Lesson 9: The Product and Quotient Rules (Section 21 slides)

Section 2.4The Product and Quotient Rules

V63.0121.021, Calculus I

New York University

October 5, 2010

Announcements

I Quiz 2 next week on §§1.5, 1.6, 2.1, 2.2I Midterm in class (covers all sections up to 2.5)

. . . . . .

Announcements

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

I Midterm in class (covers allsections up to 2.5)

V63.0121.021, Calculus I (NYU) Section 2.4 The Product and Quotient Rules October 5, 2010 2 / 41

. . . . . .

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 able touse the Product Rule forthe derivative of theproduct of two functions.

I Understand and be able touse the Quotient Rule forthe derivative of thequotient of two functions.

. . . . . .

Outline

Derivative of a ProductDerivationExamples

The Quotient RuleDerivationExamples

More derivatives of trigonometric functionsDerivative of Tangent and CotangentDerivative of Secant and Cosecant

More on the Power RulePower Rule for Negative Integers

. . . . . .

Recollection and extension

We have shown that if u and v are functions, that

(u+ v)′ = u′ + v′

(u− v)′ = u′ − v′

What about uv?

. . . . . .

Is the derivative of a product the product 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.

. . . . . .

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.

. . . . . .

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.

. . . . . .

derivatives?

.(uv)′ = u′v′?

.(uv)′ = u′v′!

. . . . . .

derivatives?

.(uv)′ = u′v′?

.(uv)′ = u′v′!

. . . . . .

Mmm...burgers

Say 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...burgers

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...burgers

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

. . . . . .

Mmm...burgers

.∆I = 5× $0.25 = $1.25?

. . . . . .

Mmm...burgers

.∆I = 5× $0.25 = $1.25?

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

. . . . . .

Money money money money

The answer depends on how much you work already and your currentwage. Suppose you work h hours and are paid w. You get a timeincrease of ∆h and a wage increase of ∆w. Income is wages timeshours, so

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

. . . . . .

A geometric argument

Draw a box:

..w .∆w

.w∆h

.∆wh

.∆w∆h

∆I = w∆h+ h∆w+∆w∆h

. . . . . .

A geometric argument

Draw a box:

..w .∆w

.w∆h

.∆wh

.∆w∆h

∆I = w∆h+ h∆w+∆w∆h

. . . . . .

Cash flow

Supose wages and hours are changing continuously over time. Over atime interval ∆t, what is the average rate of change of income?

∆I∆t

=w∆h+ h∆w+∆w∆h

= w∆h∆t

+ h∆w∆t

+∆w∆h∆t

What is the instantaneous rate of change of income?

= lim∆t→0

∆I∆t

= wdhdt

+ hdwdt

. . . . . .

Cash flow

Supose wages and hours are changing continuously over time. Over atime interval ∆t, what is the average rate of change of income?

∆I∆t

=w∆h+ h∆w+∆w∆h

= w∆h∆t

+ h∆w∆t

+∆w∆h∆t

What is the instantaneous rate of change of income?

= lim∆t→0

∆I∆t

= wdhdt

+ hdwdt

. . . . . .

Eurekamen!

We have discovered

Theorem (The Product Rule)

Let u and v be differentiable at x. Then

(uv)′(x) = u(x)v′(x) + u′(x)v(x)

in Leibniz notationddx

(uv) =dudx

· v+ udvdx

. . . . . .

Sanity Check

Example

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

Solution

(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.

Solution

(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 derivative two ways: first by direct multiplication and then bythe product rule:

[(3− x2)(x3 − x+ 1)

Solutionby direct multiplication:

[(3− x2)(x3 − x+ 1)

]FOIL=

[−x5 + 4x3 − x2 − 3x+ 3

= −5x4 + 12x2 − 2x− 3

. . . . . .

Which is better?

Example

[(3− x2)(x3 − x+ 1)

]FOIL=

[−x5 + 4x3 − x2 − 3x+ 3

= −5x4 + 12x2 − 2x− 3

. . . . . .

Which is better?

