if y to the power x = x to the power y, prove that dy/dx=y/x(y-xlogy/x-ylogx)

take logarihtm on both sides and differenciate

could u pls defferentiate it for me... i tried but couldn''t get the answer..

taking log on both sides

x lny = y lnx

diff. both sides                  let dy/dx = m

then

lny + (x/y)m = (y/x) + m lnx

m= (y/x)/(y-xlogy/x-ylogx)

y^x=x^y

xlogy = ylogx

(x/y)dy/dx + logy = y/x + logx(dy/dx)

dy/dx(x/y-logx) = y/x - logy

dy/dx = (y/x - logy)/(x/y-logx)

         = (y/x){1- (x/y)logy}/{x/y - logx}

         = y/x(y-xlogy/x-ylogx)