Section 2.4The Product and Quotient Rules

V63.0121, Calculus I

February 17–18, 2009

Announcements

I Quiz 2 is this week, covering 1.3–1.6

I Midterm I is March 4/5, covering 1.1–2.4 (today)

I ALEKS is due February 27, 11:59pm

Outline

The Product RuleDerivation of the product ruleExamples

The Quotient RuleDerivationExamples

More derivatives of trigonometric functionsDerivative of TangentDerivative of CotangentDerivative of SecantDerivative of Cosecant

More on the Power RulePower Rule for Positive Integers by InductionPower Rule for Negative Integers

Calculus

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 thederivatives?

(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 the product of thederivatives?

(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 the product of thederivatives?

(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 the product of thederivatives?

(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 the product of thederivatives?

(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...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?

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?

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?

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?

Money money money money

The answer depends on how much you work already and yourcurrent wage. Suppose you work h hours and are paid w . You geta time increase of ∆h and a wage increase of ∆w . Income iswages times hours, so

∆I = (w + ∆w)(h + ∆h)− wh

FOIL= wh + w ∆h + ∆w h + ∆w ∆h − wh

= w ∆h + ∆w h + ∆w ∆h

A geometric argument

Draw a box:

w ∆w

h

∆h

w h

w ∆h

∆w h

∆w ∆h

∆I = w ∆h + h ∆w + ∆w ∆h

A geometric argument

Draw a box:

w ∆w

h

∆h

w h

w ∆h

∆w h

∆w ∆h

∆I = w ∆h + h ∆w + ∆w ∆h

Supose wages and hours are changing continuously over time. Howdoes income change?

∆I

∆t=

w ∆h + h ∆w + ∆w ∆h

∆t

= w∆h

∆t+ h

∆w

∆t+ ∆w

∆h

∆t

SodI

dt= lim

t→0

∆I

∆t= w

dh

dt+ h

dw

dt+ 0

Theorem (The Product Rule)

Let u and v be differentiable at x. Then

(uv)′(x) = u(x)v ′(x) + u′(x)v(x)

Supose wages and hours are changing continuously over time. Howdoes income change?

∆I

∆t=

w ∆h + h ∆w + ∆w ∆h

∆t

= w∆h

∆t+ h

∆w

∆t+ ∆w

∆h

∆t

SodI

dt= lim

t→0

∆I

∆t= w

dh

dt+ h

dw

dt+ 0

Theorem (The Product Rule)

Let u and v be differentiable at x. Then

(uv)′(x) = u(x)v ′(x) + u′(x)v(x)

Supose wages and hours are changing continuously over time. Howdoes income change?

∆I

∆t=

w ∆h + h ∆w + ∆w ∆h

∆t

= w∆h

∆t+ h

∆w

∆t+ ∆w

∆h

∆t

SodI

dt= lim

t→0

∆I

∆t= w

dh

dt+ h

dw

dt+ 0

Theorem (The Product Rule)

Let u and v be differentiable at x. Then

(uv)′(x) = u(x)v ′(x) + u′(x)v(x)

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.

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.

Example

Find this derivative two ways: first by direct multiplication andthen by the product rule:

d

dx

[(3− x2)(x3 − x + 1)

]

Solutionby direct multiplication:

d

dx

[(3− x2)(x3 − x + 1)

]FOIL=

d

dx

[−x5 + 4x3 − x2 − 3x + 3

]

= −5x4 + 12x2 − 2x − 3

Example

Find this derivative two ways: first by direct multiplication andthen by the product rule:

d

dx

[(3− x2)(x3 − x + 1)

]Solutionby direct multiplication:

d

dx

[(3− x2)(x3 − x + 1)

]FOIL=

d

dx

[−x5 + 4x3 − x2 − 3x + 3

]

= −5x4 + 12x2 − 2x − 3

Example

Find this derivative two ways: first by direct multiplication andthen by the product rule:

d

dx

[(3− x2)(x3 − x + 1)

]Solutionby direct multiplication:

d

dx

[(3− x2)(x3 − x + 1)

]FOIL=

d

dx

[−x5 + 4x3 − x2 − 3x + 3

]= −5x4 + 12x2 − 2x − 3

Example

Find this derivative two ways: first by direct multiplication andthen by the product rule:

d

dx

[(3− x2)(x3 − x + 1)

]Solutionby the product rule:

dy

dx=

(d

dx(3− x2)

)(x3 − x + 1) + (3− x2)

(d

dx(x3 − x + 1)

)

