LESSON 5 DERIVATIVES OF THE INVERSE
TRIGONOMETRIC FUNCTIONS
Recall the following from trigonometry. A discuss of the inverse sine, inverse
cosine, and inverse tangent functions is given in Lesson 9 of the MATH-1330
lecture notes.
Definition The inverse sine function, denoted by sin
1
, is defined by y  sin
if and only if sin y  x , where  1  x  1 and 
   
x  [  1, 1 ] and y    , 
 2 2
Notation: Sometimes, sin
sin  1 x .
1


 y  .
2
2
x is denoted by Arc sin x .
1
That is,
That is, Arc sin x =
By the definition of the inverse sine function, we have the following
1
1. sin
1
1 
 Arc sin 
2
2
6
3
3

 Arc sin

2
2
3
3.
sin  1
5.

2 
  Arc sin
sin  1  

2 


3 
1

  Arc sin
sin

6.


2


8.
1
2. sin
sin  1 1  Arc sin 1 

2
4.
2
2

 Arc sin

2
2
4
 1
sin  1     Arc sin
 2

 1
   
6
 2

2 

  

2 
4


3 

  

2 
3

9.
1
7. sin 0  Arc sin 0  0
sin  1 (  1)  Arc sin (  1)  
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x

2
1
Definition The inverse cosine function, denoted by cos , is defined by
y  cos  1 x if and only if cos y  x , where  1  x  1 and 0  y   . That
is, x  [  1, 1 ] and y  [ 0 ,  ]
1
Notation: Sometimes, cos x is denoted by Arc cos x .
cos  1 x .
That is, Arc cos x =
By the definition of the inverse cosine function, we have the following
1
1. cos
3.
cos  1
3
3

 Arc cos

2
2
6
1
1 
 Arc cos 
2
2 3
1
2. cos
2
2

 Arc cos

2
2
4


3 
3  5
1



 
cos


Arc
cos

4.




2
2
6






2 
2  3
1



 
cos


Arc
cos

5.




2
2
4




1
 1  2
 1
6. cos     Arc cos    
3
 2
 2
8.
cos  1 1  Arc cos 1  0
1
7. cos 0  Arc cos 0 
9.

2
cos  1 (  1)  Arc cos (  1)  
1
Definition The inverse tangent function, denoted by tan , is defined by
y  tan  1 x if and only if tan y  x , where x is any real number and


   

 y  . That is, y    , 
2
2
 2 2
1
Notation: Sometimes, tan x is denoted by Arc tan x .
tan  1 x .
That is, Arc tan x =
By the definition of the inverse tangent function, we have the following
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1
1. tan
1
1

 Arc tan

6
3
3
1
2. tan 1  Arc tan 1 

3
3.
tan  1 3  Arc tan
5.
tan  1 (  1)  Arc tan (  1)  
3 
4.

4


1 
1 

tan  1  
 Arc tan  
 


6
3 
3 



4
1
6. tan (  3 )  Arc tan (  3 )  

3
1
7. tan 0  Arc tan 0  0
1
Definition The inverse cotangent function, denoted by cot , is defined by
y  cot  1 x if and only if cot y  x , where x is any real number and 0  y   .
That is, y  ( 0 ,  )
1
Notation: Sometimes, cot x is denoted by Arc cot x .
cot  1 x .
That is, Arc cot x =
By the definition of the inverse cotangent function, we have the following
1
3  Arc cot
1. cot
1
3. cot
5.
3 

6
1
2. cot 1  Arc cot 1 
1
1

 Arc cot

3
3
3
cot  1 (  1)  Arc cot (  1) 

4
1
4. cot (  3 )  Arc cot (  3 ) 
5
6
3
4


1 
1  2 
1


cot


Arc
cot


6.




3
3
3




1
7. cot 0  Arc cot 0 
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
2
1
Definition The inverse secant function, denoted by sec , is defined by

or
2
3
  
 3 
y

0
,

  y 

x

(


,

1
]

[
1
,

)
. That is,
and
 2 
  , 2  .
2
y  sec  1 x if and only if sec y  x , where x  1 and 0  y 
1
Notation: Sometimes, sec x is denoted by Arc sec x .
sec  1 x .
That is, Arc sec x =
By the definition of the inverse secant function, we have the following
1
1. sec
3.
2
2

 Arc sec

6
3
3
sec  1 2  Arc sec 2 
1
5. sec ( 
2 )  Arc sec ( 
2 

4


2 
2  7 
1


sec


Arc
sec


4.




