Answer
$\frac{ -\sqrt[3]{2^x}}{\ln(2)}2^{x} $
Work Step by Step
Given
$$y=\int_{2^{x}}^{1} \sqrt[3]{t} d t$$
Since
\begin{aligned} \frac{dy}{dx}&= \frac{d}{dx} \int_{2^{x}}^{1} \sqrt[3]{t} d t\\
&=- \frac{d}{dx} \int_{1}^{2^{x}} \sqrt[3]{t} d t\\
&= -\sqrt[3]{2^x}\frac{d}{dx}(2^x)\\
&=\frac{ -\sqrt[3]{2^x}}{\ln(2)}2^{x} \end{aligned}