CHAPTER 20 The Product and Quotient Rules

[Pages:10]CHAPTER 20

The Product and Quotient Rules

We have developed rules for the derivatives of the sum or di erence of two functions that work as follows

h

i

d

+

= 0 +0

f (x) g(x) f (x) g (x)

dx

h

i

d

?

= 0 ?0

f (x) g(x) f (x) g (x).

dx

But what about the derivative of product of two functions, or the quotient of two functions? This chapter answers these questions by deriving two new rules for

h

i

d

? f (x) g(x)

and

dx

d f (x)

. dx g(x)

These two new rules will be called the product rule and the quotient rule,

respectively.

Let's begin by deriving the product rule. Given two functionsh f (x) andi g(x), we aim to work out the derivative of their product, that is, d f (x)g(x) .

dx

By Definition 16.1, the derivative of a function F(x) is

hi

?

d

= F(z) F(x)

F(x) dx

li!m

zx

? zx

.

We are interested in the case where F(x) = f (x)g(x), which is

h

i

?

d

= f (z)g(z) f (x)g(x)

f (x)g(x) li!m

dx

zx

? zx

.

We will now work this limit out carefully. In this limit, the denominator

? z

x

approaches

zero,

so

we

have

to

get

rid

of

it

somehow.

In

the

following

computation

it

gets

absorbed

into

the

definition

of

0

f (x)

and

0

g(

x).

242

The Product and Quotient Rules

So let us begin our computation. As noted above, our first step is

h

i

?

d

= f (z)g(z) f (x)g(x)

f (x)g(x) li!m

dx

zx

? zx

.

Let's insert a little space int this expression to give ourselves room to work.

h

i

?

+

?

d

= f (z)g(z) f (x)g(z) f (x)g(z) f (x)g(x)

f (x)g(x) li!m

dx

zx

? zx

.

To

the

space

just

created,

add

zero

in

the

form

of

0

=

? f

+ (x) g( z)

f

( x) g( z).

This is an allowable step because adding in 0 doesn't alter the limit's value.

h

i

?

+

?

d

= f (z)g(z) f (x)g(z) f (x)g(z) f (x)g(x)

f (x)g(x) li!m

dx

zx

z?x

.

In the numerator, factor out g(z) from the first two terms, and f (x) from the second two, as shown below. Then split the fraction and apply limit laws:

h

i

?

?

?

?

?

+

?

d

=

f (z) f (x) g(z) f (x) g(z) g(x)

f (x)g(x) li!m

dx

z x?

?

? zx

?

=

f (z) f (x) + g(z) g(x)

li!m

zx

? g(z) f (x) ?

zx

zx

?

?

= f (z) f (x) ?

+

? g(z) g(x)

li!m

zx

? zx

li!m g(z) li!m f (x) li!m

zx

zx

zx

? zx

=0

+

0

f (x)g(x) f (x)g (x) .

In the last step we used the facts that

li!m

? f (z) f (x)

? zx

=

f 0(x),

li!m g(z) = g(x),

li!m f (x) = f (x), and

li!m

? g(z) g(x)

? zx

=

g0(x).

z

(We

x

assume

here

that

zx 0

g (x)

exists,

so,

zx

zx

by Theorem 18.1, is continuous, and hence

= .)

g(x) We have just determined that

d

? f

? (x) g(x)

lzi!mx g(z) g(x)

=

f

0

+

(x) g(x)

f

(x)

0

g

(

x).

Now

dx

that we have done this computation--and we believe it--we will never have

to do it again. It becomes our latest rule.

h

i

Rule 7 (The Product Rule)

d

=0

+

0

f (x)g(x) f (x)g(x) f (x)g (x)

dx

The derivative of a product equals the derivative of the first function times

the second function, plus the first function times the derivative of the second:

h

i

hi

hi

d

=d f (x)g(x)

? + ?d f (x) g(x) f (x)

g(x) .

dx

dx

dx

In applying the product rule to f (x)g(x), you also have to do the derivatives of f and g, using whatever rules apply to them.

