Calculus 10th Edition

The derivative is $\frac{2}{3\sqrt[3] t}-\frac{1}{3\sqrt[3] {t^2}}$.
To find the derivative of the overall function just apply the power rule$((x^n)'=nx^{n-1})$ to the individual terms as follows: $f'(t)=(t^{\frac{2}{3}}-t^{\frac{1}{3}}+4)'$ $=(t^{\frac{2}{3}})'-(t^{\frac{1}{3}})'+(4)'$ $=\frac{2}{3}t^{\frac{2}{3}-1}-\frac{1}{3}t^{\frac{1}{3}-1}+0=$ $\frac{2}{3\sqrt[3] t}-\frac{1}{3\sqrt[3] {t^2}}.$