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}
$