Section 2.3 Product & Quotient Rules and Higher-Order Derivatives
Lesson 9: The Product and Quotient Rules
-
Upload
matthew-leingang -
Category
Education
-
view
1.004 -
download
0
description
Transcript of Lesson 9: The Product and Quotient Rules
![Page 1: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/1.jpg)
. . . . . .
Section2.4TheProductandQuotientRules
V63.0121.006/016, CalculusI
February16, 2010
Announcements
I Quiz2isFebruary26, covering§§1.5–2.3I MidtermI isMarch4, covering§§1.1–2.5I OfficeHoursW 1:30–2:30, R 9–10I doget-to-know-yousurveybyThursday
![Page 2: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/2.jpg)
. . . . . .
Outline
Grader’sCorner
DerivativeofaProductDerivationExamples
TheQuotientRuleDerivationExamples
MorederivativesoftrigonometricfunctionsDerivativeofTangentandCotangentDerivativeofSecantandCosecant
MoreonthePowerRulePowerRuleforPositiveIntegersbyInductionPowerRuleforNegativeIntegers
![Page 3: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/3.jpg)
. . . . . .
Problem1.5.20
Usethetheoremsoncontinuitytoshow h(x) =sin xx+ 1
is
continuous.
SolutionByTheorem 6, f(x) = sin x and g(x) = x+ 1 arecontinuousbecause f(x) isatrigonometricfunctionand g(x) isapolynomial.
ByTheorem 4, part 5, h(x) =f(x)g(x)
iscontinuouswherever
g(x) ̸= 0.
NoteThefunction h is notarationalfunction. A rationalfunctionisthequotientoftwo polynomials.
![Page 4: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/4.jpg)
. . . . . .
Problem1.5.20
Usethetheoremsoncontinuitytoshow h(x) =sin xx+ 1
is
continuous.
SolutionByTheorem 6, f(x) = sin x and g(x) = x+ 1 arecontinuousbecause f(x) isatrigonometricfunctionand g(x) isapolynomial.
ByTheorem 4, part 5, h(x) =f(x)g(x)
iscontinuouswherever
g(x) ̸= 0.
NoteThefunction h is notarationalfunction. A rationalfunctionisthequotientoftwo polynomials.
![Page 5: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/5.jpg)
. . . . . .
Problem1.5.20
Usethetheoremsoncontinuitytoshow h(x) =sin xx+ 1
is
continuous.
SolutionByTheorem 6, f(x) = sin x and g(x) = x+ 1 arecontinuousbecause f(x) isatrigonometricfunctionand g(x) isapolynomial.
ByTheorem 4, part 5, h(x) =f(x)g(x)
iscontinuouswherever
g(x) ̸= 0.
NoteThefunction h is notarationalfunction. A rationalfunctionisthequotientoftwo polynomials.
![Page 6: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/6.jpg)
. . . . . .
Problem1.6.20
limx→∞
x3 − 2x+ 35− 2x2
= limx→∞
x3
x2· 1− 2/x2 + 3/x3
5/x2 − 2
= limx→∞
x · limx→∞
1− 2/x2 + 3/x3
5/x2 − 2
Sincethefirstfactortendsto ∞ andthesecondfactortendsto−12, theproducttendsto −∞.
Notes
I Makesurethe“lim”isthereineachstageI Donotdoarithmeticwith ∞ onpaper
![Page 7: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/7.jpg)
. . . . . .
Explanations
I Explanationsaregettingmuchbetter.I Please(continueto)formatyourpaperspresentably.
![Page 8: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/8.jpg)
. . . . . .
Outline
Grader’sCorner
DerivativeofaProductDerivationExamples
TheQuotientRuleDerivationExamples
MorederivativesoftrigonometricfunctionsDerivativeofTangentandCotangentDerivativeofSecantandCosecant
MoreonthePowerRulePowerRuleforPositiveIntegersbyInductionPowerRuleforNegativeIntegers
![Page 9: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/9.jpg)
. . . . . .
Recollectionandextension
Wehaveshownthatif u and v arefunctions, that
(u+ v)′ = u′ + v′
(u− v)′ = u′ − v′
Whatabout uv?
![Page 10: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/10.jpg)
. . . . . .
Isthederivativeofaproducttheproductofthederivatives?
..(uv)′ = u′v′?
.(uv)′ = u′v′!
Trythiswith u = x and v = x2.I Then uv = x3 =⇒ (uv)′ = 3x2.I But u′v′ = 1 · 2x = 2x.
Sowehavetobemorecareful.
![Page 11: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/11.jpg)
. . . . . .
Isthederivativeofaproducttheproductofthederivatives?
.
.(uv)′ = u′v′?
.(uv)′ = u′v′!
Trythiswith u = x and v = x2.
I Then uv = x3 =⇒ (uv)′ = 3x2.I But u′v′ = 1 · 2x = 2x.
Sowehavetobemorecareful.
![Page 12: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/12.jpg)
. . . . . .
Isthederivativeofaproducttheproductofthederivatives?
.
.(uv)′ = u′v′?
.(uv)′ = u′v′!
Trythiswith u = x and v = x2.I Then uv = x3 =⇒ (uv)′ = 3x2.
I But u′v′ = 1 · 2x = 2x.
Sowehavetobemorecareful.
![Page 13: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/13.jpg)
. . . . . .
Isthederivativeofaproducttheproductofthederivatives?
.
.(uv)′ = u′v′?