6
3
3





3
1
6. sec (  2 )  Arc sec (  2 ) 
8.
1
2  Arc sec
2. sec
2) 
4
3
5
4
1
7. sec 1  Arc sec 1  0
sec  1 (  1)  Arc sec (  1)  
1
Definition The inverse cosecant function, denoted by csc , is defined by

or
2
3
 
 3 
y

0
,

  y 

,
x

(


,

1
]

[
1
,

)
. That is,
and
.
2 
2
 2 

y  csc  1 x if and only if csc y  x , where x  1 and 0  y 
1
Notation: Sometimes, csc x is denoted by Arc csc x .
csc  1 x .
That is, Arc csc x =
By the definition of the inverse cosecant function, we have the following
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1
1. csc 2  Arc csc 2 
1
3. csc

6
1
2  Arc csc
2. csc
2
2

 Arc csc

3
3
3
1
5. csc ( 
2 )  Arc csc ( 
csc  1 (  1)  Arc csc (  1) 

4
1
4. csc (  2 )  Arc csc (  2 ) 
2) 
7
6
5
4


2 
2  4 
1


csc


Arc
csc


6.




3
3
3




8.
2 
1
7. csc 1  Arc csc 1 

2
3
2
1
Theorem If f ( x )  sin x , then f ( x ) 
1
1  x2
.
NOTE: The domain of the function f is the set { x :  1  x  1 } = [  1, 1 ] and the
domain of f  is the set { x :  1  x  1 } = (  1, 1) .
1
Proof Let y  f ( x ) . Then y  sin x . By the definition of the inverse sine
function, sin y  x , where 


 y  . Using implicit differentiation, we
2
2
obtain that y cos y  1 . Solving for y  , we have that y  

1
, where
cos y



  
cos


cos
 0 . Since cos 2 y  sin 2 y  1 for all y,
 y 


since
2
2
2
 2
2
2
2
then cos y  1  sin y  cos y   1  sin y . The angles y such that



 y 
are either in the I quadrant or IV quadrant. The cosine function is
2
2
positive in both of these quadrants. Thus, cos y 
1  sin 2 y . Since sin y  x ,
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then we have that cos y 
1  x 2 . Thus, y  
1  sin 2 y =
1
=
cos y
1
1  x2
.
1
Theorem If f ( x )  cos x , then f ( x )  
1
1  x2
.
NOTE: The domain of the function f is the set { x :  1  x  1 } = [  1, 1 ] and the
domain of f  is the set { x :  1  x  1 } = (  1, 1) .
1
Proof Let y  f ( x ) . Then y  cos x . By the definition of the inverse cosine
function, cos y  x , where 0  y   . Using implicit differentiation, we obtain
1
that (  sin y ) y   1 . Solving for y  , we have that y   
, where 0  y  
sin y
2
2
2
2
since sin 0  sin   0 . Since cos y  sin y  1 for all y, then sin y  1  cos y
 sin y   1  cos 2 y . The angles y such that 0  y   are either in the I
quadrant or II quadrant. The sine function is positive in both of these quadrants.
Thus, sin y 
=
1  cos 2 y . Since cos y  x , then we have that sin y 
1
1
1  x 2 . Thus, y   
= 
.
sin y
1  x2
1
Theorem If f ( x )  tan x , then f ( x ) 
1  cos 2 y
1
.
1  x2
NOTE: The domain of the functions f and f  is the set of all real numbers.
1
Proof Let y  f ( x ) . Then y  tan x . By the definition of the inverse tangent
function, tan y  x , where 


 y  . Using implicit differentiation, we
2
2
2
obtain that y sec y  1 . Solving for y  , we have that y  
1
. Since
sec 2 y
sec 2 y  tan 2 y  1 for all y in the domain of the secant and tangent functions, then
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sec 2 y  1  tan 2 y . Since tan y  x , then we have that sec 2 y  1  tan 2 y =
1
1
1  x 2 . Thus, y  
=
.
1  x2
sec 2 y
1
Theorem If f ( x )  cot x , then f ( x )  
1
.
1  x2
NOTE: The domain of the functions f and f  is the set of all real numbers.
1
Proof Let y  f ( x ) . Then y  cot x . By the definition of the inverse cotangent
function, cot y  x , where 0  y   . Using implicit differentiation, we obtain
1
2
that (  csc y ) y  1 . Solving for y  , we have that y   
. Since
csc 2 y
csc 2 y  cot 2 y  1 for all y in the domain of the cosecant and cotangent functions,
2
2
2
2
then csc y  1  cot y . Since cot y  x , then we have that csc y  1  cot y =
1
1

