Single Variable Calculus, Chapter 7, 7.4-2, Section 7.4-2, Problem 54

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


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