If \(g\) is the inverse of \(f\) and \(f^{\prime}(x)=\frac{1}{1+x^3}\), then \(g^{\prime}(x)\) is

\(\begin{array}{ll} & \mathrm{g}(x) \text { is inverse of function } \mathrm{f}(x) \\ & \text { i.e., } \mathrm{g}(x)=\mathrm{f}^{-1}(x) \\ & \mathrm{f}(\mathrm{g}(x))=x \\ & \text { differentiating w.r.t. } x \text {, we get } \\ & \mathrm{f}^{\prime}(\mathrm{g}(x)) \times \mathrm{g}^{\prime}(x)=1 \\ \therefore \quad & \mathrm{g}^{\prime}(x)=\frac{1}{\mathrm{f}^{\prime}(\mathrm{g}(x))}... (i)\end{array}\)
Now, \(\mathrm{f}^{\prime}(x)=\frac{1}{1+x^3}\)... [Given]
\(\therefore \quad \mathrm{f}^{\prime}(\mathrm{g}(x))=\frac{1}{1+(\mathrm{g}(x))^3}\) ... (ii)
\(\therefore \quad\) From (i) and (ii), we get
\(\begin{aligned}
& \therefore \quad \mathrm{g}^{\prime}(x)=1+(\mathrm{g}(x))^3 \\
&
\end{aligned}\)