# Math Help - Inverses

1. ## Inverses

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

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

OK, here's where I am:

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

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

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

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

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.

2. You answer is equivalent look up one line and what does

$g'(x)=$

3. Ack! Right in front of my eyes. Thanks.