e∧(x+y)+cos(xy)=0,
两边对闭游肢x求磨亏导，得
e^(x+y)*(1+y')-sin(xy)*(y+xy')=0
(e^(x+y)-xsin(xy))y'=ysin(xy)-e^(x+y)
所以
dy/轿世dx=[ysin(xy)-e^(x+y)]/[e^(x+y)-xsin(xy)]
e^(x+y)+(cosxy)=0
e^(x+y)=-cosxy
[e^(x+y)]'=[-cosxy]'
(1+y')e^(x+y)=(y+xy')sinxy
e^(x+y)-ysinxy=xy'睁念sinxy-y'e^(x+y)
y'=[e^(x+y)-ysinxy] /拍早稿袭孝 [xsinxy-e^(x+y)]