e to the x plus 3 cosine of x plus a constant c.', 'start': 38031.991, 'duration': 16.404}], 'summary': "Given g'(x) = e^x - 3sin(x), g(2π) = 5, find g(x) = e^x + 3cos(x) + c", 'duration': 35.289, 'max_score': 38013.106, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w38013106.jpg'}], 'start': 35900.748, 'title': 'Calculus techniques', 'summary': "Discusses l'hopital's rule and newton's method for evaluating limits in indeterminate forms and finding zeros of a function, as well as the concept of antiderivatives, with examples and applications.", 'chapters': [{'end': 36363.106, 'start': 35900.748, 'title': "L'hopital's rule in calculus", 'summary': "Discusses the concept of differentials, indeterminate forms, and introduces l'hopital's rule as a powerful technique for evaluating limits in indeterminate forms, with examples of 0 over 0 and infinity over infinity indeterminate forms, and emphasizes the importance of simplifying after each application of l'hopital's rule.", 'duration': 462.358, 'highlights': ["L'Hopital's rule can be applied when f and g are differentiable functions and the derivative of g is non-zero in some open interval around a, except possibly at a.", 'The limit, as x goes to a, of f of x over g of x is the same thing as the limit, as x goes to a, of f prime of x over g prime of x, provided that the second limit exists or is plus or minus infinity.', "Emphasizes the importance of simplifying after each application of L'Hopital's rule to avoid unnecessary complications and to solve the problem efficiently."]}, {'end': 36842.214, 'start': 36363.106, 'title': "L'hopital's rule for evaluating indeterminate forms", 'summary': "Explains the application of l'hopital's rule to evaluate indeterminate forms like 0 times infinity and 1 to the infinity using examples and step-by-step derivations, resulting in limits of 0 and 1 respectively.", 'duration': 479.108, 'highlights': ["L'Hopital's rule can be used to evaluate limits of the form 0 over 0 or infinity over infinity, as well as additional indeterminate forms like 0 times infinity and 1 to the infinity.", "By applying L'Hopital's rule, the limit of a product with an indeterminate form of 0 times infinity is evaluated step-by-step, resulting in a limit of 0.", "The application of L'Hopital's rule to a limit with an indeterminate form of 1 to the infinity is demonstrated using logarithms, resulting in a limit of 1."]}, {'end': 37518.103, 'start': 36842.214, 'title': "Newton's method for finding zeros", 'summary': "Explains newton's method for finding zeros of a function, with a detailed explanation of the algorithm and its application to the equation e^x = 4x, resulting in a highly accurate approximation to the zero of the function.", 'duration': 675.889, 'highlights': ["Newton's method for finding zeros of a function", 'Detailed explanation of the algorithm', 'Application to the equation e^x = 4x']}, {'end': 37802.59, 'start': 37518.103, 'title': 'Antidifferentiating and antiderivatives', 'summary': 'Discusses the concept of antidifferentiating, antiderivatives, and the power rule for anti-differentiating, highlighting the relationship between derivatives and antiderivatives and providing examples of finding antiderivatives for different functions.', 'duration': 284.487, 'highlights': ["The derivative of x to the n for any n that's not equal to negative 1 is given by x to the n plus 1, divided by n plus 1 plus a constant C.", 'Antiderivatives can be written in the form capital F of x plus c for some constant c, where capital F of X is an antiderivative for little f of x.', 'The antiderivative of x is x squared over 2, and the antiderivative of 1 is x, each with the addition of a constant C to form a more general antiderivative.', 'The chapter also explains that the derivative of a constant is 0, so the derivative of x cubed plus a constant is just going to be 3x squared, no matter what the constant is.', 'The chapter also discusses the relationship between derivatives and antiderivatives, illustrating that if two functions have the same derivative, then the distance between their antiderivatives will always stay exactly the same.']