LESSON 3 DERIVATIVES AND INTEGRALS INVOLVING THE
NATURAL AND GENERAL LOGARITHMIC FUNCTIONS
Definition The logarithmic function with base b is the function defined by
f ( x )  log b x , where b  0 and b  1 .
y
Recall that y  log b x if and only if b  x
Recall the following information about logarithmic functions:
1.
The domain of f ( x )  log b x is the set of positive real numbers. That is,
the domain of f ( x )  log b x is ( 0 ,  ) .
2.
The range of f ( x )  log b x is the set of real numbers. That is, the range of
f ( x )  log b x is (   ,  ) .
3.
The logarithmic function f ( x )  log b x and the exponential function
g ( x )  b x are inverses of one another. We studied the calculus of
exponential functions in Lesson 2.
Definition The natural logarithmic function is the logarithmic function whose base
is the irrational number e. Thus, the natural logarithmic function is the function
defined by f ( x )  log e x , where e  2.71828 . . . . . Recall that log e x  ln x .
Definition The common logarithmic function is the logarithmic function whose
base is the number 10. Thus, the common logarithmic function is the function
defined by f ( x )  log 10 x . Recall that log 10 x  log x .
Recall the following properties of logarithms:
1.
log b u r = r log b u
2.
log b u v = log b u + log b v
Copyrighted by James D. Anderson, The University of Toledo
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u
= log b u  log b v
v
3.
log b
4.
log b b  1
5.
log b 1  0
6.
b
7.
log b b x  x
8.
log a u
log
u

b
Change of Bases Formula:
log a b
log b x
 x
Theorem If f ( x )  ln x , then f ( x ) 
1
.
x
Proof Note that the domain of the function f is the set of real numbers x such
that x  0 . Let y  f ( x ) . Then y  ln x . If x  0 , then x  x . Thus,
y  ln x  ln x . If y  ln x , then by the definition of logarithm, we have that
y dy
 1 . Solving for
e y  x . Using implicit differentiation, we obtain that e
dx
dy
dy
1
dy
1
 y . Since e y  x , then
 . We have proven
, we have that
dx
dx
e
dx
x
1
that if f ( x )  ln x and x  0 , then f ( x )  . Now, we want to prove that
x
1
if f ( x )  ln x and x  0 , then f ( x )  . If x  0 , then x   x .
x
Thus, f ( x )  ln x  ln (  x ) and  x  0 . Now, using the Chain Rule and
1

f
(
x
)

D x (  x ) . Thus,
the first result proved, we obtain that
x
1
1
f ( x ) 
(  1)  .
x
x
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Theorem If y  ln u , where u  g ( x ) , then
dy
1 du


dx
u dx
du
dx
or
u
dy
1
g ( x )

g ( x ) 
.
dx
g(x)
g( x )
Proof Use the Chain Rule and the fact that D x ln x 
Theorem If f ( x )  log b x , then f ( x ) 
1
.
x
1
x ln b .
Proof Using the change of bases formula for logarithms, we have that log b x
log e x
log e b

ln x
ln b

1
ln x . Thus, f ( x )  log b x = 1 ln x . Since
ln b
ln b
ln b is a constant and D x ln x 
Theorem If y  log b
or
1
1 1
1

f
(
x
)


, then
=
x ln b .
ln b x
x
du
dy
1 du

 dx
u , where u  g ( x ) , then
dx
u ln b dx
u ln b
dy
1
g ( x )

g ( x ) 
.
dx
g ( x ) ln b
g ( x ) ln b
1
D
log
x

x
b
Proof Use the Chain Rule and the fact that
x ln b .
Examples Find the domain of the following function. Then differentiate them.
1.
=
f ( x )  ln x 2
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2
We are taking the natural logarithm of the expression x . In order for the
2
logarithm to be defined, we need the expression x to be positive. Since the
2
expression x is positive when x  0 , then the domain of the function f is
the set of real numbers x such that x  0 .
2
Since f ( x )  ln x = 2 ln x , then f ( x )  2 
Domain: { x : x  0 } or (   , 0 )  ( 0 ,  )
Answer:
f ( x ) 
2.
1
2
 .
x
x
2
x
g ( x )  ( log 2 x ) 4
We are taking the logarithm base 2 of the expression x. In order for the
logarithm to be defined, we need the expression x to be positive. Thus, the
domain of the function g is the set of positive real numbers.
g ( x )  ( log 2 x ) 4  g ( x )  4 ( log 2 x ) 3 D x log 2 x =
4 ( log 2 x ) 3
1
4 ( log 2 x )
x ln 2 =
x ln 2
3
Answer:
Domain: { x : x  0 } or ( 0 ,  )
g ( x ) 
4 ( log 2 x ) 3
x ln 2
NOTE: The set of positive real numbers is also the domain of the function
g.
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3.
s( t )  ln ( 3 t 2  16 t  12 ) 9
2
9
We are taking the natural logarithm of the expression ( 3 t  16 t  12 ) .
In order for the logarithm to be defined, we need the expression
( 3 t 2  16 t  12 ) 9 to be positive. Since the expression ( 3 t 2  16 t  12 ) 9
2
is positive when 3 t  16 t  12  0 , then we need to find the sign of the
2
expression 3 t  16 t  12 = ( t  6 ) ( 3 t  2 ) .
Sign of ( t  6 ) ( 3 t  2 ) :

