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# Lesson 10: The Chain Rule (Section 41 slides)

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The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.

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### Transcript of "Lesson 10: The Chain Rule (Section 41 slides)"

1. 1. Section 2.5 The Chain Rule V63.0121.041, Calculus I New York University October 6, 2010Announcements Quiz 2 in recitation next week (October 11-15) No class Monday, October 11 Midterm in class Monday, October 18 on §§1.1–2.5 . . . . . .
2. 2. Announcements Quiz 2 in recitation next week (October 11-15) No class Monday, October 11 Midterm in class Monday, October 18 on §§1.1–2.5 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 2 / 36
3. 3. Objectives Given a compound expression, write it as a composition of functions. Understand and apply the Chain Rule for the derivative of a composition of functions. Understand and use Newtonian and Leibnizian notations for the Chain Rule. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 3 / 36
4. 4. CompositionsSee Section 1.2 for reviewDefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog first, then f.” . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 4 / 36
5. 5. CompositionsSee Section 1.2 for reviewDefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog first, then f.” x . g . (x) g . . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 4 / 36
6. 6. CompositionsSee Section 1.2 for reviewDefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog first, then f.” x . g . (x) g . . f . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 4 / 36
7. 7. CompositionsSee Section 1.2 for reviewDefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog first, then f.” x . g . (x) f .(g(x)) g . . f . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 4 / 36
8. 8. CompositionsSee Section 1.2 for reviewDefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog first, then f.” g . (x) f .(g(x)) . ◦ g x . g . f . f . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 4 / 36
9. 9. CompositionsSee Section 1.2 for reviewDefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog first, then f.” g . (x) f .(g(x)) . ◦ g x . g . f . f .Our goal for the day is to understand how the derivative of thecomposition of two functions depends on the derivatives of theindividual functions. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 4 / 36
10. 10. OutlineHeuristics Analogy The Linear CaseThe chain ruleExamplesRelated rates of change . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 5 / 36
11. 11. Analogy Think about riding a bike. To go faster you can either: . . . . . . ..Image credit: SpringSun V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 6 / 36
12. 12. Analogy Think about riding a bike. To go faster you can either: pedal faster . . . . . . ..Image credit: SpringSun V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 6 / 36
13. 13. Analogy Think about riding a bike. To go faster you can either: pedal faster change gears . . . . . . ..Image credit: SpringSun V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 6 / 36
14. 14. Analogy Think about riding a bike. To go faster you can either: pedal faster change gears . The angular position (φ) of the back wheel depends on the position of the front sprocket (θ): R.θ . φ(θ) = r. . . . . . . ..Image credit: SpringSun V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 6 / 36
15. 15. Analogy Think about riding a bike. To go faster you can either: pedal faster change gears r . adius of front sprocket . The angular position (φ) of the back wheel depends on the position of the front sprocket (θ): R.θ . φ(θ) = r. . r . adius of back sprocket . . . . . ..Image credit: SpringSun V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 6 / 36
16. 16. Analogy Think about riding a bike. To go faster you can either: pedal faster change gears . The angular position (φ) of the back wheel depends on the position of the front sprocket (θ): R.θ . φ(θ) = r. . And so the angular speed of the back wheel depends on the derivative of this function and the speed of the front sprocket. . . . . . ..Image credit: SpringSun V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 6 / 36
17. 17. The Linear CaseQuestionLet f(x) = mx + b and g(x) = m′ x + b′ . What can you say about thecomposition? . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 7 / 36
18. 18. The Linear CaseQuestionLet f(x) = mx + b and g(x) = m′ x + b′ . What can you say about thecomposition?Answer f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 7 / 36
