Chapter 5 Derivative Rules
f ( x  h)  f (a)
, which can be used to find
h
slopes of tangent lines as well as instantaneous rates of change. Unfortunately, computing the derivative
directly from the definition can be quite tedious and overwhelming. In this chapter, we will present
several rules of differentiation that will greatly simplify the differentiation.
The derivative of a function at x is defined as f '( x)  lim
h 0
5.1 Basic Differentiation Rules
Before we start to study the rules of differentiation, we need to introduce alternate ways to represent the
derivative.
Table 5.1 Alternative Notations for the Derivative
For y  f ( x) , all of the following may be used to represent the derivative:
f '( x), y ',
dy d
,
[ f ( x)]
dx dx
The process of finding the derivative of a function is called differentiation. A function is differentiable
at x if its derivative exists at x. The following table lists the basic differentiation rules.

Table 5.1
Basic Differentiation Rules
Let C, a, and n be real numbers with a  0 . Let f(x) and g(x) be differentiable functions.
Rules
Examples
1. Constant rule: If y  C , then y '  0 .
If y   , then y '  0 .
2. Power rule: If y  xn , then y '  n  xn1 .
If y  x7 , then y '  7 x6 .
3. If y  C  f ( x) , then y '  C  f '( x) .
If y  7 x2 , then y '  7  2 x21  14 x .
4. Sum and Difference rule:
If y  f ( x)  g ( x) , then y '  f '( x)  g '( x)
If y  3x4  7 x , then y '  12 x3  7 .
5. If y  e x , then y '  e x .
If y  3e x , then y  3e x .
6. If y  a x , then y  a x ln a .
If y  3 , then y  3x ln3
x
1 3
y '  3 
x x.
If y  3  ln x , then
1
7. If y  ln x , then y '  .
x
8. If y  log a x , then y ' 
1
.
x ln a
If y  log 4 x , then y ' 
1
.
x  ln 4
Example 5.1 Find the derivatives:
1
 e2
2t 2
(a) f ( x)  5x3  7 x2.5  3x  1
(b) y  3t 2 
4
 2
x
(e) y  3  ln x  4  log 4 x
(d) y  5  e x  2  3x
(c) g ( x)  5 5 x 2 
Solutions (a)
d
d
d
d
d
[ f ( x)]  (5 x3 )  (7 x 2.5 )  (3x)  (1)
dx
dx
dx
dx
dx
d
d
d
d
=5  ( x3 )  7  ( x 2.5 )  3  ( x)  (1)
dx
dx
dx
dx
 5  3  x31  7  2.5  x 2.51  3 1  x11  0
 15 x  17.5 x  3
2
1.5
Sum and Difference rules
Rule 3
Power Rule and Rule 1
Simplify.
(b) If a power function involves negative powers, we must rewrite the function in the form of x n . Then,
1
we apply the power rule. Recall that n  a  n . Hence,
a
1
1
y  3t 2  2  e2  3t 2  t 2  e2
2t
2
dy
1
Apply the Sum, Difference, and Power rules.
 3  2  t 21  (2)t 21  0
2
dx
2
Note that e is a constant.
1
 6t  t 3  6t  3
t
(c) If a power function involves radicals, we must rewrite each radical as a fractional power. Again, we
then apply the power rule. Recall that
m
a n  a m . Hence,
n
4
 2
x
2
 5  x 5  4 x 1  2
g ( x)  5  5 x 2 
2 2
g '( x)  5  x 5 1  4  (1)  x 11  0
5
3
2
4
 2 x  5  4 x 2  3  2
5
x
x
2
4

 2
5 3
x
x
2 is a constant, it does not need to be rewritten.
Apply the Sum, Difference and Power rules.
Simplify.
(d) Using rules 5 and 6, we have
y '  5  ex  2  3x ln3 .
Note:
2  3x  6 x
(e) By rules 7 and 8, we have
dy
1
1
3
4
 3  4 
 
dx
x
x  ln 4 x x  ln 4

Applications
Since the slope of a line tangent to a curve is given by the derivative, differentiation rules can be used to
find the equation of the tangent line. The steps for deriving a tangent line from a function f(x) at x = a are
summarized in the following table.
Table 5.2 How to Find a Tangent Line
Assuming that f(x) is differentiable at x = a.
Step 1. Find the y coordinate of the point at x = a: y  f (a) (x and y are both given sometimes.)
Step 2. Find the slope of the tangent line at x = a: m  f '(a)
Step 3. Use the point-slope formula y  y0  m( x  x0 ) to find the equation of the tangent line:
y  f (a)  f '(a)( x  a)
Example 5.2
Find the equation of the line tangent to the graph of f ( x)  2x3  2x2  1 at x = 1.
Sketch the graph of f(x) and the tangent line on the same axes.
Solutions When x = 1, we can get the y coordinate by finding f(1):
y  f (1)  2 13  2 12  1  1 .
By taking the derivative, we can find a formula for the slope of a line tangent to any point of f :
d
(2 x3  2 x 2  1)
dx
 6 x2  4 x
f '( x) 
In particular, the slope of the tangent line of f at x = 1 is
m  f '(1)  6 12  4 1  2
By using the point-slope formula of the line equation with slope 2 and the point (1, 1), we have
y  1  2( x  1)
y  2x 1
The graph of f ( x)  2x3  2x2  1 and the tangent line y  2 x  1 at x = 1 can be seen in Figure 5.1.
 Figure 5.1 Graph of Example 5.2
f ( x)  2 x  1
(1, 1)

