The derivative of tan ^{-1}left(frac{sqrt{1+x | vedclass

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

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

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(B) Let $y = 	an ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}ight)$.Substitute $x = \cos 2	heta$,which implies $	heta = \frac{1}{2} \cos ^{-1} x$.Then $y = 	an ^{-1}\left(\frac{\sqrt{1+\cos 2	heta}-\sqrt{1-\cos 2	heta}}{\sqrt{1+\cos 2	heta}+\sqrt{1-\cos 2	heta}}ight)$.Using trigonometric identities $1+\cos 2	heta = 2\cos^2	heta$ and $1-\cos 2	heta = 2\sin^2	heta$,we get:$y = 	an ^{-1}\left(\frac{\sqrt{2}\cos	heta - \sqrt{2}\sin	heta}{\sqrt{2}\cos	heta + \sqrt{2}\sin	heta}ight) = 	an ^{-1}\left(\frac{1-	an	heta}{1+	an	heta}ight)$.Using the formula $	an(\frac{\pi}{4} - 	heta) = \frac{1-	an	heta}{1+	an	heta}$,we have:$y = 	an ^{-1}	an(\frac{\pi}{4} - 	heta) = \frac{\pi}{4} - 	heta = \frac{\pi}{4} - \frac{1}{2} \cos ^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 - \frac{1}{2} 	imes \left(-\frac{1}{\sqrt{1-x^2}}ight) = \frac{1}{2 \sqrt{1-x^2}}$.

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

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