3DIFFERENTIATION RULESDIFFERENTIATION RULES

3.2The Product and

Quotient Rules

DIFFERENTIATION RULES

In this section, we will learn about:

Formulas that enable us to differentiate new functions

formed from old functions by multiplication or division.

By analogy with the Sum and Difference

Rules, one might be tempted to guess—as

Leibniz did three centuries ago—that the

derivative of a product is the product of the

derivatives.

However, we can see that this guess is wrong by looking at a particular example.

THE PRODUCT RULE

Let f(x) = x and g(x) = x2.

Then, the Power Rule gives f’(x) = 1 and g’(x) = 2x.

However, (fg)(x) = x3.

So, (fg)’(x) =3 x2.

Thus, (fg)’ ≠ f’ g’.

THE PRODUCT RULE

The correct formula was discovered by

Leibniz (soon after his false start) and is

called the Product Rule.

THE PRODUCT RULE

Before stating the Product Rule, let’s see

how we might discover it.

We start by assuming that u = f(x) and

v = g(x) are both positive differentiable functions.

THE PRODUCT RULE

Then, we can interpret the product uv

as an area of a rectangle.

THE PRODUCT RULE

If x changes by an amount Δx, then

the corresponding changes in u and v

are: Δu = f(x + Δx) - f(x) Δv = g(x + Δx) - g(x)

THE PRODUCT RULE

The new value of the product, (u + Δu)

(v + Δv), can be interpreted as the area of

the large rectangle in the figure—provided

that Δu and Δv happen to be positive.

THE PRODUCT RULE

The change in the area of the rectangle is:

( ) ( )( )

the sum of the three shaded areas

uv u u v v uv

u v v u u v

Δ = + Δ + Δ −= Δ + Δ + Δ Δ=

THE PRODUCT RULE Equation 1

If we divide by ∆x, we get:

( )uv v u vu v u

x x x x

Δ Δ Δ Δ= + +Δ

Δ Δ Δ Δ

THE PRODUCT RULE

If we let ∆x → 0, we get the derivative of uv:

( )

0

0

0 0 0 0

( )( ) lim

lim

lim lim lim lim

0.

x

x

x x x x

d uvuv

dx xv u v

u v ux x x

v u vu v u

x x x

dv du dvu vdx dx dx

Δ →

Δ →

Δ → Δ → Δ → Δ →

Δ=

ΔΔ Δ Δ⎛ ⎞= + +Δ⎜ ⎟Δ Δ Δ⎝ ⎠Δ Δ Δ⎛ ⎞= + + Δ ⎜ ⎟Δ Δ Δ⎝ ⎠

= + +

THE PRODUCT RULE

Notice that ∆u → 0 as ∆x → 0 since f is differentiable and therefore continuous.

( )d dv duuv u v

dx dx dx= +

THE PRODUCT RULE Equation 2

Though we began by assuming (for the

geometric interpretation) that all quantities are

positive, we see Equation 1 is always true.

The algebra is valid whether u, v, ∆u, and ∆v are positive or negative.

So, we have proved Equation 2—known as the Product Rule—for all differentiable functions u and v.

THE PRODUCT RULE

If f and g are both differentiable, then:

In words, the Product Rule says: The derivative of a product of two functions is

the first function times the derivative of the second function plus the second function times the derivative of the first function.

[ ] [ ] [ ]( ) ( ) ( ) ( ) ( ) ( )d d df x g x f x g x g x f x

dx dx dx= +

THE PRODUCT RULE

a. If f(x) = xex, find f ’(x).

b. Find the nth derivative, f (n)(x)

THE PRODUCT RULE Example 1

By the Product Rule, we have:

THE PRODUCT RULE Example 1 a

'( ) ( )

( ) ( )

1

( 1)

x

x x

x x

x

df x xe

dxd d

x e e xdx dx

xe e

x e

=

= +

= + ⋅

= +

Using the Product Rule again, we get:

''( ) ( 1)

( 1) ( ) ( 1)

( 1) 1

( 2)

x

x x

x x

x

df x x e

dxd d

x e e xdx dx

x e e

x e

⎡ ⎤= +⎣ ⎦

= + + +

= + + ⋅

= +

THE PRODUCT RULE Example 1 b

Further applications of the Product Rule give:

4

'''( ) ( 3)

( ) ( 4)

x

x

f x x e

f x x e

= +

= +

THE PRODUCT RULE Example 1 b

In fact, each successive differentiation adds another term ex.

So:( ) ( )n xf x x n e= +

THE PRODUCT RULE Example 1 b

Differentiate the functionTHE PRODUCT RULE Example 2

( ) ( )f t t a bt= +

Using the Product Rule, we have:

THE PRODUCT RULE E. g. 2—Solution 1

( )1 21

2

'( ) ( ) ( )

( )

( ) ( 3 )

2 2

d df t t a bt a bt t

dt dt

t b a bt t

a bt a btb t

t t

−

= + + +

= ⋅ + + ⋅

+ += + =

If we first use the laws of exponents to

rewrite f(t), then we can proceed directly

without using the Product Rule.

