求由方程e^(x+y)-xy=0所确定的隐函数y=f(x)的微分dy

由已知得：e^(x+y)=xy.d e^(x+y)=dxy.e^(x+y)*d(x+y)=(ydx+xdy).e^(x+y)*(dx+dy)=ydx+xdy.e^(x+y)dx+e^(x+y) dy=ydx+xdy.[(e^(x+y)]dy-xdy=[y-e^(x+y)]dx.dy={[y-e^(x+y)]/[e^(x+y)-x]}dx.