Single Variable Calculus, Chapter 7, 7.4-1, Section 7.4-1, Problem 54

Determine $y'$ if $x^y = y^x$

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