Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 48

Find the derivative of the function $y = \sqrt{x^x}$, using log differentiation

$\begin{equation} \begin{aligned} \ln y &= \ln \sqrt{x^x}\\ \\ \ln y &= \ln x^{\frac{x}{2}}\\ \\ \ln y &= \frac{x}{2} \ln x\\ \\ \frac{d}{dx} \ln y &= \frac{x}{2} \frac{d}{dx} (\ln x) + \ln x \frac{d}{dx} \left( \frac{x}{2} \right)\\ \\ \frac{1}{y} \frac{dy}{dx} &= \frac{\cancel{x}}{2} \cdot \frac{1}{\cancel{x}} + \ln x \cdot \frac{1}{2}\\ \\ \frac{1}{y} y' &= \frac{1}{2} + \frac{1}{2} \ln x\\ \\ \frac{y'}{y} &= \frac{1+ \ln x}{2}\\ \\ y' &= y \left( \frac{1+ \ln x}{2} \right)\\ \\ y' &= \sqrt{x^x} \left( \frac{1+\ln x}{2} \right) \end{aligned} \end{equation}$