y=log left{left(frac{1+x}{1-x}right)^{1 / 4}r | vedclass

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(B) Given,$y=\frac{1}{4} \log \left(\frac{1+x}{1-x}ight)-\frac{1}{2} 	an ^{-1} x$.Using the property $\log \left(\frac{1+x}{1-x}ight) = 2 	anh ^

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

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(B) Given,$y=\frac{1}{4} \log \left(\frac{1+x}{1-x}ight)-\frac{1}{2} 	an ^{-1} x$.Using the property $\log \left(\frac{1+x}{1-x}ight) = 2 	anh ^{-1} x$,we get:$y = \frac{1}{4} (2 	anh ^{-1} x) - \frac{1}{2} 	an ^{-1} x = \frac{1}{2} 	anh ^{-1} x - \frac{1}{2} 	an ^{-1} x$.On differentiating with respect to $x$,we get:$\frac{d y}{d x} = \frac{1}{2} \frac{d}{d x}(	anh ^{-1} x) - \frac{1}{2} \frac{d}{d x}(	an ^{-1} x)$.Since $\frac{d}{d x}(	anh ^{-1} x) = \frac{1}{1-x^2}$ and $\frac{d}{d x}(	an ^{-1} x) = \frac{1}{1+x^2}$,we have:$\frac{d y}{d x} = \frac{1}{2} \left(\frac{1}{1-x^2}ight) - \frac{1}{2} \left(\frac{1}{1+x^2}ight)$.$\frac{d y}{d x} = \frac{1}{2} \left(\frac{1+x^2 - (1-x^2)}{(1-x^2)(1+x^2)}ight) = \frac{1}{2} \left(\frac{2x^2}{1-x^4}ight) = \frac{x^2}{1-x^4}$.

$y=\log \left\{\left(\frac{1+x}{1-x}\right)^{1 / 4}\right\}-\frac{1}{2} \tan ^{-1}(x)$,then $\frac{d y}{d x}$ is equal to

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