Lesson 10: The Chain Rule
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Transcript of Lesson 10: The Chain Rule
![Page 1: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/1.jpg)
. . . . . .
Section2.5TheChainRule
V63.0121.027, CalculusI
October6, 2009
Announcements
I Quiz2thisweekI Midterm inclasson§§1.1–1.4
![Page 2: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/2.jpg)
. . . . . .
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.
![Page 3: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/3.jpg)
. . . . . .
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.
![Page 4: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/4.jpg)
. . . . . .
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.
![Page 5: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/5.jpg)
. . . . . .
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.
![Page 6: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/6.jpg)
. . . . . .
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.
![Page 7: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/7.jpg)
. . . . . .
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.
![Page 8: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/8.jpg)
. . . . . .
Outline
HeuristicsAnalogyTheLinearCase
Thechainrule
Examples
Relatedratesofchange
![Page 9: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/9.jpg)
. . . . . .
Analogy
Thinkaboutridingabike. Togofasteryoucaneither:
I pedalfasterI changegears
.
.Imagecredit: SpringSun
Theangularposition(φ)ofthebackwheeldependsonthepositionofthefrontsprocket(θ):
φ(θ) =Rθ
r
Andsotheangularspeedofthebackwheeldependsonthederivativeofthisfunctionandthespeedofthefrontwheel.
![Page 10: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/10.jpg)
. . . . . .
Analogy
Thinkaboutridingabike. Togofasteryoucaneither:
I pedalfaster
I changegears
.
.Imagecredit: SpringSun
Theangularposition(φ)ofthebackwheeldependsonthepositionofthefrontsprocket(θ):
φ(θ) =Rθ
r
Andsotheangularspeedofthebackwheeldependsonthederivativeofthisfunctionandthespeedofthefrontwheel.
![Page 11: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/11.jpg)
. . . . . .
Analogy
Thinkaboutridingabike. Togofasteryoucaneither:
I pedalfasterI changegears
.
.Imagecredit: SpringSun
Theangularposition(φ)ofthebackwheeldependsonthepositionofthefrontsprocket(θ):
φ(θ) =Rθ
r
Andsotheangularspeedofthebackwheeldependsonthederivativeofthisfunctionandthespeedofthefrontwheel.
![Page 12: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/12.jpg)
. . . . . .
Analogy
Thinkaboutridingabike. Togofasteryoucaneither:
I pedalfasterI changegears
.
.Imagecredit: SpringSun
Theangularposition(φ)ofthebackwheeldependsonthepositionofthefrontsprocket(θ):
φ(θ) =Rθ
r
Andsotheangularspeedofthebackwheeldependsonthederivativeofthisfunctionandthespeedofthefrontwheel.
![Page 13: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/13.jpg)
. . . . . .
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.
![Page 14: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/14.jpg)
. . . . . .
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.
![Page 15: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/15.jpg)
. . . . . .
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.
![Page 16: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/16.jpg)
. . . . . .
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.
![Page 17: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/17.jpg)
. . . . . .
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.
![Page 18: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/18.jpg)
. . . . . .
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)
![Page 19: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/19.jpg)
. . . . . .
Outline
HeuristicsAnalogyTheLinearCase
Thechainrule
Examples
Relatedratesofchange
![Page 20: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/20.jpg)
. . . . . .
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
![Page 21: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/21.jpg)
. . . . . .
Observations
I Succinctly, thederivativeofacompositionistheproductofthederivatives
I Theonlycomplicationiswherethesederivativesareevaluated: atthesamepointthefunctionsare
I InLeibniznotation, theChainRulelookslikecancellationof(fake)fractions
.
.Imagecredit: ooOJasonOoo
![Page 22: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/22.jpg)
. . . . . .
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
![Page 23: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/23.jpg)
. . . . . .
Observations
I Succinctly, thederivativeofacompositionistheproductofthederivatives
I Theonlycomplicationiswherethesederivativesareevaluated: atthesamepointthefunctionsare
I InLeibniznotation, theChainRulelookslikecancellationof(fake)fractions
.
.Imagecredit: ooOJasonOoo
![Page 24: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/24.jpg)
. . . . . .
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.
![Page 25: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/25.jpg)
. . . . . .
Observations
I Succinctly, thederivativeofacompositionistheproductofthederivatives
I Theonlycomplicationiswherethesederivativesareevaluated: atthesamepointthefunctionsare
I InLeibniznotation, theChainRulelookslikecancellationof(fake)fractions .
