sh0 omega(dy)/(dx)=(y(x log y-y))/(x(y log x-x))" when "x^(y)

(dy)/(dx)=(y(x ln y-y))/(x(y ln x-x))

If x^(y)=y^(x) then (dy)/(dx)= ... (A) (y(x log y-y))/(x(y log x-x)) (B) (y(y log x-x))/(x(x log y-y)) (C) (y^(2)(1-log x))/(x^(2)(1-log y)) (D) (y(1-log x))/(x(1-log y))

If x^(y)y^(x),=1, prove that (dy)/(dx),=-(y(y+x log y))/(x(y log x+x))

If x^(y).y^(x)=1, prove that (dy)/(dx)=-(y(y+x log y))/(x(y log x+x))

(dy)/(dx)=(x+y)ln(x+y)-1

Solve (dy)/(dx)=(y)/(2y log y+y-x).

If y=a^(x^(x^(2)*oo)), prove that (dy)/(dx)=(y^(2)log y)/(x(1-y log x*log y))