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.

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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!