..

Sec on 2.4The Product and Quo ent Rules

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

February 23, 2011

Announcements

I Quiz 2 next week on§§1.5, 1.6, 2.1, 2.2

I Midterm March 7 on allsec ons in class (coversall sec ons up to 2.5)

Help!Free resources:

I Math Tutoring Center(CIWW 524)

I College Learning Center(schedule on Blackboard)

I TAs’ office hoursI my office hoursI each other!

Objectives

I Understand and be ableto use the Product Rulefor the deriva ve of theproduct of two func ons.

I Understand and be ableto use the Quo ent Rulefor the deriva ve of thequo ent of twofunc ons.

OutlineDeriva ve of a Product

Deriva onExamples

The Quo ent RuleDeriva onExamples

More deriva ves of trigonometric func onsDeriva ve of Tangent and CotangentDeriva ve of Secant and Cosecant

More on the Power RulePower Rule for Nega ve Integers

Recollection and extension

We have shown that if u and v are func ons, that

(u+ v)′ = u′ + v′

(u− v)′ = u′ − v′

What about uv?

Is the derivative of a product theproduct of the derivatives?

..(uv)′ = u′v′?

.(uv)′ = u′v′!

Try this with u = x and v = x2.I Then uv = x3 =⇒ (uv)′ = 3x2.I But u′v′ = 1 · 2x = 2x.

So we have to be more careful.

Is the derivative of a product theproduct of the derivatives?

.

.(uv)′ = u′v′?

.(uv)′ = u′v′!

Try this with u = x and v = x2.

I Then uv = x3 =⇒ (uv)′ = 3x2.I But u′v′ = 1 · 2x = 2x.

So we have to be more careful.

Is the derivative of a product theproduct of the derivatives?

.

.(uv)′ = u′v′?

.(uv)′ = u′v′!

Try this with u = x and v = x2.I Then uv = x3 =⇒ (uv)′ = 3x2.

I But u′v′ = 1 · 2x = 2x.So we have to be more careful.

Is the derivative of a product theproduct of the derivatives?

.

.(uv)′ = u′v′?

.(uv)′ = u′v′!

Try this with u = x and v = x2.I Then uv = x3 =⇒ (uv)′ = 3x2.I But u′v′ = 1 · 2x = 2x.

So we have to be more careful.

Is the derivative of a product theproduct of the derivatives?

.

.(uv)′ = u′v′?

.(uv)′ = u′v′!

Try this with u = x and v = x2.I Then uv = x3 =⇒ (uv)′ = 3x2.I But u′v′ = 1 · 2x = 2x.

So we have to be more careful.

Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices?

I Work longer hours.I Get a raise.

Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake?

..

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

Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices?

I Work longer hours.

I Get a raise.Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake?

..

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

Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices?

I Work longer hours.I Get a raise.

Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake?

..

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

Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices?

I Work longer hours.I Get a raise.

Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake?

..

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

Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices?

I Work longer hours.I Get a raise.

Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake?

..

.∆I = 5× $0.25 = $1.25?

.∆I = 5× $0.25 = $1.25?

.∆I = 5× $0.25 = $1.25?

Mmm...burgersSay you work in a fast-food joint. You want to make more money.What are your choices?

I Work longer hours.I Get a raise.

Say you get a 25 cent raise inyour hourly wages and work 5hours more per week. Howmuch extra money do youmake?

..

.∆I = 5× $0.25 = $1.25?

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

Money money money moneyThe answer depends on how much you work already and yourcurrent wage. Suppose you work h hours and are paid w. You get ame increase of∆h and a wage increase of∆w. Income is wagesmes hours, so

∆I = (w+∆w)(h+∆h)− whFOIL= w · h+ w ·∆h+∆w · h+∆w ·∆h− wh= w ·∆h+∆w · h+∆w ·∆h

A geometric argumentDraw a box:

..w

.∆w

.

h

.

∆h

.

wh

.

w∆h

.

∆wh

.

∆w∆h

∆I = w∆h+ h∆w+∆w∆h

A geometric argumentDraw a box:

..w

.∆w

.

h

.

∆h

.

wh

.

w∆h

.

∆wh

.

∆w∆h

∆I = w∆h+ h∆w+∆w∆h

Cash flowSupose wages and hours are changing con nuously over me. Overa me interval∆t, what is the average rate of change of income?

