Answer
$\frac{dy}{dx} = -3\sqrt[3] {t^2}$
Work Step by Step
1. Find $\frac{dy}{dt}$:
$y = 4 - t$
$dy = (0 - 1)dt$
$dy/dt = -1$
2. Find $\frac{dx}{dt}$:
$x= \sqrt[3] t = t^{1/3}$
$dx = (\frac{1}{3})t^{(\frac{1}{3}-1)}dt = \frac{t^{-\frac{2}{3}}}{3}dt = \frac{1}{3t^{\frac{2}{3}}}dt = \frac{1}{3\sqrt[3] {t^2}}dt$
$\frac{dx}{dt} = \frac{1}{3\sqrt[3] {t^2}}$
3. Using the expression: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, find the derivative:
$\frac{dy}{dx} = \frac{-1}{\frac{1}{3\sqrt[3] {t^2}}} = -3\sqrt[3] {t^2}$