}, {'end': 38501.319, 'start': 37803.05, 'title': 'Understanding antiderivatives and solving problems', 'summary': 'Covers antiderivatives, including examples of finding antiderivatives of functions, solving initial value problems, and applying antiderivatives to physical problems involving velocity and position.', 'duration': 698.269, 'highlights': ['We recognize that the antiderivative of 1 over x is just ln of the absolute value of x plus c, since the derivative of ln of the absolute value of x is 1 over x.', 'The antiderivative of a times x to the n is a times the antiderivative of x to the n, which is x to the n plus one over n plus one plus a constant C.', 'The antiderivative of f of x plus g of x is capital F of x plus capital G of x plus c, where capital F and capital G are the antiderivatives of lowercase f and lowercase g.', 'The antiderivative of e to the x minus 3 sine x is e to the x plus 3 cosine of x plus a constant c.', 'The antiderivative of f double prime of x is found by using the power rule for antiderivatives and adding a constant, c.', 'The time of impact for the tomato is approximately 5.25 seconds, and the velocity at impact is approximately -31.45 meters per second.']}], 'duration': 2600.571, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w35900748.jpg', 'highlights': ["L'Hopital's rule applies when f and g are differentiable functions and g' is non-zero at a", "L'Hopital's rule evaluates limits of 0/0, ∞/∞, 0*∞, and 1^∞, simplifying step-by-step", "Newton's method for finding zeros of a function, with detailed algorithm explanation", 'Antiderivatives: derivative of x^n is x^(n+1)/(n+1), and antiderivatives can be written as F(x)+c', "Antiderivatives of 1/x, a*x^n, f(x)+g(x), e^x-3sin(x), f''(x) using power rule, and application example"]}, {'end': 40912.011, 'segs': [{'end': 38540.998, 'src': 'embed', 'start': 38514.076, 'weight': 2, 'content': [{'end': 38518.179, 'text': "And that's all for this video on finding antiderivatives using initial conditions.", 'start': 38514.076, 'duration': 4.103}, {'end': 38520.801, 'text': 'In this video,', 'start': 38519.841, 'duration': 0.96}, {'end': 38531.55, 'text': "I'll use the mean value theorem to show that the antiderivative of 0 has to be a constant and that any two antiderivatives of the same function have to differ by a constant.", 'start': 38520.801, 'duration': 10.749}, {'end': 38540.998, 'text': 'In a previous video I stated the fact that if f is one antiderivative of a function f,', 'start': 38533.512, 'duration': 7.486}], 'summary': 'Finding antiderivatives with mean value theorem and constant differences.', 'duration': 26.922, 'max_score': 38514.076, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w38514076.jpg'}, {'end': 39142.358, 'src': 'embed', 'start': 39111.274, 'weight': 3, 'content': [{'end': 39116.617, 'text': 'The first one is point five times one, and the last one three is point five times six.', 'start': 39111.274, 'duration': 5.343}, {'end': 39123.607, 'text': 'Now if we compute this sum, we get 91 eighths which is 11.375.', 'start': 39118.138, 'duration': 5.469}, {'end': 39133.113, 'text': 'Notice from the picture that this sum of areas of rectangles is an overestimate for the area under the curve.', 'start': 39123.607, 'duration': 9.506}, {'end': 39142.358, 'text': "We can do the same sort of computation for this green picture using left end points, and we'll get an underestimate for the area under the curve.", 'start': 39134.194, 'duration': 8.164}], 'summary': 'Summing rectangles yields overestimate & underestimate for area under curve.', 'duration': 31.084, 'max_score': 39111.274, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w39111274.jpg'}, {'end': 39820.815, 'src': 'embed', 'start': 39795.683, 'weight': 0, 'content': [{'end': 39803.367, 'text': 'multiply that by my 9 halves, and I get a limit of 9, just like I expected from the previous work.', 'start': 39795.683, 'duration': 7.684}, {'end': 39812.931, 'text': 'So that was a big production, but we did successfully find the area under the curve, and it was 9.', 'start': 39804.967, 'duration': 7.964}, {'end': 39820.815, 'text': 'In this video, we approximated the area under a curve by taking the limit, as the number of rectangles goes to infinity,', 'start': 39812.931, 'duration': 7.884}], 'summary': 'Area under the curve approximated as 9 using limit and infinite rectangles', 'duration': 25.132, 'max_score': 39795.683, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w39795683.jpg'}, {'end': 40651.68, 'src': 'embed', 'start': 40630.733, 'weight': 1, 'content': [{'end': 40639.877, 'text': 'Part two of the fundamental theorem