∆I∆t

=w∆h+ h∆w+∆w∆h

∆t

= w∆h∆t

+ h∆w∆t

+∆w∆h∆t

What is the instantaneous rate of change of income?

dIdt

= lim∆t→0

∆I∆t

= wdhdt

+ hdwdt

+ 0

Cash flowSupose wages and hours are changing con nuously over me. Overa me interval∆t, what is the average rate of change of income?

∆I∆t

=w∆h+ h∆w+∆w∆h

∆t

= w∆h∆t

+ h∆w∆t

+∆w∆h∆t

What is the instantaneous rate of change of income?

dIdt

= lim∆t→0

∆I∆t

= wdhdt

+ hdwdt

+ 0

Eurekamen!We have discoveredTheorem (The Product Rule)

Let u and v be differen able at x. Then

(uv)′(x) = u(x)v′(x) + u′(x)v(x)

in Leibniz nota on

ddx

(uv) =dudx

· v+ udvdx

Sanity Check

Example

Apply the product rule to u = x and v = x2.

Solu on

(uv)′(x) = u(x)v′(x) + u′(x)v(x) = x · (2x) + 1 · x2 = 3x2

This is what we get the “normal” way.

Sanity Check

Example

Apply the product rule to u = x and v = x2.

Solu on

(uv)′(x) = u(x)v′(x) + u′(x)v(x) = x · (2x) + 1 · x2 = 3x2

This is what we get the “normal” way.

Which is better?

Example

Find this deriva ve two ways: first by direct mul plica on and thenby the product rule:

ddx

[(3− x2)(x3 − x+ 1)

]

Which is better?Example

ddx

[(3− x2)(x3 − x+ 1)

]

Solu onby direct mul plica on:

ddx

[(3− x2)(x3 − x+ 1)

]FOIL=

ddx

[−x5 + 4x3 − x2 − 3x+ 3

]

= −5x4 + 12x2 − 2x− 3

Which is better?Example

ddx

[(3− x2)(x3 − x+ 1)

]Solu onby direct mul plica on:

ddx

[(3− x2)(x3 − x+ 1)

]FOIL=

ddx

[−x5 + 4x3 − x2 − 3x+ 3

]

= −5x4 + 12x2 − 2x− 3

Which is better?Example

ddx

[(3− x2)(x3 − x+ 1)

]Solu onby direct mul plica on:

ddx

[(3− x2)(x3 − x+ 1)

]FOIL=

ddx

[−x5 + 4x3 − x2 − 3x+ 3

]= −5x4 + 12x2 − 2x− 3

Which is better?Example

ddx

[(3− x2)(x3 − x+ 1)

]Solu onby the product rule:

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

Which is better?Example

ddx

[(3− x2)(x3 − x+ 1)

]Solu onby the product rule:

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

Which is better?Example

ddx

[(3− x2)(x3 − x+ 1)

]Solu onby the product rule:

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

Which is better?Example

ddx

[(3− x2)(x3 − x+ 1)

]Solu onby the product rule:

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

Which is better?Example

ddx

[(3− x2)(x3 − x+ 1)

]Solu onby the product rule:

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

Which is better?Example

ddx

[(3− x2)(x3 − x+ 1)

]Solu onby the product rule:

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

Which is better?Example

ddx

[(3− x2)(x3 − x+ 1)

]Solu onby the product rule:

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

One moreExample

Findddx

x sin x.

Solu on

ddx

x sin x

=

(ddx

x)sin x+ x

(ddx

sin x)

= 1 · sin x+ x · cos x= sin x+ x cos x

One moreExample

Findddx

x sin x.

Solu on

ddx

x sin x =(

ddx

x)sin x+ x

(ddx

sin x)

= 1 · sin x+ x · cos x= sin x+ x cos x

One moreExample

Findddx

x sin x.

Solu on

ddx

x sin x =(

ddx

x)sin x+ x

(ddx

sin x)

= 1 · sin x+ x · cos x

= sin x+ x cos x

One moreExample

Findddx

x sin x.