Example

[(3− x2)(x3 − x+ 1)

]FOIL=

[−x5 + 4x3 − x2 − 3x+ 3

]= −5x4 + 12x2 − 2x− 3

. . . . . .

Which is better?

Example

[(3− x2)(x3 − x+ 1)

Solutionby the product rule:

(3− x2))(x3 − x+ 1) + (3− x2)

(x3 − x+ 1))

= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x− 3

. . . . . .

Which is better?

Example

[(3− x2)(x3 − x+ 1)

(3− x2))(x3 − x+ 1) + (3− x2)

(x3 − x+ 1))

= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x− 3

. . . . . .

Which is better?

Example

[(3− x2)(x3 − x+ 1)

(3− x2))(x3 − x+ 1) + (3− x2)

(x3 − x+ 1))

= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x− 3

. . . . . .

Which is better?

Example

[(3− x2)(x3 − x+ 1)

(3− x2))(x3 − x+ 1) + (3− x2)

(x3 − x+ 1))

= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x− 3

. . . . . .

Which is better?

Example

[(3− x2)(x3 − x+ 1)

(3− x2))(x3 − x+ 1) + (3− x2)

(x3 − x+ 1))

= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x− 3

. . . . . .

Which is better?

Example

[(3− x2)(x3 − x+ 1)

(3− x2))(x3 − x+ 1) + (3− x2)

(x3 − x+ 1))

= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x− 3

. . . . . .

Which is better?

Example

[(3− x2)(x3 − x+ 1)

(3− x2))(x3 − x+ 1) + (3− x2)

(x3 − x+ 1))

= (−2x)(x3 − x+ 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x− 3

. . . . . .

One more

Example

Findddx

x sin x.

Solution

x sin x

x)sin x+ x

sin x)

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

. . . . . .

One more

Example

Findddx

x sin x.

Solution

x sin x =

x)sin x+ x

sin x)

. . . . . .

One more

Example

Findddx

x sin x.

Solution

x sin x =

x)sin x+ x

sin x)

= 1 · sin x+ x · cos x

= sin x+ x cos x

. . . . . .

One more

Example

Findddx

x sin x.

Solution

x sin x =

x)sin x+ x

sin x)

. . . . . .

Mnemonic

Let u = “hi” and v = “ho”. Then

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

. . . . . .

Musical interlude

I jazz bandleader and singerI hit song “Minnie the

Moocher” featuring “hi deho” chorus

I played Curtis in The BluesBrothers

Cab Calloway1907–1994

. . . . . .

Iterating the Product Rule

Example

Use the product rule to find the derivative of a three-fold product uvw.

Solution

(uvw)′

= ((uv)w)′

.Apply the product rule

to uv and w

= (uv)′w+ (uv)w′..

to u and v

= (u′v+ uv′)w+ (uv)w′

= u′vw+ uv′w+ uvw′

So we write down the product three times, taking the derivative of eachfactor once.

. . . . . .

Example

Solution

(uvw)′

= ((uv)w)′

to uv and w

= (uv)′w+ (uv)w′..

to u and v

= (u′v+ uv′)w+ (uv)w′

. . . . . .

Example

Solution

(uvw)′ = ((uv)w)′..

to uv and w

= (uv)′w+ (uv)w′..

to u and v

= (u′v+ uv′)w+ (uv)w′

. . . . . .

Example

Solution

(uvw)′ = ((uv)w)′..

to uv and w

= (uv)′w+ (uv)w′..

to u and v

= (u′v+ uv′)w+ (uv)w′

. . . . . .

Example

Solution

(uvw)′ = ((uv)w)′..

to uv and w

= (uv)′w+ (uv)w′..

to u and v

= (u′v+ uv′)w+ (uv)w′

. . . . . .

Example

Solution

(uvw)′ = ((uv)w)′..

to uv and w

= (uv)′w+ (uv)w′..

to u and v

= (u′v+ uv′)w+ (uv)w′

. . . . . .

Example

Solution

(uvw)′ = ((uv)w)′..

to uv and w

= (uv)′w+ (uv)w′..

to u and v

= (u′v+ uv′)w+ (uv)w′

. . . . . .

