The derivative of sec ^{-1}left(frac{1}{2 x^2 | vedclass

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(D) Let $u = \sec ^{-1}\left(\frac{1}{2 x^2-1}ight)$ and $v = \sqrt{1-x^2}$.Substitute $x = \cos 	heta$.Then $u = \sec ^{-1}\left(\frac{1}{2 \cos ^

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$4$

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(D) Let $u = \sec ^{-1}\left(\frac{1}{2 x^2-1}ight)$ and $v = \sqrt{1-x^2}$.Substitute $x = \cos 	heta$.Then $u = \sec ^{-1}\left(\frac{1}{2 \cos ^2 	heta - 1}ight) = \sec ^{-1}\left(\frac{1}{\cos 2 	heta}ight) = \sec ^{-1}(\sec 2 	heta) = 2 	heta$.And $v = \sqrt{1 - \cos ^2 	heta} = \sqrt{\sin ^2 	heta} = \sin 	heta$.Now,differentiate $u$ and $v$ with respect to $	heta$:$\frac{du}{d	heta} = 2$ and $\frac{dv}{d	heta} = \cos 	heta$.Therefore,the derivative of $u$ with respect to $v$ is:$\frac{du}{dv} = \frac{du/d	heta}{dv/d	heta} = \frac{2}{\cos 	heta} = \frac{2}{x}$.At $x = \frac{1}{2}$,we have:$\left(\frac{du}{dv}ight)_{x=1/2} = \frac{2}{1/2} = 4$.

The derivative of $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ equals

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