Question

If $(x)^y = (y)^x$, then find $\frac{dy}{dx}$.

Hint:

When differentiating implicit functions involving logarithms, always apply the chain rule and product rule carefully to each term.

Answer 1

The given equation is: \[ x^y = y^x. \] Taking the natural logarithm of both sides yields: \[ \log(x^y) = \log(y^x). \] Applying the logarithmic identity $\log(a^b) = b \log(a)$, we obtain: \[ y \log x = x \log y. \] Differentiating both sides with respect to $x$ using the product rule for both $y \log x$ and $x \log y$: \[ \frac{d}{dx}(y \log x) = \frac{d}{dx}(x \log y). \] This results in: \[ \frac{dy}{dx} \log x + y \frac{1}{x} = \frac{dy}{dx} x \frac{1}{y} + \log y. \] Rearranging to group terms with $\frac{dy}{dx}$: \[ \frac{dy}{dx} \log x - \frac{dy}{dx} \frac{x}{y} = \log y - \frac{y}{x}. \] Factoring out $\frac{dy}{dx}$: \[ \frac{dy}{dx} \left( \log x - \frac{x}{y} \right) = \log y - \frac{y}{x}. \] Solving for $\frac{dy}{dx}$: \[ \frac{dy}{dx} = \frac{\log y - \frac{y}{x}}{\log x - \frac{x}{y}}. \]