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**jenkki** Given that g is the inverse function of f, where:

$\displaystyle f(x) = \int^x_0 \frac {1}{\sqrt{t^3+1}} dt$

for $\displaystyle f: [0,\infty) \rightarrow \mathbb{R}$

**I have to show that**:

$\displaystyle g''(x) = \frac{3}{2} (g(x))^2$

At first I thought that all I had to do was to use the relation:

$\displaystyle g'(y) = \frac{1}{f'(x)}$ where $\displaystyle f'(x) = \frac {1}{\sqrt{x^3+1}}$

however, this expression gives me g'(y) and g''(y) (not x). What can I do to find the g(x)'s ?