If x^y = yˣ, then find (dy)/(dx).

$x^y = y^x$ $y \log x = x \log y$ By differentiating on both sides, $y \cdot \frac{1}{x} + \log x \frac{dy}{dx} = x \cdot \frac{1}{y}\frac{dy}{dx} + \log y$ $\frac{dy}{dx}\left(\log x - \frac{x}{y}\right) = \log y - \frac{y}{x}$ $\frac{dy}{dx} = \frac{\log y - \frac{y}{x}}{\log x - \frac{x}{y}}$