.Imagecredit: ooOJasonOoo
![Page 26: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/26.jpg)
. . . . . .
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
![Page 27: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/27.jpg)
. . . . . .
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
![Page 28: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/28.jpg)
. . . . . .
Outline
HeuristicsAnalogyTheLinearCase
Thechainrule
Examples
Relatedratesofchange
![Page 29: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/29.jpg)
. . . . . .
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
![Page 30: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/30.jpg)
. . . . . .
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
![Page 31: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/31.jpg)
. . . . . .
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
![Page 32: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/32.jpg)
. . . . . .
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
![Page 33: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/33.jpg)
. . . . . .
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
![Page 34: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/34.jpg)
. . . . . .
Corollary
Corollary(ThePowerRuleCombinedwiththeChainRule)If n isanyrealnumberand u = g(x) isdifferentiable, then
ddx
(un) = nun−1dudx
.
![Page 35: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/35.jpg)
. . . . . .
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
![Page 36: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/36.jpg)
. . . . . .
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
![Page 37: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/37.jpg)
. . . . . .
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
![Page 38: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/38.jpg)
. . . . . .
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
![Page 39: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/39.jpg)
. . . . . .
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
![Page 40: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/40.jpg)
. . . . . .
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
![Page 41: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/41.jpg)
. . . . . .
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
![Page 42: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/42.jpg)
. . . . . .
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
![Page 43: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/43.jpg)
. . . . . .
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
![Page 44: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/44.jpg)
. . . . . .
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
![Page 45: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/45.jpg)
. . . . . .
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)
![Page 46: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/46.jpg)
. . . . . .
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)
![Page 47: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/47.jpg)
. . . . . .
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)
![Page 48: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/48.jpg)
. . . . . .
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)
![Page 49: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/49.jpg)
. . . . . .
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)
![Page 50: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/50.jpg)
. . . . . .
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))))))
![Page 51: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/51.jpg)
. . . . . .
Solutionto#1
ExampleFindthederivativeof y = (1− x2)10.
Solutiony′ = 10(1− x2)9(−2x) = −20x(1− x2)9
![Page 52: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/52.jpg)
. . . . . .
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
![Page 53: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/53.jpg)
. . . . . .
Solutionto#3
ExampleFindthederivativeof y = sin
√x.
Solution
y′ =ddx
sin(x1/2) = cos(x1/2)12x−1/2 =
cos(√
x)
2√x
![Page 54: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/54.jpg)
. . . . . .
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
)
![Page 55: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/55.jpg)
. . . . . .
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
![Page 56: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/56.jpg)
. . . . . .
Solutionto#6
ExampleFindthederivativeof y = tan(cos x).
Solutiony′ = sec2(cos x) · (− sin x) = − sec2(cos x) sin x
![Page 57: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/57.jpg)
. . . . . .
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
![Page 58: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/58.jpg)
. . . . . .
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))
![Page 59: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/59.jpg)
. . . . . .
Outline
HeuristicsAnalogyTheLinearCase
Thechainrule
Examples
Relatedratesofchange
![Page 60: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/60.jpg)
. . . . . .
Relatedratesofchange
QuestionTheareaofacircle, A = πr2,changesasitsradiuschanges. Iftheradiuschangeswithrespecttotime,thechangeinareawithrespecttotimeis
A.dAdr
= 2πr
B.dAdt
= 2πr +drdt
C.dAdt
= 2πrdrdt
D. notenoughinformation
.
.Imagecredit: JimFrazier
![Page 61: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/61.jpg)
. . . . . .
Relatedratesofchange
QuestionTheareaofacircle, A = πr2,changesasitsradiuschanges. Iftheradiuschangeswithrespecttotime,thechangeinareawithrespecttotimeis
A.dAdr
= 2πr
B.dAdt
= 2πr +drdt
C.dAdt
= 2πrdrdt
D. notenoughinformation
.
.Imagecredit: JimFrazier
![Page 62: Lesson 10: The Chain Rule](https://reader033.fdocuments.in/reader033/viewer/2022052410/5559e53fd8b42a39498b4ded/html5/thumbnails/62.jpg)
. . . . . .
Whathavewelearnedtoday?
I Thederivativeofacompositionistheproductofderivatives
I Insymbols:(f ◦ g)′(x) = f′(g(x))g′(x)
I Calculusislikeanonion, andnotbecauseitmakesyoucry!