The differential coefficient of tan^{-1}left( | vedclass

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(C) Let $y = 	an^{-1}\left( \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} ight)$.Put $x = \cos 2	heta$,which implies $	heta = \frac{1}{

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

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(C) Let $y = 	an^{-1}\left( \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} ight)$.Put $x = \cos 2	heta$,which implies $	heta = \frac{1}{2}\cos^{-1}x$.Substituting $x$ into the expression:$y = 	an^{-1}\left( \frac{\sqrt{1+\cos 2	heta} - \sqrt{1-\cos 2	heta}}{\sqrt{1+\cos 2	heta} + \sqrt{1-\cos 2	heta}} ight)$Using trigonometric identities $1+\cos 2	heta = 2\cos^2	heta$ and $1-\cos 2	heta = 2\sin^2	heta$:$y = 	an^{-1}\left( \frac{\sqrt{2\cos^2	heta} - \sqrt{2\sin^2	heta}}{\sqrt{2\cos^2	heta} + \sqrt{2\sin^2	heta}} ight) = 	an^{-1}\left( \frac{\cos	heta - \sin	heta}{\cos	heta + \sin	heta} ight)$Dividing numerator and denominator by $\cos	heta$:$y = 	an^{-1}\left( \frac{1 - 	an	heta}{1 + 	an	heta} ight) = 	an^{-1}(	an(\pi/4 - 	heta))$Thus,$y = \frac{\pi}{4} - 	heta = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 - \frac{1}{2} \left( -\frac{1}{\sqrt{1-x^2}} ight) = \frac{1}{2\sqrt{1-x^2}}$.

The differential coefficient of $\tan^{-1}\left( \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right)$ is

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