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3DIFFERENTIATION RULESDIFFERENTIATION RULES

3.2The Product andQuotient Rules

DIFFERENTIATION RULES

In this section, we will learn about:Formulas that enable us to differentiate new functionsformed from old functions by multiplication or division.

By analogy with the Sum and DifferenceRules, one might be tempted to guess—asLeibniz did three centuries ago—that thederivative of a product is the product of thederivatives.

However, we can see that this guess is wrongby 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 byLeibniz (soon after his false start) and iscalled the Product Rule.

THE PRODUCT RULE

Before stating the Product Rule, let’s seehow we might discover it.

We start by assuming that u = f(x) andv = g(x) are both positive differentiablefunctions.

THE PRODUCT RULE

Then, we can interpret the product uvas an area of a rectangle.

THE PRODUCT RULE

If x changes by an amount Δx, thenthe corresponding changes in u and vare:

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 ofthe large rectangle in the figure—providedthat Δu and Δv happen to be positive.

THE PRODUCT RULE

The change in the area of the rectangleis:

( ) ( )( )

the sum of the three shaded areas

uv u u v v uv

u v v u u v

! = + ! + ! "

= ! + ! + ! !

=

THE PRODUCT RULE Equation 1

If we divide by ∆x, we get:

( )uv v u vu v u

x x x x

! ! ! != + + !

! ! ! !

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 x

v u vu v ux x x

v u vu v u

x x x

dv du dvu vdx dx dx

! "

! "

! " ! " ! " ! "

!=

!

! ! !# $= + + !% &

! ! !' (

! ! !# $= + + ! % &

! ! !' (

= + +

THE PRODUCT RULE

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

( )d dv duuv u v

dx dx dx= +

THE PRODUCT RULE Equation 2

Though we began by assuming (for thegeometric interpretation) that all quantities arepositive, we see Equation 1 is always true.

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

So, we have proved Equation 2—known asthe Product Rule—for all differentiable functionsu 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 secondfunction plus the second function times the derivativeof 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

dx

d dx 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

dx

d dx e e x

dx dx

x e e

x e

! "= +# $

= + + +

= + + %

= +

THE PRODUCT RULE Example 1 b

Further applications of the Product Rulegive:

4

'''( ) ( 3)

( ) ( 4)

x

x

f x x e

f x x e

= +

= +

THE PRODUCT RULE Example 1 b

In fact, each successive differentiationadds 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 torewrite f(t), then we can proceed directlywithout using the Product Rule.

This is equivalent to the answer in Solution 1.

1 2 3 2

1 2 1 2312 2

( )

'( )

f t a t bt t at bt

f t at bt!

= + = +

= +

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 , whereg(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 dx

g xxg x g x x xg x

x

!

" # " #= = +$ % $ %

= + & = +

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

2 22 4

gf g= + = ! + =

!

We find a rule for differentiating the quotientof two differentiable functions u = f(x) andv = g(x) in much the same way that we foundthe Product Rule.

THE QUOTIENT RULE

If x, u, and v change by amounts Δx, Δu,and Δv, then the corresponding changein the quotient u / v is:

( ) ( )

( ) ( )

u u u u

v v v v

u u v u v v v u u v

v v v v v v

+ !" #! = $% &

+ !' (

+ ! $ + ! ! $ != =

+ ! + !

THE QUOTIENT RULE

So,

( )

( )

0

0

lim

lim

x

x

u vd u

dx v x

u vv ux x

v v v

! "

! "

!# $=% &

!' (

! !)

! !=+ !

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 dx

dx v v v v v

! " ! "

! "

! !# #

$ % ! != =& '+ !( )

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 numeratortimes the derivative of the denominator, all divided bythe square of the denominator.

[ ] [ ]2

( ) ( ) ( ) ( )( )

( ) ( )

d dg x f x f x g x

d f x dx dx

dx g x g x

!" #=$ % " #& ' & '

THE QUOTIENT RULE

The Quotient Rule and the otherdifferentiation formulas enable us tocompute the derivative of any rationalfunction—as the next example illustrates.

THE QUOTIENT RULE

Let2

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 tangentline 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 dx

dx 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 functionis increasing and crosses its tangent lineat (1, ½e).

THE QUOTIENT RULE Example 5

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

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

NOTE

For instance, though it is possible todifferentiate the function

using the Quotient Rule, it is much easierto perform the division first and writethe function asbefore differentiating.

23 2( )

x xF x

x

+=

1 2( ) 3 2F x x x!

= +

NOTE

Here’s a summary of the differentiationformulas 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