The derivative of a function can also be interpreted as the instantaneous rate of change of f(x). Hence,
we can use the derivative to study rates of change.
Example 5.3
The population of a fire ant colony is growing according to the function
N (t )  100 t  1000 , where t is time measured in days. What is the rate of change of the population
with respect to time when t = 4? Give units and interpret the answer.
2
Solutions
3
The rate of change of the population with respect to time is given by N '(t ) . Therefore,
d
d
3
(100 2 t 3  1000)  (100  t 2  1000)
dt
dt
3 3 2 1
1
 100  t  0  100  t 2  100 t
2
N '(t ) 
Rewrite
2
3
t 3 as t 2 .
Power rule and simplify.
When t = 4, the rate is
N '(4)  100 4  200 fire ants/day
dN
in what is called Leibniz’s notation. Leibniz’s notation
dt
is useful for determining the units of the derivative:
If N (t )  f (t ) , N '(t ) also can be written as
N '(t ) 
dN units of N

dt units of t
Since N is measured in number of the fire ants and t is measured in days, N '(t ) must be measured in
number of fire ants per day. The statement N '(4)  200 fire ants per day means that on the fourth day, the
colony is increasing in size at the rate of 200 fire ants per day. This is an instantaneous rate of change,
meaning that if the rate were remaining 200 fire ants increasing per day for a whole day on forth day,
there would be extra 200 fire ants on the fifth day.

Example 5.4 Derek purchased a new car for $20,000 dollars and its value after t years is estimated
by V (t )  20,000(0.9)t . How much was the car decreasing in value per year when it was driven out of the
show room?
Solutions
This is asking for the rate of change of the car’s value at t = 0, or V '(0) . Since
V '(t )  20,000  (0.9)t  ln(0.9) ,
we have
V '(0)  20,000  (0.9)0  ln(0.9)
 2,107.2 dollars/year
The statement V '(0)  2,107.2 dollars/year means that the value of the car decreased at a rate of $2,107.2
per year when the car was driven out of the show room.

5.2 The Product and Quotient Rule
If f(x) and g(x) are both differentiable, then the derivative of f ( x)  g ( x) can be found by the product
rule.
Table 5.2
The Product Rule
If f '( x) and g '( x) exist, and y  f ( x)  g ( x) , then
y '  f '( x)  g ( x)  f ( x)  g '( x)
Example 5.5 Find the derivative of the following functions.
(a) y  x ln x
(b) f ( x)  e x (3x4  2x2  7)
Solutions
Then
(a) y  x ln x can be thought of as y  f ( x)  g ( x) where f ( x)  x and g ( x)  ln x .
y '  f '( x)  g ( x)  f ( x)  g '( x)
 1  ln x  x 
1
x
 ln x  1
Note: In general, the derivative of the product of two functions is not given by the product of the
derivatives of the functions:
d
[ f ( x)  g ( x)]  f '( x)  g '( x)
dx
1 1
In this case, f '( x)  g '( x)  1    1  ln x .
x x
(b) f ( x)  e x (3x4  2x2  7) can be thought as f ( x)  h( x)  g ( x) where h( x)  e x and g ( x)  3x4  2 x2  7 .
Then
f '( x)  h '( x)  g ( x)  h( x)  g '( x)
 e x  (3x 4  2 x 2  7)  e x  (12 x 3  4 x)
 e x (3x 4  12 x3  2 x 2  4 x  7)

If f(x) and g(x) are both differentiable, then the derivative of
f ( x)
can be found by the quotient rule.
g ( x)
Note that we must avoid points where g(x) = 0.
Table 5.2
The Quotient Rule
If f '( x) and g '( x) exist, and y 
f ( x)
, then
g ( x)
g ( x)  f '( x)  f ( x)  g '( x)
g 2 ( x)
which is valid for any x at which g ( x)  0 .
y' 
The denominator of the Quotient Rule is easy to memorize, but the numerator is not. To remember this
expression, simply memorize the expression in this form:
 Lower   Derivative of   Higher   Derivative of 



