TS EAMCET · Maths · Differentiation
\(y=\log \left\{\left(\frac{1+x}{1-x}\right)^{1 / 4}\right\}-\frac{1}{2} \tan ^{-1}(x)\), then \(\frac{d y}{d x}\) is equal to
- A \(\frac{x}{1-x^2}\)
- B \(\frac{x^2}{1-x^4}\)
- C \(\frac{x}{1+x^4}\)
- D \(\frac{x}{1-x^4}\)
Answer & Solution
Correct Answer
(B) \(\frac{x^2}{1-x^4}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Given, } y=\log \left\{\left(\frac{1+x}{1-x}\right)^{1 / 4}\right\}-\frac{1}{2} \tan ^{-1} x \\ & \Rightarrow \quad y=\frac{1}{2} \tanh ^{-1} x-\frac{1}{2} \tan ^{-1} x\end{aligned}\) On differentiating w.r. t. \(x\), we get…
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