.(uv)′ = u′v′!
Trythiswith u = x and v = x2.I Then uv = x3 =⇒ (uv)′ = 3x2.I But u′v′ = 1 · 2x = 2x.
Sowehavetobemorecareful.
![Page 14: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/14.jpg)
. . . . . .
Isthederivativeofaproducttheproductofthederivatives?
.
.(uv)′ = u′v′?
.(uv)′ = u′v′!
Trythiswith u = x and v = x2.I Then uv = x3 =⇒ (uv)′ = 3x2.I But u′v′ = 1 · 2x = 2x.
Sowehavetobemorecareful.
![Page 15: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/15.jpg)
. . . . . .
Mmm...burgers
Sayyouworkinafast-foodjoint. Youwanttomakemoremoney.Whatareyourchoices?
I Worklongerhours.I Getaraise.
Sayyougeta25centraiseinyourhourlywagesandwork5hoursmoreperweek. Howmuchextramoneydoyoumake?
..
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
![Page 16: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/16.jpg)
. . . . . .
Mmm...burgers
Sayyouworkinafast-foodjoint. Youwanttomakemoremoney.Whatareyourchoices?
I Worklongerhours.
I Getaraise.
Sayyougeta25centraiseinyourhourlywagesandwork5hoursmoreperweek. Howmuchextramoneydoyoumake?
..
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
![Page 17: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/17.jpg)
. . . . . .
Mmm...burgers
Sayyouworkinafast-foodjoint. Youwanttomakemoremoney.Whatareyourchoices?
I Worklongerhours.I Getaraise.
Sayyougeta25centraiseinyourhourlywagesandwork5hoursmoreperweek. Howmuchextramoneydoyoumake?
..
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
![Page 18: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/18.jpg)
. . . . . .
Mmm...burgers
Sayyouworkinafast-foodjoint. Youwanttomakemoremoney.Whatareyourchoices?
I Worklongerhours.I Getaraise.
Sayyougeta25centraiseinyourhourlywagesandwork5hoursmoreperweek. Howmuchextramoneydoyoumake?
..
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
![Page 19: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/19.jpg)
. . . . . .
Mmm...burgers
Sayyouworkinafast-foodjoint. Youwanttomakemoremoney.Whatareyourchoices?
I Worklongerhours.I Getaraise.
Sayyougeta25centraiseinyourhourlywagesandwork5hoursmoreperweek. Howmuchextramoneydoyoumake?
...∆I = 5× $0.25 = $1.25?
.∆I = 5× $0.25 = $1.25?
![Page 20: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/20.jpg)
. . . . . .
Mmm...burgers
Sayyouworkinafast-foodjoint. Youwanttomakemoremoney.Whatareyourchoices?
I Worklongerhours.I Getaraise.
Sayyougeta25centraiseinyourhourlywagesandwork5hoursmoreperweek. Howmuchextramoneydoyoumake?
..
.∆I = 5× $0.25 = $1.25?
.∆I = 5× $0.25 = $1.25?
![Page 21: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/21.jpg)
. . . . . .
Moneymoneymoneymoney
Theanswerdependsonhowmuchyouwork already andyourcurrent wage. Supposeyouwork h hoursandarepaid w. Yougetatimeincreaseof ∆h andawageincreaseof ∆w. Incomeiswagestimeshours, so
∆I = (w+∆w)(h+∆h)−whFOIL= w · h+w ·∆h+∆w · h+∆w ·∆h−wh
= w ·∆h+∆w · h+∆w ·∆h
![Page 22: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/22.jpg)
. . . . . .
A geometricargument
Drawabox:
..w .∆w
.h
.∆h
.wh
.w∆h
.∆wh
.∆w∆h
∆I = w∆h+ h∆w+∆w∆h
![Page 23: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/23.jpg)
. . . . . .
A geometricargument
Drawabox:
..w .∆w
.h
.∆h
.wh
.w∆h
.∆wh
.∆w∆h
∆I = w∆h+ h∆w+∆w∆h
![Page 24: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/24.jpg)
. . . . . .
Suposewagesandhoursarechangingcontinuouslyovertime.Overatimeinterval ∆t, whatistheaveragerateofchangeofincome?
∆I∆t
=w∆h+ h∆w+∆w∆h
∆t
= w∆h∆t
+ h∆w∆t
+∆w∆h∆t
Whatistheinstantaneousrateofchangeofincome?
dIdt
= lim∆t→0
∆I∆t
= wdhdt
+ hdwdt
+ 0
![Page 25: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/25.jpg)
. . . . . .
Suposewagesandhoursarechangingcontinuouslyovertime.Overatimeinterval ∆t, whatistheaveragerateofchangeofincome?
∆I∆t
=w∆h+ h∆w+∆w∆h
∆t
= w∆h∆t
+ h∆w∆t
+∆w∆h∆t
Whatistheinstantaneousrateofchangeofincome?
dIdt
= lim∆t→0
∆I∆t
= wdhdt
+ hdwdt
+ 0
![Page 26: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/26.jpg)
. . . . . .
Eurekamen!
Wehavediscovered
Theorem(TheProductRule)Let u and v bedifferentiableat x. Then
(uv)′(x) = u(x)v′(x) + u′(x)v(x)
inLeibniznotation
ddx
(uv) =dudx
· v+ udvdx
![Page 27: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/27.jpg)
. . . . . .
ExampleApplytheproductruleto u = x and v = x2.
