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

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


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