Derivatives of Inverse
Trigonometric Function
Inverse Trig Functions

Some trig functions domains’ have to be restricted in order for them
to have an inverse function – why?
Only functions that are 1-to-1 can have inverse functions
Find

dy
dx

If

y = sin −1 ( x 3 )

dy
1
=
×( 3x 2 )
dx
3 2
1− ( x )

therefore

3x 2 )
(

dy
=
dx
3 2
1− ( x )
One more example
Find

dy
dx

dy
=
dx e x

y = sec−1 ( e x )

if

1

(e )

x 2

−1

dy
1
=
dx
e 2 x −1

×( e x )
Differentiability of Inverse Functions
If f(x) is differentiable on an interval I, one may wonder whether f-1(x) is also
differentiable? The answer to this question hinges on f'(x) being equal to 0 or
not . Indeed, if for any , then f-1(x) is also differentiable. Moreover we have

Using Leibniz's notation, the above formula becomes

which is easy to remember.
Example:
Confirm Differentiability of Inverse Function formula for the
function f (x) = x 3 +1

Solution:
y = x 3 +1
x = y3 +1

y = x −1
3

x = 3 y −1
f −1 (y) = 3 y −1

dy d  3 
=  x + 1 = 3x 2
dx dx
and

1
2
dx d  3
d
1
−
=  y − 1 =  ( y − 1) 3  = ( y − 1) 3
 dy 
 3
dy dy

(

)

2
2
dy
1
3 y−1 = 3 y−1 3 =
=3
( ) dx
dx
dy
MONOTONIC
FUNCTIONS:
Suppose that the domain of a function f is on an open interval I on which
f’(x) > 0 or on which f’(x) < 0. Then f is one-to-one, f-1(x) is differentiable
at all values of x in the range of f.
Example:
Consider the function

f (x) = x 5 + x +1

Solution:

f '(x) = 5x 4 +1
Since f’(x) > 0 on the entire domain, f(x) is
monotonic, therefore it has an inverse

.Show that f(x) is one-to=one function.
MONOTONIC
FUNCTIONS:
Suppose that the domain of a function f is on an open interval I on which
f’(x) > 0 or on which f’(x) < 0. Then f is one-to-one, f-1(x) is differentiable
at all values of x in the range of f.
Example:
Consider the function

f (x) = x 5 + x +1

Solution:

f '(x) = 5x 4 +1
Since f’(x) > 0 on the entire domain, f(x) is
monotonic, therefore it has an inverse

.Show that f(x) is one-to=one function.

4.3 derivatives of inv erse trig. functions

  • 1.
  • 2.
    Inverse Trig Functions Sometrig functions domains’ have to be restricted in order for them to have an inverse function – why? Only functions that are 1-to-1 can have inverse functions
  • 5.
    Find dy dx If y = sin−1 ( x 3 ) dy 1 = ×( 3x 2 ) dx 3 2 1− ( x ) therefore 3x 2 ) ( dy = dx 3 2 1− ( x )
  • 6.
    One more example Find dy dx dy = dxe x y = sec−1 ( e x ) if 1 (e ) x 2 −1 dy 1 = dx e 2 x −1 ×( e x )
  • 7.
    Differentiability of InverseFunctions If f(x) is differentiable on an interval I, one may wonder whether f-1(x) is also differentiable? The answer to this question hinges on f'(x) being equal to 0 or not . Indeed, if for any , then f-1(x) is also differentiable. Moreover we have Using Leibniz's notation, the above formula becomes which is easy to remember.
  • 8.
    Example: Confirm Differentiability ofInverse Function formula for the function f (x) = x 3 +1 Solution: y = x 3 +1 x = y3 +1 y = x −1 3 x = 3 y −1 f −1 (y) = 3 y −1 dy d  3  =  x + 1 = 3x 2 dx dx and 1 2 dx d  3 d 1 − =  y − 1 =  ( y − 1) 3  = ( y − 1) 3  dy   3 dy dy ( ) 2 2 dy 1 3 y−1 = 3 y−1 3 = =3 ( ) dx dx dy
  • 9.
    MONOTONIC FUNCTIONS: Suppose that thedomain of a function f is on an open interval I on which f’(x) > 0 or on which f’(x) < 0. Then f is one-to-one, f-1(x) is differentiable at all values of x in the range of f. Example: Consider the function f (x) = x 5 + x +1 Solution: f '(x) = 5x 4 +1 Since f’(x) > 0 on the entire domain, f(x) is monotonic, therefore it has an inverse .Show that f(x) is one-to=one function.
  • 10.
    MONOTONIC FUNCTIONS: Suppose that thedomain of a function f is on an open interval I on which f’(x) > 0 or on which f’(x) < 0. Then f is one-to-one, f-1(x) is differentiable at all values of x in the range of f. Example: Consider the function f (x) = x 5 + x +1 Solution: f '(x) = 5x 4 +1 Since f’(x) > 0 on the entire domain, f(x) is monotonic, therefore it has an inverse .Show that f(x) is one-to=one function.