Solution
(uv)′(x) = u(x)v′(x) + u′(x)v(x) = x · (2x) + 1 · x2 = 3x2
Thisiswhatwegetthe“normal”way.
![Page 28: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/28.jpg)
. . . . . .
ExampleApplytheproductruleto u = x and v = x2.
Solution
(uv)′(x) = u(x)v′(x) + u′(x)v(x) = x · (2x) + 1 · x2 = 3x2
Thisiswhatwegetthe“normal”way.
![Page 29: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/29.jpg)
. . . . . .
ExampleFindthisderivativetwoways: firstbydirectmultiplicationandthenbytheproductrule:
ddx
[(3− x2)(x3 − x+ 1)
]
Solutionbydirectmultiplication:
ddx
[(3− x2)(x3 − x+ 1)
]FOIL=
ddx
[−x5 + 4x3 − x2 − 3x+ 3
]
= −5x4 + 12x2 − 2x− 3
![Page 30: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/30.jpg)
. . . . . .
ExampleFindthisderivativetwoways: firstbydirectmultiplicationandthenbytheproductrule:
ddx
[(3− x2)(x3 − x+ 1)
]Solutionbydirectmultiplication:
ddx
[(3− x2)(x3 − x+ 1)
]FOIL=
ddx
[−x5 + 4x3 − x2 − 3x+ 3
]
= −5x4 + 12x2 − 2x− 3
![Page 31: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/31.jpg)
. . . . . .
ExampleFindthisderivativetwoways: firstbydirectmultiplicationandthenbytheproductrule:
ddx
[(3− x2)(x3 − x+ 1)
]Solutionbydirectmultiplication:
ddx
[(3− x2)(x3 − x+ 1)
]FOIL=
ddx
[−x5 + 4x3 − x2 − 3x+ 3
]= −5x4 + 12x2 − 2x− 3
![Page 32: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/32.jpg)
. . . . . .
ExampleFindthisderivativetwoways: firstbydirectmultiplicationandthenbytheproductrule:
ddx
[(3− x2)(x3 − x+ 1)
]Solutionbytheproductrule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x− 3
![Page 33: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/33.jpg)
. . . . . .
ExampleFindthisderivativetwoways: firstbydirectmultiplicationandthenbytheproductrule:
ddx
[(3− x2)(x3 − x+ 1)
]Solutionbytheproductrule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x− 3
![Page 34: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/34.jpg)
. . . . . .
ExampleFindthisderivativetwoways: firstbydirectmultiplicationandthenbytheproductrule:
ddx
[(3− x2)(x3 − x+ 1)
]Solutionbytheproductrule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x− 3
![Page 35: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/35.jpg)
. . . . . .
ExampleFindthisderivativetwoways: firstbydirectmultiplicationandthenbytheproductrule:
ddx
[(3− x2)(x3 − x+ 1)
]Solutionbytheproductrule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x− 3
![Page 36: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/36.jpg)
. . . . . .
ExampleFindthisderivativetwoways: firstbydirectmultiplicationandthenbytheproductrule:
ddx
[(3− x2)(x3 − x+ 1)
]Solutionbytheproductrule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x− 3
![Page 37: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/37.jpg)
. . . . . .
ExampleFindthisderivativetwoways: firstbydirectmultiplicationandthenbytheproductrule:
ddx
[(3− x2)(x3 − x+ 1)
]Solutionbytheproductrule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x− 3
![Page 38: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/38.jpg)
. . . . . .
ExampleFindthisderivativetwoways: firstbydirectmultiplicationandthenbytheproductrule:
ddx
[(3− x2)(x3 − x+ 1)
]Solutionbytheproductrule:
dydx
=
(ddx
(3− x2))(x3 − x+ 1) + (3− x2)
(ddx
(x3 − x+ 1))
= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)
= −5x4 + 12x2 − 2x− 3
![Page 39: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/39.jpg)
. . . . . .
Onemore
Example
Findddx
x sin x.
Solution
ddx
x sin x
=
(ddx
x)sin x+ x
(ddx
sin x)
= 1 · sin x+ x · cos x= sin x+ x cos x
![Page 40: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/40.jpg)
. . . . . .
Onemore
Example
Findddx
x sin x.
Solution
ddx
x sin x =(
ddx
x)sin x+ x
(ddx
sin x)
= 1 · sin x+ x · cos x= sin x+ x cos x
![Page 41: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/41.jpg)
. . . . . .
Onemore
Example
Findddx
x sin x.
Solution
ddx
x sin x =(
ddx
x)sin x+ x
(ddx
sin x)
= 1 · sin x+ x · cos x
= sin x+ x cos x
![Page 42: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/42.jpg)
. . . . . .
Onemore
Example
Findddx
x sin x.
Solution
ddx
x sin x =(
ddx
x)sin x+ x
(ddx
sin x)
= 1 · sin x+ x · cos x= sin x+ x cos x
![Page 43: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/43.jpg)
. . . . . .
Mnemonic
Let u = “hi” and v = “ho”. Then
(uv)′ = vu′ + uv′ = “hodeehiplushideeho”
![Page 44: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/44.jpg)
. . . . . .
Musicalinterlude
I jazzbandleaderandsinger
I hitsong“MinnietheMoocher”featuring“hideho”chorus
I playedCurtisin TheBluesBrothers
CabCalloway1907–1994
![Page 45: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/45.jpg)
. . . . . .
IteratingtheProductRule
ExampleUsetheproductruletofindthederivativeofathree-foldproductuvw.