= (−2x)(x3 − x + 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x − 3

Example

Find this derivative two ways: first by direct multiplication andthen by the product rule:

d

dx

[(3− x2)(x3 − x + 1)

]Solutionby the product rule:

dy

dx=

(d

dx(3− x2)

)(x3 − x + 1) + (3− x2)

(d

dx(x3 − x + 1)

)= (−2x)(x3 − x + 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x − 3

Example

Find this derivative two ways: first by direct multiplication andthen by the product rule:

d

dx

[(3− x2)(x3 − x + 1)

]Solutionby the product rule:

dy

dx=

(d

dx(3− x2)

)(x3 − x + 1) + (3− x2)

(d

dx(x3 − x + 1)

)= (−2x)(x3 − x + 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x − 3

Example

Find this derivative two ways: first by direct multiplication andthen by the product rule:

d

dx

[(3− x2)(x3 − x + 1)

]Solutionby the product rule:

dy

dx=

(d

dx(3− x2)

)(x3 − x + 1) + (3− x2)

(d

dx(x3 − x + 1)

)= (−2x)(x3 − x + 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x − 3

Example

Find this derivative two ways: first by direct multiplication andthen by the product rule:

d

dx

[(3− x2)(x3 − x + 1)

]Solutionby the product rule:

dy

dx=

(d

dx(3− x2)

)(x3 − x + 1) + (3− x2)

(d

dx(x3 − x + 1)

)= (−2x)(x3 − x + 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x − 3

Example

Find this derivative two ways: first by direct multiplication andthen by the product rule:

d

dx

[(3− x2)(x3 − x + 1)

]Solutionby the product rule:

dy

dx=

(d

dx(3− x2)

)(x3 − x + 1) + (3− x2)

(d

dx(x3 − x + 1)

)= (−2x)(x3 − x + 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x − 3

Example

Find this derivative two ways: first by direct multiplication andthen by the product rule:

d

dx

[(3− x2)(x3 − x + 1)

]Solutionby the product rule:

dy

dx=

(d

dx(3− x2)

)(x3 − x + 1) + (3− x2)

(d

dx(x3 − x + 1)

)= (−2x)(x3 − x + 1) + (3− x2)(3x2 − 1)

= −5x4 + 12x2 − 2x − 3

One more

Example

Findd

dxx sin x .

Solution

d

dxx sin x

=

(d

dxx

)sin x + x

(d

dxsin x

)= 1 · sin x + x · cos x

= sin x + x cos x

One more

Example

Findd

dxx sin x .

Solution

d

dxx sin x =

(d

dxx

)sin x + x

(d

dxsin x

)

= 1 · sin x + x · cos x

= sin x + x cos x

One more

Example

Findd

dxx sin x .

Solution

d

dxx sin x =

(d

dxx

)sin x + x

(d

dxsin x

)= 1 · sin x + x · cos x

= sin x + x cos x

One more

Example

Findd

dxx sin x .

Solution

d

dxx sin x =

(d

dxx

)sin x + x

(d

dxsin x

)= 1 · sin x + x · cos x

= sin x + x cos x

Mnemonic

Let u = “hi” and v = “ho”. Then

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

Iterating the Product Rule

Example

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

Solution

(uvw)′ = ((uv)w)′ = (uv)′w + (uv)w ′

= (u′v + uv ′)w + (uv)w ′

= u′vw + uv ′w + uvw ′

So we write down the product three times, taking the derivative ofeach factor once.

Iterating the Product Rule

Example

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

Solution

(uvw)′ = ((uv)w)′ = (uv)′w + (uv)w ′

= (u′v + uv ′)w + (uv)w ′

= u′vw + uv ′w + uvw ′

So we write down the product three times, taking the derivative ofeach factor once.

Iterating the Product Rule

Example

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

Solution

(uvw)′ = ((uv)w)′ = (uv)′w + (uv)w ′

= (u′v + uv ′)w + (uv)w ′

= u′vw + uv ′w + uvw ′

So we write down the product three times, taking the derivative ofeach factor once.

Outline

The Product RuleDerivation of the product ruleExamples

The Quotient RuleDerivationExamples

More derivatives of trigonometric functionsDerivative of TangentDerivative of CotangentDerivative of SecantDerivative of Cosecant

More on the Power RulePower Rule for Positive Integers by InductionPower Rule for Negative Integers

The Quotient Rule

What about the derivative of a quotient?

