Lesson 10: The Chain Rule

. . . . . .

Section2.5TheChainRule

V63.0121.034, CalculusI

October5, 2009

Announcements

I Quiz2thisweekI Midterm inclasson§§1.1–1.4

. . . . . .

CompositionsSeeSection1.2forreview

DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

.

.g .f.x .g(x) .f(g(x))

.f ◦ g

Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.

. . . . . .

CompositionsSeeSection1.2forreview

DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

..g

.f

.x .g(x)

.f(g(x)).f ◦ g

Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.

. . . . . .

CompositionsSeeSection1.2forreview

DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

..g .f.x .g(x)

.f(g(x)).f ◦ g

Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.

. . . . . .

CompositionsSeeSection1.2forreview

DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

..g .f.x .g(x) .f(g(x))

.f ◦ g

Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.

. . . . . .

CompositionsSeeSection1.2forreview

DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

..g .f.x .g(x) .f(g(x))

.f ◦ g

Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.

. . . . . .

CompositionsSeeSection1.2forreview

DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

..g .f.x .g(x) .f(g(x))

.f ◦ g

Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.

. . . . . .

Outline

HeuristicsAnalogyTheLinearCase

Thechainrule

Examples

Relatedratesofchange

. . . . . .

Analogy

Thinkaboutridingabike. Togofasteryoucaneither:

I pedalfasterI changegears

.

.Imagecredit: SpringSun

Theangularposition(φ)ofthebackwheeldependsonthepositionofthefrontsprocket(θ):

φ(θ) =Rθ

r

Andsotheangularspeedofthebackwheeldependsonthederivativeofthisfunctionandthespeedofthefrontwheel.

. . . . . .

Analogy

Thinkaboutridingabike. Togofasteryoucaneither:

I pedalfaster

I changegears

.

.Imagecredit: SpringSun

Theangularposition(φ)ofthebackwheeldependsonthepositionofthefrontsprocket(θ):

φ(θ) =Rθ

r

Andsotheangularspeedofthebackwheeldependsonthederivativeofthisfunctionandthespeedofthefrontwheel.

. . . . . .

Analogy

Thinkaboutridingabike. Togofasteryoucaneither:

I pedalfasterI changegears

.

.Imagecredit: SpringSun

Theangularposition(φ)ofthebackwheeldependsonthepositionofthefrontsprocket(θ):

φ(θ) =Rθ

r

Andsotheangularspeedofthebackwheeldependsonthederivativeofthisfunctionandthespeedofthefrontwheel.

. . . . . .

Analogy

Thinkaboutridingabike. Togofasteryoucaneither:

I pedalfasterI changegears

.

.Imagecredit: SpringSun

Theangularposition(φ)ofthebackwheeldependsonthepositionofthefrontsprocket(θ):

φ(θ) =Rθ

r

Andsotheangularspeedofthebackwheeldependsonthederivativeofthisfunctionandthespeedofthefrontwheel.

. . . . . .

TheLinearCase

QuestionLet f(x) = mx + b and g(x) = m′x + b′. Whatcanyousayaboutthecomposition?

Answer

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

I ThecompositionisalsolinearI Theslopeofthecompositionistheproductoftheslopesofthetwofunctions.

Thederivativeissupposedtobealocallinearizationofafunction. Sothereshouldbeananalogofthispropertyinderivatives.

. . . . . .

TheLinearCase

QuestionLet f(x) = mx + b and g(x) = m′x + b′. Whatcanyousayaboutthecomposition?

Answer

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

I ThecompositionisalsolinearI Theslopeofthecompositionistheproductoftheslopesofthetwofunctions.

Thederivativeissupposedtobealocallinearizationofafunction. Sothereshouldbeananalogofthispropertyinderivatives.

. . . . . .

TheLinearCase

QuestionLet f(x) = mx + b and g(x) = m′x + b′. Whatcanyousayaboutthecomposition?

Answer

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

I Thecompositionisalsolinear

I Theslopeofthecompositionistheproductoftheslopesofthetwofunctions.

Thederivativeissupposedtobealocallinearizationofafunction. Sothereshouldbeananalogofthispropertyinderivatives.

. . . . . .