Solution
(uvw)′
= ((uv)w)′
..
.Applytheproductrule
to uv and w
= (uv)′w+ (uv)w′..
.Applytheproductrule
to u and v
= (u′v+ uv′)w+ (uv)w′
= u′vw+ uv′w+ uvw′
Sowewritedowntheproductthreetimes, takingthederivativeofeachfactoronce.
![Page 46: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/46.jpg)
. . . . . .
IteratingtheProductRule
ExampleUsetheproductruletofindthederivativeofathree-foldproductuvw.
Solution
(uvw)′
= ((uv)w)′
..
.Applytheproductrule
to uv and w
= (uv)′w+ (uv)w′..
.Applytheproductrule
to u and v
= (u′v+ uv′)w+ (uv)w′
= u′vw+ uv′w+ uvw′
Sowewritedowntheproductthreetimes, takingthederivativeofeachfactoronce.
![Page 47: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/47.jpg)
. . . . . .
IteratingtheProductRule
ExampleUsetheproductruletofindthederivativeofathree-foldproductuvw.
Solution
(uvw)′ = ((uv)w)′..
.Applytheproductrule
to uv and w
= (uv)′w+ (uv)w′..
.Applytheproductrule
to u and v
= (u′v+ uv′)w+ (uv)w′
= u′vw+ uv′w+ uvw′
Sowewritedowntheproductthreetimes, takingthederivativeofeachfactoronce.
![Page 48: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/48.jpg)
. . . . . .
IteratingtheProductRule
ExampleUsetheproductruletofindthederivativeofathree-foldproductuvw.
Solution
(uvw)′ = ((uv)w)′..
.Applytheproductrule
to uv and w
= (uv)′w+ (uv)w′..
.Applytheproductrule
to u and v
= (u′v+ uv′)w+ (uv)w′
= u′vw+ uv′w+ uvw′
Sowewritedowntheproductthreetimes, takingthederivativeofeachfactoronce.
![Page 49: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/49.jpg)
. . . . . .
IteratingtheProductRule
ExampleUsetheproductruletofindthederivativeofathree-foldproductuvw.
Solution
(uvw)′ = ((uv)w)′..
.Applytheproductrule
to uv and w
= (uv)′w+ (uv)w′..
.Applytheproductrule
to u and v
= (u′v+ uv′)w+ (uv)w′
= u′vw+ uv′w+ uvw′
Sowewritedowntheproductthreetimes, takingthederivativeofeachfactoronce.
![Page 50: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/50.jpg)
. . . . . .
IteratingtheProductRule
ExampleUsetheproductruletofindthederivativeofathree-foldproductuvw.
Solution
(uvw)′ = ((uv)w)′..
.Applytheproductrule
to uv and w
= (uv)′w+ (uv)w′..
.Applytheproductrule
to u and v
= (u′v+ uv′)w+ (uv)w′
= u′vw+ uv′w+ uvw′
Sowewritedowntheproductthreetimes, takingthederivativeofeachfactoronce.
![Page 51: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/51.jpg)
. . . . . .
IteratingtheProductRule
ExampleUsetheproductruletofindthederivativeofathree-foldproductuvw.
Solution
(uvw)′ = ((uv)w)′..
.Applytheproductrule
to uv and w
= (uv)′w+ (uv)w′..
.Applytheproductrule
to u and v
= (u′v+ uv′)w+ (uv)w′
= u′vw+ uv′w+ uvw′
Sowewritedowntheproductthreetimes, takingthederivativeofeachfactoronce.
![Page 52: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/52.jpg)
. . . . . .
IteratingtheProductRule
ExampleUsetheproductruletofindthederivativeofathree-foldproductuvw.
Solution
(uvw)′ = ((uv)w)′..
.Applytheproductrule
to uv and w
= (uv)′w+ (uv)w′..
.Applytheproductrule
to u and v
= (u′v+ uv′)w+ (uv)w′
= u′vw+ uv′w+ uvw′
Sowewritedowntheproductthreetimes, takingthederivativeofeachfactoronce.
![Page 53: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/53.jpg)
. . . . . .
IteratingtheProductRule
ExampleUsetheproductruletofindthederivativeofathree-foldproductuvw.
Solution
(uvw)′ = ((uv)w)′..
.Applytheproductrule
to uv and w
= (uv)′w+ (uv)w′..
.Applytheproductrule
to u and v
= (u′v+ uv′)w+ (uv)w′
= u′vw+ uv′w+ uvw′
Sowewritedowntheproductthreetimes, takingthederivativeofeachfactoronce.
![Page 54: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/54.jpg)
. . . . . .
Outline
Grader’sCorner
DerivativeofaProductDerivationExamples
TheQuotientRuleDerivationExamples
MorederivativesoftrigonometricfunctionsDerivativeofTangentandCotangentDerivativeofSecantandCosecant
MoreonthePowerRulePowerRuleforPositiveIntegersbyInductionPowerRuleforNegativeIntegers
![Page 55: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/55.jpg)
. . . . . .
TheQuotientRule
Whataboutthederivativeofaquotient?
Let u and v bedifferentiablefunctionsandlet Q =uv. Then
u = Qv
If Q isdifferentiable, wehave
u′ = (Qv)′ = Q′v+Qv′
=⇒ Q′ =u′ −Qv′
v=
u′
v− u
v· v
′
v
=⇒ Q′ =(uv
)′=
u′v− uv′
v2
Thisiscalledthe QuotientRule.