+

6
+

2
3
Thus, the domain of the function s is the set of real numbers given by
2

(  ,  6)   ,   .
3

2
9
2
Since s( t )  ln ( 3 t  16 t  12 ) = 9 ln ( 3 t  16 t  12 ) , then
6 t  16
18 ( 3 t  8 )
s ( t )  9  2
=
3 t  16 t  12
3 t 2  16 t  12 .
Answer:
2

Domain: (   ,  6 )   ,  
3

s ( t ) 
18 ( 3 t  8 )
3 t 2  16 t  12
2

(


,

6
)

,


 is also the
NOTE: The set of real numbers given by
3

domain of the function s  .
NOTE: If you wanted to find the sign of the function s  , then you would
want to know when s ( t )  0 and when s  is undefined.
Copyrighted by James D. Anderson, The University of Toledo
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s ( t )  0  18 ( 3 t  8 )  0  t  
not in the domain of s nor s  .
8
. However, this number is
3
s  undefined  3 t 2  16 t  12  0  ( t  6 ) ( 3 t  2 )  0 
t   6, t 
Sign of s ( t ) :
2
. These numbers are also not in the domain of s nor s  .
3

+


6
2
3
2

Thus, the function s is increasing on the interval  ,   and is decreasing
3

on the interval (   ,  6 ) .
4.
s( t )  ln 3 t 2  16 t  12
2
We are taking the natural logarithm of the expression 3 t  16 t  12 . In
order for the logarithm to be defined, we need the expression
3 t 2  16 t  12 to be positive. The expression 3 t 2  16 t  12 is
2
2
positive when 3 t  16 t  12  0 . Since 3 t  16 t  12 =
( t  6 ) ( 3 t  2 ) , then 3 t 2  16 t  12  0 when t   6 and t 
Thus, the domain of the function s is the set of real numbers
2
2


 t : t   6 and t   or (   ,  6 )    6 ,  
3
3


given by
2

 , .
3

6 t  16
2

s
(
t
)

s
(
t
)

ln
3
t

16
t

12
Since
, then
3 t 2  16 t  12 =
2 ( 3t  8 )
3 t 2  16 t  12 .
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2
.
3
Answer:
2

t
:
t


6
and
t


 or
Domain:
3

2

2

(   ,  6 )    6,    ,  
3

3

2 ( 3t  8 )
3 t  16 t  12
s ( t ) 
2
NOTE: If you wanted to find the sign of the function s  , then you would
want to know when s ( t )  0 and when s  is undefined.
s ( t )  0  18 ( 3 t  8 )  0  t  
domain of s and s  .
8
. This number is now in the
3
s  undefined  3 t 2  16 t  12  0  ( t  6 ) ( 3 t  2 )  0 
t   6, t 
Sign of s ( t ) :
2
. These numbers are still not in the domain of s nor s  .
3


+

6

+


8
3
2
3
Thus, the function s is increasing on the set of real numbers given by
8

2

  6 ,     ,   and is decreasing on the set of real numbers given
3

3

 8 2
by (   ,  6 )    ,  .
 3 3
2
are not in the domain of the function s, then
3
8
the only critical number for the function s is t   . By the sign graph for
3
Since the numbers  6 and
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s ( t ) above, there is a local maximum occurring when t  
64
8
 8
 16    12
local maximum is s    = ln
3
3
 3
8
and the
3
=
 16

32
9
8
 16
 25 
ln 4 
 4    3  = ln 4 

  = ln 4  
 =
3
3
3
 3
 3 
 3

100
ln
.
3
5.
y 
log 3 3  5 x
We are taking the logarithm base 3 of the expression 3  5 x . In order for
the logarithm to be defined, we need the expression 3  5 x to be positive.
The expression 3  5 x is positive when 3  5 x  0 . Then
3
3  5 x  0 when x  . Thus, log 3 3  5 x is a defined as a real
5
3
x

number whenever
. The square root of log 3 3  5 x will be
5
defined as a real number (not a complex number) if and only if
log 3 3  5 x  0 . Since the base of the logarithm is three, which is
greater than one, then log 3 3  5 x  0 if and only if 3  5 x  1 . If
the base of the logarithm would have been between 0 and 1, then
3
log 3 3  5 x  0 if and only if 3  5 x  1 and x  . Thus, the
5
domain of the function y is the set of all real numbers x such that
3  5 x  1 . In order to determine this set of real numbers, we need to
solve the inequality 3  5 x  1 . Since 3  5 x  5 x  3 , then we
can solve the inequality 5 x  3  1 . Thus, 5 x  3  1 
4
2
5 x  3  1 or 5 x  3   1  x 
or x  . Thus, the domain
5
5
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of the function
2

  ,  
5

Since y 
y is the set of real numbers given by
4

,

.
 5

log 3 3  5 x
1/ 2
= [ log 3 3  5 x ] , then
dy
1
 [ log 3 3  5 x ]  1 / 2 D x log 3 3  5 x =
dx
2
1
1
5
[ log 3 3  5 x ]  1 / 2 

2
ln 3 3  5 x =
1
1
5
[ log 3 5 x  3 ]  1 / 2 

2
ln 3 5 x  3 =
5
( 5 x  3 ) ( 2 ln 3 )
Answer:
5
log 3 3  5 x
=
( 5 x  3 ) ( ln 9 )
log 3 3  5 x
2

