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7
INVERSE FUNCTIONS
INVERSE FUNCTIONS
7.6
Inverse Trigonometric
Functions
In this section, we will learn about:
Inverse trigonometric functions
and their derivatives.
INVERSE TRIGONOMETRIC FUNCTIONS
Here, we apply the ideas of Section 7.1
to find the derivatives of the so-called
inverse trigonometric functions.
INVERSE TRIGONOMETRIC FUNCTIONS
However, we have a slight difficulty
in this task.
 As the trigonometric functions are not
one-to-one, they don’t have inverse functions.
 The difficulty is overcome by restricting
the domains of these functions so that they
become one-to-one.
INVERSE TRIGONOMETRIC FUNCTIONS
Here, you can see that the sine function
y = sin x is not one-to-one.
 Use the Horizontal Line Test.
INVERSE TRIGONOMETRIC FUNCTIONS
However, here, you can see that
the function f(x) = sin x,    x  
2
is one-to-one.
2
,
INVERSE SINE FUNCTION / ARCSINE FUNCTION
The inverse function of this restricted sine
function f exists and is denoted by sin-1 or
arcsin.
 It is called
the inverse
sine function
or the arcsine
function.
INVERSE SINE FUNCTIONS
Equation 1
As the definition of an inverse function states
1
that
f ( x)  y  f ( y )  x
we have:
1
sin x  y  sin y  x and 

2
 y
 Thus, if -1 ≤ x ≤ 1, sin-1x is the number between
and  2 whose sine is x.

2
 2
INVERSE SINE FUNCTIONS
Evaluate:
1
a. sin  
2
1
1
b. tan(arcsin )
3
Example 1
INVERSE SINE FUNCTIONS
We have:
Example 1 a
1 
sin   
2 6
1
 This is because sin  / 6  1/ 2 ,
and  / 6 lies between  / 2 and  / 2 .
INVERSE SINE FUNCTIONS
Example 1 b
1
1
Let   arcsin , so sin   .
3
3
 Then, we can draw a right triangle with angle θ.
 So, we deduce from the Pythagorean Theorem
that the third side has length 9  1  2 2 .
INVERSE SINE FUNCTIONS
Example 1b
 This enables us to read from
the triangle that:
1
1
tan(arcsin )  tan  
3
2 2
INVERSE SINE FUNCTIONS
Equations 2
In this case, the cancellation equations
for inverse functions become:
1
sin (sin x)  x
1
sin(sin x)  x
for 

x

2
2
for  1  x  1
INVERSE SINE FUNCTIONS
The inverse sine function, sin-1, has
domain [-1, 1] and range   / 2,  / 2 .
INVERSE SINE FUNCTIONS
The graph is obtained from that of
the restricted sine function by reflection
about the line y = x.
INVERSE SINE FUNCTIONS
We know that:
 The sine function f is continuous,
so the inverse sine function is also continuous.
 The sine function is differentiable,
so the inverse sine function is also differentiable
(from Section 3.4).
INVERSE SINE FUNCTIONS
We could calculate the derivative of sin-1
by the formula in Theorem 7 in Section 7.1.
However, since we know that is sin-1
differentiable, we can just as easily calculate
it by implicit differentiation as follows.
INVERSE SINE FUNCTIONS
Let y = sin-1x.
 Then, sin y = x and –π/2 ≤ y ≤ π/2.
 Differentiating sin y = x implicitly with respect to x,
we obtain:
and
dy
cos y 
1
dx
dy
1

dx cos y
INVERSE SINE FUNCTIONS
Formula 3
Now, cos y ≥ 0 since –π/2 ≤ y ≤ π/2, so
cos y  1  sin y  1  x
2
Therefore
2
dy
1
1


dx cos y
1  x2
d
1
1
(sin x) 
2
dx
1 x
1  x  1
INVERSE SINE FUNCTIONS
If f(x) = sin-1(x2 – 1), find:
(a) the domain of f.
(b) f ’(x).
(c) the domain of f ’.
Example 2
INVERSE SINE FUNCTIONS
Example 2 a
Since the domain of the inverse sine
function is [-1, 1], the domain of f is:
{x | 1  x  1  1}  {x | 0  x  2}
2
2
 {x | x | 2}
   2, 2 
INVERSE SINE FUNCTIONS
Example 2 b
Combining Formula 3 with the Chain Rule,
we have:
1
d 2
f '( x) 
( x  1)
2
2 dx
1  ( x  1)


