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MHT CET · Maths · Functions

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. A \(\mathrm{e}^{\sin ^{-1} x}\)
  2. B \(\frac{1}{\sqrt{1-x^2}}\)
  3. C \(\sin ^{-1} x\)
  4. D \(\frac{\mathrm{e}^{\sin ^{-1} x}}{\sqrt{1-x^2}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{1}{\sqrt{1-x^2}}\)

Step-by-step Solution

Detailed explanation

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