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

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(B) Given,$y=\log \left\{\left(\frac{1+x}{1-x}ight)^{1 / 4}ight\}-\frac{1}{2} 	an ^{-1} x$ Using the property $\log(a^b) = b \log a$,we have $y =

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

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

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

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