Lesson 10: The Chain Rule
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The document discusses the Chain Rule in calculus, explaining the differentiation of composite functions. It outlines the essential concept that the derivative of a composition of two functions is the product of their individual derivatives. Various examples and analogies are provided to illustrate these principles and their application in finding derivatives.
![Solution to #4
Example
Find the derivative of y = (2x − 5)4 (8x2 − 5)−3
Solution
We 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
equation y′ = 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
. . . . . .](https://image.slidesharecdn.com/lesson10-thechainruleslides-100218190920-phpapp02/85/Lesson-10-The-Chain-Rule-57-320.jpg)
![Solution to #7
Example
Find the derivative of y = csc2 (sin θ).
Solution
Remember the notation:
y = csc2 (sin θ) = [csc(sin θ)]2
So
y′ = 2 csc(sin θ) · [− csc(sin θ) cot(sin θ)] · cos(θ)
= −2 csc2 (sin θ) cot(sin θ) cos θ
. . . . . .](https://image.slidesharecdn.com/lesson10-thechainruleslides-100218190920-phpapp02/85/Lesson-10-The-Chain-Rule-60-320.jpg)
Lesson 10: The Chain Rule
- 1. Section 2.5 The Chain Rule V63.0121.006/016, Calculus I February 18, 2010 Announcements Quiz 2 is February 26, covering §§1.5–2.3 Midterm is March 4, covering §§1.1–2.5 . . . . . .
- 2. Help! Free resources: recitation TAs’ office hours my office hours Math Tutoring Center (CIWW 524) College Learning Center . . . . . .
- 3. Compositions See Section 1.2 for review Definition If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “do g first, then f.” . . . . . . .
- 4. Compositions See Section 1.2 for review Definition If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “do g first, then f.” x . g . (x) g . . . . . . . .
- 5. Compositions See Section 1.2 for review Definition If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “do g first, then f.” x . g . (x) g . . f . . . . . . .
- 6. Compositions See Section 1.2 for review Definition If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “do g first, then f.” x . g . (x) f .(g(x)) g . . f . . . . . . .
- 7. Compositions See Section 1.2 for review Definition If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “do g first, then f.” f . . ◦ g x . g . (x) f .(g(x)) g . f . . . . . . .
- 8. Compositions See Section 1.2 for review Definition If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “do g first, then f.” f . . ◦ g x . g . (x) f .(g(x)) g . f . Our goal for the day is to understand how the derivative of the composition of two functions depends on the derivatives of the individual functions. . . . . . .
- 9. Outline Heuristics Analogy The Linear Case The chain rule Examples Related rates of change . . . . . .
- 10. Analogy Think about riding a bike. To go faster you can either: . . Image credit: SpringSun . . . . . .
- 11. Analogy Think about riding a bike. To go faster you can either: pedal faster . . Image credit: SpringSun . . . . . .
- 12. Analogy Think about riding a bike. To go faster you can either: pedal faster change gears . . Image credit: SpringSun . . . . . .
- 13. 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 . . . . . .
- 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 r . adius of front sprocket position of the front sprocket (θ): R.θ . φ(θ) = r. . r . adius of back sprocket . Image credit: SpringSun . . . . . .
- 15. 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 . . . . . .
- 16. The Linear Case Question Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the composition? . . . . . .
- 17. The Linear Case Question Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the composition? Answer f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b) . . . . . .
- 18. The Linear Case Question Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the composition? Answer f(g(x)) = m(m′ x + b′ ) + b = (mm′ )x + (mb′ + b) The composition is also linear . . . . . .
- 19. The Linear Case Question Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the composition? 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. . . . . . .
- 20. The Linear Case Question Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the composition? 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. So there should be an analog of this property in derivatives. . . . . . .
- 21. The Nonlinear Case See the Mathematica applet Let u = g(x) and y = f(u). Suppose x is changed by a small amount ∆x. Then ∆y ≈ f′ (y)∆u and ∆u ≈ g′ (u)∆x. So ∆y ∆y ≈ f′ (y)g′ (u)∆x =⇒ ≈ f′ (y)g′ (u) ∆x . . . . . .