4

Domain:    ,    ,  
5

5

dy
5

dx
( 5 x  3 ) ( ln 9 ) log 3 3  5 x
NOTE: The domain of the function y is the set of real numbers given by
2
dy

4

   ,    ,   and the domain of the function
is the set of real
5
dx

5

2

4



,

,




.
numbers given by
5

5

6.
y  log [ 5 9  4 w ( 2 w 4  w 2  5 ) 6 ]
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Note: log x  log 10 x
We are taking the common (base 10) logarithm of the expression
5
9  4 w ( 2 w 4  w 2  5 ) 6 . In order for the logarithm to be defined,
4
2
6
we need the expression 5 9  4 w ( 2 w  w  5 ) to be positive.
4
2
6
Since the expression ( 2 w  w  5 ) is nonnegative for all values of w,
4
2
then we just need to see if the expression 2 w  w  5 can equal zero.
4
2
This expression is quadratic in w 2 . Since 2 w  w  5 does not factor,
we can use the quadratic formula to obtain the real solution(s) if they exist.
2
Calculating the discriminant of this quadratic, we have that b  4 a c =
1  4 ( 2 ) ( 5 )   39 . Thus, the four solutions to the quadratic equation
2 w 4  w 2  5  0 are complex numbers. Thus, the expression
2 w 4  w 2  5 can not equal zero for any real number w. Thus, the
expression
5
9  4 w ( 2 w 4  w 2  5 ) 6 will be positive if the expression
9  4 w is positive. The expression 5 9  4 w is positive when
9  4 w  0 . Solving this linear inequality, we have that 9  4 w  0
9
w

when
. Thus, the domain of the function y is the interval of real
4
9

numbers given by    ,  .
4

5
4
2
6
Since log [ 5 9  4 w ( 2 w  w  5 ) ] = log 5 9  4 w +
log ( 2 w 4  w 2  5 ) 6 =
1
log ( 9  4 w ) + 6 log ( 2 w 4  w 2  5 ) ,
5
4
2
6
then y  log [ 5 9  4 w ( 2 w  w  5 ) ] =
1
log ( 9  4 w ) + 6 log ( 2 w 4  w 2  5 ) . Thus,
5
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1 1
4
8w3  2w 
y 

 6
=
ln 10  5 9  4 w
2 w 4  w 2  5 
1
ln 10

4
12 w ( 4 w 2  1) 
 5( 4 w  9 )  2 w 4  w 2  5  =



1
3 w ( 4 w 2  1) 
 5( 4 w  9 )  2 w 4  w 2  5 


9

Answer: Domain:    , 
4

4
ln 10
4
y 
ln 10
7.

1
3 w ( 4 w 2  1) 
 5( 4 w  9 )  2 w 4  w 2  5 


 2  u3
f ( u )  log 3 
2
 5 u  17



2  u3
3
We are taking the logarithm base
of the expression
. In order
5 u  17
2
2  u3
for the logarithm to be defined, we need the expression
to be
5 u  17
2  u3
positive. Thus, we need to find the sign of the expression
.
5 u  17
2  u3
Sign of
:
5 u  17


+


17
5

3 2
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Thus, the domain of the function f is the set of real numbers given by
 17 3

  , 2 .
 5

 2  u3 
 = log 3 ( 2  u 3 )  log 3 ( 5 u  17 ) ,
Since f ( u )  log 23 
2
2
 5 u  17 
  3u 2

1
5

then f ( u ) 
=
ln ( 3 / 2 )  2  u 3
5 u  17 
 3u 2

1
5

ln ( 3 / 2 )  u 3  2
5 u  17 
Answer:
 17 3

Domain:   , 2 
 5

 3u 2

1
5
f ( u ) 

ln ( 3 / 2 )  u 3  2
5 u  17 
8.
 x ( x 2  5) 3
h( x )  ln 
 3 ( 4 x  7 ) 4



x ( x 2  5) 3
We are taking the natural logarithm of the expression
. In order
( 4 x  7 ) 4/3
x ( x 2  5) 3
for the logarithm to be defined, we need the expression
to be
( 4 x  7 ) 4/3
x ( x 2  5) 3
positive. Thus, we need to find the sign of the expression
.
( 4 x  7 ) 4/3
x ( x 2  5) 3
Sign of
:
( 4 x  7 ) 4/3
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
+


5

+


7
4

0
+

5
x ( x 2  5) 3
Thus, the expression
is positive for the set of real numbers
( 4 x  7 )4 / 3

given by  

the function h
7

 5, 
4

7
 7 
    , 0   ( 5 ,  ) . Thus, the domain of
4
 4 
is the set of real numbers given by

 7 
    , 0   ( 5 , ) .