Solu on

ddx

x sin x =(

ddx

x)sin x+ x

(ddx

sin x)

= 1 · sin x+ x · cos x= sin x+ x cos x

MnemonicLet u = “hi” and v = “ho”. Then

(uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho”

Musical interlude

I jazz bandleader andsinger

I hit song “Minnie theMoocher” featuring “hide ho” chorus

I played Cur s in The BluesBrothers

Cab Calloway1907–1994

Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.

Solu on

(uvw)′

= ((uv)w)′

..

.

Apply the productrule to uv and w

= (uv)′w+ (uv)w′..

.

Apply the productrule to u and v

= (u′v+ uv′)w+ (uv)w′

= u′vw+ uv′w+ uvw′

So we write down the product three mes, taking the deriva ve of each factoronce.

Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.

Solu on

(uvw)′

= ((uv)w)′

..

.

Apply the productrule to uv and w

= (uv)′w+ (uv)w′..

.

Apply the productrule to u and v

= (u′v+ uv′)w+ (uv)w′

= u′vw+ uv′w+ uvw′

So we write down the product three mes, taking the deriva ve of each factoronce.

Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.

Solu on

(uvw)′ = ((uv)w)′..

.

Apply the productrule to uv and w

= (uv)′w+ (uv)w′..

.

Apply the productrule to u and v

= (u′v+ uv′)w+ (uv)w′

= u′vw+ uv′w+ uvw′

So we write down the product three mes, taking the deriva ve of each factoronce.

Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.

Solu on

(uvw)′ = ((uv)w)′...

Apply the productrule to uv and w

= (uv)′w+ (uv)w′..

.

Apply the productrule to u and v

= (u′v+ uv′)w+ (uv)w′

= u′vw+ uv′w+ uvw′

So we write down the product three mes, taking the deriva ve of each factoronce.

Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.

Solu on

(uvw)′ = ((uv)w)′...

Apply the productrule to uv and w

= (uv)′w+ (uv)w′..

.

Apply the productrule to u and v

= (u′v+ uv′)w+ (uv)w′

= u′vw+ uv′w+ uvw′

So we write down the product three mes, taking the deriva ve of each factoronce.

Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.

Solu on

(uvw)′ = ((uv)w)′..

.

Apply the productrule to uv and w

= (uv)′w+ (uv)w′...

Apply the productrule to u and v

= (u′v+ uv′)w+ (uv)w′

= u′vw+ uv′w+ uvw′

So we write down the product three mes, taking the deriva ve of each factoronce.

Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.

Solu on

(uvw)′ = ((uv)w)′..

.

Apply the productrule to uv and w

= (uv)′w+ (uv)w′...

Apply the productrule to u and v

= (u′v+ uv′)w+ (uv)w′

= u′vw+ uv′w+ uvw′

So we write down the product three mes, taking the deriva ve of each factoronce.

Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.

Solu on

(uvw)′ = ((uv)w)′..

.

Apply the productrule to uv and w

= (uv)′w+ (uv)w′..

.

Apply the productrule to u and v

= (u′v+ uv′)w+ (uv)w′

= u′vw+ uv′w+ uvw′

So we write down the product three mes, taking the deriva ve of each factoronce.

Iterating the Product RuleExampleUse the product rule to find the deriva ve of a three-fold product uvw.

Solu on

(uvw)′ = ((uv)w)′..

.

Apply the productrule to uv and w

= (uv)′w+ (uv)w′..

.

Apply the productrule to u and v

= (u′v+ uv′)w+ (uv)w′

= u′vw+ uv′w+ uvw′

So we write down the product three mes, taking the deriva ve of each factoronce.

OutlineDeriva ve of a Product

Deriva onExamples

The Quo ent RuleDeriva onExamples

More deriva ves of trigonometric func onsDeriva ve of Tangent and CotangentDeriva ve of Secant and Cosecant

More on the Power RulePower Rule for Nega ve Integers

The Quotient RuleWhat about the deriva ve of a quo ent?

Let u and v be differen able func ons and let Q =uv. Then

u = Qv

If Q is differen able, we have

u′ = (Qv)′ = Q′v+ Qv′

=⇒ Q′ =u′ − Qv′

v=

u′

v− u

v· v

′

v

=⇒ Q′ =(uv

)′=

u′v− uv′

v2

This is called the Quo ent Rule.

