..

Sec on 2.5The Chain Rule

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

February 23, 2011

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)

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.

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.

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.

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.

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.

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.

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.

Outline

Heuris csAnalogyThe Linear Case

The chain rule

Examples

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

I pedal fasterI change gears

..

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 sprocket

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

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

I pedal faster

I change gears

..

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 sprocket

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

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

I pedal fasterI change gears

..

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 sprocket

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

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

I pedal fasterI change gears

..

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.

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

I pedal fasterI change gears

..

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 sprocket

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

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

I pedal fasterI change gears

..

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 sprocket

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

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.

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.

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.

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.

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.

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

f′(u) ≈ ∆y∆u

=⇒ ∆y ≈ f′(u)∆u

andg′(x) ≈ ∆y

∆x=⇒ ∆u ≈ g′(x)∆x.

So∆y ≈ f′(u)g′(x)∆x =⇒ ∆y

∆x≈ f′(u)g′(x)

Outline

Heuris csAnalogyThe Linear Case

The chain rule

Examples

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

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

..

Image credit: ooOJasonOoo

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

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

..

Image credit: ooOJasonOoo

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.

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

..

Image credit: ooOJasonOoo

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

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

Outline

Heuris csAnalogyThe Linear Case

The chain rule

Examples

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

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

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.

Thenf′(u) = 1

2u−1/2, and g′(x) = 6x. So

h′(x) = 12u

−1/2(6x) = 12(3x

2 + 1)−1/2(6x) =3x√

3x2 + 1

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

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

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

.

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

Does order matter?Example

Findddx

(sin 4x) and compare it toddx

(4 sin x).

Solu onI 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

Does order matter?Example

Findddx

(sin 4x) and compare it toddx

(4 sin x).

Solu onI 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

Order matters!Example

Findddx

(sin 4x) and compare it toddx

(4 sin x).

Solu onI 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

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

Example

Let f(x) =(

3√x5 − 2+ 8

)2. Find f′(x).

Solu onddx

(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

Example

Let f(x) =(

3√x5 − 2+ 8

)2. Find f′(x).

Solu onddx

(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

Example

Let f(x) =(

3√x5 − 2+ 8

)2. Find f′(x).

Solu onddx

(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

Example

Let f(x) =(

3√x5 − 2+ 8

)2. Find f′(x).

Solu onddx

(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

Example

Let f(x) =(

3√x5 − 2+ 8

)2. Find f′(x).

Solu onddx

(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

A metaphorThink about peeling an onion:

f(x) =(

3√

x5︸︷︷︸�5

−2︸ ︷︷ ︸3√�

+8

︸ ︷︷ ︸�+8

)2

︸ ︷︷ ︸�2

..

Image credit: photobunny

f′(x) = 2(

3√x5 − 2+ 8

)13(x

5 − 2)−2/3(5x4)

Combining techniquesExample

Findddx

((x3 + 1)10 sin(4x2 − 7)

)

Solu onThe “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)

Combining techniquesExample

Findddx

((x3 + 1)10 sin(4x2 − 7)

)Solu onThe “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)

Combining techniquesExample

Findddx

((x3 + 1)10 sin(4x2 − 7)

)Solu onThe “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)

Combining techniquesExample

Findddx

((x3 + 1)10 sin(4x2 − 7)

)Solu onThe “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)

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

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

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!