 4 
5, 
 x ( x 2  5) 3 
4
2
 5 )  ln ( 4 x  7 ) ,
Since h( x )  ln 
4 / 3  = ln x  3 ln ( x
3
 (4 x  7) 
1
6x
16
1
2x
4
4

h
(
x
)


3





then
.
x
x2  5 3 4x  7 = x
x2  5
3( 4 x  7 )
Answer:

Domain:  

h( x ) 
9.
5, 
7
 7 
    , 0   ( 5 , )
4
 4 
1
6x
16
 2

x
x  5
3( 4 x  7 )
x ( x 2  5) 3
h( x )  ln
3
(4 x  7) 4
x ( x 2  5) 3
We are taking the natural logarithm of the expression
(4 x  7) 4
order for the logarithm to be defined, we need the expression
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
3
. In
x ( x 2  5) 3
3
(4 x  7) 4
x ( x 2  5) 3
to be positive. First of all, the expression
is defined as long as 4 x  7  0 . Thus, the expression
7
defined as long as x   . The defined expression
4
3
(4 x  7) 4
x ( x 2  5) 3
3
is
(4 x  7) 4
x ( x 2  5) 3
3
(4 x  7) 4
is
2
3
positive when x ( x  5 )  0  x  0 , x   5 , and x  5 .
Thus, the domain of the function h is the set of real numbers given by
7


 x : x   5 , x   , x  0 , and x  5  or
4


7

 7 
(   ,  5 )    5 ,      , 0   ( 0, 5 )  ( 5 ,  ) .
4

 4 
x ( x 2  5) 3
Since h( x )  ln
(4 x  7) 4/3
then h( x ) 
Answer:
or (   , 
2
= ln x  3 ln x  5 
4
ln 4 x  7 ,
3
1
2x
4
4
1
6x
16
 3 2
 
 2

=
.
x
x  5
3 4x  7
x
x  5
3( 4 x  7 )

Domain:  x : x  


5)  

h( x ) 
5, 
5, x  
7
, x  0 , and x 
4
7
 7 
    , 0   ( 0,
4
 4 

5 

5 )  ( 5 , )
1
6x
16
 2

x
x  5
3( 4 x  7 )
NOTE: We would do the following calculation in order to find h(  3 ) :
1
 18
16
1
9
16
10
135
32
h(  3 )   




=  
= 
=
3
4
 15
3
2
15
30
30
30
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

113
.
30
NOTE: This calculation could not have been done for the function h in
Example 7 above because  3 was not in the domain of that function.
10.
g ( t )  log 1
5
4
t 4  81
t 4  81
1
We are taking the logarithm base of the expression
5
for the logarithm to be defined, we need the expression
positive. The expression
4
t 4  81
. In order
t 4  81
4
4
t 4  81
to be
t 4  81
t 4  81
will be positive when the expression
t 4  81
4
t 4  81
is positive. Thus, we will need to find the sign of the expression
t 4  81
4
t 4  81
. In order to do this, we need to find the solutions to the
t 4  81
4
4
equations t  81  0 and t  81  0 . The four solutions to the
4
4
equation t  81  0 are complex numbers. Since t  81 =
( t 2  9 ) ( t 2  9 ) = ( t 2  9 ) ( t  3 ) ( t  3 ) , then the real solutions to
4
the equation t  81  0 are t   3 and t  3 .
Sign of
t 4  81
:
t 4  81

+

3
+

3
Thus, the domain of the function g is the set of real numbers given by
(   ,  3 )  ( 3,  ) .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Since g ( t )  log 15
4
 t 4  81 
t 4  81

log 1  4
5
t

81
t 4  81 =


1/ 4
=
 t 4  81 
1
1
 =
log 1  4
[ log 1 ( t 4  81 )  log 1 ( t 4  81 ) ] , then
5
5
5
4
4
 t  81 
 4t 3
1
1
4t 3 
g ( t )  
 4
=
4 ln ( 1 / 5 )  t 4  81
t  81 

t3 
1
1

=

 ln 5  t 4  81
t 4  81 
t3

ln 5


t 4  81
t 4  81

=
 ( t 4  81 ) ( t 4  81 )
( t 4  81) ( t 4  81 ) 

t3

ln 5


162
 162 t 3
 ( t 4  81 ) ( t 4  81 )  = ( t 4  81 ) ( t 4  81 ) ln 5


Answer:
Domain: (   ,  3 )  ( 3 ,  )
 162 t 3
g ( t )  4
( t  81 ) ( t 4  81 ) ln 5
NOTE: ln
11.
1
 ln 5  1   ln 5
5
f ( x )  log x 3  sin 3 x
3
Domain of f = { x : x  sin 3x  0 } . The three solutions to the equation
x 3  sin 3x  0 can be visualized since they are the x-coordinate of the
3
three points of intersection of the graphs of y  x and y  sin 3x . Of
course, one of the solutions is x  0 . The other two solutions could be
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
approximated using the Intermediate Value Theorem or Newton’s Method.
The approximate value of these two solutions are x   0.8377170856 and
x  0.8377170856 .
1
3 x 2  3 cos 3 x
3 ( x 2  cos 3 x )
f ( x ) 

=
ln 10
x 3  sin 3 x
( x 3  sin 3 x ) ln 10
3
Domain: { x : x  sin 3x  0 }
Answer:
3 ( x 2  cos 3 x )
f ( x ) 
( x 3  sin 3 x ) ln 10
12.
y  ln cos 7 ( 4 u 6 )
6
Domain of y = { u : cos ( 4 u )  0 } .
7
6
6
7
6
Since y  ln cos ( 4 u ) = ln [ cos ( 4 u ) ] = 7 ln cos ( 4 u ) , then
D u cos ( 4 u 6 )
168 u 5 sin ( 4 u 6 )
 [ sin ( 4 u 6 ) ] 24 u 5
dy
 7
= 7
= 
=
cos ( 4 u 6 )
cos ( 4 u 6 )
du
cos ( 4 u 6 )
 168 u 5 tan ( 4 u 6 )
6
Domain: { u : cos ( 4 u )  0 }
Answer:
dy
  168 u 5 tan ( 4 u 6 )
du
13.
y  log 3 sec 2
w
4
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Since secant is the reciprocal of cosine, then the domain of y =
{ w : cos w  0 } . Since the solution to the equation cos   0 is given