1
1  ( x  2 x  1)
4
2
2x
2x  x
2
4
2x
INVERSE SINE FUNCTIONS
Example 2 c
The domain of f ’ is:
{x | 1  x  1  1}  {x | 0  x  2}
2
2
 {x | 0  | x |  2}
 ( 2, 0)  (0, 2)
INVERSE COSINE FUNCTIONS
Equation 4
The inverse cosine function is handled
similarly.
 The restricted cosine function f(x) = cos x, 0 ≤ x ≤ π,
is one-to-one.
 So, it has an inverse function denoted by cos-1 or arccos.
cos1 x  y  cos y  x and 0  y  
INVERSE COSINE FUNCTIONS
Equation 5
The cancellation equations are:
cos (cos x)  x for 0  x  
1
1
cos(cos x)  x for  1  x  1
INVERSE COSINE FUNCTIONS
The inverse cosine function,cos-1,
has domain [-1, 1] and range [0,  ] ,
and is a continuous function.
INVERSE COSINE FUNCTIONS
Formula 6
Its derivative is given by:
d
1
1
(cos x)  
2
dx
1 x
1  x  1
 The formula can be proved by the same method
as for Formula 3.
 It is left as Exercise 17.
INVERSE TANGENT FUNCTIONS
The tangent function can be made
one-to-one by restricting it to the interval
  / 2,  / 2 .
INVERSE TANGENT FUNCTIONS
Equation 7
Thus, the inverse tangent
function is defined as
the inverse of the function
f(x) = tan x,
 / 2  x   / 2 .
 It is denoted by tan-1
or arctan.
1
tan x  y  tan y  x and 

2
 y

2
INVERSE TANGENT FUNCTIONS
E. g. 3—Solution 1
Simplify the expression
cos(tan-1x)
 Let y = tan-1x.
 Then, tan y = x and  / 2  y   / 2.
INVERSE TANGENT FUNCTIONS
E. g. 3—Solution 1
We want to find cos y.
 However, since tan y is known, it is easier
to find sec y first.
 Therefore,
sec 2 y  1  tan 2 y  1  x 2
sec y  1  x
2
(since sec y  0 for   / 2  y   / 2)
INVERSE TANGENT FUNCTIONS
E. g. 3—Solution 1
Thus,
1
cos(tan x)  cos y
1

sec y
1

2
1 x
INVERSE TANGENT FUNCTIONS
E. g. 3—Solution 2
Instead of using trigonometric identities,
it is perhaps easier to use a diagram.
 If y = tan-1x, then tan y = x.
 We can read from the figure (which illustrates
the case y > 0) that:
cos(tan 1 )
 cos y
1

1  x2
INVERSE TANGENT FUNCTIONS
The inverse tangent function,
tan-1 = arctan, has domain
range ( / 2,  / 2).
and
INVERSE TANGENT FUNCTIONS
We know that:
lim  tan x   and
x ( / 2)
 So, the lines
x   / 2
are vertical
asymptotes of
the graph of tan.
lim  tan x  
x  ( / 2)
INVERSE TANGENT FUNCTIONS
The graph of tan-1 is obtained by reflecting
the graph of the restricted tangent function
about the line y = x.
 It follows that
the lines y = π/2
and y = -π/2
are horizontal
asymptotes of
the graph of tan-1.
INVERSE TANGENT FUNCTIONS
Equations 8
This fact is expressed by these limits:
1
lim tan x 
x 
1

2
lim tan x  
x 

2
INVERSE TANGENT FUNCTIONS
Evaluate:
Example 4
 1 
lim arctan 

x 2
 x2
1

 Since
x 2
as x  2
the first equation in Equations 8 gives:
 1  
lim arctan 


x 2
 x 2 2
INVERSE TANGENT FUNCTIONS
Since tan is differentiable, tan-1 is also
differentiable.
To find its derivative, let y = tan-1x.
 Then, tan y = x.
INVERSE TANGENT FUNCTIONS
Equation 9
Differentiating that latter equation implicitly
dy
with respect to x, we have:
2
sec y
1
dx
Thus, dy 
dx
1
1
1