243

Example 20.1

Find

the

derivative

of

3

4x

ex.

This

is

a

product

3

(4x )

?

x

(e )

of

two

functions,

so

we

use

the

product

rule.

hi

??

??

d 3 x = d 3 ? x+ 3? d x

4x e

4x e 4x

e

dx

dx

dx

=

2? x+ 3? x

12x e 4x e

?

?

=

x 2+ 3

4e 3x x .

?

??

?

Example 20.2

Find the derivative of y =

2+ x3

2

x

+? 5x 7

.

This is a product of two functions, so we use the product rule.

h?

??

?i

h

i?

??

?h

i

d 2+ 2+ ?

= d 2+ ? 2+ ? + 2+ ? d 2+ ?

x 3 x 5x 7

x 3 x 5x 7 x 3

x 5x 7

dx

dx

dx

?

?? ?

=

2+ ? + 2+

+

2x x 5x 7 x 3 (2x 5)

This is the derivative, but some simplification is possible:

= 3+ 2?

+ 3+ 2+ +

2x 10x 14x 2x 5x 6x 15

= 3+ 2? + 4x 15x 8x 15 .

Here our approach was to take the derivative (using the product rule) and then simplify. Alternatively, you could first simplify (by multiplying) and then take the derivative:

h?

??

?i

h

i

d 2+ 2+ ?

= d 4+ 3? 2 + 2+ ?

x 3 x 5x 7

x 5x 7x 3x 15x 21

dx

dx

= 3+ 2? + + 4x 15x 14x 6x 15

= 3+ 2? + 4x 15x 8x 15 .

Example 20.3 Very often you'll have a choice of rules, and one choice

may lead to a simpler computation than the other. Consider finding the

derivative

of

y

=

5

3x

.

This

is

a

product,

so

you

could

use

the

product

rule:

hi

??

??

d 5 = d ? 5+ ? d 5 = ? 5+ ? 4 =

4

3x

3x 3 x

0 x 3 5x

15x .

dx

dx

dx

But it's much easier to just use the constant multiple rule:

hi

hi

d 5 = d 5 = ? 4=

4

3x 3 x 3 5x 15x .

dx

dx

244

The Product and Quotient Rules

hi Next we derive the rule for d f (x) . (You may opt to skip the derivation

dx g(x)

on a first reading and go straight to the conclusion at the bottom of this

page.) Our computation begins with the definition of the derivative and

proceed by adding zero in a clever way, as we did for the product rule.

f (z) ? f (x)

d f (x) =

g(z) g(x)

dx g(x)

li!m z x?

? zx

?

(definition of derivative)

f (z) ? f (x) ? f (x) ? f (x)

=

g(z) g(z) g(x) g(z)

li!m

zx

? zx

?

?

(insert space)

f (z) ? f (x) ? f (x) ? f (x)

=

g(z) g(z) g(x) g(z)

li!m

zx

? zx

?

?

?

?

1?

1?1

f (z) f (x)

f (x)

=

g(z)

g(x) g(z)

li!m

zx

? zx

?

?

??

?

1?

g(z) g(x)

f (z) f (x)

f (x)

= li!m

zx

g(z) ?

zx

g(x) g( z)

(add zero) (factor)

(combine fractions)

?

?

?

?

?

1 ? f (x)

?

f (z) f (x)

g(z) g(x)

=

g(z) g(x)g(z)

li!m

zx

? zx

??

?

=

f (z) f (x) 1 ? f (x) g(z) g(x)

li!m

zx

?

?

z x g(z) g(x)g(z) z x

(regroup) (split fraction)

?

?

=

f (z) f (x) ?