- 22. Outline Heuristics Analogy The Linear Case The chain rule Examples Related rates of change . . . . . .
- 23. Theorem of the day: The chain rule Theorem Let f and g be functions, with g differentiable at x and f differentiable at g(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 . . . . . .
- 24. Observations Succinctly, the derivative of a composition is the product of the derivatives . . Image credit: ooOJasonOoo . . . . . .
- 25. Theorem of the day: The chain rule Theorem Let f and g be functions, with g differentiable at x and f differentiable at g(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 . . . . . .
- 26. 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 . . . . . .
- 27. Compositions See Section 1.2 for review Definition If f and g are functions, the composition (f ◦ g)(x) = f(g(x)) means “do g first, then f.” f . . ◦ g x . g . (x) f .(g(x)) g . f . . . . . . .
- 28. 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 . . . . . .
- 29. Theorem of the day: The chain rule Theorem Let f and g be functions, with g differentiable at x and f differentiable at g(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 . . . . . .
- 30. Theorem of the day: The chain rule Theorem Let f and g be functions, with g differentiable at x and f differentiable at g(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.du. Then (x . dx du dy dy du = dx du dx . . . . . .
- 31. Outline Heuristics Analogy The Linear Case The chain rule Examples Related rates of change . . . . . .
- 32. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). . . . . . .
- 33. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). Solution First, write h as f ◦ g. . . . . . .
- 34. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). Solution √ First, write h as f ◦ g. Let f(u) = u and g(x) = 3x2 + 1. . . . . . .
- 35. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). Solution √ First, write h as f ◦ g. Let f(u) = u and g(x) = 3x2 + 1. Then f′ (u) = 2 u−1/2 , and g′ (x) = 6x. So 1 h′ (x) = 1 u−1/2 (6x) 2 . . . . . .
- 36. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). Solution √ First, write h as f ◦ g. Let f(u) = u and g(x) = 3x2 + 1. Then f′ (u) = 2 u−1/2 , and g′ (x) = 6x. So 1 3x h′ (x) = 1 u−1/2 (6x) = 1 (3x2 + 1)−1/2 (6x) = √ 2 2 3x2 + 1 . . . . . .
- 37. Corollary Corollary (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 . . . . . .
- 38. Does order matter? Example d d Find (sin 4x) and compare it to (4 sin x). dx dx . . . . . .
- 39. Does order matter? Example d d Find (sin 4x) and compare it to (4 sin x). dx dx Solution For the first, let u = 4x and y = sin(u). Then dy dy du = · = cos(u) · 4 = 4 cos 4x. dx du dx . . . . . .
- 40. Does order matter? Example d d Find (sin 4x) and compare it to (4 sin x). dx dx Solution 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 . . . . . .
- 41. Order matters! Example d d Find (sin 4x) and compare it to (4 sin x). dx dx Solution 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. Example (√ )2 . Find f′ (x). 3 Let f(x) = x5 − 2 + 8 . . . . . .
- 43. Example (√ )2 . Find f′ (x). 3 Let f(x) = x5 − 2 + 8 Solution d (√ 5 3 )2 (√ 3 ) d (√ 3 ) x −2+8 =2 x5 − 2 + 8 x5 − 2 + 8 dx dx . . . . . .
- 44. Example (√ )2 . Find f′ (x). 3 Let f(x) = x5 − 2 + 8 Solution 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 . . . . . .
- 45. Example (√ )2 . Find f′ (x). 3 Let f(x) = x5 − 2 + 8 Solution 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 . . . . . .
- 46. Example (√ )2 . Find f′ (x). 3 Let f(x) = x5 − 2 + 8 Solution 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 . . . . . .
- 47. Example (√ )2 . Find f′ (x). 3 Let f(x) = x5 − 2 + 8 Solution 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 . . . . . .
