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# Lesson 10: The Chain Rule (Section 41 handout)

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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 (Section 41 handout)

1. 1. V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010 Notes Section 2.5 The Chain Rule V63.0121.041, Calculus I New York University October 5, 2010 Announcements Quiz 2 in recitation next week (October 11-15) Midterm in class on §§1.1–2.5 Announcements Notes Quiz 2 in recitation next week (October 11-15) Midterm in class on §§1.1–2.5 V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 2 / 36 Objectives Notes 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. V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 3 / 36 1
2. 2. V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010 Compositions See Section 1.2 for review Notes 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 ◦ 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. V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 4 / 36 Outline Notes Heuristics Analogy The Linear Case The chain rule Examples Related rates of change V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 5 / 36 Analogy Notes Think about riding a bike. To go faster you can either: pedal faster change gears radius of front sprocket 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 sprocket. radius of back sprocket Image credit: SpringSun V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 6 / 36 2
3. 3. V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010 The Linear Case Notes 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. V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 7 / 36 The Nonlinear Case Notes 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 V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 8 / 36 Outline Notes Heuristics Analogy The Linear Case The chain rule Examples Related rates of change V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 9 / 36 3
4. 4. V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010 Theorem of the day: The chain rule Notes 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 dy Then du (x). dx du dy dy du = dx du dx V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 10 / 36 Observations Notes 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 V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 11 / 36 Outline Notes Heuristics Analogy The Linear Case The chain rule Examples Related rates of change V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 17 / 36 4
5. 5. V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010 Example Notes Example let h(x) = 3x 2 + 1. Find h (x). Solution √ First, write h as f ◦ g . Let f (u) = u and g (x) = 3x 2 + 1. Then −1/2 f (u) = 1 u 2 , and g (x) = 6x. So 3x h (x) = 1 u −1/2 (6x) = 2 (3x 2 + 1)−1/2 (6x) = √ 2 1 3x 2 + 1 V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 18 / 36 Corollary Notes 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 ) = nu n−1 . dx dx V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 19 / 36 Order matters! Notes Example d d Find (sin 4x) and compare it to (4 sin x). dx dx Solution V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 20 / 36 5
6. 6. V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010 Example 3 2 Notes Let f (x) = x5 − 2 + 8 . Find f (x). Solution V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 21 / 36 A metaphor Notes Think about peeling an onion: 2 3 f (x) = x 5 −2 +8 5 √ 3 +8 2 − 2)−2/3 (5x 4 ) 3 1 5 f (x) = 2 x5 − 2 + 8 3 (x Image credit: photobunny V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 22 / 36 Combining techniques Notes Example d Find (x 3 + 1)10 sin(4x 2 − 7) dx Solution V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 23 / 36 6
7. 7. V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010 Your Turn Notes Find derivatives of these functions: 1. y = (1 − x 2 )10 √ 2. y = sin x √ 3. y = sin x 4. y = (2x − 5)4 (8x 2 − 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)))))) V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 24 / 36 Solution to #1 Notes Example Find the derivative of y = (1 − x 2 )10 . Solution V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 25 / 36 Solution to #2 Notes Example √ Find the derivative of y = sin x. Solution V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 26 / 36 7
8. 8. V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010 Solution to #3 Notes Example √ Find the derivative of y = sin x. Solution V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 27 / 36 Solution to #4 Notes Example Find the derivative of y = (2x − 5)4 (8x 2 − 5)−3 Solution V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 28 / 36 Solution to #5 Notes Example z −1 Find the derivative of F (z) = . z +1 Solution V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 29 / 36 8
9. 9. V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010 Solution to #6 Notes Example Find the derivative of y = tan(cos x). Solution V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 30 / 36 Solution to #7 Notes Example Solution V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 31 / 36 Solution to #8 Notes Example Find the derivative of y = sin(sin(sin(sin(sin(sin(x)))))). Solution V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 32 / 36 9
10. 10. V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010 Outline Notes Heuristics Analogy The Linear Case The chain rule Examples Related rates of change V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 33 / 36 Related rates of change at the Deli Notes 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 per minute. Then dV dV for the machine is 5 times greater than for the deli clerk. This is dt dt explained by the A. chain rule B. product rule C. quotient Rule D. addition rule V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 34 / 36 Related rates of change in the ocean Notes Question The area of a circle, A = πr 2 , 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 V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 35 / 36 10
11. 11. V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010 Summary Notes The derivative of a composition is the product of derivatives In symbols: (f ◦ g ) (x) = f (g (x))g (x) V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 36 / 36 Notes Notes 11