of calculus is super useful because it allows us to compute integrals simply by finding antiderivatives and evaluating them.', 'start': 40630.733, 'duration': 9.144}, {'end': 40643.318, 'text': 'Finding antiderivatives tends to be really easy.', 'start': 40641.197, 'duration': 2.121}, {'end': 40648.1, 'text': 'Computing integrals using the Riemann sum definition is really hard.', 'start': 40644.098, 'duration': 4.002}, {'end': 40651.68, 'text': 'And so, because of the fundamental theorem of calculus,', 'start': 40649.318, 'duration': 2.362}], 'summary': 'Part two of the fundamental theorem of calculus simplifies integrals by finding antiderivatives easily.', 'duration': 20.947, 'max_score': 40630.733, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w40630733.jpg'}], 'start': 38514.076, 'title': 'Calculus fundamentals', 'summary': 'Covers antiderivatives, summation notation, approximating area under a curve, and the fundamental theorem of calculus. it includes properties of antiderivatives, area approximation techniques, and the concept of riemann sum. additionally, it explains the impact of increasing rectangles on accuracy and simplification of integral computation using antiderivatives, with a specific example for area under a curve being 9.', 'chapters': [{'end': 38950.376, 'start': 38514.076, 'title': 'Antiderivatives & summation notation', 'summary': 'Discusses the properties of antiderivatives, proving that any two antiderivatives of a function must differ by a constant, and reviews summation notation by providing examples and explaining how to write sums in sigma notation.', 'duration': 436.3, 'highlights': ['The antiderivative of 0 has to be a constant and any two antiderivatives of the same function have to differ by a constant, proven using the mean value theorem and the fact that if the derivative of a function is 0, the function must be a constant.', 'Reviewed summation notation using sigma notation to write sums and provided examples of evaluating sums, identifying patterns between terms, and different ways to express the sums in sigma notation.']}, {'end': 39393.261, 'start': 38952.917, 'title': 'Approximating area under a curve', 'summary': 'Discusses the approximation of the area under the curve y=x^2 using rectangles, showcasing the use of right and left endpoints to compute the sum of areas, resulting in overestimates and underestimates, and the impact of increasing the number of rectangles on the accuracy of the approximation.', 'duration': 440.344, 'highlights': ['The sum of areas of rectangles using right endpoints results in an overestimate for the area under the curve, computed as 11.375 for 6 rectangles.', 'The sum of areas of rectangles using left endpoints results in an underestimate for the area under the curve, computed as 6.875 for 6 rectangles.', 'Using 12 rectangles with right endpoints gives an overestimate of the area under the curve as 10.156, while using left endpoints gives an underestimate of 7.906.']}, {'end': 39820.815, 'start': 39393.261, 'title': 'Approximating area under a curve', 'summary': 'Explains how to approximate the area under the curve y = x squared between x = 0 and x = 3 using right and left endpoints, demonstrating the limit as the number of rectangles goes to infinity to find the exact area, which is 9.', 'duration': 427.554, 'highlights': ['The exact area under the curve y = x squared between x = 0 and x = 3 is found to be 9 using the limit as the number of rectangles goes to infinity, demonstrating the process of approximating the area using right and left endpoints.', 'The process of dividing into rectangles using right and left endpoints yields estimated areas of 9.1435 and 8.8654 respectively, providing insights into the accuracy of the approximation.', 'The width of each subinterval, or the base of each rectangle, is determined to be 3/n, leading to the estimation of area using right and left endpoints as the sum from i equals 1 to n of 3/n times 3i/n squared and 3/n times 3i-1/n squared respectively, highlighting the significance of the number of rectangles in the accuracy of the estimate.']