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

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Aicharya · Accepted Answer

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

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

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