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

If $\mathrm{f}(x)=\cos ^{-1} x, \mathrm{~g}(x)=\mathrm{e}^x$ and $\mathrm{h}(x)=\mathrm{g}(\mathrm{f}(x))$, then $\frac{\mathrm{h}^{\prime}(x)}{\mathrm{h}(x)}=$
  1. $\frac{-1}{\sqrt{1-x^2}}$
  2. $\frac{-(e)^{\left(\cos ^{-1} x\right)}}{\sqrt{1-x^2}}$
  3. $\frac{-1}{\sqrt{1-x^2}} \mathrm{e}^x$
  4. $-\sqrt{1-x^2}$

Solution

$\begin{aligned} \mathrm{f}(x) & =\cos ^{-1} x \\ \mathrm{~g}(x) & =\mathrm{e}^x \\ \mathrm{~h}(x) & =\mathrm{g}(\mathrm{f}(x)) \\ & =\mathrm{e}^{\cos ^{-1} x} \\ \mathrm{~h}^{\prime}(x) & =\mathrm{e}^{\cos ^{-1} x} \cdot \frac{-1}{\sqrt{1-x^2}} \\ \therefore \quad \frac{\mathrm{~h}^{\prime}(x)}{\mathrm{h}(x)} & =\frac{\mathrm{e}^{\cos ^{-1} x} \times \frac{-1}{\sqrt{1-x^2}}}{\mathrm{e}^{\cos ^{-1} x}}=\frac{-1}{\sqrt{1-x^2}}\end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 2)

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