If y = log1+x1-x1/4-12tan-1x, then dydx=

y=log⁡1+x1−x14−12tan−1⁡x ⇒y=14log⁡1+x1−x−12tan−1⁡x y=14log⁡(1+x)−14log⁡(1−x)−12Tan−1⁡x dydx=14(1+x)+14(1−x)−121+x2=1−x+1+x4(1+x)(1−x)−121+x2=121−x2−121+x2=x21−x4

If y = log1+x1-x1/4-12tan-1x, then dydx=

y=log⁡1+x1−x14−12tan−1⁡x ⇒y=14log⁡1+x1−x−12tan−1⁡x y=14log⁡(1+x)−14log⁡(1−x)−12Tan−1⁡x dydx=14(1+x)+14(1−x)−121+x2=1−x+1+x4(1+x)(1−x)−121+x2=121−x2−121+x2=x21−x4

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If y = $\log \{{(\frac{1 + x}{1 - x})}^{1 / 4}\} - \frac{1}{2} \tan^{- 1} (x), t h e n \frac{d y}{d x} =$

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$\frac{x}{1 - x^{2}}$

$\frac{x^{2}}{1 - x^{4}}$

$\frac{x}{1 + x^{4}}$

$\frac{x}{1 - x^{4}}$

answer is B.

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Detailed Solution

$y = \log \{{(\frac{1 + x}{1 - x})}^{\frac{1}{4}}\} - \frac{1}{2} \tan^{- 1} x \Rightarrow y = \frac{1}{4} \log \{\frac{1 + x}{1 - x})\} - \frac{1}{2} \tan^{- 1} x y = \frac{1}{4} \log (1 + x) - \frac{1}{4} \log (1 - x) - \frac{1}{2} {Tan}^{- 1} x$
$\begin{array}{l} \frac{dy}{dx} = \frac{1}{4 (1 + x)} + \frac{1}{4 (1 - x)} - \frac{1}{2 (1 + x^{2})} \\ = \frac{1 - x + 1 + x}{4 (1 + x) (1 - x)} - \frac{1}{2 (1 + x^{2})} \\ = \frac{1}{2 (1 - x^{2})} - \frac{1}{2 (1 + x^{2})} \\ = \frac{x^{2}}{1 - x^{4}} \end{array}$