frac{d}{dx} left[ tan^{-1} left( frac{sqrt{1 | vedclass

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(A) Let $y = 	an^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} ight)$.Substitute $x^2 = \cos 2	heta$,which i

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

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(A) Let $y = 	an^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} ight)$.Substitute $x^2 = \cos 2	heta$,which implies $2	heta = \cos^{-1}(x^2)$ or $	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:$y = 	an^{-1} \left( \frac{\sqrt{2} \cos 	heta + \sqrt{2} \sin 	heta}{\sqrt{2} \cos 	heta - \sqrt{2} \sin 	heta} ight) = 	an^{-1} \left( \frac{\cos 	heta + \sin 	heta}{\cos 	heta - \sin 	heta} ight)$.Dividing numerator and denominator by $\cos 	heta$:$y = 	an^{-1} \left( \frac{1 + 	an 	heta}{1 - 	an 	heta} ight) = 	an^{-1} \left( 	an \left( \frac{\pi}{4} + 	heta ight) ight) = \frac{\pi}{4} + 	heta$.Substituting back $	heta = \frac{1}{2} \cos^{-1}(x^2)$:$y = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x^2)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 + \frac{1}{2} \left( \frac{-1}{\sqrt{1 - (x^2)^2}} ight) \cdot \frac{d}{dx}(x^2) = \frac{1}{2} \left( \frac{-1}{\sqrt{1 - x^4}} ight) \cdot 2x = \frac{-x}{\sqrt{1 - x^4}}$.

$\frac{d}{dx} \left[ \tan^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \right] = $

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