( 2 n  1) 
 n 
by  
, where n is an integer, then the solutions to
2
2
( 2 n  1) 
the equation cos w  0 are given by w 
, where n is a
2
( 2 n  1)
nonnegative integer. Note that since the expression
is negative
2
when n is a negative integer. Since w is positive, then n must be a
( 2 n  1) 
( 2 n  1) 2  2
w 
 w 
nonnegative integer. Since
,
2
4

( 2 n  1) 2  2
, where n is a
then { w : cos w  0 } =  w : w 
4


nonnegative integer  .

2
Since y  log 43 sec
y  2 
w = 2 log 3 sec
4
D w sec w
1

ln ( 3 / 4)
sec w
1
 w  1 / 2 tan
ln ( 3 / 4)
Answer:
w
=
tan
=
w , then
2

ln ( 3 / 4)
[ sec
w tan
sec
w]
w
1  1/ 2
w
2
=
w
w ln ( 3 / 4)

( 2 n  1) 2  2
, where n is a nonnegative
Domain:  w : w 
4


integer 

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
y 
14.
tan
w
w ln ( 3 / 4)
g ( x )  ln ( 6 e 8 x  7 e 4 x  20 )
8x
4x
The domain of g = { x : 6 e  7 e  20  0 } . Thus, we need to solve
8x
4x
the nonlinear inequality 6 e  7 e  20  0 . Since the expression
6 e 8 x  7 e 4 x  20 is quadratic in e 4 x and factors. We have that
6 e 8 x  7 e 4 x  20 = ( 2 e 4 x  5 ) ( 3 e 4 x  4 ) . Since the expression
3 e 4 x  4 is positive for all real numbers x, then we need that the factor
2 e 4 x  5 to be positive in order for 6 e 8 x  7 e 4 x  20  0 . Thus, we
4x
just need to solve the nonlinear inequality 2 e  5  0 . We have that
5
4x
2 e 4 x  5  0  e  . Since the function y  ln x is an
2
5
4x

increasing function for all positive real numbers, then e 
2
5
5
5
1 5
ln e 4 x  ln
 4 x ln e  ln
 4 x  ln
 x  ln .
2
2
2
4 2
8x
4x
Thus, the domain of g = { x : 6 e  7 e  20  0 } =
1 5 

1 5

 x : x  ln  =  ln ,   .
4 2 

4 2

Since g ( x )  ln ( 6 e
8x
 7e
4x
6e 8x 8  7e 4x 4
 20 ) , then g ( x ) 
=
6 e 8 x  7 e 4 x  20
4 e 4 x ( 12 e 4 x  7 )
6 e 8 x  7 e 4 x  20
Answer:
1 5 

1 5

x
:
x

ln
ln
,





Domain:
or
4 2 

4 2

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
4 e 4 x ( 12 e 4 x  7 )
g ( x ) 
6 e 8 x  7 e 4 x  20
15.
y sin x 2  ln ( x 3  2 y 4 )
y sin x
2
 ln ( x
3
 2 y )  y  sin x
4
2
3 x 2  8 y 3 y
 y ( cos x ) 2 x 
x3  2y4
2
 y  ( x 3  2 y 4 ) sin x 2  2 x y ( x 3  2 y 4 ) cos x 2  3 x 2  8 y 3 y  
y ( x 3  2 y 4 ) sin x 2  8 y 3 y  3 x 2  2 x y ( x 3  2 y 4 ) cos x 2 
y [ ( x 3  2 y 4 ) sin x 2  8 y 3 ]  3 x 2  2 x y ( x 3  2 y 4 ) cos x 2 
3 x 2  2 x y ( x 3  2 y 4 ) cos x 2
y 
( x 3  2 y 4 ) sin x 2  8 y 3
3 x 2  2 x y ( x 3  2 y 4 ) cos x 2
Answer: y 
( x 3  2 y 4 ) sin x 2  8 y 3
e x  ln
3
x y  ln cos y
e x  ln
3
x y  ln cos y  e x  ln ( x y ) 1/ 3  ln cos y 
3
16.
3
ex 
3
3
3
1 1
y 
 y sin y
1
 
( ln x  ln y )  ln cos y  e x 3 x 2   
3 x
y 
cos y
3
 3x2 e x 
3
1
y

  y tan y 
3x
3y
9 x 3 y e x  y  x y   3 x y y tan y 
3
Copyrighted by James D. Anderson, The University of Toledo
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9 x 3 y e x  y  x y  3 x y y tan y 
3
9 x 3 y e x  y  y ( x  3 x y tan y ) 
3
9x3 yex  y
y 
x  3 x y tan y
3
9x3 yex  y
Answer: y 
x  3 x y tan y
3
17.
x3
sec 4 x  ln 6  7 x 2 y 5
y
x3
sec 4 x  ln 6  7 x 2 y 5  sec 4 x  3 ln x  6 ln y  7 x 2 y 5 
y
4 sec 4 x tan 4 x  3 
4 sec 4 x tan 4 x 
1
y
 6
 7 ( 2 x y 5  x 2 5 y 4 y ) 
x
y
3
6 y

