If `y^x=x^y ,t h e nfin d(dy)/(dx)dot`

Answer

`(y(x log y-y))/(x(y log x -x))`

Explanation

`y^(x)=x^(y)`
`"or "log y^(x)= log x^(y)`
`"or "x log y = y log x`
`"or "x(1)/(y)(dy)/(dx)+log y xx1 =(y)/(x)+log x (dy)/(dx)`
`"or "((x)/(y)-log x)(dy)/(dx)=(y)/(x)-log y`
`"or "(dy)/(dx)=(y/x-logy)/((x)/(y)-log x)=(y(y-x log y))/(x(x- y log x))=(y(x log y - y ))/(x(y log x -x))`