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If $y=\log \left(\frac{1+x}{1-x}\right)^{1 / 4}-\frac{1}{2} \tan ^{-1}(x)$, then $\frac{d y}{d x}$ at $x=\frac{1}{\sqrt{2}}$ equals
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Verified Answer
The correct answer is:
$\frac{2}{3}$
$$
\begin{aligned}
& \text { (d) } y=\frac{1}{4} \log \left(\frac{1+x}{1-x}\right)-\frac{1}{2} \tan ^{-1} x \\
& y=\frac{1}{4} \log (1+x)-\frac{1}{4} \log (1-x)-\frac{1}{2} \tan ^{-1} x \\
& \text { Then, } \frac{d y}{d x}=\frac{1}{4} \frac{1}{(1+x)}+\frac{1}{4} \cdot \frac{1}{(1-x)}-\frac{1}{2} \cdot \frac{1}{1+x^2} \\
& \left(\frac{d y}{d x}\right)_{x=1 / \sqrt{2}}=\frac{1}{4}\left\{\frac{1}{1+1 / \sqrt{2}}+\frac{1}{1-1 / \sqrt{2}}\right\} \\
& -\frac{1}{2} \cdot\left(\frac{1}{1+1 / 2}\right) \\
& =\frac{1}{4}\left\{\frac{\sqrt{2}(\sqrt{2}-1)+\sqrt{2}(\sqrt{2}+1)}{(\sqrt{2}+1)(\sqrt{2}-1)}\right\}-\frac{1}{3} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{1}{4}\left\{\frac{2-\sqrt{2}+2+\sqrt{2}}{2-1}\right\}-\frac{1}{3} \\
& =-\frac{1}{4}\left\{\frac{4}{1}\right\}-\frac{1}{3}=1-\frac{1}{3}=\frac{2}{3} \\
\therefore\left(\frac{d y}{d x}\right)_{x=1 / \sqrt{2}} & =\frac{2}{3}
\end{aligned}
$$
\begin{aligned}
& \text { (d) } y=\frac{1}{4} \log \left(\frac{1+x}{1-x}\right)-\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 \\
& \text { Then, } \frac{d y}{d x}=\frac{1}{4} \frac{1}{(1+x)}+\frac{1}{4} \cdot \frac{1}{(1-x)}-\frac{1}{2} \cdot \frac{1}{1+x^2} \\
& \left(\frac{d y}{d x}\right)_{x=1 / \sqrt{2}}=\frac{1}{4}\left\{\frac{1}{1+1 / \sqrt{2}}+\frac{1}{1-1 / \sqrt{2}}\right\} \\
& -\frac{1}{2} \cdot\left(\frac{1}{1+1 / 2}\right) \\
& =\frac{1}{4}\left\{\frac{\sqrt{2}(\sqrt{2}-1)+\sqrt{2}(\sqrt{2}+1)}{(\sqrt{2}+1)(\sqrt{2}-1)}\right\}-\frac{1}{3} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{1}{4}\left\{\frac{2-\sqrt{2}+2+\sqrt{2}}{2-1}\right\}-\frac{1}{3} \\
& =-\frac{1}{4}\left\{\frac{4}{1}\right\}-\frac{1}{3}=1-\frac{1}{3}=\frac{2}{3} \\
\therefore\left(\frac{d y}{d x}\right)_{x=1 / \sqrt{2}} & =\frac{2}{3}
\end{aligned}
$$
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