Question

求由方程e^(x+y)-xy=0所确定的隐函数y=f(x)的微分dy

Answer 1

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.