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}
$