TS EAMCET · Maths · Differentiation
Let \(\quad f(x)=e^x, \quad g(x)=\sin ^{-1} x \quad\) and \(h(x)=f(g(x))\), then \(\frac{h^{\prime}(x)}{h(x)}\) is equal to
- A \(\sin ^{-1} x\)
- B \(\frac{1}{\sqrt{1-x^2}}\)
- C \(-\frac{1}{\sqrt{1-x^2}}\)
- D \(e^{\sin ^{-1} x}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{1-x^2}}\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} h(x) & =f(g(x))=f\left(\sin ^{-1} x\right) \\ h(x) & =e^{\sin ^{-1} x} \\ \log h(x) & =\sin ^{-1} x \end{aligned} \] On differentiating w.r.t. \(x\), we get \[ \frac{h^{\prime}(x)}{h(x)}=\frac{1}{\sqrt{1-x^2}} \]
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