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$y=\log \left\{\left(\frac{1+x}{1-x}\right)^{1 / 4}\right\}-\frac{1}{2} \tan ^{-1}(x)$, then $\frac{d y}{d x}$ is equal to
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The correct answer is:
$\frac{x^2}{1-x^4}$
$\begin{aligned} & \text { Given, } y=\log \left\{\left(\frac{1+x}{1-x}\right)^{1 / 4}\right\}-\frac{1}{2} \tan ^{-1} x \\ & \Rightarrow \quad y=\frac{1}{2} \tanh ^{-1} x-\frac{1}{2} \tan ^{-1} x\end{aligned}$
On differentiating w.r. t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{1}{2}\left(\frac{1}{1-x^2}\right)-\frac{1}{2}\left(\frac{1}{1+x^2}\right) \\ & =\frac{1}{2}\left(\frac{1+x^2-1+x^2}{1-x^4}\right) \\ & =\frac{1}{2}\left(\frac{2 x^2}{1-x^4}\right) \\ & =\frac{x^2}{1-x^4}\end{aligned}$
On differentiating w.r. t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{1}{2}\left(\frac{1}{1-x^2}\right)-\frac{1}{2}\left(\frac{1}{1+x^2}\right) \\ & =\frac{1}{2}\left(\frac{1+x^2-1+x^2}{1-x^4}\right) \\ & =\frac{1}{2}\left(\frac{2 x^2}{1-x^4}\right) \\ & =\frac{x^2}{1-x^4}\end{aligned}$
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