frac{d}{d x}left[cos ^2left(cot ^{-1} sqrt{fr | vedclass

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Question

(D) Let $	heta = \cot ^{-1} \sqrt{\frac{2+x}{2-x}}$. Then $\cot 	heta = \sqrt{\frac{2+x}{2-x}}$.From the right-angled triangle with base $\sqrt{2+x}

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$\frac{1}{4}$

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(D) Let $	heta = \cot ^{-1} \sqrt{\frac{2+x}{2-x}}$. Then $\cot 	heta = \sqrt{\frac{2+x}{2-x}}$.From the right-angled triangle with base $\sqrt{2+x}$ and perpendicular $\sqrt{2-x}$,the hypotenuse is $\sqrt{(\sqrt{2+x})^2 + (\sqrt{2-x})^2} = \sqrt{2+x+2-x} = \sqrt{4} = 2$.Thus,$\cos 	heta = \frac{	ext{base}}{	ext{hypotenuse}} = \frac{\sqrt{2+x}}{2}$.Now,the expression becomes $\frac{d}{d x}\left[\cos ^2 	hetaight] = \frac{d}{d x}\left[\left(\frac{\sqrt{2+x}}{2}ight)^2ight]$.$= \frac{d}{d x}\left(\frac{2+x}{4}ight) = \frac{d}{d x}\left(\frac{1}{2} + \frac{x}{4}ight) = 0 + \frac{1}{4} = \frac{1}{4}$.

$\frac{d}{d x}\left[\cos ^2\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right]$ is

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