If y = operatorname{Tan}^{-1}left(frac{sqrt{1 | vedclass

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(D) Let $x^2 = \cos(2	heta)$,so $2	heta = \cos^{-1}(x^2)$,which implies $	heta = \frac{1}{2} \cos^{-1}(x^2)$.Then $\sqrt{1+x^2} = \sqrt{1+\cos(2	h

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$\frac{-x}{\sqrt{1-x^4}}$

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(D) Let $x^2 = \cos(2	heta)$,so $2	heta = \cos^{-1}(x^2)$,which implies $	heta = \frac{1}{2} \cos^{-1}(x^2)$.Then $\sqrt{1+x^2} = \sqrt{1+\cos(2	heta)} = \sqrt{2\cos^2(	heta)} = \sqrt{2}\cos(	heta)$ and $\sqrt{1-x^2} = \sqrt{1-\cos(2	heta)} = \sqrt{2\sin^2(	heta)} = \sqrt{2}\sin(	heta)$.Substituting these into the expression for $y$:$y = \operatorname{Tan}^{-1}\left(\frac{\sqrt{2}\cos(	heta) + \sqrt{2}\sin(	heta)}{\sqrt{2}\cos(	heta) - \sqrt{2}\sin(	heta)}ight) = \operatorname{Tan}^{-1}\left(\frac{\cos(	heta) + \sin(	heta)}{\cos(	heta) - \sin(	heta)}ight)$.Dividing numerator and denominator by $\cos(	heta)$:$y = \operatorname{Tan}^{-1}\left(\frac{1 + 	an(	heta)}{1 - 	an(	heta)}ight) = \operatorname{Tan}^{-1}(	an(\frac{\pi}{4} + 	heta)) = \frac{\pi}{4} + 	heta$.Substituting $	heta = \frac{1}{2} \cos^{-1}(x^2)$:$y = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x^2)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 + \frac{1}{2} \left(-\frac{1}{\sqrt{1-(x^2)^2}}ight) \cdot (2x) = -\frac{x}{\sqrt{1-x^4}}$.Thus,the correct option is $D$.

If $y = \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$ for $0 < |x| < 1$,then $\frac{dy}{dx} = $

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