Calculus: Early Transcendentals, Chapter 3, 3.1, Section 3.1, Problem 30

Differentiate the function $\displaystyle \left( \sqrt{x} + \frac{1}{\sqrt[3]{x}}\right)^2$

We have, $\displaystyle V = \left( \sqrt{x} + \frac{1}{\sqrt[3]{x}} \right) \left( \sqrt{x} + \frac{1}{\sqrt[3]{x}} \right)$
By using FOIL method, we get

$
\begin{equation}
\begin{aligned}
V &= \left( x^{\frac{1}{2}} \right) \left( x^{\frac{1}{2}} \right) + x^{\frac{1}{2}} \left( \frac{1}{x^{\frac{1}{3}}} \right)
+ x^{\frac{1}{2}} \left( \frac{1}{x^{\frac{1}{3}}} \right) + \left( \frac{1}{x^{\frac{1}{3}}} \right) \left( \frac{1}{x^{\frac{1}{3}}} \right)\\
\\
V &= x + 2 \frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}} + \frac{1}{x^{\frac{2}{3}}}\\
\\
V &= x + 2 x^{\frac{1}{2} - \frac{1}{3}} + x^{- \frac{2}{3}} = x + 2x^{\frac{1}{6}} + x^{-\frac{2}{3}}
\end{aligned}
\end{equation}
$


So,

$
\begin{equation}
\begin{aligned}
V' &= \frac{d}{dx} (x) + 2 \cdot \frac{d}{dx} \left( x^{\frac{1}{6}} \right) + \frac{d}{dx} \left( x^{-\frac{2}{3}} \right)\\
\\
&= (1) + 2 \cdot \frac{1}{6} x^{\frac{1}{6} - 1} + \left( -\frac{2}{3} \right) x^{-\frac{2}{3}-1}\\
\\
&= 1 + \frac{1}{3} x^{-\frac{5}{6}} - \frac{2}{3} x^{-\frac{5}{3}}\\
\\
&= 1 + \frac{1}{3x^{\frac{5}{6}}} - \frac{2}{3x^{\frac{5}{3}}}
\end{aligned}
\end{equation}
$