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

1. 1. Sec on 2.5 The Chain Rule V63.0121.011: Calculus I Professor Ma hew Leingang New York University February 23, 2011.
2. 2. Announcements Quiz 2 next week on §§1.5, 1.6, 2.1, 2.2 Midterm March 7 on all sec ons in class (covers all sec ons up to 2.5)
3. 3. Objectives Given a compound expression, write it as a composi on of func ons. Understand and apply the Chain Rule for the deriva ve of a composi on of func ons. Understand and use Newtonian and Leibnizian nota ons for the Chain Rule.
4. 4. CompositionsSee Section 1.2 for review Deﬁni on If f and g are func ons, the composi on (f ◦ g)(x) = f(g(x)) means “do g ﬁrst, then f.” .
5. 5. CompositionsSee Section 1.2 for review Deﬁni on If f and g are func ons, the composi on (f ◦ g)(x) = f(g(x)) means “do g ﬁrst, then f.” x g(x) g .
6. 6. CompositionsSee Section 1.2 for review Deﬁni on If f and g are func ons, the composi on (f ◦ g)(x) = f(g(x)) means “do g ﬁrst, then f.” x g(x) g . f
7. 7. CompositionsSee Section 1.2 for review Deﬁni on If f and g are func ons, the composi on (f ◦ g)(x) = f(g(x)) means “do g ﬁrst, then f.” x g(x) f(g(x)) g . f
8. 8. CompositionsSee Section 1.2 for review Deﬁni on If f and g are func ons, the composi on (f ◦ g)(x) = f(g(x)) means “do g ﬁrst, then f.” x g f g g(x) ◦. f f(g(x))
9. 9. CompositionsSee Section 1.2 for review Deﬁni on If f and g are func ons, the composi on (f ◦ g)(x) = f(g(x)) means “do g ﬁrst, then f.” x g f g g(x) ◦. f f(g(x)) Our goal for the day is to understand how the deriva ve of the composi on of two func ons depends on the deriva ves of the individual func ons.
10. 10. Outline Heuris cs Analogy The Linear Case The chain rule Examples
11. 11. Analogy Think about riding a bike. To go faster you can either: .Image credit: SpringSun
12. 12. Analogy Think about riding a bike. To go faster you can either: pedal faster .Image credit: SpringSun
13. 13. Analogy Think about riding a bike. To go faster you can either: pedal faster change gears .Image credit: SpringSun
14. 14. Analogy Think about riding a bike. To go faster you can either: pedal faster change gears . The angular posi on (φ) of the back wheel depends on the posi on of the front sprocket (θ): R.. θ φ(θ) = r..Image credit: SpringSun
15. 15. Analogy Think about riding a bike. To go faster you can either: pedal faster change gears radius of front sprocket . The angular posi on (φ) of the back wheel depends on the posi on of the front sprocket (θ): R.. θ φ(θ) = r..Image credit: SpringSun radius of back sprocket
16. 16. Analogy Think about riding a bike. To go faster you can either: pedal faster change gears . The angular posi on (φ) of the back wheel depends on the posi on of the front sprocket (θ): R.. θ φ(θ) = r.. And so the angular speed of the back wheel depends on theImage credit: SpringSun ve of this func on and the speed of the front sprocket. deriva
17. 17. The Linear Case Ques on Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the composi on?
18. 18. The Linear Case Ques on Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the composi on? Answer f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b)
19. 19. The Linear Case Ques on Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the composi on? Answer f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b) The composi on is also linear
20. 20. The Linear Case Ques on Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the composi on? Answer f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b) The composi on is also linear The slope of the composi on is the product of the slopes of the two func ons.
21. 21. The Linear Case Ques on Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the composi on? Answer f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b) The composi on is also linear The slope of the composi on is the product of the slopes of the two func ons. The deriva ve is supposed to be a local lineariza on of a func on. So there should be an analog of this property in deriva ves.
