Question

利用微分形式的不变性，求下列方程确定的隐函数y=y(x)的导数dy/dx

e^(x+y)+cos(xy)=0

Answer 1

对e^(x+y)+cos(xy)=0两物友边求微分者薯，得
d(e^(x+y)+cos(xy))=0
de^(x+y)+dcos(xy)=0
e^(x+y)*(dx+dy)-sin(xy)*(ydx+xdy)=0
e^(x+y)*dy-sin(xy)*xdy=-e^(x+y)*dx+sin(xy)*ydx
(e^(x+y)-sin(xy)*x)dy=(-e^(x+y)+sin(xy)*y)dx
dy/dx=(ysin(xy)-e^(x+y))/罩嫌槐(e^(x+y)-xsin(xy))

Answer 2

两边对x求导敏岁得：
e^(x+y)*(1+y'历卖)-sin(xy)*(y+xy'肢拿逗)=0
解得:y'=(ysin(xy)-e^(x+y))/(e^(x+y)-xcos(xy))