 14 x y 5  35 x 2 y 4 y 
x
y
4 x y sec 4 x tan 4 x  3 y  6 x y  14 x 2 y 6  35 x 3 y 5 y 
4 x y sec 4 x tan 4 x  3 y  14 x 2 y 6  35 x 3 y 5 y  6 x y 
4 x y sec 4 x tan 4 x  3 y  14 x 2 y 6  y ( 35 x 3 y 5  6 x ) 
4 x y sec 4 x tan 4 x  3 y  14 x 2 y 6
y 
35 x 3 y 5  6 x
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Examples Use logarithmic differentiation in order to differentiate the following
functions.
1.
x3 4 3x  5
y 
( x  7)2 (4 x3  9)
x3 4 3x  5
x 3 ( 3 x  5 ) 1/ 4
y 
 ln y  ln

( x  7)2 (4 x3  9)
( x  7)2 (4 x3  9)
ln y  3 ln x 
1
ln ( 3 x  5 )  2 ln ( x  7 )  ln ( 4 x 3  9 ) 
4
y
1
1
3
1
12 x 2
 3  
 2


y
x
4 3x  5
x  7
4x3  9
y
3
3
2
12 x 2





y
x
4( 3 x  5)
x  7
4x3  9
3
3
2
12 x 2 
y  y  


=
x
4
(
3
x

5
)
x

7
4 x 3  9 

x3 4 3x  5
3
3
2
12 x 2 




2
3
( x  7) (4 x  9)  x
4( 3 x  5)
x  7
4 x 3  9 
x3 4 3x  5
3
3
2
12 x 2 



Answer: y 
( x  7 ) 2 ( 4 x 3  9 )  x
4 ( 3 x  5)
x  7
4 x 3  9 
NOTE: This answer would not allow us to find the sign of y  , but it would
allow us to find the value of y  at any number in the domain of y  . For
example,
y(  1 ) 
 4 2
36 (  13 )
4 2

3
2
12 
  3  
 =

36 ( 13 )
8
6
 13 

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
3
1
12 

 
3 
 =
8
3
13 

4 2
2  936
635 4 2
117
104
288 
 635 




 =

 = 
.
36 ( 13 )  312
312
312
312 
36 ( 13 )  312 
146016
4
2.
y  x tan 3 x
y  x tan 3 x  ln y  ln x tan 3 x  ln y  ( tan 3 x ) ln x 
y
1
y
tan 3 x
 ( 3 sec 2 3 x ) ln x  ( tan 3 x ) 

 3 ( sec 2 3 x ) ln x 
y
x
y
x
tan 3 x 

 y  y  3 ( sec 2 3 x ) ln x 
=
x 

tan 3 x 

x tan 3 x  3 ( sec 2 3 x ) ln x 
x 

tan 3 x 
tan 3 x 
2
3
(
sec
3
x
)
ln
x

Answer: y  x
OR

x 
y  x tan 3 x
3.
y  ( x 3  5) 3x
2
 4
y  ( x 3  5) 3x
2
 4
 1
[ 3 x ( sec 2 3 x ) ln x  tan 3 x ]
 ln y  ln ( x 3  5 ) 3 x
2
 4
ln y  ( 3 x 2  4 ) ln ( x 3  5 ) 
y
3x2
3
2
 6 x ln ( x  5 )  ( 3 x  4 )  3

y
x  5
y
3x 2 (3x 2  4)
3
 6 x ln ( x  5 ) 

y
x3  5
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860


3x 2 (3x 2  4) 
3
y  y  6 x ln ( x  5 ) 
 =
x3  5


( x  5)
3
3x 2  4
3 x ( x  5)
3

3x 2 (3x 2  4) 
3
 6 x ln ( x  5 ) 
 =
x3  5


3x 2  4

x(3x 2  4) 
3
 2 ln ( x  5 ) 
x 3  5 

Answer: y  3 x ( x  5 )
3
3x 2  4
y  3 x ( x 3  5 ) 3 x
4.
2
 5

x(3x 2  4) 
3
OR
 2 ln ( x  5 ) 
x 3  5 

[ 2 ( x 3  5 ) ln ( x 3  5 )  x ( 3 x 2  4 ) ]
y  ( sin 2 t ) csc t
y  ( sin 2 t ) csc t  ln y  ln ( sin 2 t ) csc t  ln y  ( csc t ) ln sin 2 t 
y
2 cos 2 t
  ( csc t cot t ) ln sin 2 t  csc t 

y
sin 2 t
y
  ( csc t cot t ) ln sin 2 t  csc t  ( 2 cot 2 t ) 
y
y
  ( csc t cot t ) ln sin 2 t  2 csc t cot 2 t 
y
y  y [  ( csc t cot t ) ln sin 2 t  2 csc t cot 2 t ] =
( sin 2 t ) csc t csc t [ 2 cot 2 t  ( cot t ) ln sin 2 t ]
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
csc t
Answer: y  ( sin 2 t ) csc t [ 2 cot 2 t  ( cot t ) ln sin 2 t ]
5.
y  [ f ( x ) ] g( x )
y  [ f ( x ) ] g ( x )  ln y  ln [ f ( x ) ] g ( x )  ln y  g ( x ) ln [ f ( x ) ] 
y
f ( x )
 g ( x ) ln [ f ( x ) ]  g ( x ) 

y
f (x)
y
g ( x ) f ( x )
 g ( x ) ln [ f ( x ) ] 

y
f (x)

g ( x ) f ( x ) 
y  y  g ( x ) ln [ f ( x ) ] 
=
f ( x ) 


g ( x ) f ( x ) 
[ f ( x ) ] g ( x )  g ( x ) ln [ f ( x ) ] 
=
f ( x ) 

 f ( x ) g ( x ) ln [ f ( x ) ]
g ( x ) f ( x ) 
[ f ( x ) ] g( x ) 

