Lesson 10: The Chain Rule (Section 21 slides)

  • 660 views
Uploaded on

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. …

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.

  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads

Views

Total Views
660
On Slideshare
0
From Embeds
0
Number of Embeds
0

Actions

Shares
Downloads
22
Comments
0
Likes
0

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 1. Section 2.5 The Chain Rule V63.0121.021, Calculus I New York University October 7, 2010 Announcements Quiz 2 in recitation next week (October 11-15) Midterm in class Tuesday, october 19 on §§1.1–2.5 . . . . . .
  • 2. Announcements Quiz 2 in recitation next week (October 11-15) Midterm in class Tuesday, october 19 on §§1.1–2.5 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 2 / 36
  • 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 3 / 36
  • 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.” . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
  • 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 . . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
  • 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) g . . f . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
  • 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.” x . g . (x) f .(g(x)) g . . f . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
  • 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.” g . (x) f .(g(x)) . ◦ g x . g . f . f . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
  • 9. 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.” g . (x) f .(g(x)) . ◦ g x . g . f . 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 4 / 36
  • 10. Outline Heuristics Analogy The Linear Case The chain rule Examples Related rates of change . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 5 / 36
  • 11. Analogy Think about riding a bike. To go faster you can either: . . Image credit: SpringSun . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
  • 12. Analogy Think about riding a bike. To go faster you can either: pedal faster . . Image credit: SpringSun . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
  • 13. Analogy Think about riding a bike. To go faster you can either: pedal faster change gears . . Image credit: SpringSun . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 6 / 36
  • 17. The Linear Case Question Let f(x) = mx + b and g(x) = m′ x + b′ . What can you say about the composition? . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 7 / 36
  • 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) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 7 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 7 / 36
  • 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 7 / 36
  • 21. 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 7 / 36
  • 22. The Nonlinear Case 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 8 / 36
  • 23. Outline Heuristics Analogy The Linear Case The chain rule Examples Related rates of change . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 9 / 36
  • 24. 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 10 / 36
  • 25. Observations Succinctly, the derivative of a composition is the product of the derivatives . . Image credit: ooOJasonOoo . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 11 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 12 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 13 / 36
  • 28. 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.” g . (x) f .(g(x)) . ◦ g x . g . f . f . . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 14 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 15 / 36
  • 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) In Leibnizian notation, let y = f(u) and u = g(x). Then dy dy du = dx du dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 16 / 36
  • 31. 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(x).du Then . .  dx du dy dy du = dx du dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 16 / 36
  • 32. Outline Heuristics Analogy The Linear Case The chain rule Examples Related rates of change . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 17 / 36
  • 33. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 18 / 36
  • 34. Example Example √ let h(x) = 3x2 + 1. Find h′ (x). Solution First, write h as f ◦ g. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 18 / 36
  • 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. . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 18 / 36
  • 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) = 1 u−1/2 , and g′ (x) = 6x. So 2 h′ (x) = 1 u−1/2 (6x) 2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 18 / 36
  • 37. 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) = 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 18 / 36
  • 38. 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 19 / 36
  • 39. Does order matter? Example d d Find (sin 4x) and compare it to (4 sin x). dx dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 20 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 20 / 36
  • 41. 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 20 / 36
  • 42. 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 20 / 36
  • 43. Example (√ )2 . Find f′ (x). 3 Let f(x) = x5 − 2 + 8 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
  • 48. 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 21 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 22 / 36
  • 50. Combining techniques Example d ( 3 ) Find (x + 1)10 sin(4x2 − 7) dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 23 / 36
  • 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: . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 23 / 36
  • 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(4x − 7) + (x + 1) · 2 3 10 sin(4x − 7) 2 dx dx . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 23 / 36
  • 53. 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(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) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 23 / 36
  • 54. 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)))))) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 24 / 36
  • 55. Solution to #1 Example Find the derivative of y = (1 − x2 )10 . Solution y′ = 10(1 − x2 )9 (−2x) = −20x(1 − x2 )9 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 25 / 36
  • 56. 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 26 / 36
  • 57. Solution to #3 Example √ 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 27 / 36
  • 58. 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 28 / 36
  • 59. 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 z−1 (z + 1)2 (z + 1)3/2 (z − 1)1/2 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 29 / 36
  • 60. Solution to #6 Example Find the derivative of y = tan(cos x). Solution y′ = sec2 (cos x) · (− sin x) = − sec2 (cos x) sin x . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 30 / 36
  • 61. 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 31 / 36
  • 62. 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)) . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 32 / 36
  • 63. Outline Heuristics Analogy The Linear Case The chain rule Examples Related rates of change . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 33 / 36
  • 64. Related rates of change at the Deli Question Suppose a deli clerk can slice a stick of pepperoni (assume the tapered ends have been removed) by hand at the rate of 2 inches per minute, while a machine can slice pepperoni at the rate of 10 inches dV dV per minute. Then for the machine is 5 times greater than for dt dt the deli clerk. This is explained by the A. chain rule B. product rule C. quotient Rule D. addition rule . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 34 / 36
  • 65. Related rates of change at the Deli Question Suppose a deli clerk can slice a stick of pepperoni (assume the tapered ends have been removed) by hand at the rate of 2 inches per minute, while a machine can slice pepperoni at the rate of 10 inches dV dV per minute. Then for the machine is 5 times greater than for dt dt the deli clerk. This is explained by the A. chain rule B. product rule C. quotient Rule D. addition rule . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 34 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 35 / 36
  • 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 . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 35 / 36
  • 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! . . . . . . V63.0121.021, Calculus I (NYU) Section 2.5 The Chain Rule October 7, 2010 36 / 36