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

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Question

(A) Let $u = \sec^{-1}\left( \frac{1}{2x^2 - 1} ight) = \cos^{-1}(2x^2 - 1)$.Substitute $x = \cos 	heta$,then $u = \cos^{-1}(2\cos^2 	heta - 1) =

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let $u = \sec^{-1}\left( \frac{1}{2x^2 - 1} ight) = \cos^{-1}(2x^2 - 1)$.Substitute $x = \cos 	heta$,then $u = \cos^{-1}(2\cos^2 	heta - 1) = \cos^{-1}(\cos 2	heta) = 2	heta = 2\cos^{-1} x$.Thus,$\frac{du}{dx} = 2 	imes \left( -\frac{1}{\sqrt{1 - x^2}} ight) = -\frac{2}{\sqrt{1 - x^2}}$.Let $v = \sqrt{1 - x^2}$.Then $\frac{dv}{dx} = \frac{1}{2\sqrt{1 - x^2}} 	imes (-2x) = -\frac{x}{\sqrt{1 - x^2}}$.Now,$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-2/\sqrt{1 - x^2}}{-x/\sqrt{1 - x^2}} = \frac{2}{x}$.At $x = \frac{1}{2}$,$\frac{du}{dv} = \frac{2}{1/2} = 4$.

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

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