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

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(A) Given $y = \sin(2 \sin^{-1} x)$.Let $\sin^{-1} x = 	heta$, then $x = \sin 	heta$.So, $y = \sin(2	heta) = 2 \sin 	heta \cos 	heta$.Since $\sin

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$\frac{2 - 4x^2}{\sqrt{1 - x^2}}$

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(A) Given $y = \sin(2 \sin^{-1} x)$.Let $\sin^{-1} x = 	heta$, then $x = \sin 	heta$.So, $y = \sin(2	heta) = 2 \sin 	heta \cos 	heta$.Since $\sin 	heta = x$, then $\cos 	heta = \sqrt{1 - x^2}$.Thus, $y = 2x \sqrt{1 - x^2}$.Differentiating with respect to $x$ using the product rule:$\frac{dy}{dx} = 2 \left[ x \cdot \frac{d}{dx}(\sqrt{1 - x^2}) + \sqrt{1 - x^2} \cdot \frac{d}{dx}(x) ight]$$\frac{dy}{dx} = 2 \left[ x \cdot \frac{-2x}{2\sqrt{1 - x^2}} + \sqrt{1 - x^2} \cdot 1 ight]$$\frac{dy}{dx} = 2 \left[ \frac{-x^2 + (1 - x^2)}{\sqrt{1 - x^2}} ight] = 2 \left[ \frac{1 - 2x^2}{\sqrt{1 - x^2}} ight] = \frac{2 - 4x^2}{\sqrt{1 - x^2}}$.

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

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