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

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(B) Let $u = 	an ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}ight)$. Put $x = 	an 	heta$,then $u = 	an ^{-1}\left(\frac{\sec 	heta - 1}{	an 	heta}ig

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$\frac{1}{4}$

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(B) Let $u = 	an ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}ight)$. Put $x = 	an 	heta$,then $u = 	an ^{-1}\left(\frac{\sec 	heta - 1}{	an 	heta}ight) = 	an ^{-1}\left(\frac{1-\cos 	heta}{\sin 	heta}ight) = 	an ^{-1}\left(	an \frac{	heta}{2}ight) = \frac{	heta}{2} = \frac{1}{2} 	an ^{-1} x$. Thus,$\frac{du}{dx} = \frac{1}{2(1+x^2)}$. Let $v = 	an ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}ight)$. Put $x = \sin \phi$,then $v = 	an ^{-1}\left(\frac{2 \sin \phi \cos \phi}{1-2 \sin^2 \phi}ight) = 	an ^{-1}(	an 2\phi) = 2\phi = 2 \sin ^{-1} x$. Thus,$\frac{dv}{dx} = \frac{2}{\sqrt{1-x^2}}$. We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1}{2(1+x^2)} 	imes \frac{\sqrt{1-x^2}}{2} = \frac{\sqrt{1-x^2}}{4(1+x^2)}$. At $x=0$,$\frac{du}{dv} = \frac{\sqrt{1-0}}{4(1+0)} = \frac{1}{4}$.

The derivative of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $\tan ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right)$ at $x=0$ is

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