The Quotient RuleWhat about the deriva ve of a quo ent?Let u and v be differen able func ons and let Q =

uv. Then

u = Qv

If Q is differen able, we have

u′ = (Qv)′ = Q′v+ Qv′

=⇒ Q′ =u′ − Qv′

v=

u′

v− u

v· v

′

v

=⇒ Q′ =(uv

)′=

u′v− uv′

v2

This is called the Quo ent Rule.

The Quotient RuleWhat about the deriva ve of a quo ent?Let u and v be differen able func ons and let Q =

uv. Then

u = Qv

If Q is differen able, we have

u′ = (Qv)′ = Q′v+ Qv′

=⇒ Q′ =u′ − Qv′

v=

u′

v− u

v· v

′

v

=⇒ Q′ =(uv

)′=

u′v− uv′

v2

This is called the Quo ent Rule.

The Quotient RuleWhat about the deriva ve of a quo ent?Let u and v be differen able func ons and let Q =

uv. Then

u = Qv

If Q is differen able, we have

u′ = (Qv)′ = Q′v+ Qv′

=⇒ Q′ =u′ − Qv′

v=

u′

v− u

v· v

′

v

=⇒ Q′ =(uv

)′=

u′v− uv′

v2

This is called the Quo ent Rule.

The Quotient RuleWhat about the deriva ve of a quo ent?Let u and v be differen able func ons and let Q =

uv. Then

u = Qv

If Q is differen able, we have

u′ = (Qv)′ = Q′v+ Qv′

=⇒ Q′ =u′ − Qv′

v=

u′

v− u

v· v

′

v

=⇒ Q′ =(uv

)′=

u′v− uv′

v2

This is called the Quo ent Rule.

The Quotient RuleWhat about the deriva ve of a quo ent?Let u and v be differen able func ons and let Q =

uv. Then

u = Qv

If Q is differen able, we have

u′ = (Qv)′ = Q′v+ Qv′

=⇒ Q′ =u′ − Qv′

v=

u′

v− u

v· v

′

v

=⇒ Q′ =(uv

)′=

u′v− uv′

v2

This is called the Quo ent Rule.

The Quotient RuleWe have discoveredTheorem (The Quo ent Rule)

Let u and v be differen able at x, and v(x) ̸= 0. Thenuvis

differen able at x, and(uv

)′(x) =

u′(x)v(x)− u(x)v′(x)v(x)2

Verifying ExampleExample

Verify the quo ent rule by compu ngddx

(x2

x

)and comparing it to

ddx

(x).

Solu on

ddx

(x2

x

)=

x ddx

(x2)− x2 d

dx (x)x2

=x · 2x− x2 · 1

x2

=x2

x2= 1 =

ddx

(x)

Verifying ExampleExample

Verify the quo ent rule by compu ngddx

(x2

x

)and comparing it to

ddx

(x).

Solu on

ddx

(x2

x

)=

x ddx

(x2)− x2 d

dx (x)x2

=x · 2x− x2 · 1

x2

=x2

x2= 1 =

ddx

(x)

MnemonicLet u = “hi” and v = “lo”. Then(u

v

)′=

vu′ − uv′

v2= “lo dee hi minus hi dee lo over lo lo”

