Implicit differentiation of x^y = y^x using logarithms

Take the natural logarithm of both sides of $x^y = y^x$ to get $y\ln x = x \ln y$. This removes the variable exponents so the equation can be differentiated using the product rule instead of needing logarithmic differentiation on exponentials directly.

Differentiate both sides with respect to $x$, applying the product rule to each side: on the left, $\frac{dy}{dx}\ln x + y\cdot\frac{1}{x}$; on the right, $\ln y + x\cdot\frac{1}{y}\frac{dy}{dx}$. This gives $\frac{dy}{dx}\ln x + \frac{y}{x} = \ln y + \frac{x}{y}\frac{dy}{dx}.$

Collect all terms containing $\frac{dy}{dx}$ on one side: $\frac{dy}{dx}\left(\ln x - \frac{x}{y}\right) = \ln y - \frac{y}{x}.$ Solving for $\frac{dy}{dx}$ gives $\frac{dy}{dx} = \frac{\ln y - \dfrac{y}{x}}{\ln x - \dfrac{x}{y}},$ which can also be written as $\dfrac{dy}{dx} = \dfrac{y(x\ln y - y)}{x(y\ln x - x)}$ after clearing fractions.