If `x^y=y^x` ,then find `(dy)/(dx)`

Explanation

Taking log on both sides

`ylogx=xlogy`.


Differentiating both the sides by `uv` rule

`y/x+logx frac{dy}{dx}=x/y frac{dy}{dx}+log y`

`y/x=log y=x/yfrac{dy}{dx}-log y frac{dy}{dx}`

`frac{y-xlogy}{x}=frac{dy}{dx} frac{x-ylog x}{y}`
...