![Page 56: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/56.jpg)
. . . . . .
TheQuotientRule
Whataboutthederivativeofaquotient?
Let u and v bedifferentiablefunctionsandlet Q =uv. Then
u = Qv
If Q isdifferentiable, wehave
u′ = (Qv)′ = Q′v+Qv′
=⇒ Q′ =u′ −Qv′
v=
u′
v− u
v· v
′
v
=⇒ Q′ =(uv
)′=
u′v− uv′
v2
Thisiscalledthe QuotientRule.
![Page 57: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/57.jpg)
. . . . . .
TheQuotientRule
Whataboutthederivativeofaquotient?
Let u and v bedifferentiablefunctionsandlet Q =uv. Then
u = Qv
If Q isdifferentiable, wehave
u′ = (Qv)′ = Q′v+Qv′
=⇒ Q′ =u′ −Qv′
v=
u′
v− u
v· v
′
v
=⇒ Q′ =(uv
)′=
u′v− uv′
v2
Thisiscalledthe QuotientRule.
![Page 58: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/58.jpg)
. . . . . .
TheQuotientRule
Whataboutthederivativeofaquotient?
Let u and v bedifferentiablefunctionsandlet Q =uv. Then
u = Qv
If Q isdifferentiable, wehave
u′ = (Qv)′ = Q′v+Qv′
=⇒ Q′ =u′ −Qv′
v=
u′
v− u
v· v
′
v
=⇒ Q′ =(uv
)′=
u′v− uv′
v2
Thisiscalledthe QuotientRule.
![Page 59: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/59.jpg)
. . . . . .
TheQuotientRule
Whataboutthederivativeofaquotient?
Let u and v bedifferentiablefunctionsandlet Q =uv. Then
u = Qv
If Q isdifferentiable, wehave
u′ = (Qv)′ = Q′v+Qv′
=⇒ Q′ =u′ −Qv′
v=
u′
v− u
v· v
′
v
=⇒ Q′ =(uv
)′=
u′v− uv′
v2
Thisiscalledthe QuotientRule.
![Page 60: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/60.jpg)
. . . . . .
TheQuotientRule
Whataboutthederivativeofaquotient?
Let u and v bedifferentiablefunctionsandlet Q =uv. Then
u = Qv
If Q isdifferentiable, wehave
u′ = (Qv)′ = Q′v+Qv′
=⇒ Q′ =u′ −Qv′
v=
u′
v− u
v· v
′
v
=⇒ Q′ =(uv
)′=
u′v− uv′
v2
Thisiscalledthe QuotientRule.
![Page 61: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/61.jpg)
. . . . . .
VerifyingExample
Example
Verifythequotientrulebycomputingddx
(x2
x
)andcomparingit
toddx
(x).
Solution
ddx
(x2
x
)=
x ddx
(x2)− x2 d
dx (x)x2
=x · 2x− x2 · 1
x2
=x2
x2= 1 =
ddx
(x)
![Page 62: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/62.jpg)
. . . . . .
VerifyingExample
Example
Verifythequotientrulebycomputingddx
(x2
x
)andcomparingit
toddx
(x).
Solution
ddx
(x2
x
)=
x ddx
(x2)− x2 d
dx (x)x2
=x · 2x− x2 · 1
x2
=x2
x2= 1 =
ddx
(x)
![Page 63: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/63.jpg)
. . . . . .
Examples
Example
1.ddx
2x+ 53x− 2
2.ddx
2x+ 1x2 − 1
3.ddt
t− 1t2 + t+ 2
Answers
1. − 19(3x− 2)2
2. −2(x2 + x+ 1
)(x2 − 1)2
3.−t2 + 2t+ 3
(t2 + t+ 2)2
![Page 64: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/64.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 65: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/65.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 66: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/66.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 67: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/67.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 68: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/68.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 69: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/69.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 70: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/70.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 71: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/71.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 72: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/72.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 73: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/73.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 74: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/74.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 75: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/75.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 76: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/76.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2
= − 19(3x− 2)2
![Page 77: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/77.jpg)
. . . . . .
Solutiontofirstexample
ddx
2x+ 53x− 2
=(3x− 2) d
dx(2x+ 5)− (2x+ 5) ddx(3x− 2)
(3x− 2)2
=(3x− 2)(2)− (2x+ 5)(3)
(3x− 2)2
=(6x− 4)− (6x+ 15)
(3x− 2)2= − 19
(3x− 2)2
![Page 78: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/78.jpg)
. . . . . .
Examples
Example
1.ddx
2x+ 53x− 2
2.ddx
2x+ 1x2 − 1
3.ddt
t− 1t2 + t+ 2
Answers
1. − 19(3x− 2)2
2. −2(x2 + x+ 1
)(x2 − 1)2
3.−t2 + 2t+ 3
(t2 + t+ 2)2
![Page 79: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/79.jpg)
. . . . . .
Solutiontosecondexample
ddx
2x+ 1x2 − 1
=(x2 − 1)(2)− (2x+ 1)(2x)
(x2 − 1)2
=(2x2 − 2)− (4x2 + 2x)
(x2 − 1)2
= −2(x2 + x+ 1
)(x2 − 1)2
![Page 80: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/80.jpg)
. . . . . .