This is equivalent to the answer in Solution 1.

THE PRODUCT RULE E. g. 2—Solution 2

Example 2 shows that it is sometimes easierto simplify a product of functions than to usethe Product Rule.

In Example 1, however, the Product Rule isthe only possible method.

THE PRODUCT RULE

If , where

g(4) = 2 and g’(4) = 3, find f’(4).

THE PRODUCT RULE Example 3

( ) ( )f x xg x=

Applying the Product Rule, we get:

So,

THE PRODUCT RULE Example 3

[ ]

1 212

'( ) ( ) ( ) ( )

( )'( ) ( ) '( )

2

d d df x xg x x g x g x x

dx dx dxg x

xg x g x x xg xx

−

⎡ ⎤ ⎡ ⎤= = +⎣ ⎦ ⎣ ⎦

= + ⋅ = +

(4) 2'(4) 4 '(4) 2 3 6.5

2 22 4

gf g= + = ⋅ + =

⋅

We find a rule for differentiating the quotient

of two differentiable functions u = f(x) and

v = g(x) in much the same way that we found

the Product Rule.

THE QUOTIENT RULE

If x, u, and v change by amounts Δx, Δu,

and Δv, then the corresponding change

in the quotient u / v is:

THE QUOTIENT RULE

So,

( )

( )

0

0

lim

lim

x

x

u vd u

dx v x

u vv ux x

v v v

Δ →

Δ →

Δ⎛ ⎞=⎜ ⎟ Δ⎝ ⎠Δ Δ−

Δ Δ=+ Δ

THE QUOTIENT RULE

As ∆x → 0, ∆v → 0 also—because v = g(x)

is differentiable and therefore continuous.

Thus, using the Limit Laws, we get:

( )0 0

2

0

lim lim

lim

x x

x

u v du dvv u v u

d u x x dx dxdx v v v v v

Δ → Δ →

Δ →

Δ Δ− −

⎛ ⎞ Δ Δ= =⎜ ⎟ + Δ⎝ ⎠

THE QUOTIENT RULE

If f and g are differentiable, then:

In words, the Quotient Rule says: The derivative of a quotient is the denominator times

the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

[ ] [ ]2

( ) ( ) ( ) ( )( )

( ) ( )

d dg x f x f x g xd f x dx dx

dx g x g x

−⎡ ⎤=⎢ ⎥ ⎡ ⎤⎣ ⎦ ⎣ ⎦

THE QUOTIENT RULE

The Quotient Rule and the other

differentiation formulas enable us to

compute the derivative of any rational

function—as the next example illustrates.

THE QUOTIENT RULE

Let

2

3

2

6

x xy

x

+ −=

+

THE QUOTIENT RULE Example 4

Then,

THE QUOTIENT RULE Example 4

( ) ( ) ( ) ( )( )

( )( ) ( )( )( )

( ) ( )( )

( )

3 2 2 3

23

3 2 2

23

4 3 4 3 2

23

4 3 2

23

6 2 2 6'

6

6 2 1 2 3

6

2 12 6 3 3 6

6

2 6 12 6

6

d dx x x x x x

dx dxyx

x x x x x

x

x x x x x x

x

x x x x

x

+ + − − + − +=

+

+ + − + −=

+

+ + + − + −=

+

− − + + +=

+

Find an equation of the tangent

line to the curve y = ex / (1 + x2)

at the point (1, ½e).

THE QUOTIENT RULE Example 5

According to the Quotient Rule,

we have:

THE QUOTIENT RULE Example 5

( ) ( ) ( )( )

( ) ( )( )

( )( )

2 2

22

22

2 22 2

1 1

1

1 2 1

1 1

x x

x x x

d dx e e x

dy dx dxdx x

x e e x e x

x x

+ − +=

+

+ − −= =

+ +

So, the slope of the tangent line at

(1, ½e) is:

This means that the tangent line at (1, ½e) is horizontal and its equation is y = ½e.

1

0

x

dy

dx =

=

THE QUOTIENT RULE Example 5

In the figure, notice that the function

is increasing and crosses its tangent line

at (1, ½e).

THE QUOTIENT RULE Example 5

Don’t use the Quotient Rule every time you see a quotient.

Sometimes, it’s easier to rewrite a quotient first to put it in a form that is simpler for the purpose of differentiation.

NOTE

For instance, though it is possible to

differentiate the function

using the Quotient Rule, it is much easier

to perform the division first and write

the function as

before differentiating.

23 2( )

x xF x

x

+=

1 2( ) 3 2F x x x−= +

NOTE

Here’s a summary of the differentiation

formulas we have learned so far.

( ) ( ) ( )

( ) ( ) ( )

( )

1

'

2

0

' ' ' ' ' ' ' '

' '' ' '

n n x xd d dc x nx e e

dx dx dx

cf cf f g f g f g f g

f gf fgfg fg gf

g g

−= = =

= + = + − = −

⎛ ⎞ −= + =⎜ ⎟

⎝ ⎠

DIFFERENTIATION FORMULAS