Examples

Example

1.ddx

2x+ 53x− 2

2.ddx

sin xx2

3.ddt

1t2 + t+ 2

Answers

1. − 19(3x− 2)2

2.x cos x− 2 sin x

x3

3. − 2t+ 1(t2 + t+ 2)2

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

Solution to first exampleSolu on

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

sin xx2

3.ddt

1t2 + t+ 2

Answers

1. − 19(3x− 2)2

2.x cos x− 2 sin x

x3

3. − 2t+ 1(t2 + t+ 2)2

Solution to second exampleSolu on

ddx

sin xx2

=

x2 ddx sin x− sin x d

dxx2

(x2)2

=

x2 cos x− 2x sin xx4

=x cos x− 2 sin x

x3

Solution to second exampleSolu on

ddx

sin xx2

=x2

ddx sin x− sin x d

dxx2

(x2)2

=

x2 cos x− 2x sin xx4

=x cos x− 2 sin x

x3

Solution to second exampleSolu on

ddx

sin xx2

=x2 d

dx sin x

− sin x ddxx

2

(x2)2

=

x2 cos x− 2x sin xx4

=x cos x− 2 sin x

x3

Solution to second exampleSolu on

ddx

sin xx2

=x2 d

dx sin x− sin x

ddxx

2

(x2)2

=

x2 cos x− 2x sin xx4

=x cos x− 2 sin x

x3

Solution to second exampleSolu on

ddx

sin xx2

=x2 d

dx sin x− sin x ddxx

2

(x2)2

=

x2 cos x− 2x sin xx4

=x cos x− 2 sin x

x3

Solution to second exampleSolu on

ddx

sin xx2

=x2 d

dx sin x− sin x ddxx

2

(x2)2

=

x2 cos x− 2x sin xx4

=x cos x− 2 sin x

x3

Solution to second exampleSolu on

ddx

sin xx2

=x2 d

dx sin x− sin x ddxx

2

(x2)2

=

x2 cos x− 2x sin xx4

=x cos x− 2 sin x

x3

Solution to second exampleSolu on

ddx

sin xx2

=x2 d

dx sin x− sin x ddxx

2

(x2)2

=x2

cos x− 2x sin xx4

=x cos x− 2 sin x

x3

Solution to second exampleSolu on

ddx

sin xx2

=x2 d

dx sin x− sin x ddxx

2

(x2)2

=x2 cos x

− 2x sin xx4

=x cos x− 2 sin x

x3

Solution to second exampleSolu on

ddx

sin xx2

=x2 d

dx sin x− sin x ddxx

2

(x2)2

=x2 cos x− 2x

sin xx4

=x cos x− 2 sin x

x3

Solution to second exampleSolu on

ddx

sin xx2

=x2 d

dx sin x− sin x ddxx

2

(x2)2

=x2 cos x− 2x sin x

x4

=x cos x− 2 sin x

x3

Solution to second exampleSolu on

ddx

sin xx2

=x2 d

dx sin x− sin x ddxx

2

(x2)2

=x2 cos x− 2x sin x

x4

=x cos x− 2 sin x

x3

Solution to second exampleSolu on

ddx

sin xx2

=x2 d

dx sin x− sin x ddxx

2

(x2)2

=x2 cos x− 2x sin x

x4

=x cos x− 2 sin x

x3

Another way to do itFind the deriva ve with the product rule instead.

Solu onddx

sin xx2

=ddx

(sin x · x−2)

=

(ddx

sin x)· x−2 + sin x ·

(ddx

x−2)

= cos x · x−2 + sin x · (−2x−3)

= x−3 (x cos x− 2 sin x)

No ce the technique of factoring out the largest nega ve power,leaving posi ve powers.

Examples

Example

1.ddx

2x+ 53x− 2

2.ddx

sin xx2

3.ddt

1t2 + t+ 2

Answers

1. − 19(3x− 2)2

2.x cos x− 2 sin x

x3

3. − 2t+ 1(t2 + t+ 2)2

Solution to third exampleSolu on

ddt

1t2 + t+ 2

=(t2 + t+ 2)(0)− (1)(2t+ 1)

(t2 + t+ 2)2

= − 2t+ 1(t2 + t+ 2)2

Solution to third exampleSolu on

ddt

1t2 + t+ 2

=(t2 + t+ 2)(0)− (1)(2t+ 1)

(t2 + t+ 2)2

= − 2t+ 1(t2 + t+ 2)2

Solution to third exampleSolu on

ddt

1t2 + t+ 2

=(t2 + t+ 2)(0)− (1)(2t+ 1)

(t2 + t+ 2)2

= − 2t+ 1(t2 + t+ 2)2

A nice little takeawayFact

Let v be differen able at x, and v(x) ̸= 0. Then1vis differen able at

0, and (1v

)′= − v′

v2

Proof.

ddx

(1v

)=

v · ddx(1)− 1 · d

dxvv2

=v · 0− 1 · v′

v2= − v′

v2

Examples

Example

1.ddx

2x+ 53x− 2

2.ddx

sin xx2

3.ddt

1t2 + t+ 2

Answers

1. − 19(3x− 2)2

2.x cos x− 2 sin x

x3

3. − 2t+ 1(t2 + t+ 2)2

OutlineDeriva ve of a Product

Deriva onExamples

The Quo ent RuleDeriva onExamples

More deriva ves of trigonometric func onsDeriva ve of Tangent and CotangentDeriva ve of Secant and Cosecant