 

d  Higher term   term   Higher term   term   Lower term 

dx  Lower term 
(Lowerterm) 2
L  DH  H  DL

L2
Example 5.6 Find the derivative of the following functions.
(a) y 

ex
x3
Solutions
(b) h( x) 
(a) y 
x2  1
x2  1
ex
f ( x)
can be thought as y 
, where f ( x)  e x and g ( x)  x3 .
3
x
g ( x)
Then
y' 

g ( x)  f '( x)  f ( x)  g '( x )
g 2 ( x)
x3  e x  e x  3x 2
( x3 )2
x 2 e x ( x  3)
x6
e x ( x  3)

x4

Factor out the common factor
numerator.
x 2 e x from the
Simplify.
Note: In general, the derivative of the quotient of two functions is not given by the quotient of the
derivatives of the functions:
d  f ( x)  f '( x)

dx  g ( x)  g '( x)
In this case,
f '( x) e x
e x ( x  3)
.
 2 
g '( x) 3x
x4
(b) In this quotient, H  x 2  1 and DH  2 x , L  x 2  1 and DL  2 x . By the quotient rule, we have
L  DH  H  DL
L2
( x 2  1)  2 x  ( x 2  1)  2 x

( x 2  1) 2
h '( x) 
2 x3  2 x  2 x3  2 x
( x 2  1) 2
4x
 2
( x  1) 2

Distributive Property
Simplify.

Applications
In economics, the word “marginal” refers to an instantaneous rate of change. In other words, marginal
refers to the derivative. For example, if C(x) is the cost function, C '( x) is called the marginal cost
function. The marginal cost reflects how the cost changes in response to a one unit change in the
production. Knowing this cost is very important to management in their decision making processes.
Similarly, if R( x) and P( x) are revenue and profit functions, R '( x) and P '( x) are called the marginal
revenue and marginal profit functions.
Example 5.7
The cost to produce q items is C (q)  q 2 ( q  2) dollars. Find the marginal cost of
producing the 25th item. Interpret the answer in terms of costs.
Solutions
The marginal cost is defined as C '  q  . In this case,
C (q )  q 2 ( q  2)  q 2  (q 2  2)
1
Then, the marginal cost function is
1 1
1
C '( q )  2q  (q 2  2)  q 2  q  2
2
1 3
3
 2q 2  4q  q 2
2
5 32
 q  4q
2
Product Rule
Distributive property
Simplify.
Hence, the marginal cost of producing the 25th item is
5
3
C '(25)   (25) 2  4  25
2
5
1
  (25 2 )3  100  212.5 dollars/item
2
The statement C '(25)  212.5 dollors per item says that when 25 items have already made, the cost of
making the next item is approximately $212.5. Another way of saying this is that it costs about $212.5 to
make the 26th item.
We can also find C '(q) by first using the distributive property before differentiation:
C (q )  q 2 ( q  2)  q 2  (q 2  2)
1
 q 2  q 2  2q 2  q 2  2q 2
5
1
Then,
5 3
C '(q)  q 2  4q
2

Example 5.8 The sales (in millions of dollars) of a DVD player t years after it is put on the market
is given by
S (t ) 
100t
1  et
How fast are the sales changing at the end of the first year? At the end of the second year?
Solutions
The problems are asking for the rate at which sales are changing at the end of the first
year and at the end of second year. The rate at which sales are changing at time t is given by S '(t ) . Using
the quotient rule, we have
S '(t ) 
d  100t  L  DH  H  DL

dt 1  et 
L2

(1  et )  100  100t  et 100  100et  100t  et

(1  et ) 2
(1  et )2

100  100et (1  t )
(1  et ) 2
S (1) 
100  0
 7.233
(1  e1 )2
Therefore
S '(2) 
100  100e2 (1  2)
 9.078
(1  e2 )2
Thus, sales are increasing at a rate of $7.233 million per year at the end of the first year, and sales are
decreasing at a rate of $9.078 million per year at the end of the second year.

5.3 The Chain Rule
The derivative rules we discussed in the preceding sections can let us differentiate functions like
f ( x)  x100 and g ( x)  3x3  4x2  5 . However, we still lack the tools to find the derivative of functions
such as y  (3x3  4x2  5)100 . Functions like y  (3x3  4x2  5)100 are called composite functions.
Before we study a new rule that will let us find the derivative of a composite function, we review the
composite function first.
The composite functions
Recall that two function can be combined in various ways (sum, difference, product and quotient) to
create new functions. The composition is a special way to form a new function.
Table 5.3
Definition of Composite function
Given two function f and g, the composite function f
g is defined by
( f g )( x)  f ( g ( x))
Example 5.9 If f ( x)  x100 and g ( x)  3x3  4x2  5 , find the composite functions f g , g f
and f
f.
Solutions To get ( f g )( x)  f ( g ( x)) , replace x from f ( x)  x100 by g ( x)  3x3  4x2  5 :
( f g )( x)  f ( g ( x))  f (3x3  4x2  5)  (3x3  4x2  5)100
To get ( g f )( x)  g ( f ( x)) , replace x from g ( x)  3x3  4x2  5 by f ( x)  x100 :
( g f )( x)  g ( f ( x))  g ( x100 )  3( x100 )3  4( x100 )2  5  3x300  4x200  5
To get ( f
f )( x)  f ( f ( x)) , replace x from f ( x)  x100 by f ( x)  x100 itself:
(f
f )( x)  f ( f ( x))  f ( x100 )  ( x100 )100  x10000
Note that ( f g )( x)  ( g f )( x) , and ( f
f )( x)  f ( x)  f ( x)  [ f ( x)]2  x200