1?

f (x) ? g(z) g(x)

li!m

zx

? zx

li!m

li!m

li!m

z x g(z) z x g(x)g(z) z x

? zx

(limit laws)

= 0 1 ? f (x) 0

f (x)

g (x)

g(x) g(x)g(x)

0

0

= f (x)g(x) ? f (x)g (x)

g(x)g(x) g(x)g(x)

0

?

0

= f (x)g(x) f (x)g (x)

.

2

g(x)

(evaluate limits) (get common denominator)

(combine fractions)

We have our new rule.

Rule 8 (The Quotient Rule)

0

?

0

d f (x) = f (x)g(x) f (x)g (x)

dx g(x)

2

g(x)

245

Compare the di erences and similarities between the product rule and

the quotient rule. There are similarities, but quotient rule is more complex.

It has a minus instead of a plus, and you must divide by g(x)2.

h

i

hi

hi

Product Rule:

d

=d

? + ?d

f (x)g(x)

f (x) g(x) f (x) g(x)

dx

dx

dx

Quotient Rule:

d f (x)

dx g(x)

hi

hi

d

? ? ?d

f (x) g(x) f (x) g(x)

= dx

? ? dx

2

g(x)

Example 20.4

Find the derivative of

2

x

+

1

.

2?

xx

This function is a quotient, so we apply the quotient rule to find its derivative.

2+

h

i?

??

?h

i

d 2+ 2? ? 2+ d 2?

x 1x x x 1 x x

d x 1 = dx 2?

dx x x

?

?

?

dx

2? 2

?x ?x ?

+ 2? ? 2+

?

= (2x 0) x

x ?

x 1 (2x 1) ?

2? 2

xx

This is the derivative, but a few extra steps of algebra simplify our answer.

?

??

?

3? 2 ? 3? 2+ ?

=

2x

2x

2x x ??

2x 1

2? 2

xx

? 2? +

=

x?

2x? 2? 2

1

.

xx

5

Example 20.5 Find the derivative of x .

x

e

This function is a quotient, so we apply the quotient rule to find its derivative.

5

dx =

x

dx e

hi

hi

d

5

x?

d

5

x

xe x e

dx

dx

x2

(e )

4x? 5x = 5x e x e

.

xx

ee

This is the derivative, but some simplification is possible:

4x ?

4?

= x e (5 x) = x (5 x)

.

xx

x

ee

e

246

The Product and Quotient Rules

Example 20.6 The quotient rule can be computationally expensive, so

don't use it if you don't have to. As an example, consider di erentiating

2

3x

+4

x?5

.

This is a quotient, so you could use the quotient rule.

But the

2

denominator is constant, so a better choice would be to factor it out using

the constant multiple rule:

2+ ?

h

i

d 3x 4x 5 = 1 ? d 2 + ? = 1 + = +

3x 4x 5

(6x 4) 3x 2 .

dx

2

2 dx

2

If you used the quotient rule, your work would go like this:

2+ ?

h

i?

? hi

d 2+ ? ? + 2+ ? d

3x 4x 5 2 3x 4x 5 2

d 3x 4x 5 = dx

dx

dx

2

2

?

2?

+ ? + 2+ ? ?

= (6x 4) 2 3x 4x 5 0

4 +? = (6x 4) 2 = +

3x 2 .

4

Though it gave the same answer, the quotient rule was an overkill. Work enough exercises that you see your choices and their consequences.

Example 20.7

Find the derivative of ?10. x

We will do this two ways. First, we can simply apply the power rule.

hi

d

? 10

=?

?? 10 1

=

?

? 11

x

10x

10x .

dx

Our second step uses the quotient rule.

hi

d

? 10

=

d

1

x

10

dx

dx x

hi

hi

d

10 ? ? d 10

1x 1 x

= dx

? ? dx

2 10

x

? 10 ? 9 = 0 x 10x = ?

? 9 20

=

10x

20

x

?

? 11

10x .

Besides reminding us that a problem can be done several ways, this

example raises an important point.