=
f ( x)
f ( x ) 

 f ( x ) g ( x ) ln [ f ( x ) ]  g ( x ) f ( x ) 
[ f ( x ) ] g( x ) 
 =
f ( x)


[ f ( x ) ] g ( x )  1 [ f ( x ) g ( x ) ln [ f ( x ) ]  g ( x ) f ( x ) ]
g( x )
Answer: y  [ f ( x ) ]
 1
[ f ( x ) g( x ) ln [ f ( x ) ]  g ( x ) f ( x ) ]
Examples Determine the interval(s) on which the following functions are
increasing and decreasing, the interval(s) on which they are concave upward and
concave downward. Also, find their local maximum(s), local minimum(s), and
inflection point(s).
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
1.
f ( x )  x log 5 x  2 x
The domain of the function f is the interval ( 0 ,  ) .
f ( x )  log 5 x  x 
1
ln x
1
 2 =

 2 =
x ln 5
ln 5
ln 5
1
1
( ln x  1  2 ln 5 ) =
( ln x  1  ln 25 ) =
ln 5
ln 5
1 
x

 1
 ln
ln 5  25

f ( x ) 
1
1 1
1
( ln x  1  ln 25 )  f ( x ) 
  
ln 5
ln 5  x 
x ln 5
f ( x )  0  ln
x
x
x
 1  0  ln
 1 
 e 1 
25
25
25
x
1
25

 x 
25
e
e

Sign of f ( x ) :

0
+

25
e
 25

,   and is decreasing on
The function f is increasing on the interval 
 e

25 

0
,

.
the interval
e 

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
25
and the local
e
 ln x

 25 

 ,
x

2
f


f
(
x
)

x
log
x

2
x
minimum is
. Since
=

5
 e 
 ln 5

The function f has a local minimum occurring when x 
 ln 25 e  1


 2  =
 ln 5

1
 ln 25  ln e
2 ln 5 
ln 25 
 1  ln 25  ln e
 = 25 e  1 
 =
25 e 


ln
5
ln
5
ln
5
ln
5




 ln 25  1  ln 25 
25 
1 
25
 =
 
 = 
25 e  1 
e ln 5
ln 5
e  ln 5 


 25 
1
f
 = f ( 25 e  1 ) = 25 e
then 
 e 
Sign of f ( x ) :
+

0
The function f is concave upward on the interval ( 0 ,  ) and is concave
downward nowhere.
The function f has no inflection point.
Answer:
 25

, 
Increasing: 
 e

25 


Decreasing:  0 ,
e 

Local Maximum: None
Local Minimum: 
CU: ( 0 ,  )
CD: Nowhere
Inflection Point: None
Since D x ln x 
1
, then
x

dx

x

1
dx  ln x  c .
x
Copyrighted by James D. Anderson, The University of Toledo
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25
e ln 5
Examples Evaluate the following integrals.
1.

8
3
1
dx
x
1
is continuous on the closed interval [ 3 , 8 ] . It
x
only has a discontinuity at x  0 . Thus, we can apply the Fundamental
Theorem of Calculus.
The integrand f ( x ) 

8
3
1
dx  ln x
x
Answer: ln
2.


8
3
 ln 8  ln 3  ln
8
3
8
3
7x2
dx
4x3  5
3
Let u  4 x  5
2
Then du  12 x dx

7x2
dx = 7
4x3  5

x2
7
dx =
3
4x  5
12

12 x 2
7
dx =
3
4x  5
12
7
7
ln u  c =
ln 4 x 3  5  c
12
12
7
ln 4 x 3  5  c
Answer:
12
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

du
=
u
3.

7
5
t
dt
2
9  t
t
is continuous on the closed interval [ 5 , 7 ] .
9  t2
It only has discontinuities at t   3 and t  3 . Thus, we can apply the
Fundamental Theorem of Calculus.
The integrand f ( t ) 
2
Let u  9  t
Then du   2 t dt

7
5
t
1
dt

=
9  t2
2
1
 ln u
2

 16
 40
=

7
5
 2t
1
dt

=
9  t2
2

 40
 16
du
1
=
u
2

 16
 40
du
=
u
1
1 16
1
2
[ ln 16  ln 40 ] =
ln
ln
=
2
2
40
2
5
2
NOTE: Since u  9  t , then when t  5 , we have that u  9  25 .
Thus, the new lower limit of integration is  16 . Similarly, when t  7 , we
have that u  9  49   40 . Thus, the new upper limit of integration is
 40 .
1
2
ln
Answer:
2
5
4.

