Derivatives of Inverse Functions

I.  Quiz

II.  Homework

III.  Inverse Functions (Definition)

Let f(x) be a 1-1 function then g(x) is an inverse function of f(x) if

f(g(x)) = g(f(x)) = x

Example:

For f(x) = 2x - 1, f^-1(x) = 1/2 x +1/2

Since f(f^-1(x) ) = 2[1/2 x +1/2] - 1 = x

and f^-1(f(x)) = 1/2 [2x - 1] + 1/2 = x

IV.  The Horizontal Line Test and Roll's Theorem

Note that if f(x) is differentiable and the horizontal line test fails then f(a) = f(b) and Rolls theorem implies that there is a c such that f'(c) = 0.  A partial converse is also true:  

If f is differentiable and f'(x) is always non negative (or  always non positive) then f(x) has an inverse.

Example:

f(x) = x³ + x - 4 has an inverse since

f'(x) = 3x² + 1 which is always positive.

V.  Continuity and Differentiability of the Inverse Function 

Theorem:

1.  f cts implies that f^-1 is cts.

2.  f increasing implies that  f^-1 is increasing.

3.  f decreasing implies that f^-1 is decreasing

4.  f differentiable at c and f'(c) not 0 implies that f^-1 is differentiable at f(c).

5.  If g(x) is the inverse of  the differentiable f(x) then

g'(x) = 1/[f'(g(x))] if f'(g(x) not 0.

Proof  Since f(g(x)) = x we differentiate implicitly:

d/dx[f(g(x)) ] = d/dx [x]

Using the chain rule y = f(u),  u = g(x)

dy/dx = (dy/du)(du/dx) = (f'(u))(g'(x)) = (f'(g(x)))(g'(x))

So that

(f'(g(x)))(g'(x)) = 1  Dividing, we get:

g'(x) = 1/(f'(g(x)))

Example:  

For x > 0 Let f(x) = x² and g(x) = sqrt(x) be its inverse, then

g'(x) = 1/[2(sqrt(x)]  

Note that d/dx [sqrt(x)] = d/dx[x^1/2] = 1/2 x^-1/2  = 1/[2sqrt(x)]

Exercise:  

A)   Let f(x) = x³ + x - 4

Find d/dx[f^-1(-4)]

B)  Let f(x) = 3x⁵ + x³ + 6

Find d/dx[f^-1(10)]