Example

Solution

(uvw)′ = ((uv)w)′..

to uv and w

= (uv)′w+ (uv)w′..

to u and v

= (u′v+ uv′)w+ (uv)w′

. . . . . .

Example

Solution

(uvw)′ = ((uv)w)′..

to uv and w

= (uv)′w+ (uv)w′..

to u and v

= (u′v+ uv′)w+ (uv)w′

. . . . . .

Outline

. . . . . .

The Quotient Rule

What about the derivative of a quotient?

Let u and v be differentiable functions and let Q =uv. Then

u = Qv

If Q is differentiable, we have

u′ = (Qv)′ = Q′v+Qv′

=⇒ Q′ =u′ −Qv′

v− u

=⇒ Q′ =(uv

u′v− uv′

This is called the Quotient Rule.

. . . . . .

The Quotient Rule

What about the derivative of a quotient?Let u and v be differentiable functions and let Q =

uv. Then

u = Qv

u′ = (Qv)′ = Q′v+Qv′

=⇒ Q′ =u′ −Qv′

v− u

=⇒ Q′ =(uv

u′v− uv′

. . . . . .

The Quotient Rule

uv. Then

u = Qv

u′ = (Qv)′ = Q′v+Qv′

=⇒ Q′ =u′ −Qv′

v− u

=⇒ Q′ =(uv

u′v− uv′

. . . . . .

The Quotient Rule

uv. Then

u = Qv

u′ = (Qv)′ = Q′v+Qv′

=⇒ Q′ =u′ −Qv′

v− u

=⇒ Q′ =(uv

u′v− uv′

. . . . . .

The Quotient Rule

uv. Then

u = Qv

u′ = (Qv)′ = Q′v+Qv′

=⇒ Q′ =u′ −Qv′

v− u

=⇒ Q′ =(uv

u′v− uv′

. . . . . .

The Quotient Rule

uv. Then

u = Qv

u′ = (Qv)′ = Q′v+Qv′

=⇒ Q′ =u′ −Qv′

v− u

=⇒ Q′ =(uv

u′v− uv′

. . . . . .

The Quotient Rule

We have discovered

Theorem (The Quotient Rule)

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

at x, and (uv

)′(x) =

u′(x)v(x)− u(x)v′(x)v(x)2

. . . . . .

Verifying Example

Example

Verify the quotient rule by computingddx

)and comparing it to

Solution

(x2)− x2 d

dx (x)x2

=x · 2x− x2 · 1

x2= 1 =

. . . . . .

Verifying Example

Example

Verify the quotient rule by computingddx

)and comparing it to

Solution

(x2)− x2 d

dx (x)x2

=x · 2x− x2 · 1

x2= 1 =

. . . . . .

Mnemonic

Let u = “hi” and v = “lo”. Then(uv

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 xx3

3. − 2t+ 1(t2 + t+ 2)2

. . . . . .

Solution to first example

Solution

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

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

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

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

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

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

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

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

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

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

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

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

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

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

3. − 2t+ 1(t2 + t+ 2)2

. . . . . .

Solution to second example

Solution

sin xx2

x2 ddx sin x− sin x d

x2 cos x− 2x sin xx4

=x cos x− 2 sin x

. . . . . .

Solution

sin xx2

ddx sin x− sin x d

=x cos x− 2 sin x

. . . . . .

Solution

sin xx2

dx sin x

− sin x ddxx

=x cos x− 2 sin x

. . . . . .

Solution

sin xx2

dx sin x− sin x

=x cos x− 2 sin x

. . . . . .

Solution

sin xx2

dx sin x− sin x ddxx

=x cos x− 2 sin x

. . . . . .

Solution

sin xx2

=x cos x− 2 sin x

. . . . . .

Solution

sin xx2

=x cos x− 2 sin x

. . . . . .

Solution

sin xx2

cos x− 2x sin xx4

=x cos x− 2 sin x

. . . . . .

Solution

sin xx2

=x2 cos x

− 2x sin xx4

=x cos x− 2 sin x

. . . . . .

Solution

sin xx2

=x2 cos x− 2x

sin xx4

=x cos x− 2 sin x

. . . . . .

