Lesson 12 (Section 3.2)The Product and Quotient Rules

Math 1a

October 22, 2007

Announcements

I Midterm I: 10/24, 7-9pm. A-G in Hall A, H-Z in Hall C

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 in your hourly wages and work 5 hoursmore per week. How much extra money do you make?

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 in your hourly wages and work 5 hoursmore per week. How much extra money do you make?

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 in your hourly wages and work 5 hoursmore per week. How much extra money do you make?

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 in your hourly wages and work 5 hoursmore per week. How much extra money do you make?

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 over time. How does incomechange?

∆I

∆t=

w ∆h + h ∆w + ∆w ∆h

∆t

= w∆h

∆t+ h

∆w

∆t+ ∆w

∆h

∆t

→ wdh

dt+ h

dw

dt+ 0

as t → 0.

Theorem (The Product Rule)

Let u and v be differentiable at x. Then

(uv)′(x) = u(x)v ′(x) + u′(x)v(x)

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Supose wages and hours are changing over time. How does incomechange?

∆I

∆t=

w ∆h + h ∆w + ∆w ∆h

∆t

= w∆h

∆t+ h

∆w

∆t+ ∆w

∆h

∆t

→ wdh

dt+ h

dw

dt+ 0

as t → 0.

Theorem (The Product Rule)

Let u and v be differentiable at x. Then

(uv)′(x) = u(x)v ′(x) + u′(x)v(x)

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Example

Find this derivative two ways: first by FOIL and then by theproduct rule:

d

dx(3− x2)(x3 − x + 1).

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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

Example (Quadratic Tangent to identity function)

The curve y = ax2 + bx + c passes through the point (1, 2) and istangent to the line y = x at the point (0, 0). Find a, b, and c .

Answera = 1, b = 1, c = 0.

Example (Quadratic Tangent to identity function)

The curve y = ax2 + bx + c passes through the point (1, 2) and istangent to the line y = x at the point (0, 0). Find a, b, and c .

Answera = 1, b = 1, c = 0.

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.

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