}, {'end': 40629.372, 'start': 39822.222, 'title': 'Fundamental theorem of calculus', 'summary': 'Discusses the concept of riemann sum, accumulated area function and the fundamental theorem of calculus, stating that the derivative of the integral of a function is the original function, and the integral of the derivative is equal to the original function evaluated on the endpoints.', 'duration': 807.15, 'highlights': ['The derivative of the integral of a function is just the original function', 'The integral of the derivative is equal to the original function evaluated on the endpoints', 'Relationship between antiderivatives']}, {'end': 40912.011, 'start': 40630.733, 'title': 'Fundamental theorem of calculus', 'summary': 'Discusses the fundamental theorem of calculus, which simplifies the computation of integrals by finding antiderivatives, as demonstrated with examples involving evaluation of definite integrals.', 'duration': 281.278, 'highlights': ['The fundamental theorem of calculus allows us to compute integrals simply by finding antiderivatives and evaluating them.', 'Examples are provided to demonstrate the evaluation of definite integrals using antiderivatives.', 'The evaluation of an antiderivative at the endpoints is demonstrated using specific examples.', 'Simplification of expressions and application of the power rule in reverse are demonstrated for finding antiderivatives.']}], 'duration': 2397.935, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w38514076.jpg', 'highlights': ['The exact area under the curve y = x squared between x = 0 and x = 3 is found to be 9 using the limit as the number of rectangles goes to infinity', 'The fundamental theorem of calculus allows us to compute integrals simply by finding antiderivatives and evaluating them', 'The antiderivative of 0 has to be a constant and any two antiderivatives of the same function have to differ by a constant, proven using the mean value theorem', 'The sum of areas of rectangles using right endpoints results in an overestimate for the area under the curve, computed as 11.375 for 6 rectangles', 'The sum of areas of rectangles using left endpoints results in an underestimate for the area under the curve, computed as 6.875 for 6 rectangles']}, {'end': 42826.57, 'segs': [{'end': 40970.835, 'src': 'embed', 'start': 40942.826, 'weight': 0, 'content': [{'end': 40949.871, 'text': 'The first part of the fundamental theorem of calculus says that if f of x is a continuous function, then the function g of x,', 'start': 40942.826, 'duration': 7.045}, {'end': 40966.094, 'text': 'defined as the integral from a constant a to the variable x of f of t dt, is differentiable and has derivative equal to the original function f of x.', 'start': 40949.871, 'duration': 16.223}, {'end': 40970.835, 'text': "To prove this theorem, let's start with the limit definition of derivative.", 'start': 40966.094, 'duration': 4.741}], 'summary': 'Fundamental theorem of calculus: g(x) is differentiable and has derivative equal to f(x).', 'duration': 28.009, 'max_score': 40942.826, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w40942826.jpg'}, {'end': 41225.585, 'src': 'embed', 'start': 41192.048, 'weight': 1, 'content': [{'end': 41203.994, 'text': "We've now proved the first part of the fundamental theorem of calculus that the derivative of g exists and equals f.", 'start': 41192.048, 'duration': 11.946}, {'end': 41208.796, 'text': 'The second part of the fundamental theorem of calculus says that if f is continuous,', 'start': 41203.994, 'duration': 4.802}, {'end': 41220.741, 'text': "then the integral from a to b of f dx is equal to the antiderivative of lowercase f, which I'll denote by capital F evaluated at b,", 'start': 41208.796, 'duration': 11.945}, {'end': 41225.585, 'text': 'minus that antiderivative evaluated at a.', 'start': 41220.741, 'duration': 4.844}], 'summary': 'Proved first part of fundamental theorem of calculus, derivative of g equals f.', 'duration': 33.537, 'max_score': 41192.048, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w41192048.jpg'}, {'end': 41388.462, 'src': 'embed', 'start': 41358.656, 'weight': 3, 'content': [{'end': 41361.777, 'text': 'This video is about the substitution method for evaluating integrals.', 'start': 41358.656, 'duration': 3.121}, {'end': 41364.111, 'text': 'also known as u substitution.', 'start': 41362.25, 'duration': 1.861}, {'end': 41370.254, 'text': "As a first example, let's try to integrate 2x sine of x squared dx.", 'start': 41365.271, 'duration': 4.983}, {'end': 41377.217, 'text': 'Now sine of x squared is the composition of the function sine and the function x squared.', 'start': 41371.654, 'duration': 5.563}, {'end': 