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]