Find frac{dy}{dx} if y = sin^{-1}(2xsqrt{1-x^ | vedclass

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

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

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(A) Given $y = \sin^{-1}(2x\sqrt{1-x^2})$.Let $x = \sin 	heta$,then $	heta = \sin^{-1} x$.Since $-\frac{1}{\sqrt{2}} Then $y = \sin^{-1}(2 \sin 	heta \sqrt{1-\sin^2 	heta}) = \sin^{-1}(2 \sin 	heta \cos 	heta) = \sin^{-1}(\sin 2	heta)$.Since $-\frac{\pi}{4} Thus,$y = 2	heta = 2\sin^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}}$.

Find $\frac{dy}{dx}$ if $y = \sin^{-1}(2x\sqrt{1-x^2})$ where $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$.

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