If tan u=sqrt{frac{1-x}{1+x}} and cos v=4 x^{ | vedclass

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(A) Given $	an u=\sqrt{\frac{1-x}{1+x}}$.Let $x=\cos 	heta$,then $	heta=\cos ^{-1} x$.Substituting $x$,we get $	an u=\sqrt{\frac{1-\cos 	heta}{1+

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$\frac{1}{6}$

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(A) Given $	an u=\sqrt{\frac{1-x}{1+x}}$.Let $x=\cos 	heta$,then $	heta=\cos ^{-1} x$.Substituting $x$,we get $	an u=\sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}}=\sqrt{\frac{2 \sin ^{2} (	heta/2)}{2 \cos ^{2} (	heta/2)}}=	an (	heta/2)$.Thus,$u=\frac{	heta}{2}=\frac{1}{2} \cos ^{-1} x$.Differentiating with respect to $x$,we get $\frac{du}{dx}=\frac{1}{2} \left(-\frac{1}{\sqrt{1-x^{2}}}ight) = -\frac{1}{2 \sqrt{1-x^{2}}}$.Given $\cos v=4 x^{3}-3 x$.Substituting $x=\cos 	heta$,we get $\cos v=4 \cos ^{3} 	heta-3 \cos 	heta = \cos 3 	heta$.Thus,$v=3 	heta = 3 \cos ^{-1} x$.Differentiating with respect to $x$,we get $\frac{dv}{dx}=3 \left(-\frac{1}{\sqrt{1-x^{2}}}ight) = -\frac{3}{\sqrt{1-x^{2}}}$.Finally,$\frac{du}{dv}=\frac{du/dx}{dv/dx} = \frac{-1/(2 \sqrt{1-x^{2}})}{-3/\sqrt{1-x^{2}}} = \frac{1}{6}$.

If $\tan u=\sqrt{\frac{1-x}{1+x}}$ and $\cos v=4 x^{3}-3 x$,then $\frac{d u}{d v}=$

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