In deriving the power rule

d

n=

? n1

[x ] nx

dx

in Chapter 17, we proved it only for positive integer values of . But we

n

asserted then that it actually holds for all real values of n, positive or

negative, and that that we would see this in due time. Example 20.7 suggests

that we can use the quotient rule to show that the power rule holds for

negative integer values of n. Exercise 13 below asks for the details.

247

?

?

Example 20.8

Find

all

x

for

which

the

tangent

to

y

=

x

xe

at

x, f (x)

has

slope 0.

Solution: The slope of the tangent line is given by the derivative, so as in

the

previous

example

our

first

task

is

to

find

the

derivative

of

= y

xex.

This requires the product rule:

y

=x y xe

dy = ? x+ ? x = x + 1 e x e e (1 x).

dx

This means the tangent has slope 0

where

x+ e (1 x)

=

0.

Because

x

e

>

0

for all x, the only way we can have

x+ e (1

x)

=

0

is

if

x

=

?1.

Answer:

The

tangent

to

=x y xe

has

slope 0 at x = ?1. (See the graph on

the right.)

? 1 x

Exercises for Chapter 20

In problems 1?12 find the indicated derivative.

h?

??

?i

h pi

1. d

4+

2?

2x 3x 3x x

2.

d

x

e

x

dx hi

dx hi

3.

d

x

xe

4.

d

2x

e

(Hint:

2x

e

=

x

e

x

e

.)

5.

dx 2 + dx x

+

dx x 5

7.

x+ d ex

3+ 2+

dx x x 1

9. d

x 3+ 2+

dx x x 1

6.

dx 2 + ? d x 3px 4

dx

+ x5

8.

3+ 2+ dx x1

dx

x

10.

+ d x1

?

dx x 1

11. d

1 2+

dx x 1

x

12. d pe

dx x

13.

In Chapter 17 we proved that the power rule

d

n= [x ]

? n1

nx

works

for

positive

integer values of . Combine this fact with thedxquotient rule to show that the n

power rule also holds for negative integer values of n. (See Example 20.7.)

p

14.

Find

the

equation

of

the

tangent

line

to

the

graph

of

f

(x)

=

x

e

x at point (1, f (1)).

15.

Find the equation of the tangent line to the graph of

= p1 f (x)

at point (4, f (4)).

x

16.

Find

all

points

(x,

y)

on

the

graph

of

y

=

+ x

1 ?

where the tangent has slope 0.

x3

248

The Product and Quotient Rules

17.

Find all points (x, y) on the

graph of

= y

x

where the tangent line is horizontal.

x

e

18. Two functions f (x) and g(x) are graphed below.

Estimate

0

h (1.5)

if:

(a) h(x) = f (x)g(x).

(b)

= h(x)

f

(x) .

g(x)

(c)

= h(x)

g(x) .

f (x)

4

3

2

1

? 4

? 3

? 2

??1

y

=

f

1

(x?)

2

? 3

? 4

y

x

1234

4

3

2

1

? 4

? 3

? 2

??1

1

?2

=

? 3

y g(x)

?

4

y

x

1234

19. A function f (x) is graphed below.

Find

0

g (3)

if:

4y

3

= y f (x)

(a)

g(x)

=

2

x

f

(x).

2

(b)

= g(x)

x

.

?f (x) ?

(c)

= g(x)

f (x) 2.

2

1

? 7

? 6

? 5

? 4

? 3

? 2

??1

1

? 2

? 3

? 4

x

1234567

20.

Information about

f (x),

g(x),

f

0

(x)

and

0

g (x)

is

given

in

the

table

below.

Find

0

h (3)

if:

?

?

(a)

= h(x)

+3 xx

f (x).

x

0

123 45

? f (x) 4

? 2

01

10

0

f (x)

2

1

1

3

0.5

? 1

(b)

= h(x)

g(x) .

g(x) 10

9

74

0

? 4

f (x) (c) h(x) = f (x)g(x).

0

g (x)

0

? 0.5

? 1

? 3

? 4

? 4

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