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.. Sec on 2.5The Chain Rule

V63.0121.001: Calculus IProfessor Ma hew Leingang

New York University

February 23, 2011

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Announcements

I Quiz 2 next week on§§1.5, 1.6, 2.1, 2.2

I Midterm March 7 on allsec ons in class (coversall sec ons up to 2.5)

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ObjectivesI Given a compoundexpression, write it as acomposi on of func ons.

I Understand and applythe Chain Rule for thederiva ve of acomposi on of func ons.

I Understand and useNewtonian and Leibniziannota ons for the ChainRule.

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Notes

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Notes

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Notes

. 1.

. Sec on 2.5: The Chain Rule. V63.0121.001: Calculus I . February 23, 2011

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CompositionsSee Section 1.2 for review

Defini onIf f and g are func ons, the composi on (f ◦ g)(x) = f(g(x))means “do g first,then f.”

..g . f.x .g(x). f(g(x)).f ◦ g

Our goal for the day is to understand how the deriva ve of the composi on oftwo func ons depends on the deriva ves of the individual func ons.

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Outline

Heuris csAnalogyThe Linear Case

The chain rule

Examples

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AnalogyThink about riding a bike. Togo faster you can either:

I pedal fasterI change gears

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Image credit: SpringSun

The angular posi on (φ) of the back wheel depends on the posi onof the front sprocket (θ):

φ(θ) =R...

radius of front sprocket

θ

r...

radius of back sprocketAnd so the angular speed of the back wheel depends on thederiva ve of this func on and the speed of the front sprocket.

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Notes

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Notes

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Notes

. 2.

. Sec on 2.5: The Chain Rule. V63.0121.001: Calculus I . February 23, 2011

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The Linear CaseQues onLet f(x) = mx+ b and g(x) = m′x+ b′. What can you say about the composi on?

Answer

I f(g(x)) = m(m′x+ b′) + b = (mm′)x+ (mb′ + b)

I The composi on is also linear

I The slope of the composi on is the product of the slopes of the twofunc ons.

The deriva ve is supposed to be a local lineariza on of a func on. So thereshould be an analog of this property in deriva ves.

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The Nonlinear CaseLet u = g(x) and y = f(u). Suppose x is changed by a small amount∆x. Then

f′(y) ≈ ∆y∆x

=⇒ ∆y ≈ f′(y)∆u

andg′(y) ≈ ∆u

∆x=⇒ ∆u ≈ g′(u)∆x.

So∆y ≈ f′(y)g′(u)∆x =⇒ ∆y

∆x≈ f′(y)g′(u)

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Outline

Heuris csAnalogyThe Linear Case

The chain rule

Examples

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Notes

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Notes

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Notes

. 3.

. Sec on 2.5: The Chain Rule. V63.0121.001: Calculus I . February 23, 2011

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Theorem of the day: The chain ruleTheoremLet f and g be func ons, with g differen able at x and f differen ableat 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

dydx

=dydu

dudx

..dy��du

��dudx

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ObservationsI Succinctly, the deriva ve of acomposi on is the productof the deriva ves

I The only complica on iswhere these deriva ves areevaluated: at the same pointthe func ons are

I In Leibniz nota on, the ChainRule looks like cancella on of(fake) frac ons

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Image credit: ooOJasonOoo

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Outline

Heuris csAnalogyThe Linear Case

The chain rule

Examples

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Notes

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Notes

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Notes

. 4.

. Sec on 2.5: The Chain Rule. V63.0121.001: Calculus I . February 23, 2011

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ExampleExample

let h(x) =√3x2 + 1. Find h′(x).

Solu onFirst, write h as f ◦ g. Let f(u) =

√u and g(x) = 3x2 + 1. Then

f′(u) = 12u

−1/2, and g′(x) = 6x. So

h′(x) = 12u

−1/2(6x) = 12(3x

2 + 1)−1/2(6x) =3x√

3x2 + 1

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Corollary

Corollary (The Power Rule Combined with the Chain Rule)

If n is any real number and u = g(x) is differen able, then

ddx

(un) = nun−1dudx

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Does order matter?Example

Findddx

(sin 4x) and compare it toddx

(4 sin x).

Solu on

I For the first, let u = 4x and y = sin(u). Then

dydx

=dydu

· dudx

= cos(u) · 4 = 4 cos 4x.

I For the second, let u = sin x and y = 4u. Then

dydx

=dydu

· dudx

= 4 · cos x

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Notes

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Notes

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Notes

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. Sec on 2.5: The Chain Rule. V63.0121.001: Calculus I . February 23, 2011

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Example

Let f(x) =(

3√x5 − 2+ 8

)2. Find f′(x).

Solu on

ddx

(3√x5 − 2+ 8

)2= 2

(3√x5 − 2+ 8

) ddx

(3√x5 − 2+ 8

)= 2

(3√x5 − 2+ 8

) ddx

3√x5 − 2

= 2(

3√x5 − 2+ 8

)13(x

5 − 2)−2/3 ddx

(x5 − 2)

= 2(

3√x5 − 2+ 8

)13(x

5 − 2)−2/3(5x4)

=103x4(

3√x5 − 2+ 8

)(x5 − 2)−2/3

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A metaphorThink about peeling an onion:

f(x) =(

3√

x5︸︷︷︸�5

−2︸ ︷︷ ︸3√�

+8

︸ ︷︷ ︸�+8

)2

︸ ︷︷ ︸�2

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Image credit: photobunny

f′(x) = 2(

3√x5 − 2+ 8

)13(x

5 − 2)−2/3(5x4)

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Combining techniquesExample

Findddx

((x3 + 1)10 sin(4x2 − 7)

)Solu on

The “last” part of the func on is the product, so we apply the product rule. Eachfactor’s deriva ve requires the chain rule:

ddx

((x3 + 1)10 · sin(4x2 − 7)

)=

(ddx

(x3 + 1)10)· sin(4x2 − 7) + (x3 + 1)10 ·

(ddx

sin(4x2 − 7))

= 10(x3 + 1)9(3x2) sin(4x2 − 7) + (x3 + 1)10 · cos(4x2 − 7)(8x)

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Notes

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Notes

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Notes

. 6.

. Sec on 2.5: The Chain Rule. V63.0121.001: Calculus I . February 23, 2011

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Related rates of change in the oceanQues onThe area of a circle, A = πr2, changes as its radiuschanges. If the radius changes with respect to me,the change in area with respect to me is

A.dAdr

= 2πr

B.dAdt

= 2πr+drdt

C.dAdt

= 2πrdrdt

D. not enough informa on..

Image credit: Jim Frazier

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Summary

I The deriva ve of acomposi on is theproduct of deriva ves

I In symbols:(f ◦ g)′(x) = f′(g(x))g′(x)

I Calculus is like an onion,and not because it makesyou cry!

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Notes

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Notes

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. Sec on 2.5: The Chain Rule. V63.0121.001: Calculus I . February 23, 2011