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}
$