For a composite function y  f ( g ( x)) , we might think of f(x) as the “outer” function that acts on the
values of the “inner” function g(x). This point of view will help us to recognize a composite function.
Example 5.10 Find the inner and outer functions for the following functions so that each one can
be represented as a composite function. This process is called the Decomposition.
1
x 1
(d) y  3ln( x2  1)
(a) y  x  e x
(b) h( x) 
(c) f ( x)  e3 x2
Solutions
Decomposition of Composite Functions
( f g )( x)  f ( g ( x))
Inner Function: g(x)
Outer Function: f(x)
g ( x)  x  e x
f ( x)  x
g ( x)  x  1
f ( x) 
(c) k ( x)  e3 x 2
g ( x )  3x  2
f ( x)  e x
(d) y  3ln( x2  1)
g ( x)  x 2  1
f ( x)  3ln x
(a) y  x  e x
(b) h( x) 
1
x 1
1
x

The Chain Rule
In the above example, we observe that the outer and inner functions are much simpler functions that can
be differentiated by the derivative rules from the previous sections. The following rule describes the
relationship between the derivative of a composite function f g and derivatives of its inner and outer
functions.
Table 5.3
The Chain Rule
If f '( g ( x)) and g '( x) exist, and y  ( f g )( x)  f ( g ( x)) , then
y '  f '( g ( x))  g '( x)
The alternative form of the Chain Rule:
dy
du
and g '( x) 
. Therefore,
du
dx
dy du
y '  f '( g ( x))  g '( x) 

du dx
Let u  g ( x) , then y  f ( g ( x))  f (u) , f '( g ( x))  f '(u) 
When applying the Chain Rule, it is helpful to think of the composite function f g as having two parts:
an inner part and an outer part. Then, the derivative of f g is the product of the derivative of the outer
function (at the inner function u = g(x) ) and the derivative of the inner function at x.
Derivative of outer function f at u  g ( x)
Outer function
y '  f '( g ( x))  g '( x) 
y  f ( g ( x))  f (u)
dy du

du dx
Derivative of inner function g at x
Inner function
Example 5.11 Find the derivative of y  (3x3  4x2  5)100 .
Solution
For this function, you can consider the inner function to be g ( x)  3x3  4x2  5 .
Consequently, the outer function will be f ( x)  x100 . Note
g '( x)  9 x 2  8 x
f '( x)  100 x99
These steps can be performed mentally.
f '( g ( x))  100(3x3  4 x 2  5)99
By the Chain Rule, we have
y '  f '( g ( x))  g '( x)
 100(3x3  4 x 2  5)99  (9 x 2  8 x)
Derivative of the inner function at x
Derivative of the outer function at u = g(x)
Note that when taking the derivative of the outer
function, do not change the inner function g(x).
Let u  g ( x)  3x3  4x2  5 and y  f ( g ( x))  f (u)  u100 , then
dy
 100u 99 and
du
du
 9 x2  8x
dx
By the alternative form of the Chain Rule, we have
d
dy du
[ f ( g ( x))] 

dx
du dx
 100 u 99  (9 x 2  8 x)
y' 
Replacing u by g ( x)  3x3  4x2  5 .
 100(3 x3  4 x 2  5) 99  (9 x2  8 x)

The above example is one of the most common composite functions, y  [ g ( x)]n , where the outer function
is a power function, f ( x)  xn . The rule for differentiating such function is called the General Power
Rule and it can be considered a special case of the Chain Rule.
Table 5.4
The General Power Rule
If g '( x) exists, and y  ( f g )( x)  [ g ( x)]n , then
y '  n  [ g ( x)]n1  g '( x)
Example 5.12 Find the derivatives:
(a) y  x  e x
(b) h( x) 
1
x 1
(c) y  (1  3x)3 (2x2  3)4
Solution (a) In this function, the outer function is f ( x)  x and the inner function is g ( x)  x  e x .
By the General Power Rule, we have
y  x  ex  (x  ex ) 2
1
1
y '   ( x  e x ) 2 1  (1  e x )
2
n [g( x)]n1
g '( x)
1
1
1
 ( x  e x )  2 (1  e x )
2
1  ex
1  ex


