3.3 Differentiation Formulas In this section, we will learn: How to differentiate constant functions, power functions, polynomials, and exponential functions.

3.3Differentiation Formulas

In this section, we will learn:

How to differentiate constant functions,

power functions, polynomials, and

exponential functions.

DERIVATIVES

The constant function: f(x) = c.

The graph of this function is the

horizontal line y = c, which has slope 0.

So, we must have f’(x) = 0.

CONSTANT FUNCTION

A formal proof—from the definition of

a derivative—is also easy.

0

0 0

( )'( ) lim

lim lim0 0

h

h h

f x h f xf x

hc c

h

CONSTANT FUNCTION

In Leibniz notation, we write this rule

as follows.

( ) 0dc

dx

CONSTANT FUNCTION—DERIVATIVE

We next look at the functions

f(x) = xn, where n is a positive

integer.

POWER FUNCTIONS

If n = 1, the graph of f(x) = x is the line

y = x, which has slope 1.

So,

You can also verify Equation 1 from the definition of a derivative.

( ) 1dx

dx

POWER FUNCTIONS Equation 1

Figure 3.3.2, p. 135

We have already investigated the cases

n = 2 and n = 3.

In fact, in Section 3.2, we found that:

2 3 2( ) 2 ( ) 3d dx x x x

dx dx

POWER FUNCTIONS Equation 2

For n = 4, we find the derivative of f(x) = x4

as follows:

4 4

0 0

4 3 2 2 3 4 4

0

3 2 2 3 4

0

3 2 2 3 3

0

( ) ( ) ( )'( ) lim lim

4 6 4lim

4 6 4lim

lim 4 6 4 4

h h

h

h

h

f x h f x x h xf x

h h

x x h x h xh h x

h

x h x h xh h

h

x x h xh h x

POWER FUNCTIONS

Thus,

4 3( ) 4dx x

dx

POWER FUNCTIONS Equation 3

Comparing Equations 1, 2, and 3,

we see a pattern emerging.

It seems to be a reasonable guess that, when n is a positive integer, (d/dx)(xn) = nxn - 1.

This turns out to be true.

We prove it in two ways; the second proof uses the Binomial Theorem.

POWER FUNCTIONS

If n is a positive integer,

then1( )n nd

x nxdx

POWER RULE

The formula

can be verified simply by multiplying

out the right-hand side (or by summing

the second factor as a geometric series).

1 2 2 1( )( )n n n n n nx a x a x x a xa a

Proof 1POWER RULE

If f(x) = xn, we can use Equation 5 in

Section 3.1 for f’(a) and the previous equation

to write:

1 2 2 1

1 2 2 1

1

( ) ( )'( ) lim lim

lim( )

n n

x a x a

n n n n

x a

n n n n

n

f x f a x af a

x a x a

x x a xa a

a a a aa a

na

POWER RULE Proof 1

In finding the derivative of x4, we had to

expand (x + h)4.

Here, we need to expand (x + h)n . To do so, we use the Binomial Theorem —as follows.

0 0

( ) ( ) ( )'( ) lim lim

n n

h h

f x h f x x h xf x

h h

Proof 2POWER RULE

This is because every term except the first has h as a factor and therefore approaches 0.

1 2 2 1

0

1 2 2 1

0

1 2 2 1 1

0

( 1)2

'( ) lim

( 1)2lim

( 1)lim

2

n n n n n n

h

n n n n

h

n n n n n

h

n nx nx h x h nxh h x

f xh

n nnx h x h nxh h

hn n

nx x h nxh h nx

POWER RULE Proof 2

a.If f(x) = x6, then f’(x) = 6x5

b.If y = x1000, then y’ = 1000x999

c.If y = t4, then

d. = 3r23( )

dr

dr

Example 1POWER RULE

34dy

tdt

If c is a constant and f is a differentiable

function, then

( ) ( )d dcf x c f x

dx dx

CONSTANT MULTIPLE RULE

p. 137

Let g(x) = cf(x).

Then,0

0

0

0

( ) ( )'( ) lim

( ) ( )lim

( ) ( )lim

( ) ( )lim (Law 3 of limits)

'( )

h

h

h

h

g x h g xg x

hcf x h cf x

hf x h f x

ch

f x h f xc

hcf x

ProofCONSTANT MULTIPLE RULE

4 4

3 3

a. (3 ) 3 ( )

3(4 ) 12

b. ( ) ( 1)

( 1) ( ) 1(1) 1

d dx x

dx dx

x x

d dx x

dx dxdx

dx

NEW DERIVATIVES FROM OLD Example 2

If f and g are both differentiable,

then

( ) ( ) ( ) ( )d d df x g x f x g x

dx dx dx

SUM RULE

Let F(x) = f(x) + g(x). Then,

0

0

0

0 0

( ) ( )'( ) lim

( ) ( ) ( ) ( )lim

( ) ( ) ( ) ( )lim

( ) ( ) ( ) ( )lim lim (Law 1)

'( ) '( )

h

h

h

h h

F x h F xF x

hf x h g x h f x g x

hf x h f x g x h g x

h h

f x h f x g x h g x

h hf x g x

ProofSUM RULE

The Sum Rule can be extended to

the sum of any number of functions.