1  x 2 . Thus, y   
=
.
csc 2 y
1  x2
1
Theorem If f ( x )  sec x , then f ( x ) 
1
x x2  1
.
NOTE: The domain of the function f is the set { x : x  1 } = (   ,  1]  [1,  )
and the domain of f  is the set { x : x  1 } = (   ,  1)  (1,  ) .
1
Proof Let y  f ( x ) . Then y  sec x .
 
function, sec y  x , where y   0 ,  
 2
By the definition of the inverse secant
 3 
  , 2  . Using implicit differentiation,
1
we obtain that y  sec y tan y  1. Solving for y  , we have that y  
,
sec y tan y
    3 
y

 0,     ,
 since tan 0  tan   0 . Since sec 2 y  tan 2 y  1 for
where
2 
 2 
2
2
all y in the domain of the secant and tangent functions, then tan y  sec y  1 
    3 
tan y   sec 2 y  1 . The angles y such that y   0 ,     ,  are either in
2 
 2 
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the I quadrant or III quadrant. The tangent function is positive in both of these
quadrants. Thus, tan y 
tan y 
sec 2 y  1 =
sec 2 y  1 . Since sec y  x , then we have that
1
1
x 2  1 . Thus, y  
=
.
sec y tan y
x x2  1
1
Theorem If f ( x )  csc x , then f ( x )  
1
x x2  1
.
NOTE: The domain of the function f is the set { x : x  1 } = (   ,  1]  [1,  )
and the domain of f  is the set { x : x  1 } = (   ,  1)  (1,  ) .
1
Proof Let y  f ( x ) . Then y  csc x . By the definition of the inverse secant
    3 
function, csc y  x , where y   0 ,     ,  . Using implicit differentiation,
2 
 2 
1
we obtain that (  csc y cot y ) y  1 . Solving for y  , we have that y  
,
csc y cot y

3
    3 
 cot
 0 . Since csc 2 y  cot 2 y  1
where y   0 ,     ,  since cot
2 
2
2
 2 
for all y in the domain of the cosecant and cotangent functions, then
cot 2 y  csc 2 y  1  cot y   csc 2 y  1 . The angles y such that
    3 
y   0,     ,
 are either in the I quadrant or III quadrant. The cotangent
2 
 2 
function is positive in both of these quadrants. Thus, cot y 
csc y  x , then we have that cot y 
1
1
y  
= 
.
csc y cot y
x x2  1
csc 2 y  1 =
csc 2 y  1 . Since
x 2  1 . Thus,
Summarizing, we have the following differentiation formulas:
1.
D x ( sin  1 x ) 
1
1  x2
2.
D x ( cos  1 x )  
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1
1  x2
3.
D x ( tan  1 x ) 
1
1  x2
D x ( sec  1 x ) 
1
5.
x2  1
x
4.
D x ( cot  1 x )  
1
1  x2
D x ( csc  1 x )  
1
6.
x
x2  1
Theorem If u is a differentiable function of x, then
1
du
dx
1.
D x ( sin  1 u ) 
3.
D x ( tan  1 u ) 
1 du
1  u 2 dx
D x ( sec  1 u ) 
1
5.
1  u2
du
u u 2  1 dx
2.
4.
D x ( cot  1 u )  
1 du
1  u 2 dx
D x ( csc  1 u )  
1
6.
Proof Use the Chain Rule and the differentiation formulas above.
Examples Differentiate the following functions.
1.
f ( x )  sin  1 ( 5 x 3 )
f ( x ) 
1
1  (5x )
3 2
Answer: f ( x ) 
 D x (5x ) =
3
15 x 2
1  25 x 6
15 x 2
1  25 x 6
y  Arc cos ( 4 t )
2
2.
1
D x ( cos  1 u )  
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1  u2
du
dx
du
u u 2  1 dx
2
t 4 t ln 16
dy
1
1
t2
t2

