If y = sin^{-1}left(frac{3x}{2} - frac{x^3}{2 | vedclass

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(B) Given $y = \sin^{-1}\left(\frac{3x}{2} - \frac{x^3}{2}\right)$.We can rewrite the expression as $y = \sin^{-1}\left(3\left(\frac{x}{2}\right) - 4\

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$\frac{3}{\sqrt{4-x^2}}$

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(B) Given $y = \sin^{-1}\left(\frac{3x}{2} - \frac{x^3}{2}ight)$.We can rewrite the expression as $y = \sin^{-1}\left(3\left(\frac{x}{2}ight) - 4\left(\frac{x}{2}ight)^3ight)$.Let $\frac{x}{2} = \sin 	heta$,which implies $	heta = \sin^{-1}\left(\frac{x}{2}ight)$.Substituting this into the equation,we get $y = \sin^{-1}(3\sin 	heta - 4\sin^3 	heta)$.Using the trigonometric identity $\sin(3	heta) = 3\sin 	heta - 4\sin^3 	heta$,we have $y = \sin^{-1}(\sin 3	heta) = 3	heta$.Substituting $	heta$ back,$y = 3\sin^{-1}\left(\frac{x}{2}ight)$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 3 \cdot \frac{1}{\sqrt{1 - (x/2)^2}} \cdot \frac{1}{2}$.$\frac{dy}{dx} = \frac{3}{2 \cdot \sqrt{\frac{4-x^2}{4}}} = \frac{3}{2 \cdot \frac{\sqrt{4-x^2}}{2}} = \frac{3}{\sqrt{4-x^2}}$.

If $y = \sin^{-1}\left(\frac{3x}{2} - \frac{x^3}{2}\right)$,then $\frac{dy}{dx}$ is equal to

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