19. 19. The Linear CaseQuestionLet f(x) = mx + b and g(x) = m′ x + b′ . What can you say about thecomposition?Answer f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b) The composition is also linear . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 7 / 36
20. 20. The Linear CaseQuestionLet f(x) = mx + b and g(x) = m′ x + b′ . What can you say about thecomposition?Answer f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b) The composition is also linear The slope of the composition is the product of the slopes of the two functions. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 7 / 36
21. 21. The Linear CaseQuestionLet f(x) = mx + b and g(x) = m′ x + b′ . What can you say about thecomposition?Answer f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b) The composition is also linear The slope of the composition is the product of the slopes of the two functions.The derivative is supposed to be a local linearization of a function. Sothere should be an analog of this property in derivatives. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 7 / 36
22. 22. The Nonlinear CaseLet u = g(x) and y = f(u). Suppose x is changed by a small amount∆x. Then ∆y ≈ f′ (y)∆uand ∆u ≈ g′ (u)∆x.So ∆y ∆y ≈ f′ (y)g′ (u)∆x =⇒ ≈ f′ (y)g′ (u) ∆x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 8 / 36
23. 23. OutlineHeuristics Analogy The Linear CaseThe chain ruleExamplesRelated rates of change . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 9 / 36
24. 24. Theorem of the day: The chain ruleTheoremLet f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and (f ◦ g)′ (x) = f′ (g(x))g′ (x)In Leibnizian notation, let y = f(u) and u = g(x). Then dy dy du = dx du dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 10 / 36
25. 25. Observations Succinctly, the derivative of a composition is the product of the derivatives . . . . . . ..Image credit: ooOJasonOoo V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 11 / 36
26. 26. Theorem of the day: The chain ruleTheoremLet f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and (f ◦ g)′ (x) = f′ (g(x))g′ (x)In Leibnizian notation, let y = f(u) and u = g(x). Then dy dy du = dx du dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 12 / 36
27. 27. Observations Succinctly, the derivative of a composition is the product of the derivatives The only complication is where these derivatives are evaluated: at the same point the functions are . . . . . . ..Image credit: ooOJasonOoo V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 13 / 36
28. 28. CompositionsSee Section 1.2 for reviewDefinitionIf f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “dog first, then f.” g . (x) f .(g(x)) . ◦ g x . g . f . f . . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 14 / 36
29. 29. Observations Succinctly, the derivative of a composition is the product of the derivatives The only complication is where these derivatives are evaluated: at the same point the functions are In Leibniz notation, the Chain Rule looks like cancellation of (fake) fractions . . . . . . ..Image credit: ooOJasonOoo V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 15 / 36
30. 30. Theorem of the day: The chain ruleTheoremLet f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and (f ◦ g)′ (x) = f′ (g(x))g′ (x)In Leibnizian notation, let y = f(u) and u = g(x). Then dy dy du = dx du dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 16 / 36
31. 31. Theorem of the day: The chain ruleTheoremLet f and g be functions, with g differentiable at x and f differentiable atg(x). Then f ◦ g is differentiable at x and (f ◦ g)′ (x) = f′ (g(x))g′ (x) dy In Leibnizian notation, let y = f(u) and u = g(x).du Then . . dx du dy dy du = dx du dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 16 / 36
32. 32. OutlineHeuristics Analogy The Linear CaseThe chain ruleExamplesRelated rates of change . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 17 / 36
33. 33. ExampleExample √let h(x) = 3x2 + 1. Find h′ (x). . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 18 / 36
34. 34. ExampleExample √let h(x) = 3x2 + 1. Find h′ (x).SolutionFirst, write h as f ◦ g. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 18 / 36
35. 35. ExampleExample √let h(x) = 3x2 + 1. Find h′ (x).Solution √First, write h as f ◦ g. Let f(u) = u and g(x) = 3x2 + 1. . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 18 / 36
36. 36. ExampleExample √let h(x) = 3x2 + 1. Find h′ (x).Solution √First, write h as f ◦ g. Let f(u) = u and g(x) = 3x2 + 1. Thenf′ (u) = 1 u−1/2 , and g′ (x) = 6x. So 2 h′ (x) = 1 u−1/2 (6x) 2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 18 / 36