Solution

sin xx2

=x2 cos x− 2x sin x

=x cos x− 2 sin x

. . . . . .

Solution

sin xx2

=x cos x− 2 sin x

. . . . . .

Solution

sin xx2

=x cos x− 2 sin x

. . . . . .

Another way to do it

Find the derivative with the product rule instead.

Solution

sin xx2

(sin x · x−2

sin x)· x−2 + sin x ·

x−2)

= cos x · x−2 + sin x · (−2x−3)

= x−3 (x cos x− 2 sin x)

Notice the technique of factoring out the largest negative power,leaving positive powers.

. . . . . .

Examples

Example

1. ddx

2x+ 53x− 2

2. ddx

sin xx2

3. ddt

1t2 + t+ 2

Answers

1. − 19(3x− 2)2

3. − 2t+ 1(t2 + t+ 2)2

. . . . . .

Solution to third example

Solution

1t2 + t+ 2

=(t2 + t+ 2)(0)− (1)(2t+ 1)

(t2 + t+ 2)2

= − 2t+ 1(t2 + t+ 2)2

. . . . . .

Solution

1t2 + t+ 2

=(t2 + t+ 2)(0)− (1)(2t+ 1)

(t2 + t+ 2)2

= − 2t+ 1(t2 + t+ 2)2

. . . . . .

Solution

1t2 + t+ 2

=(t2 + t+ 2)(0)− (1)(2t+ 1)

(t2 + t+ 2)2

= − 2t+ 1(t2 + t+ 2)2

. . . . . .

A nice little takeaway

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

and (1v

)′= − v′

Proof.

v · ddx(1)− 1 · d

=v · 0− 1 · v′

v2= − v′

. . . . . .

Examples

Example

1. ddx

2x+ 53x− 2

2. ddx

sin xx2

3. ddt

1t2 + t+ 2

Answers

1. − 19(3x− 2)2

3. − 2t+ 1(t2 + t+ 2)2

. . . . . .

Outline

. . . . . .

Derivative of Tangent

Example

Findddx

Solution

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

. . . . . .

Example

Findddx

Solution

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

. . . . . .

Example

Findddx

Solution

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

. . . . . .

Example

Findddx

Solution

tan x =ddx

(sin xcos x

=cos2 x+ sin2 x

cos2 x

cos2 x= sec2 x

. . . . . .

Example

Findddx

Solution

tan x =ddx

(sin xcos x

=cos2 x+ sin2 x

cos2 x=

1cos2 x

= sec2 x

. . . . . .

Example

Findddx

Solution

tan x =ddx

(sin xcos x

=cos2 x+ sin2 x

cos2 x=

1cos2 x

= sec2 x

. . . . . .

Derivative of Cotangent

Example

Findddx

Answer

cot x = − 1sin2 x

= − csc2 x

Solution

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

Answer

cot x = − 1sin2 x

= − csc2 x

Solution

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

Answer

cot x = − 1sin2 x

= − csc2 x

Solution

cot x =ddx

(cos xsin x

sin x · (− sin x)− cos x · cos xsin2 x

sin2 x= − 1

sin2 x= − csc2 x

. . . . . .

Example

Findddx

Answer

cot x = − 1sin2 x

= − csc2 x

Solution

cot x =ddx

(cos xsin x

sin2 x

= − 1sin2 x

= − csc2 x

. . . . . .

Example

Findddx

Answer

cot x = − 1sin2 x

= − csc2 x

Solution

cot x =ddx

(cos xsin x

sin2 x= − 1

sin2 x

= − csc2 x

. . . . . .

Example

Findddx

Answer

cot x = − 1sin2 x

= − csc2 x

Solution

cot x =ddx

(cos xsin x

sin2 x= − 1

sin2 x= − csc2 x

. . . . . .

Derivative of Secant

Example

Findddx

Solution

sec x =ddx

=cos x · 0− 1 · (− sin x)

cos2 x

=sin xcos2 x

cos x· sin xcos x

= sec x tan x

. . . . . .

Example

Findddx

Solution

sec x =ddx

=cos x · 0− 1 · (− sin x)

cos2 x

=sin xcos2 x

cos x· sin xcos x

= sec x tan x

. . . . . .