Solutiontosecondexample
ddx
2x+ 1x2 − 1
=(x2 − 1)(2)− (2x+ 1)(2x)
(x2 − 1)2
=(2x2 − 2)− (4x2 + 2x)
(x2 − 1)2
= −2(x2 + x+ 1
)(x2 − 1)2
![Page 81: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/81.jpg)
. . . . . .
Solutiontosecondexample
ddx
2x+ 1x2 − 1
=(x2 − 1)(2)− (2x+ 1)(2x)
(x2 − 1)2
=(2x2 − 2)− (4x2 + 2x)
(x2 − 1)2
= −2(x2 + x+ 1
)(x2 − 1)2
![Page 82: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/82.jpg)
. . . . . .
Examples
Example
1.ddx
2x+ 53x− 2
2.ddx
2x+ 1x2 − 1
3.ddt
t− 1t2 + t+ 2
Answers
1. − 19(3x− 2)2
2. −2(x2 + x+ 1
)(x2 − 1)2
3.−t2 + 2t+ 3
(t2 + t+ 2)2
![Page 83: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/83.jpg)
. . . . . .
Solutiontothirdexample
ddt
t− 1t2 + t+ 2
=(t2 + t+ 2)(1)− (t− 1)(2t+ 1)
(t2 + t+ 2)2
=(t2 + t+ 2)− (2t2 − t− 1)
(t2 + t+ 2)2
=−t2 + 2t+ 3(t2 + t+ 2)2
![Page 84: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/84.jpg)
. . . . . .
Solutiontothirdexample
ddt
t− 1t2 + t+ 2
=(t2 + t+ 2)(1)− (t− 1)(2t+ 1)
(t2 + t+ 2)2
=(t2 + t+ 2)− (2t2 − t− 1)
(t2 + t+ 2)2
=−t2 + 2t+ 3(t2 + t+ 2)2
![Page 85: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/85.jpg)
. . . . . .
Solutiontothirdexample
ddt
t− 1t2 + t+ 2
=(t2 + t+ 2)(1)− (t− 1)(2t+ 1)
(t2 + t+ 2)2
=(t2 + t+ 2)− (2t2 − t− 1)
(t2 + t+ 2)2
=−t2 + 2t+ 3(t2 + t+ 2)2
![Page 86: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/86.jpg)
. . . . . .
Examples
Example
1.ddx
2x+ 53x− 2
2.ddx
2x+ 1x2 − 1
3.ddt
t− 1t2 + t+ 2
Answers
1. − 19(3x− 2)2
2. −2(x2 + x+ 1
)(x2 − 1)2
3.−t2 + 2t+ 3
(t2 + t+ 2)2
![Page 87: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/87.jpg)
. . . . . .
Mnemonic
Let u = “hi” and v = “lo”. Then(uv
)′=
vu′ − uv′
v2= “lodeehiminushideelooverlolo”
![Page 88: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/88.jpg)
. . . . . .
Outline
Grader’sCorner
DerivativeofaProductDerivationExamples
TheQuotientRuleDerivationExamples
MorederivativesoftrigonometricfunctionsDerivativeofTangentandCotangentDerivativeofSecantandCosecant
MoreonthePowerRulePowerRuleforPositiveIntegersbyInductionPowerRuleforNegativeIntegers
![Page 89: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/89.jpg)
. . . . . .
DerivativeofTangent
Example
Findddx
tan x
Solution
ddx
tan x =ddx
(sin xcos x
)
=cos x · cos x− sin x · (− sin x)
cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
![Page 90: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/90.jpg)
. . . . . .
DerivativeofTangent
Example
Findddx
tan x
Solution
ddx
tan x =ddx
(sin xcos x
)
=cos x · cos x− sin x · (− sin x)
cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
![Page 91: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/91.jpg)
. . . . . .
DerivativeofTangent
Example
Findddx
tan x
Solution
ddx
tan x =ddx
(sin xcos x
)=
cos x · cos x− sin x · (− sin x)cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
![Page 92: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/92.jpg)
. . . . . .
DerivativeofTangent
Example
Findddx
tan x
Solution
ddx
tan x =ddx
(sin xcos x
)=
cos x · cos x− sin x · (− sin x)cos2 x
=cos2 x+ sin2 x
cos2 x
=1
cos2 x= sec2 x
![Page 93: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/93.jpg)
. . . . . .
DerivativeofTangent
Example
Findddx
tan x
Solution
ddx
tan x =ddx
(sin xcos x
)=
cos x · cos x− sin x · (− sin x)cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
![Page 94: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/94.jpg)
. . . . . .
DerivativeofTangent
Example
Findddx
tan x
Solution
ddx
tan x =ddx
(sin xcos x
)=
cos x · cos x− sin x · (− sin x)cos2 x
=cos2 x+ sin2 x
cos2 x=
1cos2 x
= sec2 x
![Page 95: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/95.jpg)
. . . . . .
DerivativeofCotangent
Example
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
Solution
ddx
cot x =ddx
(cos xsin x
)
=sin x · (− sin x)− cos x · cos x
sin2 x
=− sin2 x− cos2 x
sin2 x= − 1
sin2 x= − csc2 x
![Page 96: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/96.jpg)
. . . . . .
DerivativeofCotangent
Example
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
Solution
ddx
cot x =ddx
(cos xsin x
)
=sin x · (− sin x)− cos x · cos x
sin2 x
=− sin2 x− cos2 x
sin2 x= − 1
sin2 x= − csc2 x
![Page 97: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/97.jpg)
. . . . . .
