Single Variable Calculus, Chapter 7, 7.3-2, Section 7.3-2, Problem 38

Differentiate $\displaystyle g (x) = \sqrt{x} e^x$


$
\begin{equation}
\begin{aligned}

g'(x) =& \frac{d}{dx} (\sqrt{x} e^x)
\\
\\
g'(x) =& \sqrt{x} \frac{d}{dx} (e^x) + (e^x) \frac{d}{dx} (\sqrt{x})
\\
\\
g'(x) =& (x)^{\frac{1}{2}} \frac{d}{dx} (e^x) + (e^x) \frac{d}{dx} (x)^{\frac{1}{2}}
\\
\\
g'(x) =& (x)^{\frac{1}{2}} e^x + (e^x) \left( \frac{1}{2} \right) (x)^{\frac{-1}{2}}
\\
\\
g'(x) =& x^{\frac{1}{2}} e^x + \frac{e^x}{2 \sqrt{x}}
\\
\\
g'(x) =& \sqrt{x} e^x + \frac{e^x}{2 \sqrt{x}}
\\
\\
& \text{or}
\\
\\
g'(x) =& \sqrt{x} e^x + \frac{e^x}{2 \sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}}
\\
\\
g'(x) =& \sqrt{x} e^x + \frac{\sqrt{x} e^x}{2x}

\end{aligned}
\end{equation}
$