If f(x) = operatorname{Tan}^{-1} left[ frac{s | vedclass

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(B) Let $x^2 = \cos(2	heta)$,where $2	heta \in (0, \pi)$,so $	heta = \frac{1}{2} \cos^{-1}(x^2)$.Then $\sqrt{1+x^2} = \sqrt{1+\cos(2	heta)} = \sqr

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

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(B) Let $x^2 = \cos(2	heta)$,where $2	heta \in (0, \pi)$,so $	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 $f(x)$:$f(x) = \operatorname{Tan}^{-1} \left[ \frac{\sqrt{2}\cos	heta + \sqrt{2}\sin	heta}{\sqrt{2}\cos	heta - \sqrt{2}\sin	heta} ight] = \operatorname{Tan}^{-1} \left[ \frac{\cos	heta + \sin	heta}{\cos	heta - \sin	heta} ight] = \operatorname{Tan}^{-1} \left[ \frac{1 + 	an	heta}{1 - 	an	heta} ight] = \operatorname{Tan}^{-1} [	an(\frac{\pi}{4} + 	heta)] = \frac{\pi}{4} + 	heta$.Substituting $	heta$ back: $f(x) = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x^2)$.Differentiating with respect to $x$:$f'(x) = 0 + \frac{1}{2} \cdot \frac{-1}{\sqrt{1-(x^2)^2}} \cdot \frac{d}{dx}(x^2) = \frac{-1}{2\sqrt{1-x^4}} \cdot 2x = \frac{-x}{\sqrt{1-x^4}}$.

If $f(x) = \operatorname{Tan}^{-1} \left[ \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \right]$ for $0 < |x| < 1$,then $f'(x) =$

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