frac{d}{dx} left( tan^{-1} left( frac{sqrt{1 | vedclass

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

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

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(B) Let $y = 	an^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} ight)$.Substitute $x = 	an 	heta$,then $	heta = 	an^{-1} x$.$y = 	an^{-1} \left( \frac{\sqrt{1 + 	an^2 	heta} - 1}{	an 	heta} ight) = 	an^{-1} \left( \frac{\sec 	heta - 1}{	an 	heta} ight)$.Using trigonometric identities,$\frac{\sec 	heta - 1}{	an 	heta} = \frac{1 - \cos 	heta}{\sin 	heta} = \frac{2 \sin^2(	heta/2)}{2 \sin(	heta/2) \cos(	heta/2)} = 	an(	heta/2)$.Thus,$y = 	an^{-1}(	an(	heta/2)) = \frac{	heta}{2} = \frac{1}{2} 	an^{-1} x$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1 + x^2} = \frac{1}{2(1 + x^2)}$.

$\frac{d}{dx} \left( \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right) \right)$ is equal to

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