MHT CET · Maths · Differentiation
If \(\mathrm{f}(x)=\cos ^{-1} x, \mathrm{~g}(x)=\mathrm{e}^x\) and \(\mathrm{h}(x)=\mathrm{g}(\mathrm{f}(x))\), then \(\frac{\mathrm{h}^{\prime}(x)}{\mathrm{h}(x)}=\)
- A \(\frac{-1}{\sqrt{1-x^2}}\)
- B \(\frac{-(e)^{\left(\cos ^{-1} x\right)}}{\sqrt{1-x^2}}\)
- C \(\frac{-1}{\sqrt{1-x^2}} \mathrm{e}^x\)
- D \(-\sqrt{1-x^2}\)
Answer & Solution
Correct Answer
(A) \(\frac{-1}{\sqrt{1-x^2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \mathrm{f}(x) & =\cos ^{-1} x \\ \mathrm{~g}(x) & =\mathrm{e}^x \\ \mathrm{~h}(x) & =\mathrm{g}(\mathrm{f}(x)) \\ & =\mathrm{e}^{\cos ^{-1} x} \\ \mathrm{~h}^{\prime}(x) & =\mathrm{e}^{\cos ^{-1} x} \cdot \frac{-1}{\sqrt{1-x^2}} \\ \therefore \quad \frac{\mathrm{~h}^{\prime}(x)}{\mathrm{h}(x)} & =\frac{\mathrm{e}^{\cos ^{-1} x} \times \frac{-1}{\sqrt{1-x^2}}}{\mathrm{e}^{\cos ^{-1} x}}=\frac{-1}{\sqrt{1-x^2}}\end{aligned}\)
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