More on the Power RulePower Rule for Nega ve Integers

Derivative of TangentExample

Findddx

tan x

Solu on

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

Derivative of TangentExample

Findddx

tan x

Solu on

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

Derivative of TangentExample

Findddx

tan x

Solu on

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

Derivative of TangentExample

Findddx

tan x

Solu on

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

Derivative of TangentExample

Findddx

tan x

Solu on

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

Derivative of TangentExample

Findddx

tan x

Solu on

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

Derivative of CotangentExample

Findddx

cot x

Derivative of CotangentExample

Findddx

cot x

Answer

ddx

cot x = − 1sin2 x

= − csc2 x

Derivative of CotangentExample

Findddx

cot x

Solu on

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

Derivative of CotangentExample

Findddx

cot x

Solu on

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

Derivative of CotangentExample

Findddx

cot x

Solu on

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

Derivative of CotangentExample

Findddx

cot x

Solu on

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

Derivative of SecantExample

Findddx

sec x

Solu on

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

Derivative of SecantExample

Findddx

sec x

Solu on

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

Derivative of SecantExample

Findddx

sec x

Solu on

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

Derivative of SecantExample

Findddx

sec x

Solu on

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

Derivative of SecantExample

Findddx

sec x

Solu on

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

Derivative of SecantExample

Findddx

sec x

Solu on

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

Derivative of CosecantExample

Findddx

csc x

Derivative of CosecantExample

Findddx

csc x

Answer

ddx

csc x = − csc x cot x

Derivative of CosecantExample

Findddx

csc x

Solu on

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

Derivative of CosecantExample

Findddx

csc x

Solu on

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

Derivative of CosecantExample

Findddx

csc x

Solu on

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

Derivative of CosecantExample

Findddx

csc x

Solu on

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: Derivatives oftrigonometric functions

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 Func ons come in pairs(sin/cos, tan/cot, sec/csc)

I Deriva ves of pairs followsimilar pa erns, withfunc ons andco-func ons switchedand an extra sign.

OutlineDeriva ve of a Product

Deriva onExamples

The Quo ent RuleDeriva onExamples

More deriva ves of trigonometric func onsDeriva ve of Tangent and CotangentDeriva ve of Secant and Cosecant

More on the Power RulePower Rule for Nega ve Integers

Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem

ddx

x−n = (−n)x−n−1

for posi ve integers n.

Proof.

ddx

x−n =ddx

1xn

= −ddxx

n

(xn)2= −nxn−1

x2n= −nxn−1−2n = −nx−n−1

Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem

ddx

x−n = (−n)x−n−1

for posi ve integers n.

Proof.

ddx

x−n =ddx

1xn

= −ddxx

n

(xn)2= −nxn−1

x2n= −nxn−1−2n = −nx−n−1

Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem

ddx

x−n = (−n)x−n−1

for posi ve integers n.

Proof.

ddx

x−n =ddx

1xn

= −ddxx

n

(xn)2= −nxn−1

x2n= −nxn−1−2n = −nx−n−1

Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem

ddx

x−n = (−n)x−n−1

for posi ve integers n.

Proof.

ddx

x−n =ddx

1xn

= −ddxx

n

(xn)2

= −nxn−1

x2n= −nxn−1−2n = −nx−n−1

Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem

ddx

x−n = (−n)x−n−1

for posi ve integers n.

Proof.

ddx

x−n =ddx

1xn

= −ddxx

n

(xn)2= −nxn−1

x2n

= −nxn−1−2n = −nx−n−1

Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem

ddx

x−n = (−n)x−n−1

for posi ve integers n.

Proof.

ddx

x−n =ddx

1xn

= −ddxx

n

(xn)2= −nxn−1

x2n= −nxn−1−2n

= −nx−n−1

Power Rule for Negative IntegersWe will use the quo ent rule to proveTheorem

ddx

x−n = (−n)x−n−1

for posi ve integers n.

Proof.

ddx

x−n =ddx

1xn

= −ddxx

n

(xn)2= −nxn−1

x2n= −nxn−1−2n = −nx−n−1

SummaryI The Product Rule: (uv)′ = u′v+ uv′

I The Quo ent Rule:(uv

)′=

vu′ − uv′

v2I Deriva ves of tangent/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 The Power Rule is true for all whole number powers, includingnega ve powers:

ddx

xn = nxn−1