The derivative of operatorname{Sec}^{-1}left( | vedclass

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(D) Let $u = \operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}\right)$ and $v = \sqrt{1-x^2}$.Using the identity $\operatorname{Sec}^{-1}(\frac{1}{z}) = \

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$4$

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(D) Let $u = \operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}ight)$ and $v = \sqrt{1-x^2}$.Using the identity $\operatorname{Sec}^{-1}(\frac{1}{z}) = \cos^{-1}(z)$,we have $u = \cos^{-1}(2x^2-1)$.Let $x = \cos 	heta$,then $u = \cos^{-1}(2\cos^2 	heta - 1) = \cos^{-1}(\cos 2	heta) = 2	heta = 2\cos^{-1}x$.Thus,$\frac{du}{dx} = 2 	imes (-\frac{1}{\sqrt{1-x^2}}) = -\frac{2}{\sqrt{1-x^2}}$.Now,$v = \sqrt{1-x^2}$,so $\frac{dv}{dx} = \frac{1}{2\sqrt{1-x^2}} 	imes (-2x) = -\frac{x}{\sqrt{1-x^2}}$.We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-2/\sqrt{1-x^2}}{-x/\sqrt{1-x^2}} = \frac{2}{x}$.At $x = \frac{1}{2}$,$\frac{du}{dv} = \frac{2}{1/2} = 4$.

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

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