If y = log {left( frac{1 + x}{1 - x} right)^{ | vedclass

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(A) Given $y = \log {\left( \frac{1 + x}{1 - x} ight)^{1/4}} - \frac{1}{2}{	an ^{ - 1}}x$. Using properties of logarithms,$y = \frac{1}{4} [\log(1

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

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(A) Given $y = \log {\left( \frac{1 + x}{1 - x} ight)^{1/4}} - \frac{1}{2}{	an ^{ - 1}}x$. Using properties of logarithms,$y = \frac{1}{4} [\log(1 + x) - \log(1 - x)] - \frac{1}{2}{	an ^{ - 1}}x$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{4} \left[ \frac{1}{1 + x} - \frac{1}{1 - x}(-1) ight] - \frac{1}{2} \left( \frac{1}{1 + x^2} ight)$ $\frac{dy}{dx} = \frac{1}{4} \left[ \frac{1}{1 + x} + \frac{1}{1 - x} ight] - \frac{1}{2(1 + x^2)}$ $\frac{dy}{dx} = \frac{1}{4} \left[ \frac{1 - x + 1 + x}{(1 + x)(1 - x)} ight] - \frac{1}{2(1 + x^2)}$ $\frac{dy}{dx} = \frac{1}{4} \left[ \frac{2}{1 - x^2} ight] - \frac{1}{2(1 + x^2)}$ $\frac{dy}{dx} = \frac{1}{2(1 - x^2)} - \frac{1}{2(1 + x^2)}$ $\frac{dy}{dx} = \frac{1}{2} \left[ \frac{(1 + x^2) - (1 - x^2)}{(1 - x^2)(1 + x^2)} ight]$ $\frac{dy}{dx} = \frac{1}{2} \left[ \frac{2x^2}{1 - x^4} ight] = \frac{x^2}{1 - x^4}$.

If $y = \log {\left( \frac{1 + x}{1 - x} \right)^{1/4}} - \frac{1}{2}{\tan ^{ - 1}}x,$ then $\frac{dy}{dx} = $

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