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

$\displaystyle \displaystyle g''(x) = - \frac{f''(g(x))}{[f'(g(x))]^3}$

OK, here's where I am:

$\displaystyle f^{-1}(x) = g(x)$

$\displaystyle \displaystyle g'(x) = \frac {1}{f'(g(x))}$

$\displaystyle \displaystyle g''x = \frac{f'(g(x))\cdot(0) - (1)(f''(g(x))(g'(x))}{[f'(g(x))]^2}$

$\displaystyle \displaystyle g''(x) = \frac{-f''(g(x))(g'(x))}{[f'(g(x))]^2}$

So I'm close to the answer, but I don't see how to take it from here. Have I made any mistakes thus far or am I just missing the next step? (or both).

Thanks.