4 t 3  5t 2  6
dt
8t

4 t 3  5t 2  6
1
dt =
( 4 t 2  5 t  6 t  1 ) dt =

8t
8

1  4t3
5t 2


6
ln
t
  c
8  3
2

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

1  4t3
5t 2


6
ln
t
Answer:
  c
8  3
2

5.

2
6
5  7x
dx
3 x2
5  7x
is continuous on the closed interval [  6 ,  2 ] .
3x 2
It only has a discontinuity at x  0 . Thus, we can apply the Fundamental
Theorem of Calculus.
The integrand f ( x ) 

2
6
5  7x
1
dx =
2
3x
3

2
6
(5 x
2
 7x
1
1
) dx =
3
2
 5x1


7
ln
x
=
 1


 6
2
15
1 5

 5

   7 ln x 
=     7 ln 2     7 ln 6   =
3x
3 2
 6
 6

5
1 5
 15




7
ln
2


7
ln
6



7
ln
2

7
ln
6
=
 6

 =
6
3  3

1
3
1
3
1 5
1 5
6
5



7
ln
2

7
ln
6

7
(
ln
6

ln
2
)

7
ln
=
=
=
 3


3  3
3  3
2 
1 5
1  5  21 ln 3 
5  21 ln 3


7
ln
3
=
=


3  3
3 
3
9
Answer:
6.

5  21 ln 3
9
dw
w ( ln w ) 3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Let u  ln w
1
du

dw
Then
w

dw
w ( ln w ) 3 =

1
 c
2 ( ln w ) 2

du
=
u3
u
3
1
u2
 c =
 c = 
du =
2u 2
2
1

 c
Answer:
2 ( ln w ) 2
7.

dt
64
4
6t
log 4 t
4
1
The integrand f ( t ) 
6t
4
log 4 t
is continuous on the closed interval
[ 4 , 64 ] . It only has a discontinuity at t  0 . Thus, we can apply the
Fundamental Theorem of Calculus.
Let u  log 4 t
1
dt
Then du 
t ln 4

dt
64
4
ln 4
6
6t

log 4 t
4
du
3
1
4
u
1
= 6
ln 4
=
6


dt
64
4
t
3
 1/ 4
1
4
log 4 t
1
ln
4

=
6

dt
64
4
t ln 4 4 log 4 t
3
u

ln 4  4 3 / 4 
2 ln 4 3 / 4
u
du =
u
=

6  3
9
1
Copyrighted by James D. Anderson, The University of Toledo
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=

3
1
=
( 4 27  1 ) ln 16
2 ln 4 3 / 4
2 ln 4 4
(3
 1) =
( 27  1 ) =
9
9
9
NOTE: Since u  log 4 t , then when t  4 , we have that u  log 4 4  1 .
Thus, the new lower limit of integration is 1. Similarly, when t  64 , we
have that u  log 4 64  3 . Thus, the new upper limit of integration is 3.
( 4 27  1 ) ln 16
Answer:
9
8.

csc 2 
d
8  3 cot 
Let u  8  3 cot 
2
Then du   3 csc  d
1
csc 2 
d


 8  3cot  = 3



du
1

ln u  c =
=
3
u
1
ln 8  3 cot   c
3
Answer: 
9.

 3 csc 2 
1
d = 
8  3 cot 
3
1
ln 8  3 cot   c
3
4e 2x
dx
e 2x  4
2x
Let u  e  4
2x
Then du  2 e dx
Copyrighted by James D. Anderson, The University of Toledo
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4e 2x
e 2x
4
dx
4
dx
=  2x
=
e 2x  4
e  4
2


2e 2x
du
dx
2
= 
=
e 2x  4
u
2 ln u  c = 2 ln e 2 x  4  c = ln ( e 2 x  4 ) 2  c
2x
2
Answer: ln ( e  4 )  c
3 / 4
10.

/6
8 sin 5 w
dw
3  2 cos 5 w
The integrand f ( w ) 
8 sin 5 w
is continuous on the closed interval
3  2 cos 5 w
  3 
 6 , 4  . It does not have any discontinuities. Thus, we can apply the
Fundamental Theorem of Calculus.
Let u  3  2 cos 5 w
Then du   2 (  sin 5 w ) 5 dw = 10 sin 5 w dw
3 / 4

=
/6
4
5

3 / 4
8 sin 5 w
sin 5 w
8
dw = 8 
dw =
 / 6 3  2 cos 5 w
10
3  2 cos 5 w
3
3
4 3 
ln
5
3 
2
3
du
4
=  ln u
u
5

3 
2
3 
3
=
4
[ ln ( 3 
5
3 / 4

/6
10 sin 5 w
dw
3  2 cos 5 w
2 )  ln ( 3 
2
3
Copyrighted by James D. Anderson, The University of Toledo
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3 )] =
NOTE: Since u  3  2 cos 5 w , then when w 
u  3  2 cos

3 
5

  3 
3

2
=

2 
6

3 . Thus, the new lower
3
, we have that
4
7 
15
7

2 cos
 = 3  2 cos
= 3  2 cos  2  
4 
4
4

2 
  3  2
. Thus, the new upper limit of integration is
2 
limit of integration is 3 
u  3 

, we have that
6
3 . Similarly, when w 

3  2


3  2.
4 3 
ln
Answer:
5
3 
2
3
Copyrighted by James D. Anderson, The University of Toledo
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