Lesson 9: The Product and Quotient Rules

. . . . . .

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


is

continuous.




g(x) ̸= 0.


. . . . . .

Problem1.5.20


is

continuous.




g(x) ̸= 0.


. . . . . .

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





. . . . . .

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.

. . . . . .


.

.(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.


. . . . . .


.

.(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.


. . . . . .


.

.(uv)′ = u′v′?

.(uv)′ = u′v′!



. . . . . .


.

.(uv)′ = u′v′?

.(uv)′ = u′v′!



. . . . . .

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


I Worklongerhours.

I Getaraise.


..

.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?

. . . . . .

Mmm...burgers




..

.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?

. . . . . .

Mmm...burgers




..

.∆I = 5× $0.25 = $1.25?.∆I = 5× $0.25 = $1.25?

. . . . . .

Mmm...burgers




...∆I = 5× $0.25 = $1.25?

.∆I = 5× $0.25 = $1.25?

. . . . . .

Mmm...burgers




..

.∆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

. . . . . .


ddx

[(3− x2)(x3 − x+ 1)

]Solutionbydirectmultiplication:

ddx

[(3− x2)(x3 − x+ 1)

]FOIL=

ddx

[−x5 + 4x3 − x2 − 3x+ 3

]

= −5x4 + 12x2 − 2x− 3

. . . . . .


ddx

[(3− x2)(x3 − x+ 1)

]Solutionbydirectmultiplication:

ddx

[(3− x2)(x3 − x+ 1)

]FOIL=

ddx

[−x5 + 4x3 − x2 − 3x+ 3

]= −5x4 + 12x2 − 2x− 3

. . . . . .


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

. . . . . .


ddx

[(3− x2)(x3 − x+ 1)


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

. . . . . .


ddx

[(3− x2)(x3 − x+ 1)


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

. . . . . .


ddx

[(3− x2)(x3 − x+ 1)


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

. . . . . .


ddx

[(3− x2)(x3 − x+ 1)


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

. . . . . .


ddx

[(3− x2)(x3 − x+ 1)


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

. . . . . .


ddx

[(3− x2)(x3 − x+ 1)


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)


. . . . . .

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)


. . . . . .

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′..


to u and v

= (u′v+ uv′)w+ (uv)w′

= u′vw+ uv′w+ uvw′

Sowewritedowntheproductthreetimes, takingthederivativeofeachfactoronce.

. . . . . .



Solution

(uvw)′

= ((uv)w)′

..


to uv and w

= (uv)′w+ (uv)w′..


to u and v

= (u′v+ uv′)w+ (uv)w′



. . . . . .



Solution

(uvw)′ = ((uv)w)′..


to uv and w

= (uv)′w+ (uv)w′..


to u and v

= (u′v+ uv′)w+ (uv)w′



. . . . . .



Solution

(uvw)′ = ((uv)w)′..


to uv and w

= (uv)′w+ (uv)w′..


to u and v

= (u′v+ uv′)w+ (uv)w′



. . . . . .



Solution

(uvw)′ = ((uv)w)′..


to uv and w

= (uv)′w+ (uv)w′..


to u and v

= (u′v+ uv′)w+ (uv)w′



. . . . . .



Solution

(uvw)′ = ((uv)w)′..


to uv and w

= (uv)′w+ (uv)w′..


to u and v

= (u′v+ uv′)w+ (uv)w′



. . . . . .



Solution

(uvw)′ = ((uv)w)′..


to uv and w

= (uv)′w+ (uv)w′..


to u and v

= (u′v+ uv′)w+ (uv)w′



. . . . . .



Solution

(uvw)′ = ((uv)w)′..


to uv and w

= (uv)′w+ (uv)w′..


to u and v

= (u′v+ uv′)w+ (uv)w′



. . . . . .



Solution

(uvw)′ = ((uv)w)′..


to uv and w

= (uv)′w+ (uv)w′..


to u and v

= (u′v+ uv′)w+ (uv)w′



. . . . . .

Outline

Grader’sCorner





. . . . . .

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



u = Qv


u′ = (Qv)′ = Q′v+Qv′

=⇒ Q′ =u′ −Qv′

v=

u′

v− u

v· v

′

v

=⇒ Q′ =(uv

)′=

u′v− uv′

v2


. . . . . .

TheQuotientRule



u = Qv


u′ = (Qv)′ = Q′v+Qv′

=⇒ Q′ =u′ −Qv′

v=

u′

v− u

v· v

′

v

=⇒ Q′ =(uv

)′=

u′v− uv′

v2


. . . . . .

TheQuotientRule



u = Qv


u′ = (Qv)′ = Q′v+Qv′

=⇒ Q′ =u′ −Qv′

v=

u′

v− u

v· v

′

v

=⇒ Q′ =(uv

)′=

u′v− uv′

v2


. . . . . .

