If $\mathrm{f}(x)=\mathrm{e}^x, \mathrm{~g}(x)=\sin ^{-1} x$ and $\mathrm{h}(x)=\mathrm{f}(\mathrm{g}(x))$,…

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
  1. $\mathrm{e}^{\sin ^{-1} x}$
  2. $\frac{1}{\sqrt{1-x^2}}$
  3. $\sin ^{-1} x$
  4. $\frac{\mathrm{e}^{\sin ^{-1} x}}{\sqrt{1-x^2}}$

Solution

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

Asked in: MHT CET 2023 (10 May Shift 1)

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