2
2
2
sec y 1  tan y 1  x
d
1
1
(tan x ) 
2
dx
1 x
INVERSE TRIG. FUNCTIONS
Equations 10
The remaining inverse trigonometric
functions are not used as frequently and
are summarized as follows.
INVERSE TRIG. FUNCTIONS
Equations 10
1
y  csc x (| x |  1)  csc y  x
and
y  (0,  / 2]  ( ,3 / 2]
1
y  sec x (| x |  1)  sec y  x
and
y  [0,  / 2)  [ ,3 / 2)
1
y  cot x ( x  )  cot y  x
and
y  (0,  )
INVERSE TRIG. FUNCTIONS
The choice of intervals for y in
the definitions of csc-1 and sec-1
is not universally agreed upon.
INVERSE TRIG. FUNCTIONS
For instance, some authors use
y  0,  / 2   / 2,   in the definition
of sec-1.
 You can see from the graph of the secant function
that both this
choice and the
one in Equations
10 will work.
DERIVATIVES OF INVERSE TRIG. FUNCTIONS
In the following table, we collect
the differentiation formulas for all
the inverse trigonometric functions.
 The proofs of the formulas for the derivatives of
csc-1, sec-1, and cot-1 are left as Exercises 19–21.
DERIVATIVES
Table 11
d
1
1
(sin x ) 
dx
1 x 2
d
1
1
(csc x )  
dx
x x2  1
d
1
1
(cos x )  
2
dx
1 x
d
1
1
(sec x ) 
2
dx
x x 1
d
1
1
(tan x ) 
dx
1 x 2
d
1
1
(cot x )  
dx
1 x 2
DERIVATIVES
Each of these formulas can be
combined with the Chain Rule.
 For instance, if u is a differentiable function
of x, then
d
1 du
1
(sin u ) 
dx
1  u 2 dx
and
d
1 du
1
(tan u ) 
dx
1  u 2 dx
DERIVATIVES
Differentiate:
1
a. y 
1
sin x
b. f ( x )  x arctan x
Example 5
DERIVATIVES
Example 5 a
1
y
1
sin x
dy
d
1
1

(sin x )
dx dx
1
2 d
1
 (sin x )
(sin x )
dx
1

1
2
2
(sin x ) 1  x
DERIVATIVES
Example 5 b
f ( x )  x arctan x
f '( x )  x
1
1
2
(
x
2
1 ( x )
1/ 2
)  arctan x
x

 arctan x
2(1  x )
INVERSE TRIG. FUNCTIONS
Example 6
Prove the identity
tan x  cot x   / 2
1
1
 Although calculus is not needed to prove this,
the proof using calculus is quite simple.
INVERSE TRIG. FUNCTIONS
Example 6
If f(x) = tan-1x + cot-1x , then
1
1
f '( x ) 

0
2
2
1 x
1 x
for all values of x.
 Therefore f(x) = C, a constant.
INVERSE TRIG. FUNCTIONS
Example 6
To determine the value of C,
we put x = 1.
 Then,
1
1
C  f (1)  tan x  cot 1 
 Thus, tan-1x + cot-1x = π/2.

4


4


2
INTEGRATION FORMULAS
Equations 12 & 13
Each of the formulas in Table 11 gives
rise to an integration formula.
 The two most useful of these are:

1
1 x
2
dx  sin1 x  C
1
1
dx

tan
x C
 x2  1
INTEGRATION FORMULAS
Find:

1
1/ 4
1 4x
0
 If we write
Example 7

1/ 4
0
2
1
1 4x
2
dx
dx  
1/ 4
0
1
1  (2 x )
the integral resembles Equation 12 and
the substitution u = 2x is suggested.
 This gives du = 2dx; so dx = du/2.
2
dx
INTEGRATION FORMULAS
Example 7
When x = 0, u = 0.
When x = ¼, u = ½.
Thus,

1/ 4
0
1
1 4x
2
dx 
1/ 2
1
2 0

du
1 u
2
1/ 2
 sin u 
0
1
2
1
 21 sin1  21   sin1 0 
1 

  
2 6 12
INTEGRATION FORMULAS
Evaluate:
Example 8
1
dx
 x 2  a2
 To make the given integral more like Equation 13,
we write:
dx
 x 2  a2  
dx
1
dx
 2 2
2
 a x 
2 x
a  2  1
 2  1
a

a 
 This suggests that we substitute u = x/a.
INTEGRATION FORMULAS
Example 8
Then, du = dx/a, dx = a du, and
dx
1 a du
 x 2  a2  a2  u 2  1
1 du
  2
a u 1
1
1
 tan u  C
a
INTEGRATION FORMULAS
E.g. 8—Formula 14
Thus, we have the formula
1
1
1  x 
dx

tan

C


2
2
 x a
a
a
INTEGRATION FORMULAS
Find:
Example 9
x
dx
 x4  9
 We substitute u = x2 because then du = 2x dx
and we can use Equation 14 with a = 3:
x
1 du
1 1
1  u 
 x 4  9 dx  2  u 2  9  2  3 tan  3   C
2


1
x
1
 tan    C
6
 3 