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Let $x = \sin\theta$. Then $\theta = \sin^{-1}x$. Since $x \in [-\frac{1}{2}, \frac{1}{2}]$, $\theta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$.
Substitute $x = \sin\theta$ into the expression $3x - 4x^3$:\r\n$$3x - 4x^3 = 3\sin\theta - 4\sin^3\theta = \sin(3\theta)$$\r\nTherefore, $y = \sin^{-1}(\sin(3\theta))$.
Since $\theta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$, we have $3\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Therefore, $\sin^{-1}(\sin(3\theta)) = 3\theta$.\r\nSo, $y = 3\theta = 3\sin^{-1}x$.
Now, differentiate $y$ with respect to $x$:\r\n$$\frac{dy}{dx} = \frac{d}{dx}(3\sin^{-1}x) = 3\frac{d}{dx}(\sin^{-1}x) = 3\cdot\frac{1}{\sqrt{1-x^2}}$$\r\nThus, $\frac{dy}{dx} = \frac{3}{\sqrt{1-x^2}}$.
Final Answer: $\frac{3}{\sqrt{1-x^2}}$
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