1
2( x  e x ) 2 2 x  e x
Rewrite the outer function as ( )
1
2
General Power Rule
Simply
1
Rewrite ( ) as
2
(b)
h( x ) 
1
 ( x  1) 1
x 1
h'(x)  ( 1)  ( x  1) 2  (1) 
Rewrite
1
(1  x) 2
General Power Rule
We could also use the Quotient Rule to differentiate this function,
h'(x) 
(1  x 2 )  0  1  (1)
1

2
(1  x)
(1  x) 2
but it is much easier to use the Chain Rule. In general, do not use the Quotient Rule for functions of the
1
form
. Instead, rewrite the function as [ g ( x)] n and use the Chain Rule instead.
n
[ g ( x)]
(c) For this function, we need to apply the Product Rule before applying the Chain Rule.
d
d
Apply the Product Rule.
(2 x 2  3) 4  (2 x 2  3) 4  (1  3 x)3
dx
dx
 (1  3 x)3  4(2 x 2  3)3  4 x  (2 x 2  3) 4  3(1  3 x) 2  ( 3) Apply the General Power Rule for derivatives.
y '  (1  3 x)3 
 16 x(1  3 x)3 (2 x 2  3)3  9(2 x 2  3) 4 (1  3 x) 2
Simplify.
 (1  3 x) (2 x  3) [16 x(1  3 x)  9(2 x  3)]
Factor the common factor: (1  3x)2 (2 x2  3)3
 (1  3 x) 2 (2 x 2  3)3 (16 x  48 x 2  18 x 2  27)
Apply the Distributive property.
 (1  3 x) (2 x  3) (16 x  66 x  27)
Simplify.
2
2
2
2
3
3
2
2

Recall that the derivative of e x is simply e x . Can we apply the Chain Rule to functions of the
form y  e g ( x ) ? Using the Chain Rule, we find the following formulas:
Table 5.5
The Chain For Exponential functions
If g '( x) exists, and y  e g ( x ) , then
y '  e g ( x )  g '( x)
If g '( x) exists, and y  a g ( x ) with a  0, a  1 , then
y '  a g ( x )  ln a  g '( x)
Note that the derivative of y  e g ( x ) is a special case of the derivative of y  a g ( x ) , because ln e  1 .
Example 5.13
(a) k ( x)  e3 x 2
Find the derivatives:
(b) g ( x)  3  32 e
x
 3 x 1
(c) y 
x2  1
ex
2
Solution (a) In this function, g ( x)  3x  2 and g '( x)  3 . By the Chain Rule, we have
k '( x)  e g ( x )  g '( x)
 e3 x  2  3  3e3 x  2
The derivative of inner function g ( x)
The derivative of the outer function
(b) Let f ( x)  2e x  3x  1 , then f '( x)  2e x  3 . We have
eu , where u  g ( x)
g '( x)  a f ( x )  ln a  f '( x)
 3  32e
x
 3 x 1
 3  ln 3  32e
(c) Rewrite the function as y 
( x 2  1)
ex
2
1
x
ln 3  (2e x  3)
 3 x 1
(2e x  3)
2
.
1
1
d 2
d 2
( x  1) 2  ( x 2  1) 2  e x
dx
dx
y' 
2
(e x ) 2
2
2
1
1
1
e x  ( x 2  1)  2  2 x  ( x 2  1) 2  e x  2 x
2

2
(e x ) 2
ex 
2
e x  x  ( x 2  1)  2 1  2( x 2  1) 

2
2
ex  ex
x(1  2 x 2  2)  x(2 x 2  1)
 2
 2
2
1
( x  1) 2 e x
ex x2  1
2
Apply the Quotient Rule.
Apply the Chain Rule to each term.
1
Factor out the common factor: e x  x  ( x 2  1) 
2
1
2
Simplify.
This function can also be differentiated by the Product Rule.
y
x2  1
e
x2
 x 2  1  e x  ( x 2  1)  e x
2
2
2
2
1
1
1
y '  ( x 2  1) 2 1  2 x  e  x  ( x 2  1) 2  e  x  (2 x)
2
 ( x 2  1)  2  x  e  x  ( x 2  1) 2  e  x  2 x
2
1
 ( x  1)
2

 12
 xe
x(1  2 x 2  2)
( x 2  1) 2  e x
1
2
 x2

1
[1  2( x  1)]
2
 x(2 x 2  1)
x2  1  ex
2
Rewrite the function as a product.
Apply the Product Rule.
2
Simplify.
Factor out the common factor: e x  x  ( x 2  1) 
2
1
2
Simplify.
The Product Rule is easier to remember. Hence, whenever a function can be written as a product it is
more convenient to use the Product Rule than the Quotient Rule.