DerivativeofCotangent
Example
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
Solution
ddx
cot x =ddx
(cos xsin x
)=
sin x · (− sin x)− cos x · cos xsin2 x
=− sin2 x− cos2 x
sin2 x= − 1
sin2 x= − csc2 x
![Page 98: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/98.jpg)
. . . . . .
DerivativeofCotangent
Example
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
Solution
ddx
cot x =ddx
(cos xsin x
)=
sin x · (− sin x)− cos x · cos xsin2 x
=− sin2 x− cos2 x
sin2 x
= − 1sin2 x
= − csc2 x
![Page 99: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/99.jpg)
. . . . . .
DerivativeofCotangent
Example
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
Solution
ddx
cot x =ddx
(cos xsin x
)=
sin x · (− sin x)− cos x · cos xsin2 x
=− sin2 x− cos2 x
sin2 x= − 1
sin2 x
= − csc2 x
![Page 100: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/100.jpg)
. . . . . .
DerivativeofCotangent
Example
Findddx
cot x
Answer
ddx
cot x = − 1sin2 x
= − csc2 x
Solution
ddx
cot x =ddx
(cos xsin x
)=
sin x · (− sin x)− cos x · cos xsin2 x
=− sin2 x− cos2 x
sin2 x= − 1
sin2 x= − csc2 x
![Page 101: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/101.jpg)
. . . . . .
DerivativeofSecant
Example
Findddx
sec x
Solution
ddx
sec x =ddx
(1
cos x
)
=cos x · 0− 1 · (− sin x)
cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
![Page 102: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/102.jpg)
. . . . . .
DerivativeofSecant
Example
Findddx
sec x
Solution
ddx
sec x =ddx
(1
cos x
)
=cos x · 0− 1 · (− sin x)
cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
![Page 103: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/103.jpg)
. . . . . .
DerivativeofSecant
Example
Findddx
sec x
Solution
ddx
sec x =ddx
(1
cos x
)=
cos x · 0− 1 · (− sin x)cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
![Page 104: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/104.jpg)
. . . . . .
DerivativeofSecant
Example
Findddx
sec x
Solution
ddx
sec x =ddx
(1
cos x
)=
cos x · 0− 1 · (− sin x)cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
![Page 105: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/105.jpg)
. . . . . .
DerivativeofSecant
Example
Findddx
sec x
Solution
ddx
sec x =ddx
(1
cos x
)=
cos x · 0− 1 · (− sin x)cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
![Page 106: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/106.jpg)
. . . . . .
DerivativeofSecant
Example
Findddx
sec x
Solution
ddx
sec x =ddx
(1
cos x
)=
cos x · 0− 1 · (− sin x)cos2 x
=sin xcos2 x
=1
cos x· sin xcos x
= sec x tan x
![Page 107: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/107.jpg)
. . . . . .
DerivativeofCosecant
Example
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
Solution
ddx
csc x =ddx
(1
sin x
)
=sin x · 0− 1 · (cos x)
sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
![Page 108: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/108.jpg)
. . . . . .
DerivativeofCosecant
Example
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
Solution
ddx
csc x =ddx
(1
sin x
)
=sin x · 0− 1 · (cos x)
sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
![Page 109: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/109.jpg)
. . . . . .
DerivativeofCosecant
Example
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
Solution
ddx
csc x =ddx
(1
sin x
)=
sin x · 0− 1 · (cos x)sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
![Page 110: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/110.jpg)
. . . . . .
DerivativeofCosecant
Example
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
Solution
ddx
csc x =ddx
(1
sin x
)=
sin x · 0− 1 · (cos x)sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
![Page 111: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/111.jpg)
. . . . . .
DerivativeofCosecant
Example
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
Solution
ddx
csc x =ddx
(1
sin x
)=
sin x · 0− 1 · (cos x)sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
![Page 112: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/112.jpg)
. . . . . .
DerivativeofCosecant
Example
Findddx
csc x
Answer
ddx
csc x = − csc x cot x
Solution
ddx
csc x =ddx
(1
sin x
)=
sin x · 0− 1 · (cos x)sin2 x
= − cos xsin2 x
= − 1sin x
· cos xsin x
= − csc x cot x
![Page 113: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/113.jpg)
. . . . . .
Recap: Derivativesoftrigonometricfunctions
y y′
sin x cos x
cos x − sin x
tan x sec2 x
cot x − csc2 x
sec x sec x tan x
csc x − csc x cot x
I Functionscomeinpairs(sin/cos, tan/cot, sec/csc)
I Derivativesofpairsfollowsimilarpatterns,withfunctionsandco-functionsswitchedandanextrasign.
![Page 114: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/114.jpg)
. . . . . .
Outline
Grader’sCorner
DerivativeofaProductDerivationExamples
TheQuotientRuleDerivationExamples
MorederivativesoftrigonometricfunctionsDerivativeofTangentandCotangentDerivativeofSecantandCosecant
MoreonthePowerRulePowerRuleforPositiveIntegersbyInductionPowerRuleforNegativeIntegers
![Page 115: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/115.jpg)
. . . . . .
PowerRuleforPositiveIntegersbyInductionTheoremLet n beapositiveinteger. Then
ddx
xn = nxn−1
Proof.Byinductionon n. Wecanshowittobetruefor n = 1 directly.
Supposeforsome n thatddx
xn = nxn−1. Then
ddx
xn+1
=ddx
(x · xn)
=
(ddx
x)xn + x
(ddx
xn)
= 1 · xn + x · nxn−1 = (n+ 1)xn
![Page 116: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/116.jpg)
. . . . . .
