Transcript Sec. 5.8 Inverse Trig Functions and Differentiation

Inverse Trig Functions and
Differentiation
By
Dr. Julia Arnold
Since a function must pass the horizontal line
test to have an inverse function, the trig
functions, being periodic, have to have their
domains restricted in order to pass the
horizontal line test.
For example: Let’s look at the graph of sin x
on [-2pi, 2 pi].
y
1.0
-6.0
-5.0
-4.0
-3.0
-2.0
-1.0
-1.0
x
1.02.03.04.05.06.07.0
Flunks the
horizontal
line test.
By restricting the domain from [- pi/2, pi/2] we produce a
portion of the sine function which will pass the horizontal
line test and go from [-1,1].
y
1.0
-1.0
1.0
x
The inverse sine function
is written as y = arcsin(x)
which means that sin(y)=x.
Thus y is an angle and x is
a number. y
1.0
-1.0
Y = sinx
-1.0
1.0
-1.0
x
Y = arcsin x
We must now do this for each of the other trig functions:
y
3.0
2.0
1.0
2.0
x
3.0
1.0
-1.0
1.0
x
2.0
-1.0
Y = cos x on [0, pi]
Range [-1,1]
-2.0
Y= arccos (x) on [-1,1]
range [0, pi]
arctan(x) has a range of (- pi/2, pi/2)
arccot(x) has a range of (0, pi)
arcsec(x) has a range of [0, pi], y

pi/2
arccsc(x) has a range of [- pi/2, pi/2],
y

0
Evaluate without a calculator:
1
arcsin
2
1
x
2
Step 1: Set equal to x
arcsin
Step 2: Rewrite as
1
sin x 
2
Step 3: Since the inverse is only defined in
quadrants 1 & 4 for sin we are looking for an angle in
the 4th quadrant whose value is -1/2.
The value must be - pi/6.
Evaluate without a calculator:
arccos0
Step 1: Set equal to x
Step 2: Rewrite as
arccos0  x
cos x  0
Step 3: The inverse is only defined in quadrants 1 &
2 for cos so we are looking for the angle whose
value is 0.
The value must be pi/2.
Evaluate without a calculator:
 
arctan  3   x
arctan 3
tan x  3
Step 3: The inverse is only defined in
quadrants 1 & 4 for tan so we are
looking for the angle whose tan value
is sqr(3).
The tan 60 = sqr(3) or x = pi/3
Evaluate with a calculator:
arcsin .3
arcsin .3  x
Check the mode setting on your calculator. Radian
should be highlighted.
Press 2nd function sin .3 ) Enter.
The answer is .3046
Inverse Properties
If -1 < x < 1 and - pi/2 < y < pi/2 then
sin(arcsin x)=x and arcsin(siny)=y
If - pi/2 < y < pi/2 then
tan(arctan x)=x and arctan(tany)=y
If -1 < x < 1 and 0 < y < pi/2 or pi/2 < y < pi then
sec(arcsec x)=x and arcsec(secy)=y
On the next slide we will see how these properties
are applied
Inverse Properties Examples
Solve for x: arctan(2 x  3) 

4
If - pi/2 < y < pi/2 then
tan(arctan x)=x and arctan(tany)=y
Thus:
tan  arctan(2 x  3)   tan
2x  3  1
2x  4
x2

4
Inverse Properties Examples
If y  arcsin( x)
where 0 < y < pi/2 find cos y
Solution: For this problem we use the right triangle
1
x
1 x2
Cos(y) =
Sin(y) = x, thus the opp side must be x
and the hyp must be 1, so sin y = x
By the pythagorean theorem, this
makes
2
1

x
the bottom side
1 x2
Inverse Properties Examples
If
 5
y  arc sec 

 2 
find tan y
Solution: For this problem we use the right triangle
5
1
By the pythagorean theorem, this
makes
the opp side
2
2
tan(y) =
1
2
5  2  5 4  1 1
2
Derivatives of the Inverse Trig Functions
d arcsin( u )
u

dx
1 u2
d arccos( u )
 u

dx
1 u2
d arctan( u )
u

dx
1 u2
d arc cot(u )  u

dx
1 u2
d arc sec(u )
u

dx
u u 2 1
d arc csc(u )
 u

dx
u u 2 1
Examples Using the Derivatives of the Inverse Trig
Functions
Find the derivative of
arcsin 2x
Note: u’ = du/dx
d arcsin( u )
u

dx
1 u2
Let u = 2x
du/dx = 2
d  arcsin(2 x) 
dx

2
1  2x
2

2
1  4 x2
Examples Using the Derivatives of the Inverse Trig
Functions
Note: u’ = du/dx
Find the derivative of
arctan 3x
d arctan( u )
u

dx
1 u2
Let u = 3x
du/dx = 3
d  arctan(3x) 
dx
3

1  9x2
Examples Using the Derivatives of the Inverse Trig
Functions
Find the derivative of
d arc sec(u )
u

dx
u u 2 1
2x
arc sec(e )
2x
u

e
Let
du
 2e 2 x
dx
d arc sec(u )
2e 2 x
2


2
4x
2x
2x
dx
e
1
e
e
1
 
Some more examples:
Let arctan   3   x
 5 

 3  
3
tan x 
find sec  arctan     sec( x)
5
 5 

Solution: Use the right triangle on the coordinate graph
Now using the triangle we can
find sec x after we find the hyp.
 32  52
34
 9  25  34
34
sec( x) 
5
Some more examples:
Write the expression in algebraic form
Let arctan 3 x   y
secarctan 3x 
then tan y  3 x
Solution: Use the right triangle
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 more examples:
Find the derivative of:
1
2
f ( x)  arctan x  arctan x 
Let u = x
du
1

dx 2 x
1
f ( x) 
2 x
1
 x
2
1

2 x 1  x 
1
2
Please let us know if this presentation
has been beneficial.
Thanks.