The derivative of y=tan ^{-1}left(frac{sqrt{1 | vedclass

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(B) Given,$y=	an ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}ight)$.Let $x=	an 	heta$,then $	heta = 	an ^{-1} x$.Substituting $x$ in the expression:$y =

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$\frac{1}{2(1+x^2)}$

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(B) Given,$y=	an ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}ight)$.Let $x=	an 	heta$,then $	heta = 	an ^{-1} x$.Substituting $x$ in the expression:$y = 	an ^{-1}\left(\frac{\sqrt{1+	an ^2 	heta}-1}{	an 	heta}ight)$$y = 	an ^{-1}\left(\frac{\sec 	heta-1}{	an 	heta}ight)$$y = 	an ^{-1}\left(\frac{\frac{1}{\cos 	heta}-1}{\frac{\sin 	heta}{\cos 	heta}}ight) = 	an ^{-1}\left(\frac{1-\cos 	heta}{\sin 	heta}ight)$Using trigonometric identities $1-\cos 	heta = 2\sin ^2(	heta/2)$ and $\sin 	heta = 2\sin(	heta/2)\cos(	heta/2)$:$y = 	an ^{-1}\left(\frac{2\sin ^2(	heta/2)}{2\sin(	heta/2)\cos(	heta/2)}ight) = 	an ^{-1}(	an(	heta/2)) = 	heta/2$.Since $	heta = 	an ^{-1} x$,we have $y = \frac{1}{2} 	an ^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2} = \frac{1}{2(1+x^2)}$.Hence,option $B$ is correct.

The derivative of $y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ is equal to

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