Find frac{dy}{dx},if y = sin^{-1}x + sin^{-1} | vedclass

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(A) Given $y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}$.Let $x = \sin	heta$,then $	heta = \sin^{-1}x$. Since $-1 \le x \le 1$,we have $-\frac{\pi}{2} \le

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(A) Given $y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}$.Let $x = \sin	heta$,then $	heta = \sin^{-1}x$. Since $-1 \le x \le 1$,we have $-\frac{\pi}{2} \le 	heta \le \frac{\pi}{2}$.Then $\sqrt{1-x^2} = \sqrt{1-\sin^2	heta} = \sqrt{\cos^2	heta} = |\cos	heta|$.Since $-\frac{\pi}{2} \le 	heta \le \frac{\pi}{2}$,$\cos	heta \ge 0$,so $|\cos	heta| = \cos	heta$.Thus,$y = \sin^{-1}(\sin	heta) + \sin^{-1}(\cos	heta) = 	heta + \sin^{-1}(\sin(\frac{\pi}{2} - 	heta))$.For $x \in [0, 1]$,$0 \le 	heta \le \frac{\pi}{2}$,so $y = 	heta + (\frac{\pi}{2} - 	heta) = \frac{\pi}{2}$.For $x \in [-1, 0]$,$-\frac{\pi}{2} \le 	heta $y = 	heta + \sin^{-1}(\sin(\frac{\pi}{2} + 	heta))$. Since $\frac{\pi}{2} + 	heta \in [0, \pi]$,we use the identity $\sin^{-1}(\sin A) = \pi - A$ for $A \in [\frac{\pi}{2}, \pi]$.$y = 	heta + \pi - (\frac{\pi}{2} + 	heta) = \frac{\pi}{2}$.In both cases,$y = \frac{\pi}{2}$.Therefore,$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2}) = 0$.

Find $\frac{dy}{dx}$,if $y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}$,where $-1 \le x \le 1$.

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