Lesson 10: The Chain Rule (slides)
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The document discusses the chain rule in calculus, emphasizing its application to the composition of functions. It includes definitions, analogies, and examples to illustrate how the derivative of a composition can be expressed as the product of the derivatives of individual functions. Observations highlight the importance of evaluating derivatives at the appropriate points in the composition.
Lesson 10: The Chain Rule (slides)
- 1. Sec on 2.5 The Chain Rule V63.0121.011: Calculus I Professor Ma hew Leingang New York University February 23, 2011 .
- 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. 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. Compositions See 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. Compositions See 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. Compositions See 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. Compositions See 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. Compositions See 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. Compositions See 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. Outline Heuris cs Analogy The Linear Case The chain rule Examples
- 11. Analogy Think about riding a bike. To go faster you can either: . Image credit: SpringSun
- 12. Analogy Think about riding a bike. To go faster you can either: pedal faster . Image credit: SpringSun
- 13. Analogy Think about riding a bike. To go faster you can either: pedal faster change gears . Image credit: SpringSun
- 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. 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. 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 the Image credit: SpringSun ve of this func on and the speed of the front sprocket. deriva
- 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. 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. 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. 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. 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. 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. Outline Heuris cs Analogy The Linear Case The chain rule Examples
- 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. Observations Succinctly, the deriva ve of a composi on is the product of the deriva ves . Image credit: ooOJasonOoo
- 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. 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. Compositions See 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. 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 ons Image credit: ooOJasonOoo
- 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. 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. Outline Heuris cs Analogy The Linear Case The chain rule Examples
- 33. Example Example √ let h(x) = 3x2 + 1. Find h′ (x).
- 34. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). Solu on First, write h as f ◦ g.
- 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. 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. 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. 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. Does order matter? Example d d Find (sin 4x) and compare it to (4 sin x). dx dx
- 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. 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. 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. Example (√ )2 x − 2 + 8 . Find f′ (x). 3 5 Let f(x) =
- 44. Example (√ )2 x − 2 + 8 . Find f′ (x). 3 5 Let 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. Example (√ )2 x − 2 + 8 . Find f′ (x). 3 5 Let 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. Example (√ )2 x − 2 + 8 . Find f′ (x). 3 5 Let 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. Example (√ )2 x − 2 + 8 . Find f′ (x). 3 5 Let 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. Example (√ )2 x − 2 + 8 . Find f′ (x). 3 5 Let 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. 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 (x Image credit: photobunny
- 50. Combining techniques Example d ( 3 ) Find (x + 1)10 sin(4x2 − 7) dx
- 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. 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. 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. 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 on Image credit: Jim Frazier
- 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 on Image credit: Jim Frazier
- 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!