TheLinearCase

QuestionLet f(x) = mx + b and g(x) = m′x + b′. Whatcanyousayaboutthecomposition?

Answer

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

I ThecompositionisalsolinearI Theslopeofthecompositionistheproductoftheslopesofthetwofunctions.

Thederivativeissupposedtobealocallinearizationofafunction. Sothereshouldbeananalogofthispropertyinderivatives.

. . . . . .

TheLinearCase

QuestionLet f(x) = mx + b and g(x) = m′x + b′. Whatcanyousayaboutthecomposition?

Answer

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

I ThecompositionisalsolinearI Theslopeofthecompositionistheproductoftheslopesofthetwofunctions.

Thederivativeissupposedtobealocallinearizationofafunction. Sothereshouldbeananalogofthispropertyinderivatives.

. . . . . .

TheNonlinearCaseSeetheMathematicaapplet

Let u = g(x) and y = f(u). Suppose x ischangedbyasmallamount ∆x. Then

∆y ≈ f′(y)∆u

and∆u ≈ g′(u)∆x.

So∆y ≈ f′(y)g′(u)∆x =⇒ ∆y

∆x≈ f′(y)g′(u)

. . . . . .

Outline

HeuristicsAnalogyTheLinearCase

Thechainrule

Examples

Relatedratesofchange

. . . . . .

Theoremoftheday: Thechainrule

TheoremLet f and g befunctions, with g differentiableat x and fdifferentiableat g(x). Then f ◦ g isdifferentiableat x and

(f ◦ g)′(x) = f′(g(x))g′(x)

InLeibniziannotation, let y = f(u) and u = g(x). Then

dydx

=dydu

dudx

..dy

��du��dudx

. . . . . .

Observations

I Succinctly, thederivativeofacompositionistheproductofthederivatives

I Theonlycomplicationiswherethesederivativesareevaluated: atthesamepointthefunctionsare

I InLeibniznotation, theChainRulelookslikecancellationof(fake)fractions

..Imagecredit: ooOJasonOoo

. . . . . .

Theoremoftheday: Thechainrule

TheoremLet f and g befunctions, with g differentiableat x and fdifferentiableat g(x). Then f ◦ g isdifferentiableat x and

(f ◦ g)′(x) = f′(g(x))g′(x)

InLeibniziannotation, let y = f(u) and u = g(x). Then

dydx

=dydu

dudx

..dy

��du��dudx

. . . . . .

Observations

I Succinctly, thederivativeofacompositionistheproductofthederivatives

I Theonlycomplicationiswherethesederivativesareevaluated: atthesamepointthefunctionsare

I InLeibniznotation, theChainRulelookslikecancellationof(fake)fractions

..Imagecredit: ooOJasonOoo

. . . . . .

CompositionsSeeSection1.2forreview

DefinitionIf f and g arefunctions, the composition (f ◦ g)(x) = f(g(x)) means“do g first, then f.”

..g .f.x .g(x) .f(g(x))

.f ◦ g

Ourgoalforthedayistounderstandhowthederivativeofthecompositionoftwofunctionsdependsonthederivativesoftheindividualfunctions.

. . . . . .

Observations

I Succinctly, thederivativeofacompositionistheproductofthederivatives

I Theonlycomplicationiswherethesederivativesareevaluated: atthesamepointthefunctionsare

I InLeibniznotation, theChainRulelookslikecancellationof(fake)fractions .

.Imagecredit: ooOJasonOoo

. . . . . .

Theoremoftheday: Thechainrule

TheoremLet f and g befunctions, with g differentiableat x and fdifferentiableat g(x). Then f ◦ g isdifferentiableat x and

(f ◦ g)′(x) = f′(g(x))g′(x)

InLeibniziannotation, let y = f(u) and u = g(x). Then

dydx

=dydu

dudx

..dy

��du��dudx

. . . . . .

Theoremoftheday: Thechainrule

TheoremLet f and g befunctions, with g differentiableat x and fdifferentiableat g(x). Then f ◦ g isdifferentiableat x and

(f ◦ g)′(x) = f′(g(x))g′(x)

InLeibniziannotation, let y = f(u) and u = g(x). Then

dydx

=dydu

dudx

..dy

��du��dudx

. . . . . .

