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

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
  1. $\frac{1}{\sqrt{1-x^2}}$
  2. $\left(1-x^2\right)^2$
  3. $\frac{1}{1-x^2}$
  4. $\left(1-x^2\right)$

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 $\begin{aligned} \mathrm{h}^{\prime}(x) & =\mathrm{e}^{\sin ^{-1} x} \cdot \frac{\mathrm{~d}}{\mathrm{~d} x}\left(\sin ^{-1} x\right) \\ & =\mathrm{e}^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}} \end{aligned}$
Now, $\frac{\mathrm{h}^{\prime}(x)}{\mathrm{h}(x)}=\frac{\mathrm{e}^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}}{\mathrm{e}^{\sin ^{-1} x}}=\frac{1}{\sqrt{1-x^2}}$ $\therefore \quad\left(\frac{h^{\prime}(x)}{h(x)}\right)^2=\frac{1}{1-x^2}$

Asked in: MHT CET 2024 (09 May Shift 2)

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