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