lesson 9: the product and quotient rules
DESCRIPTION
The product rule or Leibniz rule is the rule by which we can differentiate products of functions!TRANSCRIPT
. . . . . .
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
. . . . . .
Outline
Grader’sCorner
DerivativeofaProductDerivationExamples
TheQuotientRuleDerivationExamples
MorederivativesoftrigonometricfunctionsDerivativeofTangentandCotangentDerivativeofSecantandCosecant
MoreonthePowerRulePowerRuleforPositiveIntegersbyInductionPowerRuleforNegativeIntegers
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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
. . . . . .
Explanations
I Explanationsaregettingmuchbetter.I Please(continueto)formatyourpaperspresentably.
. . . . . .
Outline
Grader’sCorner
DerivativeofaProductDerivationExamples
TheQuotientRuleDerivationExamples
MorederivativesoftrigonometricfunctionsDerivativeofTangentandCotangentDerivativeofSecantandCosecant
MoreonthePowerRulePowerRuleforPositiveIntegersbyInductionPowerRuleforNegativeIntegers
. . . . . .
Recollectionandextension
Wehaveshownthatif u and v arefunctions, that
(u+ v)′ = u′ + v′
(u− v)′ = u′ − v′
Whatabout uv?
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
Mmm...burgers
Sayyouworkinafast-foodjoint. Youwanttomakemoremoney.Whatareyourchoices?
I Worklongerhours.I Getaraise.
Sayyougeta25centraiseinyourhourlywagesandwork5hoursmoreperweek. Howmuchextramoneydoyoumake?
..
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
. . . . . .
Mmm...burgers
Sayyouworkinafast-foodjoint. Youwanttomakemoremoney.Whatareyourchoices?
I Worklongerhours.
I Getaraise.
Sayyougeta25centraiseinyourhourlywagesandwork5hoursmoreperweek. Howmuchextramoneydoyoumake?
..
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
. . . . . .
Mmm...burgers
Sayyouworkinafast-foodjoint. Youwanttomakemoremoney.Whatareyourchoices?
I Worklongerhours.I Getaraise.
Sayyougeta25centraiseinyourhourlywagesandwork5hoursmoreperweek. Howmuchextramoneydoyoumake?
..
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
. . . . . .
Mmm...burgers
Sayyouworkinafast-foodjoint. Youwanttomakemoremoney.Whatareyourchoices?
I Worklongerhours.I Getaraise.
Sayyougeta25centraiseinyourhourlywagesandwork5hoursmoreperweek. Howmuchextramoneydoyoumake?
..
.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?
. . . . . .
Mmm...burgers
Sayyouworkinafast-foodjoint. Youwanttomakemoremoney.Whatareyourchoices?
I Worklongerhours.I Getaraise.
Sayyougeta25centraiseinyourhourlywagesandwork5hoursmoreperweek. Howmuchextramoneydoyoumake?
...∆I = 5× $0.25 = $1.25?
.∆I = 5× $0.25 = $1.25?
. . . . . .
Mmm...burgers
Sayyouworkinafast-foodjoint. Youwanttomakemoremoney.Whatareyourchoices?
I Worklongerhours.I Getaraise.
Sayyougeta25centraiseinyourhourlywagesandwork5hoursmoreperweek. Howmuchextramoneydoyoumake?
..
.∆I = 5× $0.25 = $1.25?
.∆I = 5× $0.25 = $1.25?
. . . . . .
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
. . . . . .
A geometricargument
Drawabox:
..w .∆w
.h
.∆h
.wh
.w∆h
.∆wh
.∆w∆h
∆I = w∆h+ h∆w+∆w∆h
. . . . . .
A geometricargument
Drawabox:
..w .∆w
.h
.∆h
.wh
.w∆h
.∆wh
.∆w∆h
∆I = w∆h+ h∆w+∆w∆h
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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.
. . . . . .
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.
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
Mnemonic
Let u = “hi” and v = “ho”. Then
(uv)′ = vu′ + uv′ = “hodeehiplushideeho”
. . . . . .
Musicalinterlude
I jazzbandleaderandsinger
I hitsong“MinnietheMoocher”featuring“hideho”chorus
I playedCurtisin TheBluesBrothers
CabCalloway1907–1994
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
Outline
Grader’sCorner
DerivativeofaProductDerivationExamples
TheQuotientRuleDerivationExamples
MorederivativesoftrigonometricfunctionsDerivativeofTangentandCotangentDerivativeofSecantandCosecant
MoreonthePowerRulePowerRuleforPositiveIntegersbyInductionPowerRuleforNegativeIntegers
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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)
. . . . . .
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)
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
Mnemonic
Let u = “hi” and v = “lo”. Then(uv
)′=
vu′ − uv′
v2= “lodeehiminushideelooverlolo”
. . . . . .
Outline
Grader’sCorner
DerivativeofaProductDerivationExamples
TheQuotientRuleDerivationExamples
MorederivativesoftrigonometricfunctionsDerivativeofTangentandCotangentDerivativeofSecantandCosecant
MoreonthePowerRulePowerRuleforPositiveIntegersbyInductionPowerRuleforNegativeIntegers
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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.
. . . . . .
Outline
Grader’sCorner
DerivativeofaProductDerivationExamples
TheQuotientRuleDerivationExamples
MorederivativesoftrigonometricfunctionsDerivativeofTangentandCotangentDerivativeofSecantandCosecant
MoreonthePowerRulePowerRuleforPositiveIntegersbyInductionPowerRuleforNegativeIntegers
. . . . . .
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
. . . . . .
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
. . . . . .
PrincipleofMathematicalInduction
.
.Suppose S(1) istrue and S(n + 1)is true wheneverS(n) is true. ThenS(n) is true for alln.
.
.Imagecredit: KoolSkatkat
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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