f(x)=∫0x dt1+t3 and g(x) be the inverse of f(x). Then the value of  4g′′(x)(g(x))2  is __.

# $f\left(x\right)={\int }_{0}^{x} \frac{dt}{\sqrt{1+{t}^{3}}}$ and g(x) be the inverse of $f\left(x\right)$. Then the value of  is __.

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### Solution:

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

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 $\frac{dx}{dy}=\sqrt{1+{x}^{3}}$

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

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