22. 22. The Nonlinear Case Let u = g(x) and y = f(u). Suppose x is changed by a small amount ∆x. Then ∆y f′ (u) ≈ =⇒ ∆y ≈ f′ (u)∆u ∆u and ∆y g′ (x) ≈ =⇒ ∆u ≈ g′ (x)∆x. ∆x So ∆y ∆y ≈ f′ (u)g′ (x)∆x =⇒ ≈ f′ (u)g′ (x) ∆x
23. 23. Outline Heuris cs Analogy The Linear Case The chain rule Examples
24. 24. Theorem of the day: The chain rule Theorem Let f and g be func ons, with g diﬀeren able at x and f diﬀeren able at g(x). Then f ◦ g is diﬀeren able at x and (f ◦ g)′ (x) = f′ (g(x))g′ (x) In Leibnizian nota on, let y = f(u) and u = g(x). Then dy dy du = dx du dx
25. 25. Observations Succinctly, the deriva ve of a composi on is the product of the deriva ves .Image credit: ooOJasonOoo
26. 26. Theorem of the day: The chain rule Theorem Let f and g be func ons, with g diﬀeren able at x and f diﬀeren able at g(x). Then f ◦ g is diﬀeren able at x and (f ◦ g)′ (x) = f′ (g(x))g′ (x) In Leibnizian nota on, let y = f(u) and u = g(x). Then dy dy du = dx du dx
27. 27. Observations Succinctly, the deriva ve of a composi on is the product of the deriva ves The only complica on is where these deriva ves are evaluated: at the same point the func ons are .Image credit: ooOJasonOoo
28. 28. CompositionsSee Section 1.2 for review Deﬁni on If f and g are func ons, the composi on (f ◦ g)(x) = f(g(x)) means “do g ﬁrst, then f.” x g f g g(x) ◦. f f(g(x))
29. 29. Observations Succinctly, the deriva ve of a composi on is the product of the deriva ves The only complica on is where these deriva ves are evaluated: at the same point the func ons are In Leibniz nota on, the Chain Rule looks like cancella on of . (fake) frac onsImage credit: ooOJasonOoo
30. 30. Theorem of the day: The chain rule Theorem Let f and g be func ons, with g diﬀeren able at x and f diﬀeren able at g(x). Then f ◦ g is diﬀeren able at x and (f ◦ g)′ (x) = f′ (g(x))g′ (x) In Leibnizian nota on, let y = f(u) and u = g(x). Then dy dy du = dx du dx
31. 31. Theorem of the day: The chain rule Theorem Let f and g be func ons, with g diﬀeren able at x and f diﬀeren able at g(x). Then f ◦ g is diﬀeren able at x and (f ◦ g)′ (x) = f′ (g(x))g′ (x) In Leibnizian nota on, let y = f(u) and u = g(x). Then dy .du dy dy du dx du = dx du dx
32. 32. Outline Heuris cs Analogy The Linear Case The chain rule Examples
33. 33. Example Example √ let h(x) = 3x2 + 1. Find h′ (x).
34. 34. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). Solu on First, write h as f ◦ g.
35. 35. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). Solu on √ First, write h as f ◦ g. Let f(u) = u and g(x) = 3x2 + 1.
36. 36. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). Solu on √ First, write h as f ◦ g. Let f(u) = u and g(x) = 3x2 + 1. Then f′ (u) = 1 u−1/2 , and g′ (x) = 6x. So 2 h′ (x) = 1 u−1/2 (6x) 2
37. 37. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). Solu on √ First, write h as f ◦ g. Let f(u) = u and g(x) = 3x2 + 1. Then f′ (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
38. 38. Corollary Corollary (The Power Rule Combined with the Chain Rule) If n is any real number and u = g(x) is diﬀeren able, then d n du (u ) = nun−1 . dx dx
39. 39. Does order matter? Example d d Find (sin 4x) and compare it to (4 sin x). dx dx
