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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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    That is the definition of an inverse function.2011-11-03

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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.