Thread: Inverses and Second Derivatives

Problem:
If f is a one-to-one, twice differentiable function with inverse function g, show that:

g''(x) = -( f''(g(x)) ) / ( [f'(g(x))]^3 )

and deduce that if f is increasing and concave upward, then its inverse function is concave downward.
___________________________

My work so far:
For the first part, I replaced all of the g's with f^-1 (I won't show that here since it would be really cluttered).
Then I took the equation (f^-1)'(x) = 1 / (f'(f^-1(x))) and changed it to
f'(f^-1(x)) = 1 / ((f^-1)'(x)) and plugged that into the bottom of the problem's equation. After a little simplifying, I got
(f^-1)''(x) = (-f''(f^-1(x))) * ((f^-1)'(x))^3

I'm not sure where to go from here. What can I do about the second derivatives? Thank you!

Originally Posted by uberbandgeek6

Problem:
If f is a one-to-one, twice differentiable function with inverse function g, show that:

g''(x) = -( f''(g(x)) ) / ( [f'(g(x))]^3 )

and deduce that if f is increasing and concave upward, then its inverse function is concave downward.
___________________________

My work so far:
For the first part, I replaced all of the g's with f^-1 (I won't show that here since it would be really cluttered).
Then I took the equation (f^-1)'(x) = 1 / (f'(f^-1(x))) and changed it to
f'(f^-1(x)) = 1 / ((f^-1)'(x)) and plugged that into the bottom of the problem's equation. After a little simplifying, I got
(f^-1)''(x) = (-f''(f^-1(x))) * ((f^-1)'(x))^3

I'm not sure where to go from here. What can I do about the second derivatives? Thank you!

Part I

Remember the identity that for inverses, we have

.

Taking the derivative, we get

.

Divide by to get:

.

Take the derivative again, yielding

.

Remember that and substitute:

.

And then simplify to yield

.

Part II

Now, when we say that is increasing and concave upward, we are essentially saying that

for all . Since , then

.

We can use that information with our previous equations to infer that .