Download Inverse Trig. Functions & Differentiation

Inverse Trig.
Functions &
Differentiation
Section 5.8
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 FUNCTIONS
Equation 1
• As the definition of an inverse function
states
1
f ( x)  y  f ( y )  x
• that
• we have:
• sin 1 x  y  sin y  x and    y  
2
• Thus, if -1 ≤ x ≤ 1, sin-1x is the number
between  2 and  2 whose sine is x.
2
INVERSE SINE FUNCTIONS
• Evaluate:
1

• a. sin  
2
1
1
• b. tan(arcsin )
3
Example 1
Solve.
1.
1
y  arcsin
2
3.
3

sin  arctan 
4


3
2. arccos  
 2 


Solve.

3
2. arccos  
 2 


1
1. y  arcsin
2
1
sin   QI
sin y 
2 ref  
y

cos   QII

ref 
6
6
 3

3. sin  arctan   x
4

3
  arctan
4
3
5
tan  
4

tan   QI
3

 cos x
2
5
x
6
6
4
QI
3
x
sin    x
opp
x
hyp
3
x
5
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 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
•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:
dy
cos y 
1
dx
dy
1
and

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


2
dx cos y
1 x
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).
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 COSINE FUNCTIONS
Equation 4
•The inverse cosine function is handled
similarly. cos x  y  cos y  x and 0  y  
1
– 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.
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.
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 Trig. Functions
None of the 6 basic
trig. functions has
an inverse unless
you restrict their
domains.
Function
Domain
Range
y = arcsin x -1< x < 1
I & IV
y = arccos x -1< x < 1
I & II
y = arctan x  < x <  I & IV
y= arccot x  < x <  I & I
y = arcsec x
I & II
x 1
y = arccsc
I
&
IV
x 1
The Inverse Trigonometric Functions
Graphs of six inverse trigonometric functions :
The Inverse Trigonometric Functions
Graphs of six inverse trigonometric functions :
Inverse Properties
f (f
–1(x))
= x and f
–1(f
(x)) = x
Remember that the trig.
functions have inverses only in
restricted domains.
DERIVATIVES
d
1
1
(sin x ) 
• dx
1 x 2
Table 11
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 of Inverse Trig.
Functions
Let u be a
differentiable
function of x.
'
d
u
arcsin u  
2
dx
1 u
d
u '
arccos u  
2
dx
1 u
d
u'
arctan u  
2
dx
1 u
d
u
arc cot u  
2
dx
1 u
'
'
d
u
arc sec u  
2
dx
u u 1
d
u
arc csc u  
2
dx
u u 1
'
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 )
Find each derivative with respect to x.
2. f  x   arc sec  2x 
3.
y  x arctan x
Find each derivative with respect to x.
dy
2. f  x   arc sec  2x 
sec y tan y
2
dx
sec y  2 x
dy
 2cos y cot y
2x
dx
2
4x  1

1
 1 
 2
y



2
 2x   4 x  1 
1
dy
1

dx x 4 x 2  1
3. y  x arctan x
y  x
  arctan x
dy
d
x

d
2
1
dx
dx
tan  x sec 
dx
 1 
 x
 arctan x
2 
2 d
1 x 2
x
1 x
1
 1 x 
dx

x

 arctan x
d
1
2
1

1 x
2
dx 1  x


Find each derivative with respect to the given variable.
4. h  t   sin  arccos t 
5. f  x   2arcsin  x  1
Find each derivative with respect to the given variable.
4. h  t   sin  arccos t 
d
 sin
1
cos  t
dt
1
1 t 2

t
1  t 2 d

1
1 dt
d
1

dt
1 t 2
5. f  x   2arcsin  x  1
cos 
sin  x  1
1
x 1

1   x  1
2
h  t   sin  
d
1
dx
d
1

dx cos 
d
1

2
dx
1   x  1
d
h '  t   cos  
dt
t
1 
 

2
1
1 t 

t
1 t 2
f  x   2
d
f 'x  2
dx

2
1   x  1
2
Example 1
Differenti ate arcsin( 2 x)
arcsin( 2 x) 

1
1  (2 x)
2
1 4x2
2
2
Example 2
x
Differenti ate arctan 
 2

x
 
arctan 
 2
1
1

2 2
x
1   
 2
1

 x2 
21  
4

2

4  x2
Example 2
x
Differenti ate arctan 
 2

x
 
arctan 
 2
1
1

2 2
x
1   
 2
1

 x2 
21  
4

2

4  x2
Example 2
x
Differenti ate arctan 
 2

x
 
arctan 
 2
1
1

2 2
x
1   
 2
1

 x2 
21  
4

2

4  x2
Example 2
x
Differenti ate arctan 
 2

x
 
arctan 
 2
1
1

2 2
x
1   
 2
1

 x2 
21  
4

2

4  x2
Example 2
Differenti ate xarccos x
 x arccos x 
 arccos x  x
 arccos x 
1
1 x2
x
1 x2
Example 2
Differenti ate xarccos x
 x arccos x 
 arccos x  x
 arccos x 
1
1 x2
x
1 x2
Example 2
Differenti ate xarccos x
 x arccos x 
 arccos x  x
 arccos x 
1
1 x2
x
1 x2
Some homework examples:
Write the expression in algebraic form
Let arctan 3x   y
then tan y  3x
Solution: Use the right triangle
secarctan 3x
Now using the triangle we can
find the hyp.
1  9x2
3x
y
1
12  3x   1  9 x 2
2
1  9x2
sec y 
 1  9x2
1
Some homework examples:
Find the derivative of:
f ( x)  arctan x  arctan x 
1
2
1
2
Let u = x
du
1

dx 2 x
1
f ( x) 
2 x
1
 x
2
1

2 x 1  x 
Example

d
sec 1 x 2  x
dx

ux x
2
du
 2x 1
dx
1
du


2
u u  1 dx

2x 1
x x
2
x
2
 x 1
2
Example
d
tan 1 sin x 
dx
u  sin x
du
 cos x
dx
1 du


2
1  u dx
cos x

1  sin 2 x
Find an equation for the line
1
tangent to the graph of y  cot x at
x = -1
d
cot 1 x
dx
At x = -1
1

1 x2
1
1

2
1  (1)
2
Slope of
tangent line
When x = -1, y =
3  1
x  1
y

4
2
3
4