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

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(B) Let $x = \sin 	heta$, where $	heta = \sin^{-1} x$.Then $\sqrt{1 - x^2} = \sqrt{1 - \sin^2 	heta} = \cos 	heta$.The expression becomes $y = 	a

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

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(B) Let $x = \sin 	heta$, where $	heta = \sin^{-1} x$.Then $\sqrt{1 - x^2} = \sqrt{1 - \sin^2 	heta} = \cos 	heta$.The expression becomes $y = 	an^{-1} \left[ \frac{\sin 	heta - \cos 	heta}{\sin 	heta + \cos 	heta} ight]$.Divide numerator and denominator by $\cos 	heta$: $y = 	an^{-1} \left[ \frac{	an 	heta - 1}{	an 	heta + 1} ight]$.Using $	an(\frac{\pi}{4} - 	heta) = \frac{1 - 	an 	heta}{1 + 	an 	heta}$, we have $y = 	an^{-1} [-	an(\frac{\pi}{4} - 	heta)] = 	an^{-1} [	an(	heta - \frac{\pi}{4})]$.Thus, $y = 	heta - \frac{\pi}{4} = \sin^{-1} x - \frac{\pi}{4}$.Differentiating with respect to $x$, $\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1} x) - 0 = \frac{1}{\sqrt{1 - x^2}}$.

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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