If y = sin^{-1}(sqrt{1 - x^2}),then dy/dx = | vedclass

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(C) Given $y = \sin^{-1}(\sqrt{1 - x^2})$.Let $\sqrt{1 - x^2} = \sin 	heta$,which implies $1 - x^2 = \sin^2 	heta$.Then $x^2 = 1 - \sin^2 	heta = \

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

Vedclass · Answer

(C) Given $y = \sin^{-1}(\sqrt{1 - x^2})$.Let $\sqrt{1 - x^2} = \sin 	heta$,which implies $1 - x^2 = \sin^2 	heta$.Then $x^2 = 1 - \sin^2 	heta = \cos^2 	heta$,so $x = \cos 	heta$ (assuming $x > 0$ for the principal value range).Thus,$	heta = \cos^{-1} x$.Substituting this back into the expression for $y$,we get $y = \sin^{-1}(\sin 	heta) = 	heta = \cos^{-1} x$.Now,differentiating $y = \cos^{-1} x$ with respect to $x$,we get:$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}$.

If $y = \sin^{-1}(\sqrt{1 - x^2})$,then $dy/dx = $

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