Let \(\mathrm{f}(x)=\mathrm{e}^x, \mathrm{~g}(x)=\sin ^{-1} x\) and \(\mathrm{h}(x)=\mathrm{f}(\mathrm{g}(x))\), then \(\left(\frac{\mathrm{h}^{\prime}(x)}{\mathrm{h}(x)}\right)^2\) is equal to

\(\begin{aligned}
\mathrm{h}(x) & =\mathrm{f}(\mathrm{~g}(x)) \\
& =\mathrm{f}\left(\sin ^{-1} x\right) \\
\therefore \quad \mathrm{h}(x) & =\mathrm{e}^{\sin ^{-1} x}
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
\mathrm{h}^{\prime}(x) & =\mathrm{e}^{\sin ^{-1} x} \cdot \frac{\mathrm{~d}}{\mathrm{~d} x}\left(\sin ^{-1} x\right) \\
& =\mathrm{e}^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}
\end{aligned}\)
Now, \(\frac{\mathrm{h}^{\prime}(x)}{\mathrm{h}(x)}=\frac{\mathrm{e}^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}}{\mathrm{e}^{\sin ^{-1} x}}=\frac{1}{\sqrt{1-x^2}}\)
\(\therefore \quad\left(\frac{h^{\prime}(x)}{h(x)}\right)^2=\frac{1}{1-x^2}\)