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If $\mathrm{f}(x)=\mathrm{e}^x, \mathrm{~g}(x)=\sin ^{-1} x$ and $\mathrm{h}(x)=\mathrm{f}(\mathrm{g}(x))$, then $\frac{\mathrm{h}^{\prime}(x)}{\mathrm{h}(x)}$ is
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{1-x^2}}$
$\begin{aligned}
\mathrm{h}(x) & =\mathrm{f}(\mathrm{g}(x)) \\
& =\mathrm{f}\left(\sin ^{-1} x\right) \\
\therefore \quad \mathrm{h}(x) & =\mathrm{e}^{\sin ^{-1} x}
\end{aligned}$
Differentiating w.r.t. $x$, we get
$\begin{aligned}
& \begin{aligned}
\mathrm{h}^{\prime}(x) & =\mathrm{e}^{\sin ^{-1} x} \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left(\sin ^{-1} x\right) \\
& =\mathrm{e}^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}
\end{aligned} \\
& \text { Now, } \frac{\mathrm{h}^{\prime}(x)}{\mathrm{h}(x)}=\frac{\mathrm{e}^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}}{\mathrm{e}^{\sin ^{-1} x}}=\frac{1}{\sqrt{1-x^2}}
\end{aligned}$
\mathrm{h}(x) & =\mathrm{f}(\mathrm{g}(x)) \\
& =\mathrm{f}\left(\sin ^{-1} x\right) \\
\therefore \quad \mathrm{h}(x) & =\mathrm{e}^{\sin ^{-1} x}
\end{aligned}$
Differentiating w.r.t. $x$, we get
$\begin{aligned}
& \begin{aligned}
\mathrm{h}^{\prime}(x) & =\mathrm{e}^{\sin ^{-1} x} \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left(\sin ^{-1} x\right) \\
& =\mathrm{e}^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}
\end{aligned} \\
& \text { Now, } \frac{\mathrm{h}^{\prime}(x)}{\mathrm{h}(x)}=\frac{\mathrm{e}^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}}{\mathrm{e}^{\sin ^{-1} x}}=\frac{1}{\sqrt{1-x^2}}
\end{aligned}$
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