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

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(A) Let $y = 	an^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}ight)$.Substitute $x = 	an 	heta$,so $	heta = 	an^{-1} x$.Then $y = 	an^{-1}\left(\frac{\sq

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

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

The derivative of $\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $x$ is:

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