 Dt ( 4 ) = 
 4 ( ln 4 ) 2 t = 
2
t2 2
2t 2
dt
1  16 t
1  (4 )
14
2t
2 t
t
NOTE: 4  ( 4 )  16
2
2
2
2
dy
t 4 t ln 16
Answer: dt  
2
1  16 t
3.
h( u )  cot  1 ( 6  e 4 u )
1
 e 4u 4
4u
h( u )  
 Du (6  e ) = 
=
1  ( 6  e 4u ) 2
1  36  12 e 4 u  e 8 u
4e 4u
37  12 e 4 u  e 8 u
4e 4u
Answer: h( u ) 
37  12 e 4 u  e 8 u
4.
 3
y  6 x 5 Arc tan  2
 7x





 4
 3 
1
 6  3
5

y  6  5 x Arc tan  2   x 
  x 
9  7
 =
 7x 

1


49 x 4
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860


 3
6  5 x 4 Arc tan  2

 7x




 3
6  5 x 4 Arc tan  2

 7x



 
 7  1




6x2

9  =


49 x 4  

 
 7  1




6x2
49 x 4 

4
9  49 x  =



49 x 4 


 3
6  5 x 4 Arc tan 
2
 7x



294 x 6
 

7 ( 49 x 4  9 )  =


 3
6 x  5 Arc tan 
2
 7x



294 x 2
 

7 ( 49 x 4  9 ) 

4

 3


y

6
x
5
Arc
tan

Answer:
2
 7x

4
5.


294 x 2
 

7 ( 49 x 4  9 ) 

g ( t )  ln sec  1 3 8  t 3
g ( t ) 
D t sec  1 3 8  t 3
sec  1 3 8  t 3
1
D t sec  1 3 8  t 3 =
3
1
3
8  t3
(8  t 3 ) 2/3
8  t3
(8  t 3 ) 2/3  1
 D t ( 8  t 3 ) 1/ 3 =
1
 ( 8  t 3 )  2 / 3 (  3t 2 ) =
1 3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860


1
( 8  t 3 ) 1/ 3
(8  t 3 ) 2/3
t2

3 2/3 =
 1 (8  t )
t2
t2
(8  t 3 ) (8  t 3 ) 2/3  1
g ( t ) 
=
( t 3  8) (8  t 3 ) 2 / 3  1
D t sec  1 3 8  t 3
1
=
sec  1 3 8  t 3
sec  1 3 8  t 3

t2
( t 3  8) (8  t 3 ) 2 / 3  1
t2
( t 3  8 ) ( 8  t 3 ) 2 / 3  1 sec  1 3 8  t 3
Answer: g ( t ) 
6.
y  csc
1
t2
( t 3  8 ) ( 8  t 3 ) 2 / 3  1 sec  1 3 8  t 3
x3
4
 x3 

 D x 
 3
6
=
4
x
x


1
4
16
dy
 
dx
x3
4
1
3

x
=  x
1
6
( x  16 )
4
16
dy
Answer: dx  
x
3
x 6  16
1
1 6
( x  16 )
16
= 
3x 2

4 =
12
x
x 6  16
12
x 6  16
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
=
Example Find the equation (in point-slope form) of the tangent line to the graph of
x
y

Arc
cos
the function
at the point for which x   3 .
2
 

(

3
,
y
(

3
)
)


3
,


Tangent Point:
6 

3
y(  3 )  Arc cos

2

6
1

 D x  (  x ) 1/ 2 

=
2



x


1  

 4 
1

  (  x )  1/ 2 (  1) 
x 4
 =
1 
4
1
y  
1
1
(4  x)
4

1
x
4
m tan  y (  3 ) 
Answer: y 

6
1
1
2 3

3
6
= 1
2

4  x

1
1
4
x
1
=
2  x(4  x)
3
6
( x  3)
Example If s( t )  Arc cot ( 4 t  3 ) is the position function which gives the
position (in meters) of a particle at time t (in hours), then find (a) the instantaneous
velocity and (b) the instantaneous acceleration of the particle at time t  2 hours.
v ( t )  s ( t )  
1
4

4


=
1  ( 4 t  3) 2
1  ( 4 t  3) 2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
v( 2 )  
4
4
2
 
