Upcoming SlideShare
×

# Lesson 10: The Chain Rule

1,790 views
1,626 views

Published on

The chain rule helps us find a derivative of a composition of functions. It turns out that it's the product of the derivatives of the composed functions.

Published in: Education, Technology
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
1,790
On SlideShare
0
From Embeds
0
Number of Embeds
17
Actions
Shares
0
68
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Lesson 10: The Chain Rule

1. 1. Section 2.5 The Chain Rule V63.0121.034, Calculus I October 5, 2009 Announcements Quiz 2 this week Midterm in class on §§1.1–1.4 . . . . . .
2. 2. 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.” . . . . . . .
3. 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.” x . g . (x) g . . . . . . . .
4. 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 . . f . . . . . . .
5. 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) f .(g(x)) g . . f . . . . . . .
6. 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.” f . . ◦ g x . g . (x) f .(g(x)) g . f . . . . . . .
7. 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 . 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. . . . . . .
8. 8. Outline Heuristics Analogy The Linear Case The chain rule Examples Related rates of change . . . . . .
9. 9. Analogy Think about riding a bike. To go faster you can either: . . Image credit: SpringSun . . . . . .
10. 10. Analogy Think about riding a bike. To go faster you can either: pedal faster . . Image credit: SpringSun . . . . . .
11. 11. Analogy Think about riding a bike. To go faster you can either: pedal faster change gears . . Image credit: SpringSun . . . . . .
12. 12. 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): Rθ 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 wheel. . Image credit: SpringSun . . . . . .
13. 13. The Linear Case Question Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the composition? . . . . . .
14. 14. 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) . . . . . .
15. 15. 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 . . . . . .
16. 16. 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. . . . . . .
17. 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) 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. . . . . . .
18. 18. 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 . . . . . .
19. 19. Outline Heuristics Analogy The Linear Case The chain rule Examples Related rates of change . . . . . .
20. 20. 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 . . . . . .
21. 21. Observations Succinctly, the derivative of a composition is the product of the derivatives . Image credit: o . . . . . . .
22. 22. 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 . . . . . .
23. 23. 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: o . . . . . . .
24. 24. 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 . . . . . . .
25. 25. 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) . Image credit: o fractions . . . . . . .
26. 26. 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 . . . . . .
27. 27. 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 . . . . . .
28. 28. Outline Heuristics Analogy The Linear Case The chain rule Examples Related rates of change . . . . . .
29. 29. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). . . . . . .
30. 30. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). Solution First, write h as f ◦ g. . . . . . .
31. 31. 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. . . . . . .
32. 32. 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 . . . . . .
33. 33. 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 . . . . . .
34. 34. 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 . . . . . .
35. 35. Does order matter? Example d d Find (sin 4x) and compare it to (4 sin x). dx dx . . . . . .
36. 36. 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 . . . . . .
37. 37. 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 . . . . . .
38. 38. 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 . . . . . .
39. 39. Example (√ )2 . Find f′ (x). 3 Let f(x) = x5 − 2 + 8 . . . . . .
40. 40. 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 . . . . . .
41. 41. 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 . . . . . .
42. 42. 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 . . . . . .
43. 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 (√ ) 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 . . . . . .
44. 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 (√ ) 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 . . . . . .
45. 45. A metaphor Think about peeling an onion: (√ )2 3 5 f(x) = x −2 +8 5 √ 3 +8 . Image credit: p . 2 (√ ) f′ (x) = 2 x5 − 2 + 8 1 (x5 − 2)−2/3 (5x4 ) 3 3 . . . . . .
46. 46. Combining techniques Example d ( 3 ) Find (x + 1)10 sin(4x2 − 7) dx . . . . . .
47. 47. 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: . . . . . .
48. 48. 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 . . . . . .
49. 49. 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) . . . . . .
50. 50. 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)))))) . . . . . .
51. 51. Solution to #1 Example Find the derivative of y = (1 − x2 )10 . Solution y′ = 10(1 − x2 )9 (−2x) = −20x(1 − x2 )9 . . . . . .
52. 52. 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 . . . . . .
53. 53. 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 . . . . . .
54. 54. 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 . . . . . .
55. 55. 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 . . . . . .
56. 56. Solution to #6 Example Find the derivative of y = tan(cos x). Solution y′ = sec2 (cos x) · (− sin x) = − sec2 (cos x) sin x . . . . . .
57. 57. 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 . . . . . .
58. 58. 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)) . . . . . .
59. 59. Outline Heuristics Analogy The Linear Case The chain rule Examples Related rates of change . . . . . .
60. 60. 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 . Image dA dr . B. = 2π r + dt dt dA dr C. = 2π r dt dt D. not enough information . . . . . .
61. 61. 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 . Image dA dr . B. = 2π r + dt dt dA dr C. = 2π r dt dt D. not enough information . . . . . .
62. 62. 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! . . . . . .