## Calculus with Applications (10th Edition)

$$\dfrac{dy}{dx}= \frac{2}{3} x^{-4 / 3}$$
Since $y=\dfrac{-2}{\sqrt[3]{x}} = -2 x^{-1 / 3}$ Using the rule $\frac{d}{dx}x^n=nx^{n-1},$ then \begin{align*} \dfrac{dy}{dx}&= (-2)\frac{-1}{3}x^{-4/3}\\ &=\frac{2}{3} x^{-4 / 3}\end{align*}
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