1
. How can we apply the Chain Rule to functions of the
x
form y  ln g ( x) ? To apply the Chain Rule to the function of the form y  ln g ( x) , we use the
following formulas:
Recall that the derivative of ln x is

Table 5.6
The Chain Rule for Logarithmic Functions
If g '( x) exists, and y  ln g ( x) , then
f '( x) 
1
g '( x)
 g '( x) 
g ( x)
g ( x)
If g '( x) exists, and y  log a g ( x) with a  0, a  1 , then
f '( x) 
1
g '( x)
 g '( x) 
g ( x)ln a
g ( x)ln a
Note that the derivative of y  ln g ( x) is a special case of the derivative of y  log a g ( x) because
ln e  1 .
Example 5.14
(a) y  log10 (5 x  3)
Determine the derivative for the following functions.
(c) f ( x)  ln
(b) k (t )  ln t 4  2t
x2  1
for x  1
x 1
Solution (a) In this function, g ( x)  5x  3 and g '( x)  5 , then by the Chain Rule,
y' 
1
1
5
 g '( x) 
5 
g ( x)ln a
(5x  3)ln10
(5x  3)ln10
The derivative of the inner function g(x)
The derivative of the outer function log a u , where u  g ( x)
(b) Here, we need to use The Chain Rule twice. Let g (t )  t 4  2t , then
g (t )  (t 4  2t ) 2
1
1
g '(t )  (t 4  2t )  2 (4t 3  2)
2
1
1
 (t 4  2t ) 2  2  (2t 3  1)
2
2t 3  1

t 4  2t
1
Now, applying the Chain Rule again, we have
k '(t ) 

1
 g '(t )
g (t )
1
t 4  2t

2t 3  1
t 4  2t

2t 3  1
t 4  2t
The General Power Rule
Factor out the common factor
Simplify
We can also find k '(t ) by first rewriting the function using the logarithmic property, and then taking the
derivative:
1
1
k (t )  ln t 4  2t  ln(t 4  2t ) 2  ln(t 4  2t )
2
3
3
1 4t  2 2t  1
k '(t )   4

2 t  2t t 4  2t
(c) if g ( x) 
Recall that: ln M p  p  ln M
Apply the Chain Rule for ln(t 4  2t )
x2  1
, then we use the Quotient Rule to find g '( x) .
x 1
f '( x) 
1
 g '( x)
g ( x)
1
( x  1)  (2 x)  ( x 2  1)  1

x2  1
( x  1) 2
x 1
x  1 2x2  2x  x2  1
 2

x 1
( x  1) 2


Apply the quotient rule for g(x).
Simplify
x2  2x  1
( x 2  1)( x  1)
Another way of finding f '( x) is by rewriting the function using the logarithmic property, and then taking
the derivative:
x2  1
 ln( x 2  1)  ln( x  1)
x 1
2x
1
f '( x)  2

x 1 x 1
f ( x)  ln
Recall that: ln M  ln M  ln N
N
Apply the Chain Rule for each term.
To get the same answer, we need to combine the result as follows:
2x
1
2 x( x  1)  ( x 2  1)
x2  2x  1



x2  1 x  1
( x 2  1)( x  1)
( x 2  1)( x  1)



1
x2
Example 5.15 Suppose that the demand equation of a product is given by p( x)  e 18 , where x is
measured in thousands. Find the marginal revenue function and find where the marginal revenue is zero.
Solution Recall that the revenue function is defined as R( x)  x  p( x)  x  e 
revenue function is R '( x) . Thus,
1 x2
18
and the marginal
R '( x)  1  e

1 2
x
18
 xe

1 2
x
18
 (
1
2 x)
18
1
 x2
1
 x 2  e 18
9
1 2
 x
1
 e 18 (1  x 2 )
9
e

1 2
x
18
Apply the product Rule and the Chain Rule.
Simplify.
Factor out the common factor
To find where the marginal revenue is zero, let R '( x)  0 . Thus

1 2
x
18
1
(1  x 2 )  0
9
1 2
 x
1
1
e 18 (1  x)(1  x)  0
3
3
e
Recall that e x is never zero no matter what the value of x is. Thus, the marginal revenue will be zero only
if x = 3 or x = -3. For this case, x = -3 doesn’t make sense. Therefore, the marginal revenue will be zero if
three thousand products are made.

Example 5.16
Find where the tangent line to the graph y  ln( x4  x3  1) is horizontal.
Solution The tangent line will be horizontal when the slope is zero, that is, at points where the
derivative is zero. Applying the Chain Rule, we have
y' 
4 x3  3x 2 x 2 (4 x  3)

0
x 4  x3  1 x 4  x3  1
Since the only time that a fraction can equal zero is when the numerator equals zero, x2 (4x  3)  0 . By
solving this equation, we find that the graph has two horizontal tangent lines at x = 0 and x = ¾.