41383.359, 'text': 'And notice that the function x squared has derivative 2x, which is sitting right here in the integrand.', 'start': 41377.977, 'duration': 5.382}, {'end': 41388.462, 'text': "I'm going to make the substitution u equals x squared.", 'start': 41384.64, 'duration': 3.822}], 'summary': 'Video on u substitution method for evaluating integrals, exemplified by integrating 2x sine of x squared dx.', 'duration': 29.806, 'max_score': 41358.656, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w41358656.jpg'}, {'end': 41882.207, 'src': 'embed', 'start': 41849.388, 'weight': 4, 'content': [{'end': 41863.199, 'text': 'Plugging in those bounds, I get ln of e squared quantity squared over 2, minus ln of e squared over 2, which evaluates to 2 squared over 2,', 'start': 41849.388, 'duration': 13.811}, {'end': 41865.401, 'text': 'minus 1 over 2, which is again 1 half.', 'start': 41863.199, 'duration': 2.202}, {'end': 41873.025, 'text': 'This video gave some examples of u-substitution to evaluate integrals.', 'start': 41869.144, 'duration': 3.881}, {'end': 41882.207, 'text': "This method works great in examples like this one, where there's a chunk that you can call u, whose derivative,", 'start': 41874.545, 'duration': 7.662}], 'summary': 'Demonstrated u-substitution for evaluating integrals, resulting in 1/2 as the final answer.', 'duration': 32.819, 'max_score': 41849.388, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w41849388.jpg'}, {'end': 42232.724, 'src': 'embed', 'start': 42175.207, 'weight': 5, 'content': [{'end': 42182.309, 'text': 'Notice the similarity between the formula for the average value of a function and the formula for the average value of a list of numbers.', 'start': 42175.207, 'duration': 7.102}, {'end': 42194.016, 'text': 'The integral for the function corresponds to the summation sign for the list of numbers and the length of the interval b minus a for the function corresponds to n,', 'start': 42183.05, 'duration': 10.966}, {'end': 42196.277, 'text': 'the number of numbers in the list of numbers.', 'start': 42194.016, 'duration': 2.261}, {'end': 42205.74, 'text': "Now let's work an example for the function g of x equals 1 over 1 minus 5x on the interval from 2 to 5.", 'start': 42197.437, 'duration': 8.303}, {'end': 42220.334, 'text': 'We know that the average value of g is given by the integral from 2 to 5 of 1 over 1 minus 5x dx divided by the length of that interval.', 'start': 42205.74, 'duration': 14.594}, {'end': 42224.497, 'text': "I'm going to use u substitution to integrate.", 'start': 42222.316, 'duration': 2.181}, {'end': 42228.08, 'text': "So I'm going to set u equal to 1 minus 5x.", 'start': 42224.938, 'duration': 3.142}, {'end': 42232.724, 'text': 'So du is negative 5 dx.', 'start': 42228.661, 'duration': 4.063}], 'summary': 'The formula for the average value of a function is similar to that of a list of numbers, demonstrated through an example of g(x) = 1/(1-5x) on the interval [2, 5].', 'duration': 57.517, 'max_score': 42175.207, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w42175207.jpg'}], 'start': 40912.691, 'title': 'Calculus fundamentals', 'summary': 'Covers the fundamental theorem of calculus, u-substitution method, definite integrals, and the average value of a function. it demonstrates key concepts and methods, including examples, and proofs, providing a comprehensive understanding of the topics.', 'chapters': [{'end': 41124.016, 'start': 40912.691, 'title': 'Fundamental theorem of calculus', 'summary': 'Explains the fundamental theorem of calculus, proving both parts, where the first part states that if f of x is a continuous function, then g of x, defined as the integral from a to x of f of t dt, is differentiable and has derivative equal to the original function f of x.', 'duration': 211.325, 'highlights': ['The first part of the fundamental theorem of calculus states that if f of x is a continuous function, then the function g of x, defined as the integral from a constant a to the variable x of f of t dt, is differentiable and has derivative equal to the original function f of x.', 'The derivative g prime of x by definition is the limit as h goes to zero of g of x plus h minus g of x over h.', 'The integral from x to x plus h can be closely approximated by a skinny rectangle with height f of x and width h, leading to the limit as h goes to 0 of f of x times h over h, which is just f of x.']