- 48. A metaphor Think about peeling an onion: (√ )2 3 5 f(x) = x −2 +8 5 √ 3 +8 . 2 (√ ) f′ (x) = 2 x5 − 2 + 8 1 (x5 − 2)−2/3 (5x4 ) 3 3 . Image credit: photobunny . . . . . .
- 49. Combining techniques Example d ( 3 ) Find (x + 1)10 sin(4x2 − 7) dx . . . . . .
- 50. Combining techniques Example d ( 3 ) Find (x + 1)10 sin(4x2 − 7) dx Solution The “last” part of the function is the product, so we apply the product rule. Each factor’s derivative requires the chain rule: . . . . . .
- 51. Combining techniques Example d ( 3 ) Find (x + 1)10 sin(4x2 − 7) dx Solution The “last” part of the function is the product, so we apply the product rule. Each factor’s derivative requires the chain rule: d ( 3 ) (x + 1)10 · sin(4x2 − 7) dx ( ) ( ) d 3 d = (x + 1)10 · sin(4x2 − 7) + (x3 + 1)10 · sin(4x2 − 7) dx dx . . . . . .
- 52. Combining techniques Example d ( 3 ) Find (x + 1)10 sin(4x2 − 7) dx Solution The “last” part of the function is the product, so we apply the product rule. Each factor’s derivative requires the chain rule: d ( 3 ) (x + 1)10 · sin(4x2 − 7) dx ( ) ( ) d 3 d = (x + 1)10 · sin(4x2 − 7) + (x3 + 1)10 · sin(4x2 − 7) dx dx = 10(x3 + 1)9 (3x2 ) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x) . . . . . .
- 53. Your Turn Find 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)))))) . . . . . .
- 54. Solution to #1 Example Find the derivative of y = (1 − x2 )10 . Solution y′ = 10(1 − x2 )9 (−2x) = −20x(1 − x2 )9 . . . . . .
- 55. Solution to #2 Example √ 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 . . . . . .
- 56. Solution to #3 Example √ Find the derivative of y = sin x. Solution (√ ) ′d 1 /2 1/2 1 −1/2 cos x y = sin(x ) = cos(x ) 2 x = √ dx 2 x . . . . . .
- 57. Solution to #4 Example Find the derivative of y = (2x − 5)4 (8x2 − 5)−3 Solution We 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 equation y′ = 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 . . . . . .
- 58. Solution to #5 Example √ z−1 Find the derivative of F(z) = . z+1 Solution ( )−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 = 2 z−1 (z + 1) (z + 1)3/2 (z − 1)1/2 . . . . . .
- 59. Solution to #6 Example Find the derivative of y = tan(cos x). Solution y′ = sec2 (cos x) · (− sin x) = − sec2 (cos x) sin x . . . . . .
- 60. Solution to #7 Example Find the derivative of y = csc2 (sin θ). Solution Remember the notation: y = csc2 (sin θ) = [csc(sin θ)]2 So y′ = 2 csc(sin θ) · [− csc(sin θ) cot(sin θ)] · cos(θ) = −2 csc2 (sin θ) cot(sin θ) cos θ . . . . . .
- 61. Solution to #8 Example Find the derivative of y = sin(sin(sin(sin(sin(sin(x)))))). Solution Relax! It’s just a bunch of chain rules. All of these lines are multiplied together. 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)) . . . . . .
- 62. Outline Heuristics Analogy The Linear Case The chain rule Examples Related rates of change . . . . . .
- 63. Related rates of change Question The area of a circle, A = π r2 , changes as its radius changes. If the radius changes with respect to time, the change in area with respect to time is dA A. = 2π r dr dA dr . B. = 2π r + dt dt dA dr C. = 2π r dt dt D. not enough information . Image credit: Jim Frazier . . . . . .
- 64. Related rates of change Question The area of a circle, A = π r2 , changes as its radius changes. If the radius changes with respect to time, the change in area with respect to time is dA A. = 2π r dr dA dr . B. = 2π r + dt dt dA dr C. = 2π r dt dt D. not enough information . Image credit: Jim Frazier . . . . . .
- 65. What have we learned today? 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! . . . . . .