If the function \(\mathrm{f}(x)=x^3+\mathrm{e}^{\frac{x}{2}}\) and \(\mathrm{g}(x)=\mathrm{f}^{-1}(x)\), then the value of \(g^{\prime}(1)\) is

\(\begin{array}{ll} & \mathrm{f}(x)=x^3+\mathrm{e}^{\frac{x}{2}} \\ & \Rightarrow \mathrm{f}^{\prime}(x)=3 x^2+\frac{\mathrm{e}^{\frac{x}{2}}}{2} \\ & \text { Given that } \mathrm{g}(x)=\mathrm{f}^{-1}(x) \\ \therefore \quad & \mathrm{gof}(x)=x \\ \therefore \quad & \mathrm{f}(x))=x \\ & \text { Differentiating w.r.t. } x, \text { we get } \\ & \mathrm{g}^{\prime}(\mathrm{f}(x)) \mathrm{f}^{\prime}(x)=1 \\ & \text { for } x=0, \text { we get } \\ & \mathrm{g}^{\prime}(\mathrm{f}(0)) \cdot \mathrm{f}^{\prime}(0)=1 \\ \therefore \quad & \mathrm{~g}^{\prime}(1)=\frac{1}{\mathrm{f}^{\prime}(0)}=\frac{1}{0+\frac{1}{2}}=2\end{array}\)