}, {'end': 41356.995, 'start': 41124.016, 'title': 'Fundamental theorem of calculus', 'summary': 'Explains the fundamental theorem of calculus, proving that the derivative of g exists and equals f, and that the integral from a to b of f dx is equal to the antiderivative of f evaluated at b minus that antiderivative evaluated at a.', 'duration': 232.979, 'highlights': ['The derivative of g exists and equals f, proved by the first part of the Fundamental Theorem of Calculus.', 'The integral from a to b of f dx is equal to the antiderivative of f evaluated at b minus that antiderivative evaluated at a, as per the second part of the Fundamental Theorem of Calculus.', 'The intermediate value theorem states that the intermediate value between the minimum and maximum value of f has to be achieved as f of c for some c in the interval, applicable to continuous functions.']}, {'end': 41644.935, 'start': 41358.656, 'title': 'U-substitution method for integrals', 'summary': 'Discusses the u-substitution method for evaluating integrals, demonstrating the process through examples including the integration of 2x sine of x squared dx and e to the 7x dx, highlighting the key steps and results.', 'duration': 286.279, 'highlights': ['Demonstrated the u-substitution method for integrating 2x sine of x squared dx, showing the process of making the substitution u = x squared, finding du, and replacing the integrand to solve the integral.', 'Explained the process of substituting back in after obtaining the solution in terms of u, ensuring that the final answer is in terms of the original variable x.', 'Illustrated the verification process of the final answer by taking the derivative and confirming that it matches the original integrand, emphasizing the use of the chain rule.', 'Presented another example of u-substitution for integrating e to the 7x dx, demonstrating the steps of making the substitution, finding du, and obtaining the final answer.', 'Encouraged viewers to verify the obtained answers by taking derivatives, highlighting the use of the chain rule in the verification process.']}, {'end': 42005.416, 'start': 41646.976, 'title': 'U-substitution for definite integrals', 'summary': 'Explains the u-substitution method for calculating definite integrals, demonstrating two different approaches to deal with the bounds of integration, resulting in the evaluation of the integral from e to e squared of ln(x)/x dx as 1/2.', 'duration': 358.44, 'highlights': ['The integral from e to e squared of ln x over x dx is evaluated using u-substitution method, resulting in the value of 1/2.', 'The chapter presents two methods for dealing with the bounds of integration, either addressing them during the substitution process or after evaluating the indefinite integral, both leading to the same result of 1/2.', "The explanation of u-substitution's foundation based on the chain rule and its practical application in evaluating integrals is provided, emphasizing the recognition of u as g(x) and du as g'(x) dx."]}, {'end': 42196.277, 'start': 42005.416, 'title': 'Average value of a function', 'summary': 'Explains the concept of average value of a function, demonstrating how to estimate it using sample points and ultimately defining it as the integral on the interval from a to b divided by the length of the interval.', 'duration': 190.861, 'highlights': ['The integral for the function corresponds to the summation sign for the list of numbers and the length of the interval b minus a for the function corresponds to n, the number of numbers in the list of numbers.', 'The average value of the function is given by the integral on the interval from a to b divided by the length of the interval.', 'The approximation gets better as the number of sample points n gets bigger and bigger.']}, {'end': 42826.57, 'start': 42197.437, 'title': 'Finding average value and mean value theorem', 'summary': 'Explains how to find the average value of a function using integration and demonstrates the mean value theorem for integrals, showing that any continuous function must achieve its average value on an interval, and provides two proofs of the theorem.', 'duration': 629.133, 'highlights': ['The video covers the process of using u substitution to integrate the function g of x equals 1 over 1 minus 5x on the interval from 2 to 5, finding the average value of g to be approximately -0.0654.', 'It is demonstrated that g achieves its average value over the interval from 2 to 5, as predicted by the fact that a continuous function must achieve every value between its minimum and maximum, known as the mean value theorem for integrals.', 'Two proofs of the mean value theorem for integrals are provided, one using the intermediate value theorem and the other as a corollary to the regular mean value theorem for functions, showing that the mean value theorem for integrals is equivalent to the mean value theorem for functions where the function is an integral.']