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

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(C) Given $y = 	an^{-1} \left[ \frac{x - \sqrt{1 - x^2}}{x + \sqrt{1 - x^2}} ight]$.Let $x = \cos 	heta$,then $	heta = \cos^{-1} x$.Substituting

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$\frac{1}{\sqrt{1 - x^2}}$

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(C) Given $y = 	an^{-1} \left[ \frac{x - \sqrt{1 - x^2}}{x + \sqrt{1 - x^2}} ight]$.Let $x = \cos 	heta$,then $	heta = \cos^{-1} x$.Substituting $x = \cos 	heta$ into the expression:$y = 	an^{-1} \left[ \frac{\cos 	heta - \sin 	heta}{\cos 	heta + \sin 	heta} ight]$Divide numerator and denominator by $\cos 	heta$:$y = 	an^{-1} \left[ \frac{1 - 	an 	heta}{1 + 	an 	heta} ight]$Using the formula $	an(\frac{\pi}{4} - 	heta) = \frac{1 - 	an 	heta}{1 + 	an 	heta}$:$y = 	an^{-1} \left[ 	an \left( \frac{\pi}{4} - 	heta ight) ight] = \frac{\pi}{4} - 	heta$Substitute back $	heta = \cos^{-1} x$:$y = \frac{\pi}{4} - \cos^{-1} x$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - \cos^{-1} x ight) = 0 - \left( -\frac{1}{\sqrt{1 - x^2}} ight) = \frac{1}{\sqrt{1 - x^2}}$.

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

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