Question

If x^y . y^x , then prove that `"dy"/"dx" = y/x((x log y - y)/(y log x - x))` 

Answer 1

Given x^y . y^x 

Taking logarithm on both sides we get,

y log x = x log y

Differentiating with respect to 'x' we get,

`y*1/x + log x * "dy"/"dx" = x * 1/y "dy"/"dx" + log y(1)`

`=> y/x + log x ("dy"/"dx") = x/y "dy"/"dx" + log y`

`=> "dy"/"dx" (log x - x/y) = log y - y/x`

`=> "dy"/"dx" ((y log x - x)/y) = (x log y - y)/x`

`=> "dy"/"dx" = (x log y - y)/x xx y/(y log x - x)`

`=> "dy"/"dx" = y/x((x log y - y)/(y log x - x))`

Hence proved.