For instance, using this theorem twice, we get:

( ) ' ( ) '

( ) ' '

' ' '

f g h f g h

f g h

f g h

EXTENDED SUM RULE

If f and g are both differentiable,

then

( ) ( ) ( ) ( )d d df x g x f x g x

dx dx dx

DIFFERENCE RULE

8 5 4 3

8 5 4

3

7 4 3 2

7 4 3 2

( 12 4 10 6 5)

12 4

10 6 5

8 12 5 4 4 10 3 6 1 0

8 60 16 30 6

dx x x x x

dxd d dx x x

dx dx dxd d dx x

dx dx dx

x x x x

x x x x

NEW DERIVATIVES FROM OLD Example 3

Find the points on the curve

y = x4 - 6x2 + 4

where the tangent line is horizontal.

NEW DERIVATIVES FROM OLD Example 4

Horizontal tangents occur where

the derivative is zero.

We have:

Thus, dy/dx = 0 if x = 0 or x2 – 3 = 0, that is, x = ± .

4 2

3 2

( ) 6 ( ) (4)

4 12 0 4 ( 3)

dy d d dx x

dx dx dx dx

x x x x

Solution: Example 4

3

So, the given curve has horizontal tangents

when x = 0, , and - .

The corresponding points are (0, 4), ( , -5), and (- , -5).

Solution: Example 4

33

3 3

Figure 3.3.3, p. 139

The equation of motion of a particle is

s = 2t3 - 5t2 + 3t + 4, where s is measured

in centimeters and t in seconds.

Find the acceleration as a function of time.

What is the acceleration after 2 seconds?

NEW DERIVATIVES FROM OLD Example 5

The velocity and acceleration are:

The acceleration after 2s is: a(2) = 14 cm/s2

2( ) 6 10 3

( ) 12 10

dsv t t t

dtdv

a t tdt

Solution: Example 5

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

Let F(x) = f(x)g(x).

Then,

Proof

0

0

( ) ( )'( ) lim

( ) ( ) ( ) ( )lim

h

h

F x h F xF x

hf x h g x h f x g x

h

THE PRODUCT RULE

0

0

0 0 0 0

'( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )lim

( ) ( ) ( ) ( )lim ( ) ( )

( ) ( ) ( ) ( )lim ( ) lim lim ( ) lim

( ) '( ) ( ) '( )

h

h

h h h h

F x

f x h g x h f x h g x f x h g x f x g x

h

g x h g x f x h f xf x h g x

h h

g x h g x f x h f xf x h g x

h h

f x g x g x f x

ProofTHE PRODUCT RULE

Find F’(x) if F(x) = (6x3)(7x4).

By the Product Rule, we have:

Example 6

3 4 4 3

3 3 4 2

6 6

6

'( ) (6 ) (7 ) (7 ) (6 )

(6 )(28 ) (7 )(18 )

168 126

294

d dF x x x x x

dx dx

x x x x

x x

x

THE PRODUCT RULE

If h(x) = xg(x) and it is known that

g(3) = 5 and g’(3) = 2, find h’(3).

Applying the Product Rule, we get:

Therefore,

Example 7

'( ) [ ( )] [ ( )] ( ) [ ]

'( ) ( )

d d dh x xg x x g x g x x

dx dx dxxg x g x

THE PRODUCT RULE

'(3) 3 '(3) (3) 3 2 5 11 h g g

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

Let F(x) = f(x)/g(x).

Then,

THE QUOTIENT RULE Proof

0

0

0

( ) ( )'( ) lim

( ) ( )( ) ( )

lim

( ) ( ) ( ) ( )lim

( ) ( )

h

h

h

F x h F xF x

hf x h f xg x h g x

hf x h g x f x g x h

hg x h g x

We can separate f and g in that expression

by subtracting and adding the term f(x)g(x)

in the numerator:

Proof

0

0

'( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )lim

( ) ( )

( ) ( ) ( ) ( )( ) ( )

lim( ) ( )

h

h

F x

f x h g x f x g x f x g x f x g x h

hg x h g x

f x h f x g x h g xg x f x

h hg x h g x

THE QUOTIENT RULE

Again, g is continuous by Theorem 4 in Section 3.2.

Hence,

Proof

0 0 0 0

0 0

2

( ) ( ) ( ) ( )lim ( ) lim lim ( ) lim

lim ( ) lim ( )

( ) '( ) ( ) '( )

[ ( )]

h h h h

h h

f x h f x g x h g xg x f x

h hg x h g x

g x f x f x g x

g x

0lim ( ) ( )

hg x h g x

THE QUOTIENT RULE

The theorems of this section show

that:

Any polynomial is differentiable on .

Any rational function is differentiable on

its domain.

