If $y=\log \left(\frac{1+x}{1-x} ight)^{1 / 4}-\frac{1}{2} an ^{-1}(x)$, then $\frac{d y}{d x}$ at $x=\frac{1}{\sqrt{2}}$ equals

Question

If $y=\log \left(\frac{1+x}{1-x}ight)^{1 / 4}-\frac{1}{2} 	an ^{-1}(x)$, then $\frac{d y}{d x}$ at $x=\frac{1}{\sqrt{2}}$ equals

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { (d) } y=\frac{1}{4} \log \left(\frac{1+x}{1-x}ight)-\frac{1}{2} 	an ^{-1} x \ & y=\frac{1}{4} \log (1+x)-\frac{1}{4} \log (1-x)-\frac{1}{2} 	an ^{-1} x \ & 	ext { 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}ight)_{x=1 / \sqrt{2}}=\frac{1}{4}\left\{\frac{1}{1+1 / \sqrt{2}}+\frac{1}{1-1 / \sqrt{2}}ight\} \ & -\frac{1}{2} \cdot\left(\frac{1}{1+1 / 2}ight) \ & =\frac{1}{4}\left\{\frac{\sqrt{2}(\sqrt{2}-1)+\sqrt{2}(\sqrt{2}+1)}{(\sqrt{2}+1)(\sqrt{2}-1)}ight\}-\frac{1}{3} \ & \end{aligned} $ $ \begin{aligned} & =\frac{1}{4}\left\{\frac{2-\sqrt{2}+2+\sqrt{2}}{2-1}ight\}-\frac{1}{3} \ & =-\frac{1}{4}\left\{\frac{4}{1}ight\}-\frac{1}{3}=1-\frac{1}{3}=\frac{2}{3} \ 	herefore\left(\frac{d y}{d x}ight)_{x=1 / \sqrt{2}} & =\frac{2}{3} \end{aligned} $

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…

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