Outline

HeuristicsAnalogyTheLinearCase

Thechainrule

Examples

Relatedratesofchange

. . . . . .

Example

Examplelet h(x) =

√3x2 + 1. Find h′(x).

SolutionFirst, 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

. . . . . .

Example

Examplelet h(x) =

√3x2 + 1. Find h′(x).

SolutionFirst, 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

. . . . . .

Example

Examplelet h(x) =

√3x2 + 1. Find h′(x).

SolutionFirst, 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

. . . . . .

Example

Examplelet h(x) =

√3x2 + 1. Find h′(x).

SolutionFirst, 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

. . . . . .

Example

Examplelet h(x) =

√3x2 + 1. Find h′(x).

SolutionFirst, 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(ThePowerRuleCombinedwiththeChainRule)If n isanyrealnumberand u = g(x) isdifferentiable, then

ddx

(un) = nun−1dudx

.

. . . . . .

Doesordermatter?

Example

Findddx

(sin 4x) andcompareittoddx

(4 sin x).

Solution

I Forthefirst, let u = 4x and y = sin(u). Then

dydx

=dydu

· dudx

= cos(u) · 4 = 4 cos 4x.

I Forthesecond, let u = sin x and y = 4u. Then

dydx

=dydu

· dudx

= 4 · cos x

. . . . . .

Doesordermatter?

Example

Findddx

(sin 4x) andcompareittoddx

(4 sin x).

Solution

I Forthefirst, let u = 4x and y = sin(u). Then

dydx

=dydu

· dudx

= cos(u) · 4 = 4 cos 4x.

I Forthesecond, let u = sin x and y = 4u. Then

dydx

=dydu

· dudx

= 4 · cos x

. . . . . .

Doesordermatter?

Example

Findddx

(sin 4x) andcompareittoddx

(4 sin x).

Solution

I Forthefirst, let u = 4x and y = sin(u). Then

dydx

=dydu

· dudx

= cos(u) · 4 = 4 cos 4x.

I Forthesecond, let u = sin x and y = 4u. Then

dydx

=dydu

· dudx

= 4 · cos x

. . . . . .

Ordermatters!

Example

Findddx

(sin 4x) andcompareittoddx

(4 sin x).

Solution

I Forthefirst, let u = 4x and y = sin(u). Then

dydx

=dydu

· dudx

= cos(u) · 4 = 4 cos 4x.

I Forthesecond, 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).

Solution

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)

=103

x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

. . . . . .

Example

Let f(x) =(

3√

x5 − 2 + 8)2. Find f′(x).

Solution

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)

=103

x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

. . . . . .

Example

Let f(x) =(

3√

x5 − 2 + 8)2. Find f′(x).

Solution

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)

=103

x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

. . . . . .

Example

Let f(x) =(

3√

x5 − 2 + 8)2. Find f′(x).

Solution

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)

=103

x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

. . . . . .

Example

Let f(x) =(

3√

x5 − 2 + 8)2. Find f′(x).

Solution

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)

=103

x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

. . . . . .

Example

Let f(x) =(

3√

x5 − 2 + 8)2. Find f′(x).

Solution

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)

=103

x4(

3√

x5 − 2 + 8)

(x5 − 2)−2/3

. . . . . .

A metaphor

Thinkaboutpeelinganonion:

f(x) =

(3√

x5︸︷︷︸�5

−2

︸ ︷︷ ︸3√�

+8

︸ ︷︷ ︸�+8

)2

︸ ︷︷ ︸�2

..Imagecredit: photobunny

f′(x) = 2(

3√

x5 − 2 + 8)

13(x

5 − 2)−2/3(5x4)

. . . . . .

Combiningtechniques

Example

Findddx

((x3 + 1)10 sin(4x2 − 7)

)

SolutionThe“last”partofthefunctionistheproduct, soweapplytheproductrule. Eachfactor’sderivativerequiresthechainrule:

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)

. . . . . .

Combiningtechniques

Example

Findddx

((x3 + 1)10 sin(4x2 − 7)

)SolutionThe“last”partofthefunctionistheproduct, soweapplytheproductrule. Eachfactor’sderivativerequiresthechainrule:

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)

. . . . . .

