2/29/2012
Math 103 – Rimmer
3.9Inverse Trig.
Functions
f ( x ) = arcsin x
or f ( x ) = sin −1 x
Math 103 – Rimmer
3.9Inverse Trig.
Functions
 −π π 
Definition : y = arcsin x is the number in 
,
for which sin y = x
 2 2 
 3
Find arcsin 
 .
 2 
 −1 
Find arcsin   .
 2 
Find arcsin ( −1) .
1
2/29/2012
What is the derivative of y = arcsin x?
from before:
f ( x ) = sin x
1
′
( f −1 ) ( x ) = f ′ f −1 ( x )
f ′( x) =
(
)
( f )′ ( x ) =
−1
( f )′ ( x ) =
−1
( f )′ ( x ) =
−1
f −1 ′ ( x ) =
( )
Math 103 – Rimmer
3.9Inverse Trig.
Functions
f −1 ( x ) = arcsin x
f −1 ′ x = ?
( )( )
since cos x = 1 − sin 2 x , we have
this is better written as
remember sin ( arcsin x ) = x
y = arcsin x
y′ =
Let y = arcsin ( 3x ) .
Math 103 – Rimmer
3.9Inverse Trig.
Functions
 3
Find y ′ 
 .
 6 
2
2/29/2012
Math 103 – Rimmer
3.9Inverse Trig.
Functions
f ( x ) = arccos x or f ( x ) = cos −1 x
Math 103 – Rimmer
3.9Inverse Trig.
Functions
Definition : y = arccos x is the number in [ 0, π ] for which cos y = x
 −1 
Find arccos   .
 2 
 3
Find arccos 
 .
 2 
Find arccos ( −1) .
3
2/29/2012
What is the derivative of y = arccos x?
from before:
f ( x ) = cos x
1
′
f −1 ( x ) =
f ′( x) =
f ′ f −1 ( x )
( )
( f )′ ( x ) =
−1
( f )′ ( x ) =
−1
( f )′ ( x ) =
−1
f −1 ′ ( x ) =
(
)
Math 103 – Rimmer
3.9Inverse Trig.
Functions
f −1 ( x ) = arccos x
f −1 ′ x = ?
( )( )
since sin x = 1 − cos 2 x , we have
this is better written as
remember cos ( arccos x ) = x
y = arccos x
( )
y′ =
Fall 2010
Math 103 – Rimmer
3.9Inverse Trig.
Functions
4
2/29/2012
f ( x ) = arctan x or f ( x ) = tan x
−1
Math 103 – Rimmer
3.9Inverse Trig.
Functions
lim arctan x =
x→− ∞
lim arctan x =
x →∞
Math 103 – Rimmer
3.9Inverse Trig.
Functions
 −π π 
Definition : y = arctan x is the number in 
,
for which tan y = x
 2 2 
Find arctan ( −1) .
Find arctan
( 3 ).
 −1 
Find arctan 
.
 3
5
2/29/2012
What is the derivative of y = arctan x?
from before:
f ( x ) = tan x
1
′
f −1 ( x ) =
f ′( x) =
f ′ f −1 ( x )
( )
(
)
( f )′ ( x ) =
−1
( f )′ ( x ) =
−1
( f )′ ( x ) =
−1
f −1 ′ ( x ) =
( )
f −1 ( x ) = arctan x
f −1 ′ x = ?
( )( )
since sec2 x = 1 + tan 2 x, we have
this is better written as
remember tan ( arctan x ) = x
y = arctan x
y′ =
Why is the word arc used?
Math 103 – Rimmer
3.9Inverse Trig.
Functions
1
1 + x2
Math 103 – Rimmer
3.9Inverse Trig.
Functions
arclentgh s = rθ
On the unit circle r = 1,so s = θ
6