37. 37. ExampleExample √let h(x) = 3x2 + 1. Find h′ (x).Solution √First, write h as f ◦ g. Let f(u) = u and g(x) = 3x2 + 1. Thenf′ (u) = 1 u−1/2 , and g′ (x) = 6x. So 2 3x h′ (x) = 1 u−1/2 (6x) = 1 (3x2 + 1)−1/2 (6x) = √ 2 2 3x2 + 1 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 18 / 36
38. 38. CorollaryCorollary (The Power Rule Combined with the Chain Rule)If n is any real number and u = g(x) is differentiable, then d n du (u ) = nun−1 . dx dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 19 / 36
39. 39. Does order matter?Example d dFind (sin 4x) and compare it to (4 sin x). dx dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 20 / 36
40. 40. Does order matter?Example d dFind (sin 4x) and compare it to (4 sin x). dx dxSolution For the first, let u = 4x and y = sin(u). Then dy dy du = · = cos(u) · 4 = 4 cos 4x. dx du dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 20 / 36
41. 41. Does order matter?Example d dFind (sin 4x) and compare it to (4 sin x). dx dxSolution For the first, let u = 4x and y = sin(u). Then dy dy du = · = cos(u) · 4 = 4 cos 4x. dx du dx For the second, let u = sin x and y = 4u. Then dy dy du = · = 4 · cos x dx du dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 20 / 36
42. 42. Order matters!Example d dFind (sin 4x) and compare it to (4 sin x). dx dxSolution For the first, let u = 4x and y = sin(u). Then dy dy du = · = cos(u) · 4 = 4 cos 4x. dx du dx For the second, let u = sin x and y = 4u. Then dy dy du = · = 4 · cos x dx du dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 20 / 36
43. 43. Example (√ )2 . Find f′ (x). 3Let f(x) = x5 − 2 + 8 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 21 / 36
44. 44. Example (√ )2 . Find f′ (x). 3Let f(x) = x5 − 2 + 8Solution d (√ 5 3 )2 (√ 3 ) d (√ 3 ) x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8 dx dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 21 / 36
45. 45. Example (√ )2 . Find f′ (x). 3Let f(x) = x5 − 2 + 8Solution d (√ 5 3 )2 (√ 3 ) d (√ 3 ) x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8 dx dx (√ ) d√ 3 3 =2 x5 − 2 + 8 x5 − 2 dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 21 / 36
46. 46. Example (√ )2 . Find f′ (x). 3Let f(x) = x5 − 2 + 8Solution d (√ 5 3 )2 (√ 3 ) d (√ 3 ) x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8 dx dx (√ ) d√ 3 3 =2 x5 − 2 + 8 x5 − 2 dx (√ ) d x5 − 2 + 8 1 (x5 − 2)−2/3 (x5 − 2) 3 =2 3 dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 21 / 36
47. 47. Example (√ )2 . Find f′ (x). 3Let f(x) = x5 − 2 + 8Solution d (√ 5 3 )2 (√ 3 ) d (√ 3 ) x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8 dx dx (√ ) d√ 3 3 =2 x5 − 2 + 8 x5 − 2 dx (√ ) d x5 − 2 + 8 1 (x5 − 2)−2/3 (x5 − 2) 3 =2 3 (√ ) dx =2 3 x 5 − 2 + 8 1 (x5 − 2)−2/3 (5x4 ) 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 21 / 36
48. 48. Example (√ )2 . Find f′ (x). 3Let f(x) = x5 − 2 + 8Solution d (√ 5 3 )2 (√ 3 ) d (√ 3 ) x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8 dx dx (√ ) d√ 3 3 =2 x5 − 2 + 8 x5 − 2 dx (√ ) d x5 − 2 + 8 1 (x5 − 2)−2/3 (x5 − 2) 3 =2 3 (√ ) dx =2 3 x 5 − 2 + 8 1 (x5 − 2)−2/3 (5x4 ) 3 (√ 10 4 3 5 ) = x x − 2 + 8 (x5 − 2)−2/3 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 21 / 36
49. 49. A metaphor Think about peeling an onion: (√ )2 3 f(x) = x 5 −2 +8 5 √ 3 +8 . (√ ) 2 f′ (x) = 2 x5 − 2 + 8 1 (x5 − 2)−2/3 (5x4 ) 3 3 . . . . . ..Image credit: photobunny V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 22 / 36
50. 50. Combining techniquesExample d ( 3 )Find (x + 1)10 sin(4x2 − 7) dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 23 / 36
51. 51. Combining techniquesExample d ( 3 )Find (x + 1)10 sin(4x2 − 7) dxSolutionThe “last” part of the function is the product, so we apply the productrule. Each factor’s derivative requires the chain rule: . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 23 / 36
52. 52. Combining techniquesExample d ( 3 )Find (x + 1)10 sin(4x2 − 7) dxSolutionThe “last” part of the function is the product, so we apply the productrule. Each factor’s derivative requires the chain rule: d ( 3 ) (x + 1)10 · sin(4x2 − 7) dx ( ) ( ) d 3 d = (x + 1) 10 · sin(4x − 7) + (x + 1) · 2 3 10 sin(4x − 7) 2 dx dx . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 23 / 36
53. 53. Combining techniquesExample d ( 3 )Find (x + 1)10 sin(4x2 − 7) dxSolutionThe “last” part of the function is the product, so we apply the productrule. Each factor’s derivative requires the chain rule: d ( 3 ) (x + 1)10 · sin(4x2 − 7) dx ( ) ( ) d 3 d = (x + 1) 10 · sin(4x − 7) + (x + 1) · 2 3 10 sin(4x − 7) 2 dx dx = 10(x3 + 1)9 (3x2 ) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 23 / 36
