If u=tan ^{-1}left(frac{sqrt{1+x^{2}}-1}{x}ri | vedclass

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

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$\frac{1}{4}$

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(A) Given $u=	an ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}ight)$.Let $x=	an 	heta$,then $	heta=	an ^{-1} x$.$u=	an ^{-1}\left(\frac{\sec 	heta-1}{	an 	heta}ight) = 	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)) = \frac{	heta}{2} = \frac{1}{2} 	an ^{-1} x$.Thus,$\frac{d u}{d x} = \frac{1}{2(1+x^{2})}$.Given $v=	an ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}ight)$.Let $x=\sin 	heta$,then $	heta=\sin ^{-1} x$.$v=	an ^{-1}\left(\frac{2 \sin 	heta \cos 	heta}{1-2 \sin ^{2} 	heta}ight) = 	an ^{-1}\left(\frac{\sin 2 	heta}{\cos 2 	heta}ight) = 	an ^{-1}(	an 2 	heta) = 2 	heta = 2 \sin ^{-1} x$.Thus,$\frac{d v}{d x} = \frac{2}{\sqrt{1-x^{2}}}$.Then $\frac{d u}{d v} = \frac{d u/d x}{d v/d x} = \frac{1}{2(1+x^{2})} 	imes \frac{\sqrt{1-x^{2}}}{2} = \frac{\sqrt{1-x^{2}}}{4(1+x^{2})}$.At $x=0$,$\frac{d u}{d v} = \frac{\sqrt{1-0}}{4(1+0)} = \frac{1}{4}$.

If $u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ and $v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$,then $\frac{d u}{d v}$ at $x=0$ is

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