The derivative of tan ^{-1}left[frac{x}{1+sqr | vedclass

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(C) Let $y = 	an ^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}ight)$ and $u = \sec ^{-1}\left(\frac{1}{2 x^2-1}ight)$.Substitute $x = \sin 	heta$,then $\s

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$\frac{-1}{4}$

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(C) Let $y = 	an ^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}ight)$ and $u = \sec ^{-1}\left(\frac{1}{2 x^2-1}ight)$.Substitute $x = \sin 	heta$,then $\sqrt{1-x^2} = \cos 	heta$.$y = 	an ^{-1}\left(\frac{\sin 	heta}{1+\cos 	heta}ight) = 	an ^{-1}\left(\frac{2 \sin(	heta/2) \cos(	heta/2)}{2 \cos^2(	heta/2)}ight) = 	an ^{-1}(	an(	heta/2)) = \frac{	heta}{2}$.Now,$u = \sec ^{-1}\left(\frac{1}{2 \sin^2 	heta - 1}ight) = \sec ^{-1}\left(\frac{1}{-(1-2 \sin^2 	heta)}ight) = \sec ^{-1}\left(\frac{1}{-\cos 2 	heta}ight) = \sec ^{-1}(-\sec 2 	heta) = \sec ^{-1}(\sec(\pi - 2 	heta)) = \pi - 2 	heta$.Since $x = \sin 	heta$,$	heta = \sin^{-1} x$.Thus,$y = \frac{1}{2} \sin^{-1} x$ and $u = \pi - 2 \sin^{-1} x$.Then $\frac{dy}{dx} = \frac{1}{2 \sqrt{1-x^2}}$ and $\frac{du}{dx} = \frac{-2}{\sqrt{1-x^2}}$.Therefore,$\frac{dy}{du} = \frac{dy/dx}{du/dx} = \frac{1/(2 \sqrt{1-x^2})}{-2/\sqrt{1-x^2}} = -\frac{1}{4}$.

The derivative of $\tan ^{-1}\left[\frac{x}{1+\sqrt{1-x^2}}\right]$ with respect to $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ is

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