If u=sin ^{-1}left(frac{2 x}{1+x^2}right) and | vedclass

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(C) Given $u = \sin^{-1}\left(\frac{2x}{1+x^2}ight)$.Using the substitution $x = 	an 	heta$,we have $u = \sin^{-1}(\sin 2	heta) = 2	heta = 2	an

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(C) Given $u = \sin^{-1}\left(\frac{2x}{1+x^2}ight)$.Using the substitution $x = 	an 	heta$,we have $u = \sin^{-1}(\sin 2	heta) = 2	heta = 2	an^{-1}x$.Therefore,$\frac{du}{dx} = \frac{2}{1+x^2}$.Given $v = 	an^{-1}\left(\frac{2x}{1-x^2}ight)$.Using the substitution $x = 	an 	heta$,we have $v = 	an^{-1}(	an 2	heta) = 2	heta = 2	an^{-1}x$.Therefore,$\frac{dv}{dx} = \frac{2}{1+x^2}$.Now,$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{2/(1+x^2)}{2/(1+x^2)} = 1$.

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

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