40. 40. Does order matter? Example d d Find (sin 4x) and compare it to (4 sin x). dx dx Solu on For the ﬁrst, let u = 4x and y = sin(u). Then dy dy du = · = cos(u) · 4 = 4 cos 4x. dx du dx
41. 41. Does order matter? Example d d Find (sin 4x) and compare it to (4 sin x). dx dx Solu on 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
42. 42. Order matters! Example d d Find (sin 4x) and compare it to (4 sin x). dx dx Solu on 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
43. 43. Example (√ )2 x − 2 + 8 . Find f′ (x). 3 5Let f(x) =
44. 44. Example (√ )2 x − 2 + 8 . Find f′ (x). 3 5Let f(x) =Solu on d (√ 5 )2 (√ ) d (√ ) x −2+8 =2 x −2+8 x −2+8 3 3 5 3 5 dx dx
45. 45. Example (√ )2 x − 2 + 8 . Find f′ (x). 3 5Let f(x) =Solu on d (√ 5 )2 (√ ) d (√ ) x −2+8 =2 x −2+8 x −2+8 3 3 5 3 5 dx dx (√ ) d√ x −2+8 x −2 3 5 3 5 =2 dx
46. 46. Example (√ )2 x − 2 + 8 . Find f′ (x). 3 5Let f(x) =Solu on d (√ 5 )2 (√ ) d (√ ) x −2+8 =2 x −2+8 x −2+8 3 3 5 3 5 dx dx (√ ) d√ x −2+8 x −2 3 5 3 5 =2 dx (√ ) d x − 2 + 8 1 (x5 − 2)−2/3 (x5 − 2) 3 5 =2 3 dx
47. 47. Example (√ )2 x − 2 + 8 . Find f′ (x). 3 5Let f(x) =Solu on d (√ 5 )2 (√ ) d (√ ) x −2+8 =2 x −2+8 x −2+8 3 3 5 3 5 dx dx (√ ) d√ x −2+8 x −2 3 5 3 5 =2 dx (√ ) d x − 2 + 8 1 (x5 − 2)−2/3 (x5 − 2) 3 5 =2 3 (√ ) dx x − 2 + 8 1 (x5 − 2)−2/3 (5x4 ) 3 5 =2 3
48. 48. Example (√ )2 x − 2 + 8 . Find f′ (x). 3 5Let f(x) =Solu on d (√ 5 )2 (√ ) d (√ ) x −2+8 =2 x −2+8 x −2+8 3 3 5 3 5 dx dx (√ ) d√ x −2+8 x −2 3 5 3 5 =2 dx (√ ) d x − 2 + 8 1 (x5 − 2)−2/3 (x5 − 2) 3 5 =2 3 (√ ) dx x − 2 + 8 1 (x5 − 2)−2/3 (5x4 ) 3 5 =2 3 (√ 10 4 3 5 ) = x x − 2 + 8 (x5 − 2)−2/3 3
49. 49. A metaphor Think about peeling an onion: (√ )2 3 f(x) = x5 −2 +8 5 √ 3 +8 . 2 (√ ) ′ −2+8 1 5 − 2)−2/3 (5x4 ) 3 f (x) = 2 x5 3 (xImage credit: photobunny
50. 50. Combining techniques Example d ( 3 ) Find (x + 1)10 sin(4x2 − 7) dx
51. 51. Combining techniques Example d ( 3 ) Find (x + 1)10 sin(4x2 − 7) dx Solu on The “last” part of the func on is the product, so we apply the product rule. Each factor’s deriva ve requires the chain rule:
52. 52. Combining techniques Example d ( 3 ) Find (x + 1)10 sin(4x2 − 7) dx Solu on The “last” part of the func on is the product, so we apply the product rule. Each factor’s deriva ve 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
53. 53. Combining techniques Example d ( 3 ) Find (x + 1)10 sin(4x2 − 7) dx Solu on The “last” part of the func on is the product, so we apply the product rule. Each factor’s deriva ve 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)
54. 54. Related rates of change in the ocean Ques on The area of a circle, A = πr2 , changes as its radius changes. If the radius changes with respect to me, the change in area with respect to me is dA A. = 2πr dr dA dr B. = 2πr + dt dt dA dr C. = 2πr dt dt . D. not enough informa onImage credit: Jim Frazier
55. 55. Related rates of change in the ocean Ques on The area of a circle, A = πr2 , changes as its radius changes. If the radius changes with respect to me, the change in area with respect to me is dA A. = 2πr dr dA dr B. = 2πr + dt dt dA dr C. = 2πr dt dt . D. not enough informa onImage credit: Jim Frazier
56. 56. Summary The deriva ve of a composi on is the product of deriva ves In symbols: (f ◦ g)′ (x) = f′ (g(x))g′ (x) Calculus is like an onion, and not because it makes you cry!