Let $\quad f(x)=e^x, \quad g(x)=\sin ^{-1} x \quad$ and $h(x)=f(g(x))$, then $\frac{h^{\prime}(x)}{h(x)}$ is equal to

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Aicharya · Accepted Answer

We have, $ \begin{aligned} h(x) & =f(g(x))=f\left(\sin ^{-1} xight) \ h(x) & =e^{\sin ^{-1} x} \ \log h(x) & =\sin ^{-1} x \end{aligned} $ On differentiating w.r.t. $x$, we get $ \frac{h^{\prime}(x)}{h(x)}=\frac{1}{\sqrt{1-x^2}} $

Let $\quad f(x)=e^x, \quad g(x)=\sin ^{-1} x \quad$ and $h(x)=f(g(x))$, then $\frac{h^{\prime}(x)}{h(x)}$ is…

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