THE QUOTIENT RULE

Let

2

3

2

6

x xy

x

THE QUOTIENT RULE Example 8

Then,

THE QUOTIENT RULE Example 8

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

The figure shows the graphs of the function of

Example 8 and its derivative. Notice that, when y grows rapidly (near -2),

y’ is large. When y grows

slowly, y’ is near 0.

Figures:

Figure 3.3.4, p. 141

Example 8

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:

It is possible to differentiate the function

using the Quotient Rule. However, it is much easier to perform the division

first and write the function as

before differentiating.

23 2( )

x x

F xx

1 2( ) 3 2 F x x x

NOTE

If n is a positive integer,

then

GENERAL POWER FUNCTIONS

1( ) n ndx nx

dx

GENERAL POWER FUNCTIONS Proof

2

1

2

11 2

2

1

(1) 1 ( )1( )

( )

0 1

n n

nn n

n n

n

nn n

n

n

d dx xd d dx dxx

dx dx x x

x nx

x

nxnx

x

nx

a. If y = 1/x, then

b.

GENERAL POWER FUNCTIONS Example 9

1 22

1( )

dy dx x

dx dx x

3 43

4

66 ( ) 6( 3)

18

d dt t

dt t dt

t

So far, we know that the Power Rule

holds if the exponent n is a positive or

negative integer.

If n = 0, then x0 = 1, which we know has a derivative of 0.

Thus, the Power Rule holds for any integer n.

POWER RULE – integer version

What if the exponent is a fraction?

In Example 3 in Section 3.2, we found that:

This can be written as:

1

2

dx

dx x

1 2 1 212

dx x

dx

FRACTIONS

This shows that the Power Rule is true

even when n = ½.

In fact, it also holds for any real number n,

as we will prove in Chapter 7.

FRACTIONS

If n is any real number,

then1( )n nd

x nxdx

POWER RULE—GENERAL VERSION

a. If f(x) = xπ, then f ’(x) = πxπ-1.

b.

Example 10POWER RULE

3 2

2/3

(2 /3) 123

5/323

1

( )

Let

Then

yx

dy dx

dx dx

x

x

Differentiate the function

Here, a and b are constants. It is customary in mathematics to use letters near

the beginning of the alphabet to represent constants and letters near the end of the alphabet to represent variables.

PRODUCT RULE Example 11

( ) ( )f t t a bt

Using the Product Rule, we have:

PRODUCT RULE E. g. 11—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.

1 2 3 2

1 2 1 2312 2

( )

'( )

f t a t bt t at bt

f t at bt

LAWS OF EXPONENTS E. g. 11—Solution 2

They also enables us to find normal

lines.

The normal line to a curve C at a point P is the line through P that is perpendicular to the tangent line at P.

In the study of optics, one needs to consider the angle between a light ray and the normal line to a lens.

NORMAL LINES

Find equations of the tangent line

and normal line to the curve

at the point (1, ½).

Example 12TANGENT AND NORMAL LINES

2/(1 ) y x x

According to the Quotient Rule,

we have:

Example 12Solution: derivative

2 2

2 2

2

2 2

2 2 2

2 2 2 2

(1 ) ( ) (1 )

(1 )

1(1 ) (2 )

2(1 )

(1 ) 4 1 3

2 (1 ) 2 (1 )

d dx x x xdy dx dx

dx x

x x xxx

x x x

x x x x

So, the slope of the tangent line at (1, ½) is:

We use the point-slope form to write an equation of the tangent line at (1, ½):

31 14 4 41 ( 1) or y x y x

Solution: tangent line Example 12

2

2 21

1 3 1 1

42 1(1 1 )

x

dy

dx

The slope of the normal line at (1, ½) is

the negative reciprocal of -¼, namely 4.

Thus, an equation of the normal line is:

712 24( 1) 4or y x y x

Solution: normal line Example 12

The curve and its tangent and normal

lines are graphed in the figure.

Figures: Example 12

© Thomson Higher Education

Figure 3.3.5, p. 144

At what points on the hyperbola xy = 12

is the tangent line parallel to the line

3x + y = 0?

Since xy = 12 can be written as y = 12/x, we have:

TANGENT LINE Example 13

1 22

1212 ( ) 12( )

dy dx x

dx dx x

Let the x-coordinate of one of the points

in question be a.

Then, the slope of the tangent line at that point is 12/a2.

This tangent line will be parallel to the line 3x + y = 0, or y = -3x, if it has the same slope, that is, -3.

Solution: Example 13

Equating slopes, we get:

Therefore, the required points are: (2, 6) and (-2, -6)

Solution: Example 13

22

123 4 2or or a a

a

The hyperbola and the tangents are

shown in the figure.

Figure: Example 13

© Thomson Higher Education

Figure 3.3.6, p. 144

Here’s a summary of the differentiation

formulas we have learned so far.

1

'

2

0

' ' ' ' ' ' ' '

' '' ' '

n nd dc x nx

dx dx

cf cf f g f g f g f g

f gf fgfg fg gf

g g

DIFFERENTIATION FORMULAS