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

Answer 1

对e^(x+y)+cos(xy)=0两边求微分,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))/(e^(x+y)-xsin(xy))