}], 'duration': 1913.879, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/HfACrKJ_Y2w/pics/HfACrKJ_Y2w40912691.jpg', 'highlights': ['The first part of the fundamental theorem of calculus states that if f of x is a continuous function, then the function g of x, defined as the integral from a constant a to the variable x of f of t dt, is differentiable and has derivative equal to the original function f of x.', 'The derivative of g exists and equals f, proved by the first part of the Fundamental Theorem of Calculus.', 'The integral from a to b of f dx is equal to the antiderivative of f evaluated at b minus that antiderivative evaluated at a, as per the second part of the Fundamental Theorem of Calculus.', 'Demonstrated the u-substitution method for integrating 2x sine of x squared dx, showing the process of making the substitution u = x squared, finding du, and replacing the integrand to solve the integral.', 'The integral from e to e squared of ln x over x dx is evaluated using u-substitution method, resulting in the value of 1/2.', 'The average value of the function is given by the integral on the interval from a to b divided by the length of the interval.', 'The video covers the process of using u substitution to integrate the function g of x equals 1 over 1 minus 5x on the interval from 2 to 5, finding the average value of g to be approximately -0.0654.']}], 'highlights': ['The fundamental theorem of calculus allows us to compute integrals simply by finding antiderivatives and evaluating them', 'The video covers the process of using u substitution to integrate the function g of x equals 1 over 1 minus 5x on the interval from 2 to 5, finding the average value of g to be approximately -0.0654', 'The integral from a to b of f dx is equal to the antiderivative of f evaluated at b minus that antiderivative evaluated at a, as per the second part of the Fundamental Theorem of Calculus', 'The first part of the fundamental theorem of calculus states that if f of x is a continuous function, then the function g of x, defined as the integral from a constant a to the variable x of f of t dt, is differentiable and has derivative equal to the original function f of x', 'The sum of areas of rectangles using left endpoints results in an underestimate for the area under the curve, computed as 6.875 for 6 rectangles', 'The sum of areas of rectangles using right endpoints results in an overestimate for the area under the curve, computed as 11.375 for 6 rectangles', 'The exact area under the curve y = x squared between x = 0 and x = 3 is found to be 9 using the limit as the number of rectangles goes to infinity', 'The derivative of g exists and equals f, proved by the first part of the Fundamental Theorem of Calculus', 'The average value of the function is given by the integral on the interval from a to b divided by the length of the interval', 'The integral from e to e squared of ln x over x dx is evaluated using u-substitution method, resulting in the value of 1/2', 'Demonstrated the u-substitution method for integrating 2x sine of x squared dx, showing the process of making the substitution u = x squared, finding du, and replacing the integrand to solve the integral', 'The antiderivative of 0 has to be a constant and any two antiderivatives of the same function have to differ by a constant, proven using the mean value theorem', "L'Hopital's rule applies when f and g are differentiable functions and g' is non-zero at a", "L'Hopital's rule evaluates limits of 0/0, ∞/∞, 0*∞, and 1^∞, simplifying step-by-step", "Newton's method for finding zeros of a function, with detailed algorithm explanation", 'Antiderivatives: derivative of x^n is x^(n+1)/(n+1), and antiderivatives can be written as F(x)+c', "Antiderivatives of 1/x, a*x^n, f(x)+g(x), e^x-3sin(x), f''(x) using power rule, and application example"]}