设y=arctan(x-y),则dy=
设F(x,y)=y-arctan(x-y)=0；
则dy/dx=-(∂F/∂x)/(∂F/∂y)=-{-1/[1+(x-y)²]}/{1+1/[1+(x-y)²]}=1/[2+(x-y)²]
故dy=dx/[2+(x-y)²]