Example

Findddx

Solution

sec x =ddx

cos x · 0− 1 · (− sin x)cos2 x

=sin xcos2 x

cos x· sin xcos x

= sec x tan x

. . . . . .

Example

Findddx

Solution

sec x =ddx

=sin xcos2 x

cos x· sin xcos x

= sec x tan x

. . . . . .

Example

Findddx

Solution

sec x =ddx

=sin xcos2 x

cos x· sin xcos x

= sec x tan x

. . . . . .

Example

Findddx

Solution

sec x =ddx

=sin xcos2 x

cos x· sin xcos x

= sec x tan x

. . . . . .

Derivative of Cosecant

Example

Findddx

Answer

csc x = − csc x cot x

Solution

csc x =ddx

=sin x · 0− 1 · (cos x)

sin2 x

= − cos xsin2 x

= − 1sin x

· cos xsin x

= − csc x cot x

. . . . . .

Example

Findddx

Answer

Solution

csc x =ddx

=sin x · 0− 1 · (cos x)

sin2 x

= − cos xsin2 x

= − 1sin x

· cos xsin x

= − csc x cot x

. . . . . .

Example

Findddx

Answer

Solution

csc x =ddx

sin x · 0− 1 · (cos x)sin2 x

= − cos xsin2 x

= − 1sin x

· cos xsin x

= − csc x cot x

. . . . . .

Example

Findddx

Answer

Solution

csc x =ddx

= − cos xsin2 x

= − 1sin x

· cos xsin x

= − csc x cot x

. . . . . .

Example

Findddx

Answer

Solution

csc x =ddx

= − cos xsin2 x

= − 1sin x

· cos xsin x

= − csc x cot x

. . . . . .

Example

Findddx

Answer

Solution

csc x =ddx

= − cos xsin2 x

= − 1sin x

· cos xsin x

= − csc x cot x

. . . . . .

Recap: Derivatives of trigonometric 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 Functions come in pairs(sin/cos, tan/cot, sec/csc)

I Derivatives of pairs followsimilar patterns, withfunctions and co-functionsswitched and an extra sign.

. . . . . .

Outline

. . . . . .

Power Rule for Negative Integers

We will use the quotient rule to prove

Theorem

x−n = (−n)x−n−1

for positive integers n.

Proof.

x−n =ddx

= −ddxx

= −nxn−1

x2n= −nxn−1−2n = −nx−n−1

. . . . . .

Theorem

x−n = (−n)x−n−1

Proof.

x−n =ddx

= −ddxx

= −nxn−1

x2n= −nxn−1−2n = −nx−n−1

. . . . . .

Theorem

x−n = (−n)x−n−1

Proof.

x−n =ddx

= −ddxx

= −nxn−1

x2n= −nxn−1−2n = −nx−n−1

. . . . . .

Theorem

x−n = (−n)x−n−1

Proof.

x−n =ddx

= −ddxx

= −nxn−1

x2n= −nxn−1−2n = −nx−n−1

. . . . . .

Theorem

x−n = (−n)x−n−1

Proof.

x−n =ddx

= −ddxx

= −nxn−1

= −nxn−1−2n = −nx−n−1

. . . . . .

Theorem

x−n = (−n)x−n−1

Proof.

x−n =ddx

= −ddxx

= −nxn−1

x2n= −nxn−1−2n

= −nx−n−1

. . . . . .

Theorem

x−n = (−n)x−n−1

Proof.

x−n =ddx

= −ddxx

= −nxn−1

x2n= −nxn−1−2n = −nx−n−1

. . . . . .

Summary

I The Product Rule: (uv)′ = u′v+ uv′

I The Quotient Rule:(uv

vu′ − uv′

v2I Derivatives of tangent/cotangent, secant/cosecant

tan x = sec2 xddx

sec x = sec x tan x

cot x = − csc2 xddx

I The Power Rule is true for all whole number powers, includingnegative powers:

xn = nxn−1

Lesson 9: The Product and Quotient Rules (Section 21 slides)

quotient rules v63

nyu section

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