Let u and v be differentiable functions and let Q =u

v. Then

u = Qv

If Q is differentiable, we have

u′ = (Qv)′ = Q ′v + Qv ′

=⇒ Q ′ =u′ − Qv ′

v=

u′

v− u

v· v ′

v

=⇒ Q ′ =(u

v

)′=

u′v − uv ′

v2

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 =u

v. Then

u = Qv

If Q is differentiable, we have

u′ = (Qv)′ = Q ′v + Qv ′

=⇒ Q ′ =u′ − Qv ′

v=

u′

v− u

v· v ′

v

=⇒ Q ′ =(u

v

)′=

u′v − uv ′

v2

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 =u

v. Then

u = Qv

If Q is differentiable, we have

u′ = (Qv)′ = Q ′v + Qv ′

=⇒ Q ′ =u′ − Qv ′

v=

u′

v− u

v· v ′

v

=⇒ Q ′ =(u

v

)′=

u′v − uv ′

v2

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 =u

v. Then

u = Qv

If Q is differentiable, we have

u′ = (Qv)′ = Q ′v + Qv ′

=⇒ Q ′ =u′ − Qv ′

v=

u′

v− u

v· v ′

v

=⇒ Q ′ =(u

v

)′=

u′v − uv ′

v2

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 =u

v. Then

u = Qv

If Q is differentiable, we have

u′ = (Qv)′ = Q ′v + Qv ′

=⇒ Q ′ =u′ − Qv ′

v=

u′

v− u

v· v ′

v

=⇒ Q ′ =(u

v

)′=

u′v − uv ′

v2

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 =u

v. Then

u = Qv

If Q is differentiable, we have

u′ = (Qv)′ = Q ′v + Qv ′

=⇒ Q ′ =u′ − Qv ′

v=

u′

v− u

v· v ′

v

=⇒ Q ′ =(u

v

)′=

u′v − uv ′

v2

This is called the Quotient Rule.

Verifying Example

Example

Verify the quotient rule by computingd

dx

(x2

x

)and comparing it

tod

dx(x).

Solution

d

dx

(x2

x

)=

x ddx

(x2

)− x2 d

dx (x)

x2

=x · 2x − x2 · 1

x2

=x2

x2= 1 =

d

dx(x)

Verifying Example

Example

Verify the quotient rule by computingd

dx

(x2

x

)and comparing it

tod

dx(x).

Solution

d

dx

(x2

x

)=

x ddx

(x2

)− x2 d

dx (x)

x2

=x · 2x − x2 · 1

x2

=x2

x2= 1 =

d

dx(x)

Examples

Example

1.d

dx

2x + 5

3x − 2

2.d

dx

2x + 1

x2 − 1

3.d

dt

t − 1

t2 + t + 2

Answers

1. − 19

(3x − 2)2

2. −2

(x2 + x + 1

)(x2 − 1)2

3.−t2 + 2t + 3

(t2 + t + 2)2

Solution to first example

d

dx

2x + 5

3x − 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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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 example

d

dx

2x + 5

3x − 2=

(3x − 2) ddx (2x + 5)− (2x + 5) d

dx (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.d

dx

2x + 5

3x − 2

2.d

dx

2x + 1

x2 − 1

3.d

dt

t − 1

t2 + t + 2

Answers

1. − 19

(3x − 2)2

2. −2

(x2 + x + 1

)(x2 − 1)2

3.−t2 + 2t + 3

(t2 + t + 2)2

Solution to second example

d

dx

2x + 1

x2 − 1

=(x2 − 1)(2)− (2x + 1)(2x)

(x2 − 1)2

=(2x2 − 2)− (4x2 + 2x)

(x2 − 1)2

= −2

(x2 + x + 1

)(x2 − 1)2

Solution to second example

d

dx

2x + 1

x2 − 1=

(x2 − 1)(2)− (2x + 1)(2x)

(x2 − 1)2

=(2x2 − 2)− (4x2 + 2x)

(x2 − 1)2

= −2

(x2 + x + 1

)(x2 − 1)2

Solution to second example

d

dx

2x + 1

x2 − 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.d

dx

2x + 5

3x − 2

2.d

dx

2x + 1

x2 − 1

3.d

dt

t − 1

t2 + t + 2

Answers

1. − 19

(3x − 2)2

2. −2

(x2 + x + 1

)(x2 − 1)2

3.−t2 + 2t + 3

(t2 + t + 2)2

Solution to third example

d

dt

t − 1

t2 + 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

Solution to third example

d

dt

t − 1

t2 + 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

Solution to third example

d

dt

t − 1

t2 + 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.d

dx

2x + 5

3x − 2

2.d

dx

2x + 1

x2 − 1

3.d

dt

t − 1

t2 + 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(u

v

)′=

vu′ − uv ′

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

Outline

The Product RuleDerivation of the product ruleExamples

The Quotient RuleDerivationExamples

More derivatives of trigonometric functionsDerivative of TangentDerivative of CotangentDerivative of SecantDerivative of Cosecant

