Aim: How do we find the derivative of inverse functions?
Theorem: The derivative of an inverse function: Let f be a function that is differentiable on an
interval [a, b]. if f has an inverse function g ( x)  f 1 ( x) , then g is differentiable at any x
1
1
, f '( g ( x))  0 or g '(b) 
, f '(a )  0
g '( x) 
f '( g ( x))
f '(a)
Do Now: a) Given the function f ( x)  5x3  x  8 has an inverse f 1 ( x) . Find ( f 1 ) '(8)
f '( x)  15 x 2  1
8  5 x3  x  8
x0
( f 1 ) '(8) 
1
1
15(0)  1
b) Given the function f ( x)  2 x5  x3  1 has the inverse function g ( x)  f
f '( x)  10 x 4  3x 2
4  2 x5  x3  1
x 1
g '(4) 
1
( x) . Find g '(4)
1
1

10(1)  3(1) 13
We will use the derivative of an inverse function Theorem to get the derivative of
sin 1 x,cos 1 x, and tan 1 x .
x  tan 1 ( y )
tan x  y
sec 2 xdx  1dy
dx
1
1
1



2
2
dy sec x 1  tan x
1 y2
f ( x)  sin x
f 1 ( x)  g ( x)  arcsin x
1
1

f '( g ( x)) cos( g ( x))
1
1


1  sin 2 ( g ( x))
1  x2
( f 1 ) '( x) 
f ( x)  cos x
f 1 ( x)  g ( x)  arccos x
1
1

f '( g ( x))  sin( g ( x))
1
1


1  cos 2 ( g ( x))
1  x2
( f 1 )' ( x) 
EX1: Differentiate each function
a) y  arcsin( x)  x
dy
1

1
dx
1  x2
b) y  arctan(e x )
Inside: u  e x
outside: tan 1 u
dy
1
ex
x

e

dx 1  (e x )2
1  e2 x
c) y  arcsin( x )
d) f ( x)  x arcsin( x 2 )
Inside: u  x
outside: sin 1 u
f’(x) = sin-1(x2)+2x2/SQRT(1-x4)
dy
1
1

 ( x) 1/2
dx
1  ( x )2 2

1
1

( 1  x )2 x 2 x  x 2
e) y 
arccos(3 x)
x
1
f) f ( x)  x arctan x  ln(1  4 x 2 )
4


1

 3  x  cos 1 (3 x) 1
2

dy  1  (3 x)


dx
x2

1  3
= 2
 cos 1 (3x) 
x  1  9 x2

1
1 1

 2x
2
1  x 4 1  4 x2
x
x
= tan 1 x 

2
1  x 2(1  4 x 2 )
f '( x)  1tan 1 x  x
1 
EX2: Find the line tangent to the graph of y  3x arcsin x at ( , ) .
2 4
y  3x arcsin x
dy
1
 3arcsin x  3x
dx
1  x2
dy
1.5
|x 0.5  3arcsin 0.5 
 3.303
dx
0.75
y

1
 3.303( x  )
4
2
Aim: How do we find the derivative of inverse functions?
Theorem: The derivative of an inverse function: Let f be a function that is differentiable on an
interval [a, b]. if f has an inverse function g ( x)  f 1 ( x) , then g is differentiable at any x
1
1
, f '( g ( x))  0 or g '(b) 
, f '(a )  0
g '( x) 
f '( g ( x))
f '(a)
Do Now: a) Given the
function f ( x)  5x3  x  8 . Find ( f 1 ) '(8)
b) Given the function f ( x)  2 x5  x3  1 has the
inverse function g ( x)  ( f 1 ) '( x) . Find g '(4)
We will use the derivative of an inverse function Theorem to derive the derivative of
arcsin x, arccos x, arctan x .
f ( x)  sin x
f 1 ( x)  g ( x)  arcsin x
EX1: Differentiate each function
a) y  arcsin( x)  x
f ( x)  cos x
b) y  arctan(e x )
f 1 ( x)  g ( x)  arccos x
c) y  arcsin( x )
e) y 
arccos(3 x)
x
d) f ( x)  x arcsin( x 2 )
1
f) f ( x)  x arctan x  ln(1  4 x 2 )
4
1 
EX2: Find the line tangent to the graph of y  3x arcsin x at ( , ) .
2 4