For f(x,y)=x2y+xy2f(x, y) = x^2 y + xy^2f(x,y)=x2y+xy2, compute both mixed partial derivatives fxy(x,y)f_{xy}(x, y)fxy(x,y) and fyx(x,y)f_{yx}(x, y)fyx(x,y). Why are they equal?
fxy=fyx=2x+2yf_{xy} = f_{yx} = 2x + 2yfxy=fyx=2x+2y; they are equal because both exist and are continuous
fxy=2x+2yf_{xy} = 2x + 2yfxy=2x+2y and fyx=x+yf_{yx} = x + yfyx=x+y; they differ because the domain is not simply connected
fxy=2x+2yf_{xy} = 2x + 2yfxy=2x+2y and fyx=2y+2x=2x+2yf_{yx} = 2y + 2x = 2x + 2yfyx=2y+2x=2x+2y; they are equal by Schwarz's theorem (Clairaut's theorem)
They cannot both be computed from the definition