$f\left(x\right)={\int }_0^x \frac{dt}{\sqrt{1+t^3}}$ and g(x) be the inverse of $f\left(x\right)$. Then the value of  $4\frac{g^{\prime \prime }\left(x\right)}{\left(g\right(x){)}^2}$is __.

Solution:

$y=f\left(x\right)\Rightarrow x=f^{-1}\left(y\right)\Rightarrow x=g\left(y\right)$

Given $y=f\left(x\right)={\int }_0^x \frac{dt}{\sqrt{1+t^3}}$

$\frac{dy}{dx}=\frac{1}{\sqrt{1+x^3}}$ or $\frac{dx}{dy}=\sqrt{1+x^3}$

$\begin{array}{l}{g}^{\mathrm{\prime }}\left(y\right)=\sqrt{1+g^3\left(y\right)}\\ g^{\prime \prime }\left(y\right)=\frac{3g^2\left(y\right)g^{\mathrm{\prime }}\left(y\right)}{2\sqrt{1+g^3\left(y\right)}}\\ \therefore 2g^{\prime \prime }\left(y\right)=3g^2\left(y\right)\frac{g^{\mathrm{\prime }}\left(y\right)}{\sqrt{1+g^3\left(y\right)}}=3g^2\left(y\right)\frac{\sqrt{1+g^3\left(y\right)}}{\sqrt{1+g^3\left(y\right)}}=3g^2\left(y\right)\end{array}$

or $2g^{\prime \prime }\left(y\right)=3g^2\left(y\right)$