CHAPTER 20 The Product and Quotient Rules

CHAPTER 20

The Product and Quotient Rules

e have developed rules for the derivatives of the sum or di erence of

W two functions that work as follows

h

i

+

= 0 +0

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

h

i

?

= 0 ?0

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

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

? Dx f (x) g(x) and

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, Dx f (x)g(x) .

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

hi

?

= F(z) F(x)

Dx F(x)

li!m

zx

? zx

.

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

h

i

?

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

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

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

0

0

computation it gets absorbed into the definition of f (x) and g (x).

244

The Product and Quotient Rules

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

h

i

?

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

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

zx

? zx

.

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

h

i

?

+

?

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

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

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

?

+

?

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

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

zx

? zx

.

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

?

?

?

?

?

+

?

=

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

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

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) .

Ilzi!nmxtfh(ex)l=asft(xs)t,eapnwd elziu!msxegd(zz)t??hxge(x)fa=cgts0(xt)h. a(tWelzi!maxssf (uzz)m??xfe(xh) e=ref

0

(x),

that

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

zx 0

g (x) exists, so,

=

by

Theorem

18.1,

g(x)

is

continuous,

a?nd

hence?

li!m g(z)

z

=

x0

g(x).) +

0

We have just determined that Dx f (x)g(x) f (x)g(x) f (x)g (x). Now

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)

=0

+

0

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

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

=

?+?

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

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.

245

Example 20.1

3x

Find the derivative of 4x e .

3? x This is a product (4x ) (e ) of two functions, so we use the product rule.

h

i

??

??

3x =

3 ? x+ 3?

x

Dx 4x e

Dx 4x e 4x Dx e

=

2? x+ 3? x

12x e 4x e

?

?

=

x 2+ 3

4e 3x x .

?

??

?

Example 20.2

= 2+ Find the derivative of y x 3

2+ ? x 5x 7 .

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

h?

??

?i

h

i?

??

?h

i

2+ 2+ ?

=

2+ ? 2+ ? + 2+ ?

2+ ?

Dx x 3 x 5x 7

Dx x 3 x 5x 7 x 3 Dx x 5x 7

?

?? ?

=

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

2+ 2+ ?

=

4+ 3? 2 + 2+ ?

Dx x 3 x 5x 7

Dx x 5x 7x 3x 15x 21

= 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 =5

derivative of y 3x . This is a product, so you could use the product rule:

hi

??

??

5=

? 5+ ?

5 = ? 5+ ? 4 =

4

Dx 3x

Dx 3 x 3 Dx x

0 x 3 5x

15x .

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

hi

hi

5=

5 = ? 4=

4

Dx 3x

3Dx x

3 5x

15x .

246

The Product and Quotient Rules

hi

f (x)

Next we derive the rule for Dx

. (You may opt to skip the derivation

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)

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

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

Dx

2

g(x)

g(x)

247

Compare the di erences and similarities between the product rule and

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

2

It has a minus insteadh of a pluis, and yhou mi ust divide by gh(x) . i

=

?+?

Product Rule: Dx f (x)g(x) Dx f (x) g(x) f (x) Dx g(x)

f (x)

Quotient Rule: Dx g(x)

hi

hi

???

=

Dx f (x)

g(x) ?

f (x) Dx g(x) ?

2

g(x)

Example 20.4

2+ x1 Find the derivative of 2 ? .

xx

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

2+ x1

Dx 2 ? xx

=

h

i?

??

?h

i

2+ 2? ? 2+

2?

Dx x 1 x x x 1 Dx x x

?

?

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?

1 .

2? 2

xx

5

Example 20.5

x Find the derivative of .

x

e

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

hi

hi

5 x? 5

x

5

x

= Dx x e

x Dx e

Dx

x

e

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

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