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

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

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`y^(x) = x^(y)`
`rArr "log" (y^(x)) = "log"(x^(y))`
`rArr x"log"y = y"log"x`
Differentiate both sides w.r.t.x
`x * (d)/(dx) "log" y + "log" y * (d)/(dx)x`
`= y * (d)/(dx)"log"x + "log"x * (d)/(dx)y`
`rArr (x)/(y)(dy)/(dx) + "log" y * 1 = (y)/(x) + "log"x (dy)/(dx)`
`rArr ((x)/(y) - "log"x)(dy)/(dx) = (y)/(x) - "log" y`
`rArr (x-y "log"x)/(y) (dy)/(dx) = (y-x "log"y)/(x)`
`rArr (dy)/(dx) = (y)/(x) * (y-x"log"y)/(x-y"log"x)`
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