Derivatives of Inverse
Functions
Continuity and Differentiability of Inverse
Functions:
Let f be a function whose domain is an interval I. If
f possesses an inverse, then the following statements
are true.
1. If f is continuous on its domain, then its inverse is
continuous on its domain
2. If f is increasing on its domain, then its inverse is
increasing on its domain.
3. If f is decreasing on its domain, then its inverse is
decreasing on its domain.
4. If f is differentiable at c and f (c) isn t zero, then its
inverse is differentiable at f (c).
The Derivative of an Inverse Function:
If f is differentiable on an interval and possesses
an inverse function,g, then the derivative of g is
given by …..
1
g '( x) =
, of _ course_ f '( g ( x)) ≠ 0
f '( g ( x))
u sin g _ the_ f −1 _ notation_ for _ the_ inverse...
−1
( f )'( x) =
1
−1
f '( f ( x))
, of _ course_ f '( f −1 ( x)) ≠ 0
Let s see if we can find the formula for the derivative
of the inverse function…………..
Remember the Alternate Limit Definition of the Derivative ?
g '(c) = lim x →c
g ( x ) − g ( c)
x−c
Let:
y = g ( x )....... and ...... a = g (c)
Since... f _ and _ g _ are _ inverse _ functions:
f ( y ) = x....... and ..... f (a ) = c
If _ _ x → c....., then _ _ g ( x ) → g (c)
In terms of the inverse, then ….
g( x) → g(c)..... same_ as..... y → a
g( x) → g(c)..... same_ as..... y → a
And if y approaches a, we can also say…..
If ..... y → a.... then.... f ( y ) → f (a )
Remember where we started?
g '(c) = lim x →c
g ( x ) − g ( c)
x−c
g '(c) = lim y→a
g '(c) = lim y→a
This can be now be written as …….
y−a
f ( y ) − f (a )
Some algebra manipulation…
1
1
1
1
=
=
f ( y ) − f (a ) f '(a ) g '(c) =
f '(a ) f '( g (c))
y−a
1
g '( x) =
, of _ course_ f '( g ( x)) ≠ 0
f '( g ( x))
Put into practice…………
Let _ f ( x) = x 3 + x
and let g be the inverse function of f. Find
g (2).
1. Find the derivative of f (x).
f '( x) = 3x + 1
2
2. Substitute into the equation for the derivative of g.
g '( x ) =
1
2
3 g( x) + 1
3. Now, evaluate it for an input of 2
g '(2) =
1
2
3 g ( 2) + 1
Let _ f ( x) = x + x
3
and let g be the inverse function of f. Find
g (2).
3. Now, evaluate it for an input of 2
g ' ( 2) =
1
3 g ( 2)
2
+1
4. What s g (2) equal to ?
Since g is the inverse of f, it has to be where the y-value
of f equals 2
2 = x3 + x
by _ inspection _ we _ see _ x = 1
f _ has_ po int_ (1,2)
so, g _ has _ po int_ (2,1)
g(2) = 1
1
1
g'(2) =
=
2
3 1 +1 4
g is the inverse of the function f. Find g (4).
f ( x) = x 2 , where_ x ≥ 0
1. Remember the definition.
1
g '( x) =
, of _ course_ f '( g ( x)) ≠ 0
f '( g ( x))
2. Find the derivative of f
f '( x ) = 2 x
3. Plug info into the equation
1
g'(4) =
2 g (4)
g is the inverse of the function f. Find g (4).
f ( x) = x 2
3. Plug info into the equation
1
g'(4) =
2 g (4)
4. Find g (4)
f ( x) = 4
x2 = 4
x = 2 → pt (2,4)
so, g (4) = 2
5. Put in equation.
1
1
g'(4) =
=
22 4
One more example to show that the answer is not
always ¼!!!
Find the slope of the inverse at the point (2,1) for the
function…
3
f ( x) = x + 2 x − 1
1. Recall that the slope of the curve equals the
derivative s value at that x-value.
slope_ of _ g _ at _ (2,1) = g '(2)
2. Remember equation.
1
g '( x) =
, of _ course_ f '( g ( x)) ≠ 0
f '( g ( x))
Find the slope of the inverse at the point (2,1) for the
function…
3
f ( x) = x + 2 x − 1
2. Remember equation.
1
g '( x) =
, of _ course_ f '( g ( x)) ≠ 0
f '( g ( x))
3. Find the derivative of f
4. Plug into equation
g '(2) =
1
2
3 g ( 2) + 2
f '( x) = 3x + 2
2
5. We know g (2) = 1
1
1
g'(2) =
=
2
31 +2 5