Transcript Sec. 5.8 Inverse Trig Functions and Differentiation

Sec. 5.8 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 [-2, 2 ].
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 [- /2, /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, ]
Range [-1,1]
-2.0
Y= arccos (x) on [-1,1]
range [0, ]
arctan(x) has a range of (- /2, /2)
arccot(x) has a range of (0, )
arcsec(x) has a range of [0, ] y

/2
arccsc(x) has a range of [- /2, /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 - /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 /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 3.
The tan 60 = 3 or x = /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 - /2 < y < /2 then
sin(arcsin x)=x and arcsin(siny)=y
If - /2 < y < /2 then
tan(arctan x)=x and arctan(tany)=y
If -1 < x < 1 and 0 < y < /2 or /2 < y <  then
sec(arcsec x)=x and arcsec(secy)=y
On the next slide we will see how these properties
are applied
Inverse Properties Examples
4
Solve for x: arctan( 2 x  3)  
 This is y
If - /2 < y < /2 then
tan(arctan x)=x and arctan(tany)=y
Thus:

tan arctan( 2 x  3)   tan
4
2x  3  1
2x  4
x2
Inverse Properties Examples
If y  arcsin( x) where 0 < y < /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
 2 

y  arc sec

If
 5  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
1  2 x 
1  4x2
dx
2


d arcsin( u )
2
2
Examples Using the Derivatives of the Inverse Trig
Functions
Find the derivative of
arctan 3x
Note: u’ = du/dx
d arctan( u )
u

dx
1 u2
Let u = 3x
du/dx = 3
dx
1  (3x) 2 1  9 x 2


d arctan( u )
3
3
Examples Using the Derivatives of the Inverse Trig
Functions
Find the derivative of
 
arc sec e 2 x
d arc sec(u )
u

dx
u u 2 1
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
 
For your next test make a 4x6 notecard and copy
all of the derivatives summarized on page 382.
Some homework examples:
22. Page 383
3

Let
arctan 
x
 5 

  3 
sec arctan 
3
 
tan 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 homework examples:
24. Page 383
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 homework examples:
48. Page 384
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
On Wednesday there will be
Quiz 3B on the homework
for 5.1, 5.2, 5.3, 5.4, 5.5, 5.8
While you are in BB, please go to Communication
then to discussion board and answer the question I
have posed. Thanks.
This is where
I’ve been: in
Orlando, Fl
at the Swan Hotel
on the Disney
property.