Section 3.2The Product and Quotient Rules

Math 1a

February 22, 2008

Announcements

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I Midterm I Friday 2/29 in class (up to §3.2)

Outline

The Product RuleDerivation of the product ruleExamples

The Quotient RuleDerivationExamples

More on the Power RulePower Rule for nonnegative integers by inductionPower 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 it u′v ′?

Is the derivative of a product the product of thederivatives?

NO!

Try this with u = x and v = x2. Then uv = x3, so (uv)′ = 3x2.But u′v ′ = 1(2x) = 2x .So we have to be more careful.

Is the derivative of a product the product of thederivatives?

NO!Try this with u = x and v = x2.

Then uv = x3, so (uv)′ = 3x2.But u′v ′ = 1(2x) = 2x .So we have to be more careful.

Is the derivative of a product the product of thederivatives?

NO!Try this with u = x and v = x2. Then uv = x3, so (uv)′ = 3x2.But u′v ′ = 1(2x) = 2x .

So we have to be more careful.

Is the derivative of a product the product of thederivatives?

NO!Try this with u = x and v = x2. Then uv = x3, so (uv)′ = 3x2.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 FOIL and then by theproduct rule:

d

dx

[(3− x2)(x3 − x + 1)

]

Solution

(i) by FOIL:

d

dx

[(3− x2)(x3 − x + 1)

] FOIL=

d

dx

[−x5 + 4x3 − x2 − 3x + 3

]= −5x4 + 12x2 − 2x − 3

(ii) by 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 FOIL and then by theproduct rule:

d

dx

[(3− x2)(x3 − x + 1)

]Solution

(i) by FOIL:

d

dx

[(3− x2)(x3 − x + 1)

] FOIL=

d

dx

[−x5 + 4x3 − x2 − 3x + 3

]= −5x4 + 12x2 − 2x − 3

(ii) by 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

dxxex

Answery ′ = ex + xex

One more

Example

Findd

dxxex

Answery ′ = ex + xex

Mnemonic

Let u = “ho” and v = “hi”. Then

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

Outline

The Product RuleDerivation of the product ruleExamples

The Quotient RuleDerivationExamples

More on the Power RulePower Rule for nonnegative integers by inductionPower Rule for negative integers

The Quotient Rule

What about the derivative of a quotient?

Let u and v be differentiable 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− uv ′

v2

=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 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− uv ′

v2

=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 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− uv ′

v2

=u′v − uv ′

v2

This is called the Quotient Rule.

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

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

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

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

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 on the Power RulePower Rule for nonnegative integers by inductionPower Rule for negative integers

Power Rule for nonnegative integers by induction

TheoremLet n be a positive integer. Then

d

dxxn = nxn−1

Proof.By induction on n. We have shown it to be true for n = 1.

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 nonnegative integers by induction

TheoremLet n be a positive integer. Then

d

dxxn = nxn−1

Proof.By induction on n. We have shown it to be true for n = 1.

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