$x^{y} = y^{x}$

This curve is symmetric about the line $y=x$ (switching the x's and y's results in the same equation). This symmetry combined with its differentiability imply that the derivative $\frac{dy}{dx}$ of this curve is $-1$ at the crossing point (since it must be perpendicular to $y=x$ at the crossing point).

Starting with $x^y=y^x$ , take the ln of both sides: $y\ln x=x\ln y$ .

Next we differentiate both sides:

$\frac{dy}{dx}\cdot\ln x+y\cdot\frac1x=\ln y+x\cdot\frac1y\frac{dy}{dx}$

Finally we substitute $y=x$ and $\frac{dy}{dx}=-1$ :

$-\ln x+1=\ln x-1$

Solving for x, we get that $x=e$ .