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

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(A) Given $y = \sin(2\sin^{-1}x)$.Let $x = \sin	heta$,then $	heta = \sin^{-1}x$.Substituting this into the expression for $y$:$y = \sin(2	heta) = 2

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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 $x = \sin	heta$,then $	heta = \sin^{-1}x$.Substituting this into the expression for $y$:$y = \sin(2	heta) = 2\sin	heta\cos	heta$.Since $\sin	heta = x$,we have $\cos	heta = \sqrt{1 - \sin^2	heta} = \sqrt{1 - x^2}$.Thus,$y = 2x\sqrt{1 - x^2}$.Now,differentiate $y$ 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]$$\frac{dy}{dx} = 2 \left[ x \cdot \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) + \sqrt{1 - x^2} \cdot 1 ight]$$\frac{dy}{dx} = 2 \left[ \frac{-x^2}{\sqrt{1 - x^2}} + \sqrt{1 - x^2} 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} = $

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