PowerRuleforPositiveIntegersbyInductionTheoremLet n beapositiveinteger. Then
ddx
xn = nxn−1
Proof.Byinductionon n.
Wecanshowittobetruefor n = 1 directly.
Supposeforsome n thatddx
xn = nxn−1. Then
ddx
xn+1
=ddx
(x · xn)
=
(ddx
x)xn + x
(ddx
xn)
= 1 · xn + x · nxn−1 = (n+ 1)xn
![Page 117: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/117.jpg)
. . . . . .
PrincipleofMathematicalInduction
.
.Suppose S(1) istrue and S(n + 1)is true wheneverS(n) is true. ThenS(n) is true for alln.
.
.Imagecredit: KoolSkatkat
![Page 118: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/118.jpg)
. . . . . .
PowerRuleforPositiveIntegersbyInductionTheoremLet n beapositiveinteger. Then
ddx
xn = nxn−1
Proof.Byinductionon n. Wecanshowittobetruefor n = 1 directly.
Supposeforsome n thatddx
xn = nxn−1. Then
ddx
xn+1
=ddx
(x · xn)
=
(ddx
x)xn + x
(ddx
xn)
= 1 · xn + x · nxn−1 = (n+ 1)xn
![Page 119: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/119.jpg)
. . . . . .
PowerRuleforPositiveIntegersbyInductionTheoremLet n beapositiveinteger. Then
ddx
xn = nxn−1
Proof.Byinductionon n. Wecanshowittobetruefor n = 1 directly.
Supposeforsome n thatddx
xn = nxn−1. Then
ddx
xn+1 =ddx
(x · xn)
=
(ddx
x)xn + x
(ddx
xn)
= 1 · xn + x · nxn−1 = (n+ 1)xn
![Page 120: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/120.jpg)
. . . . . .
PowerRuleforPositiveIntegersbyInductionTheoremLet n beapositiveinteger. Then
ddx
xn = nxn−1
Proof.Byinductionon n. Wecanshowittobetruefor n = 1 directly.
Supposeforsome n thatddx
xn = nxn−1. Then
ddx
xn+1 =ddx
(x · xn)
=
(ddx
x)xn + x
(ddx
xn)
= 1 · xn + x · nxn−1 = (n+ 1)xn
![Page 121: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/121.jpg)
. . . . . .
PowerRuleforPositiveIntegersbyInductionTheoremLet n beapositiveinteger. Then
ddx
xn = nxn−1
Proof.Byinductionon n. Wecanshowittobetruefor n = 1 directly.
Supposeforsome n thatddx
xn = nxn−1. Then
ddx
xn+1 =ddx
(x · xn)
=
(ddx
x)xn + x
(ddx
xn)
= 1 · xn + x · nxn−1 = (n+ 1)xn
![Page 122: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/122.jpg)
. . . . . .
PowerRuleforNegativeIntegersUsethequotientruletoprove
Theorem
ddx
x−n = (−n)x−n−1
forpositiveintegers n.
Proof.
ddx
x−n =ddx
1xn
=xn · d
dx1− 1 · ddxx
n
x2n
=0− nxn−1
x2n
= −nx−n−1
![Page 123: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/123.jpg)
. . . . . .
PowerRuleforNegativeIntegersUsethequotientruletoprove
Theorem
ddx
x−n = (−n)x−n−1
forpositiveintegers n.
Proof.
ddx
x−n =ddx
1xn
=xn · d
dx1− 1 · ddxx
n
x2n
=0− nxn−1
x2n
= −nx−n−1
![Page 124: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/124.jpg)
. . . . . .
PowerRuleforNegativeIntegersUsethequotientruletoprove
Theorem
ddx
x−n = (−n)x−n−1
forpositiveintegers n.
Proof.
ddx
x−n =ddx
1xn
=xn · d
dx1− 1 · ddxx
n
x2n
=0− nxn−1
x2n
= −nx−n−1
![Page 125: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/125.jpg)
. . . . . .
PowerRuleforNegativeIntegersUsethequotientruletoprove
Theorem
ddx
x−n = (−n)x−n−1
forpositiveintegers n.
Proof.
ddx
x−n =ddx
1xn
=xn · d
dx1− 1 · ddxx
n
x2n
=0− nxn−1
x2n
= −nx−n−1
![Page 126: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/126.jpg)
. . . . . .
PowerRuleforNegativeIntegersUsethequotientruletoprove
Theorem
ddx
x−n = (−n)x−n−1
forpositiveintegers n.
Proof.
ddx
x−n =ddx
1xn
=xn · d
dx1− 1 · ddxx
n
x2n
=0− nxn−1
x2n= −nx−n−1
![Page 127: Lesson 9: The Product and Quotient Rules](https://reader034.fdocuments.us/reader034/viewer/2022051323/54926015b47959744d8b45c4/html5/thumbnails/127.jpg)
. . . . . .
Whathavewelearnedtoday?
I TheProductRule: (uv)′ = u′v+ uv′
I TheQuotientRule:(uv
)′=
vu′ − uv′
v2I Derivativesoftangent/cotangent, secant/cosecant
ddx
tan x = sec2 xddx
sec x = sec x tan x
ddx
cot x = − csc2 xddx
csc x = − csc x cot x
I ThePowerRuleistrueforallwholenumberpowers,includingnegativepowers:
ddx
xn = nxn−1