5.4 Higher-Order Derivatives
The derivative f ' of a function f is also a function; hence, where it exists, we can also find the derivative
of f ' . We call f ' the first order derivative of f , and the derivative of f ' , denoted as f '' , as the
second order derivative of f. Continuing in this manner, we are led to consider the third, fourth, and
higher order derivatives of f whenever they exist. The following table represents the notations for the
higher order derivatives.

Table 5.7
Notations for the Higher-Order Derivatives
Name
Notation
The First-Order Derivative
f ', y ',
dy d
,
[ f ( x)], D1 f ( x)
dx dx
The Second-Order Derivative
f '', y '',
d2y d2
,
[ f ( x)], D 2 f ( x)
dx 2 dx 2
The Third-Order Derivative
f ''', y ''',
d3y d3
,
[ f ( x)], D3 f ( x)
dx3 dx3
The nth -Order Derivative ( n  4 )
f (n) , y (n) ,
dny dn
,
[ f ( x)], D n f ( x)
dx n dx n
Assume y  f ( x) and f ', f '', f ''', ... f ( n) exist.
Example 5.17
Find all derivatives of all orders of the function f ( x)  x4  2x3  5x2  10x  8 .
Solution We have
f '( x)  4 x 3  6 x 2  10 x  10
f ''( x)  12 x 2  12 x  10
f '''( x)  24 x  12
f (4) ( x)  24
f (5) ( x)  0
In fact f ( n) ( x)  0 , for all n  5 .
Example 5.18

Find the fourth derivative of the function y  3 x .
Solution We have
y3 xx
1
3
1 1
1 2
y '  x 3 1  x  3
3
3
1  2 2
2 5
y ''      x  3 1   x  3
3  3
9
2  5 5
10  8 3
y '''       x  3 1 
x
9  3
27
y (4) 
10  8   8 3 1
80 11
80
80
  x
  x 3  

11
3
27  3 
81
81x
81  3 x11

Application
What exactly does the second derivative f '' tell us? Just as the first derivative f ' measures the
instantaneous rate of change of the function f , the second derivative of f measures the rate of change of
the first derivative f ' of the function f; in other words, the rate of change of the rate of change of f.
Acceleration
In physics, the second derivative f '' can be interpreted as acceleration. Recall that if s  s (t ) is the
position function of an object that moves in a straight line, then the instantaneous rate of change of the
position of the object is called the velocity of the object at time t:
v(t )  s '(t ) 
ds
dt
The instantaneous rate of change of velocity with respect to time is called the acceleration a(t) of the
object. Thus, the acceleration function is the derivative of the velocity function, and hence, is the second
derivative of the position function:
a(t )  v '(t )  s ''(t ) 
Example 5.19
d 2s
dt 2
The vertical position of a ball thrown from the top of a lighthouse is given by
s(t )  16t 2  96t  1000
where s is measured in feet and t is measured in seconds. Find the velocity and acceleration of the ball 5
seconds after it is thrown.
Solution The velocity v and acceleration a of the ball at any time t are given by
ds
 32t  96
dt
d 2s
a(t )  v '(t )  2  32 ft/sec2
dt
v(t ) 
and
Therefore, the velocity and the acceleration of the ball 5 seconds after it is thrown is
v(5)  32(5)  96  64 ft/sec
a(5)  32 ft/sec2
This means that after 5 seconds the ball is falling at a speed of 64 ft/sec and the acceleration of the ball is
a constant 32 ft/sec2 . The negative sign indicates that the acceleration is also directed downward. We
should not be surprised by the constant acceleration since we know there is only one force acting on such
a following object, gravity, which is constant.

