Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 39

Differentiate $\displaystyle y = \sqrt[3]{t} (t^2 + t + t^{-1})$



$\begin{equation} \begin{aligned} y' =& (t)^{\frac{1}{3}} (t^2 + t + t^{-1}) && \text{Expand the equation} \\ \\ y' =& t^{\frac{7}{3}} + t^{\frac{4}{3}} + t^{\frac{-2}{3}} && \text{} \\ \\ y' =& \frac{d}{dt} (t^{\frac{7}{3}}) + \frac{d}{dt} (t^{\frac{4}{3}}) + \frac{d}{dt} (t^{\frac{-2}{3}}) && \text{Apply Power Rule} \\ \\ y' =& \frac{7}{3} t^{\frac{4}{3}} + \frac{4}{3} t^{\frac{1}{3}} - \frac{2}{3} t^{\frac{-5}{3}} && \text{Simplify the equation} \\ \\ y' =& \frac{7}{3} t^{\frac{4}{3}} + \frac{4}{3} t^{\frac{1}{3}} - \frac{2}{3t^{\frac{5}{3}}} && \text{Get the LCD} \\ \\ y' =& \frac{7 t^{\frac{9}{3}} + 4 t^{\frac{6}{3}} - 2}{3 t^{\frac{5}{3}}} && \text{Simplify the equation} \\ \\ y' =& \frac{7t^3 + 4t^2-2}{3t^{\frac{5}{3}}} && \text{} \\ \\ \end{aligned} \end{equation}$