Answer
$y'=\frac{lny-\frac{y}{x}}{lnx-\frac{x}{y}}$
or
$y'=\frac{y(xlny-y)}{x(ylnx-x)}$
Work Step by Step
Given: $x^{y}=y^{x}$
Use logarithmic properties $ln(x^{y})=ylnx$.
$ylnx=xlny$
Simplify the given function using both product rule and chain rule of differentiation.
$\frac{y}{x}+lnx\frac{dy}{dx}=\frac{x}{y}\frac{dy}{dx}+lny$
or
$\frac{y}{x}+lnxy'=\frac{x}{y}y'+lny$
Hence, $y'=\frac{lny-\frac{y}{x}}{lnx-\frac{x}{y}}$
or
$y'=\frac{y(xlny-y)}{x(ylnx-x)}$