More on the Power RulePower Rule for Positive Integers by InductionPower Rule for Negative Integers

Derivative of Tangent

Example

Findd

dxtan x

Solution

d

dxtan x =

d

dx

(sin x

cos x

)

=cos x · sin x − sin x · (− sin x)

cos2 x

=cos2 x + sin2 x

cos2 x=

1

cos2 x= sec2 x

Derivative of Tangent

Example

Findd

dxtan x

Solution

d

dxtan x =

d

dx

(sin x

cos x

)

=cos x · sin x − sin x · (− sin x)

cos2 x

=cos2 x + sin2 x

cos2 x=

1

cos2 x= sec2 x

Derivative of Tangent

Example

Findd

dxtan x

Solution

d

dxtan x =

d

dx

(sin x

cos x

)=

cos x · sin x − sin x · (− sin x)

cos2 x

=cos2 x + sin2 x

cos2 x=

1

cos2 x= sec2 x

Derivative of Tangent

Example

Findd

dxtan x

Solution

d

dxtan x =

d

dx

(sin x

cos x

)=

cos x · sin x − sin x · (− sin x)

cos2 x

=cos2 x + sin2 x

cos2 x

=1

cos2 x= sec2 x

Derivative of Tangent

Example

Findd

dxtan x

Solution

d

dxtan x =

d

dx

(sin x

cos x

)=

cos x · sin x − sin x · (− sin x)

cos2 x

=cos2 x + sin2 x

cos2 x=

1

cos2 x

= sec2 x

Derivative of Tangent

Example

Findd

dxtan x

Solution

d

dxtan x =

d

dx

(sin x

cos x

)=

cos x · sin x − sin x · (− sin x)

cos2 x

=cos2 x + sin2 x

cos2 x=

1

cos2 x= sec2 x

Derivative of Cotangent

Example

Findd

dxcot x

Answer

d

dxcot x = − 1

sin2 x= − csc2 x

Derivative of Cotangent

Example

Findd

dxcot x

Answer

d

dxcot x = − 1

sin2 x= − csc2 x

Derivative of Secant

Example

Findd

dxsec x

Solution

d

dxsec x =

d

dx

(1

cos x

)

=cos x · 0− 1 · (− sin x)

cos2 x

=sin x

cos2 x=

1

cos x· sin x

cos x= sec x tan x

Derivative of Secant

Example

Findd

dxsec x

Solution

d

dxsec x =

d

dx

(1

cos x

)

=cos x · 0− 1 · (− sin x)

cos2 x

=sin x

cos2 x=

1

cos x· sin x

cos x= sec x tan x

Derivative of Secant

Example

Findd

dxsec x

Solution

d

dxsec x =

d

dx

(1

cos x

)=

cos x · 0− 1 · (− sin x)

cos2 x

=sin x

cos2 x=

1

cos x· sin x

cos x= sec x tan x

Derivative of Secant

Example

Findd

dxsec x

Solution

d

dxsec x =

d

dx

(1

cos x

)=

cos x · 0− 1 · (− sin x)

cos2 x

=sin x

cos2 x

=1

cos x· sin x

cos x= sec x tan x

Derivative of Secant

Example

Findd

dxsec x

Solution

d

dxsec x =

d

dx

(1

cos x

)=

cos x · 0− 1 · (− sin x)

cos2 x

=sin x

cos2 x=

1

cos x· sin x

cos x

= sec x tan x

Derivative of Secant

Example

Findd

dxsec x

Solution

d

dxsec x =

d

dx

(1

cos x

)=

cos x · 0− 1 · (− sin x)

cos2 x

=sin x

cos2 x=

1

cos x· sin x

cos x= sec x tan x

Derivative of Cosecant

Example

Findd

dxcsc x

Answer

d

dxcsc x = − csc x cot x

Derivative of Cosecant

Example

Findd

dxcsc x

Answer

d

dxcsc 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 pairsfollow similar patterns,with functions andco-functions switchedand an extra sign.

