If y = tan^2 left( cos^{-1} sqrt{frac{1+x^2}{ | vedclass

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Question

(C) Let $	heta = \cos^{-1} \sqrt{\frac{1+x^2}{2}}$. Then $\cos 	heta = \sqrt{\frac{1+x^2}{2}}$.Squaring both sides, $\cos^2 	heta = \frac{1+x^2}{2}

Vedclass · Accepted Answer

$-\frac{4x}{(1+x^2)^2}$

Vedclass · Answer

(C) Let $	heta = \cos^{-1} \sqrt{\frac{1+x^2}{2}}$. Then $\cos 	heta = \sqrt{\frac{1+x^2}{2}}$.Squaring both sides, $\cos^2 	heta = \frac{1+x^2}{2}$.Using the identity $	an^2 	heta = \sec^2 	heta - 1 = \frac{1}{\cos^2 	heta} - 1$, we get:$y = \frac{1}{\cos^2 	heta} - 1 = \frac{1}{\frac{1+x^2}{2}} - 1 = \frac{2}{1+x^2} - 1 = \frac{2 - (1+x^2)}{1+x^2} = \frac{1-x^2}{1+x^2}$.Now, differentiate $y$ with respect to $x$ using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} ight) = \frac{v u' - u v'}{v^2}$:$\frac{dy}{dx} = \frac{(1+x^2)(-2x) - (1-x^2)(2x)}{(1+x^2)^2} = \frac{-2x - 2x^3 - 2x + 2x^3}{(1+x^2)^2} = \frac{-4x}{(1+x^2)^2}$.Thus, the correct option is $C$.

If $y = \tan^2 \left( \cos^{-1} \sqrt{\frac{1+x^2}{2}} \right)$, then $\frac{dy}{dx} = $

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