# Inverse and Derivative of $g'(x)=(1+x^{3})^{-1/2}$ Question

I am stuck on the following question:

Suppose that $g$ is differentiable with derivative $g'(x)=(1+x^{3})^{-1/2}$. Show that the inverse function $h=g^{-1}$ satisfies $h''(x)=\frac{3}{2}[h(x)]^{2}$.

I feel like I've missed something in my notes. I need to find $g^{-1}(x)$ I suppose. I've been searching for theorems relating to inverses and derivatives in order to find $g^{-1}(x)$ from $g'(x)$.

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We have $g(h(x))=x$. Differentiate, using the Chain Rule. We get $h'(x)g'(h(x))=1$. That gets you $h'(x)$ in terms of $h(x)$. Differentiate both sides of this relationship. – André Nicolas Nov 3 '11 at 6:57
Where do you get $g(h(x))=x$ from? – Malthus Nov 3 '11 at 7:11
That is the definition of an inverse function. – user7530 Nov 3 '11 at 7:14

Note that $g'(x)$ is undefined when $x\le -1$. So we are only interested in values of $x$ greater than $-1$. Since $g'(x)>0$ for $x>-1$, our function $g$ is increasing in the interval $(-1,\infty)$. Thus, with a suitable restriction on the domain, an inverse function $h$ exists.
For all suitable $x$, we have $g(h(x))=x$, and for all suitable $x$, we have $h(g(x))=x$, by the definition of inverse function.
We use the identity $g(h(x))=x$. Differentiate both sides with respect to $x$, using the Chain Rule. We obtain $$h'(x)g'(h(x))=1.$$ But we are told that $g'(u)=(1+u^3)^{-1/2}$. It follows that $$h'(x)(1+[h(x)]^3)^{-1/2}=1.$$ More simply, $$h'(x)=(1+[h(x)]^3)^{1/2}.$$ To find $h''(x)$, differentiate. We get, by the Chain Rule, applied twice, $$h''(x)=3[h(x)]^2h'(x)(1/2)(1+[h(x)]^3)^{-1/2}.$$ The term $h'(x)(1+[h(x)]^3)^{-1/2}$ sort of hidden in the right-hand side is equal to $1$. It follows that $$h''(x)=\frac{3}{2}[h(x)]^2.$$
Comment: One might imagine integrating $g'(x)$ to find $g(x)$, and then finding the inverse function $h(x)$ explicitly. However, the function $(1+x^3)^{-1/2}$ does not have an elementary antiderivative, so that approach will not succeed.