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

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(B) Let $u = 	an^{-1} \sqrt{\frac{1-x}{1+x}}$ and $v = \sin^{-1} x$.Substitute $x = \cos 	heta$,where $	heta = \cos^{-1} x$.Then $\sqrt{\frac{1-x}{

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$-\frac{1}{2}$

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(B) Let $u = 	an^{-1} \sqrt{\frac{1-x}{1+x}}$ and $v = \sin^{-1} x$.Substitute $x = \cos 	heta$,where $	heta = \cos^{-1} x$.Then $\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}} = \sqrt{\frac{2\sin^2(	heta/2)}{2\cos^2(	heta/2)}} = 	an(	heta/2)$.So,$u = 	an^{-1}(	an(	heta/2)) = \frac{	heta}{2} = \frac{1}{2} \cos^{-1} x$.Since $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$,we have $u = \frac{1}{2} (\frac{\pi}{2} - v) = \frac{\pi}{4} - \frac{1}{2} v$.Now,differentiate $u$ with respect to $v$:$\frac{du}{dv} = \frac{d}{dv} (\frac{\pi}{4} - \frac{1}{2} v) = -\frac{1}{2}$.

The derivative of $\tan^{-1} \sqrt{\frac{1-x}{1+x}}$ with respect to $\sin^{-1}x$ is -

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