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

We are given that, ${y^x} = {x^y}$
By taking log on both sides , we get
x log y $=$ y log x …(i)

Differentiating (i) on both sides w.r.t. x, we get

$x\cfrac{d}{{dx}}\log y + \log y \cdot 1 = y\cfrac{d}{{dx}}\log x + \log x\cfrac{{dy}}{{dx}}$

$\Rightarrow$ $x \cdot \cfrac{1}{y} \cdot \cfrac{{dy}}{{dx}} + \log y = \cfrac{y}{x} + \log x \cdot \cfrac{{dy}}{{dx}}$

$\Rightarrow$ $\cfrac{{dy}}{{dx}}\left[ {\log x - \cfrac{x}{y}} \right] = \log y - \cfrac{y}{x} \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{y(x\log y - y)}}{{x(y\log x - x)}}$