## Calculus: Early Transcendentals 8th Edition

$y'=\frac{lny-\frac{y}{x}}{lnx-\frac{x}{y}}$ or $y'=\frac{y(xlny-y)}{x(ylnx-x)}$
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 $y'=\frac{y(xlny-y)}{x(ylnx-x)}$