Combiningtechniques

Example

Findddx

((x3 + 1)10 sin(4x2 − 7)

)SolutionThe“last”partofthefunctionistheproduct, soweapplytheproductrule. Eachfactor’sderivativerequiresthechainrule:

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)

. . . . . .

Combiningtechniques

Example

Findddx

((x3 + 1)10 sin(4x2 − 7)

)SolutionThe“last”partofthefunctionistheproduct, soweapplytheproductrule. Eachfactor’sderivativerequiresthechainrule:

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)

. . . . . .

YourTurn

Findderivativesofthesefunctions:

1. y = (1− x2)10

2. y =√sin x

3. y = sin√x

4. y = (2x− 5)4(8x2 − 5)−3

5. F(z) =

√z− 1z + 1

6. y = tan(cos x)

7. y = csc2(sin θ)

8. y = sin(sin(sin(sin(sin(sin(x))))))

. . . . . .

Solutionto#1

ExampleFindthederivativeof y = (1− x2)10.

Solutiony′ = 10(1− x2)9(−2x) = −20x(1− x2)9

. . . . . .

Solutionto#2

ExampleFindthederivativeof y =

√sin x.

SolutionWriting

√sin x as (sin x)1/2, wehave

y′ = 12 (sin x)−1/2 (cos x) =

cos x

2√sin x

. . . . . .

Solutionto#3

ExampleFindthederivativeof y = sin

√x.

Solution

y′ =ddx

sin(x1/2) = cos(x1/2)12x−1/2 =

cos(√

x)

2√x

. . . . . .

Solutionto#4

ExampleFindthederivativeof y = (2x− 5)4(8x2 − 5)−3

SolutionWeneedtousetheproductruleandthechainrule:

y′ = 4(2x− 5)3(2)(8x2 − 5)−3 + (2x− 5)4(−3)(8x2 − 5)−4(16x)

Therestisabitofalgebra, usefulifyouwantedtosolvetheequation 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

)

. . . . . .

Solutionto#5

Example

Findthederivativeof F(z) =

√z− 1z + 1

.

Solution

y′ =12

(z− 1z + 1

)−1/2 ((z + 1)(1) − (z− 1)(1)

(z + 1)2

)=

12

(z + 1z− 1

)1/2 (2

(z + 1)2

)=

1(z + 1)3/2(z− 1)1/2

. . . . . .

Solutionto#6

ExampleFindthederivativeof y = tan(cos x).

Solutiony′ = sec2(cos x) · (− sin x) = − sec2(cos x) sin x

. . . . . .

Solutionto#7

ExampleFindthederivativeof y = csc2(sin θ).

SolutionRememberthenotation:

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

. . . . . .

Solutionto#8

ExampleFindthederivativeof y = sin(sin(sin(sin(sin(sin(x)))))).

SolutionRelax! It’sjustabunchofchainrules. Alloftheselinesaremultipliedtogether.

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))

. . . . . .

Outline

HeuristicsAnalogyTheLinearCase

Thechainrule

Examples

Relatedratesofchange

. . . . . .

Relatedratesofchange

QuestionTheareaofacircle, A = πr2,changesasitsradiuschanges. Iftheradiuschangeswithrespecttotime,thechangeinareawithrespecttotimeis

A.dAdr

= 2πr

B.dAdt

= 2πr +drdt

C.dAdt

= 2πrdrdt

D. notenoughinformation

..Imagecredit: JimFrazier

. . . . . .

Relatedratesofchange

QuestionTheareaofacircle, A = πr2,changesasitsradiuschanges. Iftheradiuschangeswithrespecttotime,thechangeinareawithrespecttotimeis

A.dAdr

= 2πr

B.dAdt

= 2πr +drdt

C.dAdt

= 2πrdrdt

D. notenoughinformation

..Imagecredit: JimFrazier

. . . . . .

Whathavewelearnedtoday?

I Thederivativeofacompositionistheproductofderivatives

I Insymbols:(f ◦ g)′(x) = f′(g(x))g′(x)

I Calculusislikeanonion, andnotbecauseitmakesyoucry!