Consider the following for the two (02) items that follow: Let y =...

Let $x = 3\sin\theta$. Then $\theta = \sin^{-1} \left(\dfrac{x}{3}\right).<br><br>Substituting$x = 3\sin$\theta into the given expression:<br><br>$y = \sin^{-1} $\left(3\sin\theta - \dfrac{4(3\sin\theta)^3}{27}\right)$

y = \sin^{-1} \left(3\sin\theta - \dfrac{4(27\sin^3\theta)}{27}\right)<br><br>y = \sin^{-1}(3\sin$\theta -$4\sin^3$\theta)<br><br>Using the trigonometric identity$\sin(3$\theta) =$3\sin$\theta -$4\sin^3$\theta:<br><br>$y = \sin^{-1}(\sin 3$\theta)<br><br>Assuming principal values, we get:<br><br>$y = 3$\theta<br><br>$y = 3\sin^{-1} \left(\dfrac{x}{3}\right)$<br><br>Differentiating with respect to$x$using the chain rule:<br><br>$\dfrac{dy}{dx} = 3 \cdot \dfrac{1}{\sqrt{1 - \left(\dfrac{x}{3}\right)^2}}\cdot \dfrac{d}{dx}\left(\dfrac{x}{3}\right)

\dfrac{dy}{dx} =$3 \cdot$\dfrac{1}{\sqrt{1 - \dfrac{x^2}{9}}} \cdot \dfrac{1}{3}$$<br><br>$$\dfrac{dy}{dx} = \dfrac{1}{\sqrt{\dfrac{9 - x²}{9}}}$$<br><br>$$\dfrac{dy}{dx} = \dfrac{3}{\sqrt{9 - x²}}$$<br><br>Answer:$$\dfrac{3}{\sqrt{9-x²}}