1.
Derivatives of Inverse Functions – Notes (1)
ex / Given: f ( x)  y  e x and let g ( x)  f 1( x)
(a) If (a, b)  f , find f '(a) and g '(b) .
Recall f '( x)  e x and ea  b , so f '(a)  e a  b .
From the Derivative of an Inverse Theorem, we have: f '(a) 
1
g '(b)
d (ln x) 1

dx
x
(b) Use implicit differentiation to find the derivative of the inverse of y = ex.
Here we’ll use an old trick exchanging the x & y to find the inverse.
So we begin with: x  e y . Now take the derivative w.r.t. ‘x’.
So g '(b)  b1 . Recall:
d
dx
 x   dxd  e y   1  e y  dy
dx
(chain rule)
Substituting x for ey (or keeping in mind that y = ln x)…
1 dy
1 d

or
  ln x 
x dx
e y dx
2.
ex / Given: f ( x)  y  Sin x and let g ( x)  f 1( x)  Sin -1 x
(a) If
 6 , 12   f , find f ' 6  and g ' 12 .
Recall f '( x)  cos x with D f    2 , 2  and cos
1
2 2 3


So f ' 6  23 and g ' 12 
3
3
3
 
 
 6  
3
2
 
2
(b) Use implicit differentiation to get the derivative of the inverse of y  Sin x
Exchange the ‘x’ and the ‘y’ to get: x = Sin y. Take the derivative w.r.t. x:
dy
1
dy
. Now we pull out that ‘Rt Triangle Trick Trick’!
1   cos y   dx 

dx cos y
If sin y = x, what is cos y = ? (See figure below) cos y =
1  x2
dy/dx = 1/cos y or


1 x 2
1
d
1
Sin 1x 
dx
1  x2
=
1  x2
Derivatives of Inverse Functions – Notes (2)
3. Given f ( x)  y  Tan x, D   2 , 2 with 4 ,1  f ( x) with f 1 ( x)  g ( x)  Tan 1x ,
(a) Find f '
d
dx
 4  and g'1


 Tan x   sec2 x and cos  4  
f'
4

 sec2
4  


2
2
 
2
2

1
2
 sec
 4  
 2 and then we’ll use:
1
2
(b) Use implicit differentiation to find
2
f ' a  
1
g '(b)
to get: g ' 1 
d
dx
 Tan x .
1
To find the inverse function for y = Tan x, we exchange the ‘x’ and the ‘y’ to get:
x  Tan y and then we take the derivative w.r.t. ‘x’ to get:
1
dy
d
1  sec2 y  dx  dx
Tan 1x 
 cos 2 y and then we use the rt  trig trick:
2
sec y




So if Tan y = x, then
1
1
cos y 
 cos2 y 
1  x2
1  x2
1
d
Hence dx
Tan 1x 
1  x2
1  x2


4. Derivatives of the other Inverse Trig Functions
1
1
d
d
Cot 1x 
and dx
Cos1x 
dx
1  x2
1  x2
both of which can be proven by taking the derivative w.r.t. ‘x’ of the trig identities:
Sin 1x + Cos1x = 2 and Tan 1x + Cot 1x = 2




We don’t worry too much about Inverse Secant and Inverse Cosecant since we can
Always use the following identities involving Inverse Cosine and Inverse Sine:
Sec1x = Cos 1 1x and Csc1x = Sin 1 1x
By the way… this doesn’t quite work for Inverse Cotangent, since:
Tan 1 1
,x  0
x
1
Cot x  
1
Tan 1x   , x  0
ex/ Cot 11  Tan 1 11  4 but… ex/ Cot 1  1  34 and Tan 1  11      4    34