Concavity
The second derivative f '' can be used to study the shape of the graph of a function. For example both
f ( x)  x2 and g ( x)  x are increasing in the interval (0, ∞), but there is a fundamental difference in the way
they are increasing. The slopes of the lines tangent to the graph of f ( x)  x2 are increasing as we move from
left to right. Meanwhile, the slopes of lines tangent to the graph g ( x)  x are decreasing as we move from
left to right. See the following graphs.
 Figure 5.2 Graphs of f ( x)  x2 and g ( x)  x
Slope is increasing means
f ' is increasing. We say
f is concave up.
Slope is decreasing means
f ' is decreasing. We say f
is concave down.
The word concavity is used to describe the shape of a function. From the above graphs, we have the
following definition of concavity.
Table 5.8
Definition of Concavity
We say that the graph of f is concave up on the interval (a, b), if f ' is increasing on (a, b).
We say that the graph of f is concave down on the interval (a, b), if f ' is decreasing on (a, b).
A point where the graph of f changes concavity is called a point of inflection.
Recall that f ' tells us where f is increasing and where it is decreasing. The derivative of f ' , which is
f '' , tells us where f ' is increasing and where it is decreasing. We can use the following second
derivative test for concavity.
Table 5.9
Test for Concavity
If f ''  0 on the interval (a, b) ( f ' is increasing) , then f is concave up on (a, b).
If f ''  0 on the interval (a, b) ( f ' is decreasing) , then f is concave down on (a, b).
Concavity is very important for sketching the graph of a function, which we discussed it in Chapter 6.
Here we will concentrate on its application to economics problems.
Example 5.20
The total sales (in thousands) of a new video game is given by
59
s(t ) 
1  64e.4t
where t is the time in months. Where is the graph of s (t ) concave up and where is it concave down? Are
there any inflection points of s (t ) ? Explain what this means.
Solution The given expression s(t) is called the logistic function. A logistic function is always
increasing. Its graph is concave up at first, then concave down, and finally levels off at a horizontal
asymptote. (See Figure 5.3). In this case, the point of inflection (also called the point of diminishing),
where the concavity changes, is somewhere around 10 weeks. To find out exactly where the curve change
the concavity, we need the second derivative of s(t).
 Figure 5.3 Graph of the logistic function
Maximum potential sales
Concave down. The rate of sales is decreasing.
Less units are being sold each moth.
Point of inflection point. The rate of
sales stop increasing.
Concave up. The rate of sales is increasing.
More units are being sold each month.
When the curve is concave up, we know f ''  0 ( f ' is increasing), and when the curve is concave down,
f ''  0 ( f ' is decreasing). Therefore, the curve changes concavity when f ''  0 or when f '' does not
exist. Since
s(t ) 
59
 59(1  64e.4t )1
1  64e.4t
We can use the chain rule or quotient rule to find the first derivative of s(t):
s '(t )  59  (1)  (1  64e.4t ) 2  64e.4t  (.4)
 1510.4(1  64e.4t )2 e.4t

1510.4e.4t
(1  64e.4t )2
Using the quotient rule and the chain rule, we have
s ''(t ) 
(1  64e .4t ) 2  1510.4e .4t (.4)  1510.4e .4t  2(1  64e .4t )  64e .4t  (.4)
(1  64e .4t )4

.4(1  64e .4t )  1510.4e .4t [(1  64e .4t )  128e .4t ]
(1  64e .4t )4

604.16e .4t (64e .4t  1)
(1  64e .4t )3
The denominator in s '' is never zero, so there is no point at which s '' is undefined. To find where the
s ''  0 , we only need to set the numerator equals to zero:
604.16e.4t (64e.4t  1)  0
The factor 604.16e.4t can never be equal to zero, so we have (64e.4t  1)  0 . Now, solve for t:
64e .4t  1
1
e .4t 
64
 1 
 .4t  ln  
 64 
ln  164 
t
 10.4 months
.4
The exponential function with base e.
Taking the nature logarithm
Thus, the curve of the sales function changes the concavity around 10.4 months. This means when the
new game just appears on the market, the sales increase rapidly, more units are sold each month than the
previous month until the middle of the 11th month. Then the sales slow down and fewer units are sold
each month than the previous month. The inflection point is the time when the rate of sales stops
increasing. Eventually, most people who want the new game have already bought it, the sales will stop
and the 59 million units should be the maximum potential sales of the new game.

Sample Quiz
Find derivatives for problems 1- 5:
1. f ( x)  3x 2  5 x 
2 2
e
x
2. y  3  3x  log3 (3x  5)
3. g (t )  (t 2  5)3 (2t  1)4
4. f ( z )  ln
z2
z2  1
z0
5. y  3x 2  e2 x
6. Find the equation of the tangent line to the graph of f ( x) 
x 5
at the point at which x = 0.
2x 1
7. The cost function for a product is given by C( x)  .01x3  x2  50 x0.3  1000 , where x is the number of
items produced and C(x) is the cost in dollars to produce x items. How much would it cost to produce the
101st item?
8. A drug concentration curve is given by C(t )  20te0.2t , with C in mg/ml and t in minutes. What is the
rate of the concentration 4 minutes after it was administered?
9. Suppose the demand equation for a product is given by p 
1
10. The deer population in a wild park is approximated by P(t ) 
400
with t measured in years
1  9e0.3t
x2  1
sold and p is the price in dollars. Find the marginal revenue function.
, where x is the number of items
since 1980. When does the deer population stop growing?
Answers to Sample Quiz
1. 6 x 
5
2 x

2
x2
2. 3  3x ln 3 
3
(3x  5)ln 3
3x  e2 x
z 
1
4. 2   2

 z z 1
5.
7. $100.6
8. 1.797 mg/ml/min
10. After t = 7.324 years or in 1987
3x 2  e2 x
3. 2(t 2  5)2 (2t  1)3 (10t 2  3t  20)
6. y  9 x  5
9. R '( x) 
1
x2  1

x2
( x 2  1)3