Outline

The Product RuleDerivation of the product ruleExamples

The Quotient RuleDerivationExamples

More derivatives of trigonometric functionsDerivative of TangentDerivative of CotangentDerivative of SecantDerivative of Cosecant

More on the Power RulePower Rule for Positive Integers by InductionPower Rule for Negative Integers

Power Rule for Positive Integers by Induction

TheoremLet n be a positive integer. Then

d

dxxn = nxn−1

Proof.By induction on n. We can show it to be true for n = 1 directly.

Suppose for some n thatd

dxxn = nxn−1. Then

d

dxxn+1

=d

dx(x · xn)

=

(d

dxx

)xn + x

(d

dxxn

)= 1 · xn + x · nxn−1 = (n + 1)xn

Power Rule for Positive Integers by Induction

TheoremLet n be a positive integer. Then

d

dxxn = nxn−1

Proof.By induction on n.

We can show it to be true for n = 1 directly.

Suppose for some n thatd

dxxn = nxn−1. Then

d

dxxn+1

=d

dx(x · xn)

=

(d

dxx

)xn + x

(d

dxxn

)= 1 · xn + x · nxn−1 = (n + 1)xn

Principle of Mathematical Induction

Suppose S(1) istrue and S(n + 1)is true when-ever S(n) is true.Then S(n) is truefor all n.

Image credit: Kool Skatkat

Power Rule for Positive Integers by Induction

TheoremLet n be a positive integer. Then

d

dxxn = nxn−1

Proof.By induction on n. We can show it to be true for n = 1 directly.

Suppose for some n thatd

dxxn = nxn−1. Then

d

dxxn+1

=d

dx(x · xn)

=

(d

dxx

)xn + x

(d

dxxn

)= 1 · xn + x · nxn−1 = (n + 1)xn

Power Rule for Positive Integers by Induction

TheoremLet n be a positive integer. Then

d

dxxn = nxn−1

Proof.By induction on n. We can show it to be true for n = 1 directly.

Suppose for some n thatd

dxxn = nxn−1. Then

d

dxxn+1 =

d

dx(x · xn)

=

(d

dxx

)xn + x

(d

dxxn

)= 1 · xn + x · nxn−1 = (n + 1)xn

Power Rule for Positive Integers by Induction

TheoremLet n be a positive integer. Then

d

dxxn = nxn−1

Proof.By induction on n. We can show it to be true for n = 1 directly.

Suppose for some n thatd

dxxn = nxn−1. Then

d

dxxn+1 =

d

dx(x · xn)

=

(d

dxx

)xn + x

(d

dxxn

)

= 1 · xn + x · nxn−1 = (n + 1)xn

Power Rule for Positive Integers by Induction

TheoremLet n be a positive integer. Then

d

dxxn = nxn−1

Proof.By induction on n. We can show it to be true for n = 1 directly.

Suppose for some n thatd

dxxn = nxn−1. Then

d

dxxn+1 =

d

dx(x · xn)

=

(d

dxx

)xn + x

(d

dxxn

)= 1 · xn + x · nxn−1 = (n + 1)xn

Power Rule for Negative Integers

Use the quotient rule to prove

Theorem

d

dxx−n = (−n)x−n−1

for positive integers n.

Proof.

d

dxx−n =

d

dx

1

xn

=xn · d

dx 1− 1 · ddx xn

x2n

=0− nxn−1

x2n

= −nx−n−1

Power Rule for Negative Integers

Use the quotient rule to prove

Theorem

d

dxx−n = (−n)x−n−1

for positive integers n.

Proof.

d

dxx−n =

d

dx

1

xn

=xn · d

dx 1− 1 · ddx xn

x2n

=0− nxn−1

x2n

= −nx−n−1

Power Rule for Negative Integers

Use the quotient rule to prove

Theorem

d

dxx−n = (−n)x−n−1

for positive integers n.

Proof.

d

dxx−n =

d

dx

1

xn

=xn · d

dx 1− 1 · ddx xn

x2n

=0− nxn−1

x2n

= −nx−n−1

Power Rule for Negative Integers

Use the quotient rule to prove

Theorem

d

dxx−n = (−n)x−n−1

for positive integers n.

Proof.

d

dxx−n =

d

dx

1

xn

=xn · d

dx 1− 1 · ddx xn

x2n

=0− nxn−1

x2n

= −nx−n−1

Power Rule for Negative Integers

Use the quotient rule to prove

Theorem

d

dxx−n = (−n)x−n−1

for positive integers n.

Proof.

d

dxx−n =

d

dx

1

xn

=xn · d

dx 1− 1 · ddx xn

x2n

=0− nxn−1

x2n= −nx−n−1