TheQuotientRule



u = Qv


u′ = (Qv)′ = Q′v+Qv′

=⇒ Q′ =u′ −Qv′

v=

u′

v− u

v· v

′

v

=⇒ Q′ =(uv

)′=

u′v− uv′

v2


. . . . . .

TheQuotientRule



u = Qv


u′ = (Qv)′ = Q′v+Qv′

=⇒ Q′ =u′ −Qv′

v=

u′

v− u

v· v

′

v

=⇒ Q′ =(uv

)′=

u′v− uv′

v2


. . . . . .

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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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

. . . . . .


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





. . . . . .

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

)=


=cos2 x+ sin2 x

cos2 x

=1

cos2 x= sec2 x

. . . . . .

DerivativeofTangent

Example

Findddx

tan x

Solution

ddx

tan x =ddx

(sin xcos x

)=


=cos2 x+ sin2 x

cos2 x=

1cos2 x

= sec2 x

. . . . . .

DerivativeofTangent

Example

Findddx

tan x

Solution

ddx

tan x =ddx

(sin xcos 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

. . . . . .


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= − 1

sin2 x= − csc2 x

. . . . . .


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= − 1

sin2 x= − csc2 x

. . . . . .


Example

Findddx

cot x

Answer

ddx

cot x = − 1sin2 x

= − csc2 x

Solution

ddx

cot x =ddx

(cos xsin x

)=



sin2 x

= − 1sin2 x

= − csc2 x

. . . . . .


Example

Findddx

cot x

Answer

ddx

cot x = − 1sin2 x

= − csc2 x

Solution

ddx

cot x =ddx

(cos xsin x

)=



sin2 x= − 1

sin2 x

= − csc2 x

. . . . . .


Example

Findddx

cot x

Answer

ddx

cot x = − 1sin2 x

= − csc2 x

Solution

ddx

cot x =ddx

(cos xsin 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

)=


=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

)=


=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

)=


=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

. . . . . .


Example

Findddx

csc x

Answer

ddx


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

. . . . . .


Example

Findddx

csc x

Answer

ddx


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

. . . . . .


Example

Findddx

csc x

Answer

ddx


Solution

ddx

csc x =ddx

(1

sin x

)=


= − cos xsin2 x

= − 1sin x

· cos xsin x

= − csc x cot x

. . . . . .


Example

Findddx

csc x

Answer

ddx


Solution

ddx

csc x =ddx

(1

sin x

)=


= − cos xsin2 x

= − 1sin x

· cos xsin x

= − csc x cot x

. . . . . .


Example

Findddx

csc x

Answer

ddx


Solution

ddx

csc x =ddx

(1

sin 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





. . . . . .

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

. . . . . .


ddx

xn = nxn−1

Proof.Byinductionon n.

Wecanshowittobetruefor n = 1 directly.


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

http://flickr.com/photos/kool_skatkat/

. . . . . .


ddx

xn = nxn−1



xn = nxn−1. Then

ddx

xn+1

=ddx

(x · xn)

=

(ddx

x)xn + x

(ddx

xn)

= 1 · xn + x · nxn−1 = (n+ 1)xn

. . . . . .


ddx

xn = nxn−1



xn = nxn−1. Then

ddx

xn+1 =ddx

(x · xn)

=

(ddx

x)xn + x

(ddx

xn)

= 1 · xn + x · nxn−1 = (n+ 1)xn

. . . . . .


ddx

xn = nxn−1



xn = nxn−1. Then

ddx

xn+1 =ddx

(x · xn)

=

(ddx

x)xn + x

(ddx

xn)

= 1 · xn + x · nxn−1 = (n+ 1)xn

. . . . . .


ddx

xn = nxn−1



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

. . . . . .


Theorem

ddx

x−n = (−n)x−n−1


Proof.

ddx

x−n =ddx

1xn

=xn · d

dx1− 1 · ddxx

n

x2n

=0− nxn−1

x2n

= −nx−n−1

. . . . . .


Theorem

ddx

x−n = (−n)x−n−1


Proof.

ddx

x−n =ddx

1xn

=xn · d

dx1− 1 · ddxx

n

x2n

=0− nxn−1

x2n

= −nx−n−1

. . . . . .


Theorem

ddx

x−n = (−n)x−n−1


Proof.

ddx

x−n =ddx

1xn

=xn · d

dx1− 1 · ddxx

n

x2n

=0− nxn−1

x2n

= −nx−n−1

. . . . . .


Theorem

ddx

x−n = (−n)x−n−1


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


I ThePowerRuleistrueforallwholenumberpowers,includingnegativepowers:

ddx

xn = nxn−1

Lesson 9: The Product and Quotient Rules

Education