54. 54. Your TurnFind derivatives of these functions: 1. y = (1 − x2 )10 √ 2. y = sin x √ 3. y = sin x 4. y = (2x − 5)4 (8x2 − 5)−3 √ z−1 5. F(z) = z+1 6. y = tan(cos x) 7. y = csc2 (sin θ) 8. y = sin(sin(sin(sin(sin(sin(x)))))) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 24 / 36
55. 55. Solution to #1ExampleFind the derivative of y = (1 − x2 )10 .Solutiony′ = 10(1 − x2 )9 (−2x) = −20x(1 − x2 )9 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 25 / 36
56. 56. Solution to #2Example √Find the derivative of y = sin x.Solution √Writing sin x as (sin x)1/2 , we have cos x y′ = 1 2 (sin x)−1/2 (cos x) = √ 2 sin x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 26 / 36
57. 57. Solution to #3Example √Find the derivative of y = sin x.Solution (√ ) ′ d 1/2 1 −1/2 cos x y = 1/2 sin(x ) = cos(x ) 2 x = √ dx 2 x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 27 / 36
58. 58. Solution to #4ExampleFind the derivative of y = (2x − 5)4 (8x2 − 5)−3SolutionWe need to use the product rule and the chain rule: y′ = 4(2x − 5)3 (2)(8x2 − 5)−3 + (2x − 5)4 (−3)(8x2 − 5)−4 (16x)The rest is a bit of algebra, useful if you wanted to solve the equationy′ = 0: [ ] y′ = 8(2x − 5)3 (8x2 − 5)−4 (8x2 − 5) − 6x(2x − 5) ( ) = 8(2x − 5)3 (8x2 − 5)−4 −4x2 + 30x − 5 ( ) = −8(2x − 5)3 (8x2 − 5)−4 4x2 − 30x + 5 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 28 / 36
59. 59. Solution to #5Example √ z−1Find the derivative of F(z) = . z+1Solution ( )−1/2 ( ) 1 z−1(z + 1)(1) − (z − 1)(1) y′ = 2 z+1 (z + 1)2 ( )1/2 ( ) 1 z+1 2 1 = = 2 z−1 (z + 1)2 (z + 1)3/2 (z − 1)1/2 . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 29 / 36
60. 60. Solution to #6ExampleFind the derivative of y = tan(cos x).Solutiony′ = sec2 (cos x) · (− sin x) = − sec2 (cos x) sin x . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 30 / 36
61. 61. Solution to #7ExampleFind the derivative of y = csc2 (sin θ).SolutionRemember the notation: y = csc2 (sin θ) = [csc(sin θ)]2So y′ = 2 csc(sin θ) · [− csc(sin θ) cot(sin θ)] · cos(θ) = −2 csc2 (sin θ) cot(sin θ) cos θ . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 31 / 36
62. 62. Solution to #8ExampleFind the derivative of y = sin(sin(sin(sin(sin(sin(x)))))).SolutionRelax! It’s just a bunch of chain rules. All of these lines are multipliedtogether. y′ = cos(sin(sin(sin(sin(sin(x)))))) · cos(sin(sin(sin(sin(x))))) · cos(sin(sin(sin(x)))) · cos(sin(sin(x))) · cos(sin(x)) · cos(x)) . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 32 / 36
63. 63. OutlineHeuristics Analogy The Linear CaseThe chain ruleExamplesRelated rates of change . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 33 / 36
64. 64. Related rates of change at the DeliQuestionSuppose a deli clerk can slice a stick of pepperoni (assume thetapered ends have been removed) by hand at the rate of 2 inches perminute, while a machine can slice pepperoni at the rate of 10 inches dV dVper minute. Then for the machine is 5 times greater than for dt dtthe deli clerk. This is explained by theA. chain ruleB. product ruleC. quotient RuleD. addition rule . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 34 / 36
65. 65. Related rates of change at the DeliQuestionSuppose a deli clerk can slice a stick of pepperoni (assume thetapered ends have been removed) by hand at the rate of 2 inches perminute, while a machine can slice pepperoni at the rate of 10 inches dV dVper minute. Then for the machine is 5 times greater than for dt dtthe deli clerk. This is explained by theA. chain ruleB. product ruleC. quotient RuleD. addition rule . . . . . . V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 34 / 36
66. 66. Related rates of change in the ocean Question The area of a circle, A = πr2 , changes as its radius changes. If the radius changes with respect to time, the change in area with respect to time is dA A. = 2πr dr dA dr B. = 2πr + dt dt . dA dr C. = 2πr dt dt D. not enough information . . . . . ..Image credit: Jim Frazier V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 35 / 36
67. 67. Related rates of change in the ocean Question The area of a circle, A = πr2 , changes as its radius changes. If the radius changes with respect to time, the change in area with respect to time is dA A. = 2πr dr dA dr B. = 2πr + dt dt . dA dr C. = 2πr dt dt D. not enough information . . . . . ..Image credit: Jim Frazier V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 35 / 36
68. 68. Summary The derivative of a composition is the product of derivatives In symbols: (f ◦ g)′ (x) = f′ (g(x))g′ (x) Calculus is like an onion, and not because it makes you cry! . . . . . .V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 6, 2010 36 / 36
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