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

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Aicharya · Accepted Answer

Given $g(x)$ is inverse of the function $f(x) \& f^{\prime}(x)$ $=\frac{1}{\mathrm{~h}(\mathrm{x})}$ Now, $g(x)=f^{-1}(x) \Rightarrow f(g(x))=x$ Differentiating w.r. to $x$. $\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{g}(\mathrm{x})) \cdot \mathrm{g}^{\prime}(\mathrm{x})=1 \Rightarrow \mathrm{g}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{f}^{\prime}(\mathrm{g}(\mathrm{x}))} \ & =\frac{1}{(1 / \mathrm{h}(\mathrm{g}(\mathrm{x})))}=\mathrm{h}(\mathrm{g}(\mathrm{x})) \end{aligned}$

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

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