 
1  25
26
13
Answer for (a): 
2 m
13 hr
4
2 1
v
(
t
)


Since
, then
2 =  4 [1  ( 4 t  3 ) ]
1  ( 4 t  3)
32 ( 4 t  3 )
a( t )  v ( t )  s( t )  4 [1  ( 4 t  3 ) 2 ]  2 2 ( 4 t  3 ) 4 =
[1  ( 4 t  3 ) 2 ] 2
a( 2 ) 
32 ( 5 )
8(5)
32 ( 5 )
16 ( 5 )
32 ( 5 )
40
=
=
=
=
2 =
2
( 1  25 )
13  13
26  26
13  26
26
169
Answer for (b):
40 m
169 hr 2
 1 3 x  12
y

tan
Examples Determine the interval(s) on which the function
x 2  9 is
increasing and decreasing. Then find the local maximum(s) and local minimum(s)
of the function.
y 
 3 x  12 
 2


D
x
2
x

9
 3 x  12 

 =

1   2
 x  9
1
1
3 ( x 2  9 )  ( 3 x  12 ) 2 x

=
( 3 x  12 ) 2
( x 2  9) 2
1 
( x 2  9) 2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
1
3 x 2  27  6 x 2  24 x

=
( 3 x  12 ) 2
( x 2  9) 2
1 
( x 2  9) 2
1
 3 x 2  24 x  27
 3( x 2  8 x  9 )

=
( 3 x  12 ) 2
( x 2  9) 2
( x 2  9 ) 2  ( 3 x  12 ) 2
1 
( x 2  9) 2
In order to find the sign of y  , we need to know when y   0 and when y  is
undefined.
Since y  is undefined when the denominator of y  , which is the expression
( x 2  9 ) 2  ( 3 x  12 ) 2 , is equal to zero. The expression
( x 2  9 ) 2  ( 3 x  12 ) 2 is the sum of the two nonnegative expressions
( x 2  9 ) 2 and ( 3 x  12 ) 2 . Thus, in order for the expression
( x 2  9 ) 2  ( 3 x  12 ) 2 to be equal to zero, we must have that ( x 2  9 ) 2  0
2
2
2
2
and ( 3 x  12 )  0 . Since ( x  9 )  0  x  9  0 , then solutions to
2
2
the equation ( x  9 )  0 are x   3 . However, neither of these numbers is
2
2
a solution to the equation ( 3 x  12 )  0 . Since ( 3 x  12 )  0 
3 x  12  0 , then solution to the equation ( 3 x  12 ) 2  0 is x   4 .
2
2
However, this number is not a solution to the equation ( x  9 )  0 . Thus, the
denominator of y  does not equal zero.
y   0  x 2  8 x  9  0 . Since the quadratic does not factor, we will use
the quadratic formula to solve this equation. Thus, x 
8 
4 ( 16  9 )
2
Sign of y  :
8  2
=
7
2

= 4 
4 
7

+

8 

7
4 
7
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
64  4 ( 1) 9
2
=
The function y is increasing on the interval (  4 
7 , 4 
The function y is decreasing on the set (   ,  4 
7 )  ( 4 
There is a local maximum occurring when x   4 
is y(  4 
tan  1
1
7 ) = tan
3
7
14  8
y(  4 
7 , ) .
7 . The local maximum
7 ) . There is a local minimum occurring when x   4 
and the local minimum is y(  4 
y(  4 
7 ).
7
7 ).
3(  4 
7 )  12
( 4 
7 )2  9
1
= tan
 12  3
16  8
7  12
7  7  9
=


3 7
3 7
1 
 =  tan  1
tan

=
 8 7  14 
7
8 7  14


1
7 ) = tan
3(  4 
7 )  12
( 4 
7 )2  9
1
= tan
 12  3
16  8
7  12
7  7  9
=


3 7
3 7
 =  tan  1
tan  1  
 14  8 7 
14  8 7


Answer:
Increasing: (  4 
7 , 4 
Decreasing: (   ,  4 
1
Local Maximum: tan
7)
7 )  ( 4 
3
7
14  8
7
7 , )
1
or  tan
3
8
7
7  14


3 7
3 7
1 
1

tan


tan
Local Minimum:
 14  8 7  or
14  8 7


Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860