z = f(y/x,x+2y)
的兩端求微分,得
dz = f1*[(xdy-ydx)/x?]+f2*(dx+2dy)
= [-(y/x?)f1+f2]dx+[(1/x)f1+2*f2]dy,
得到
Dz/Dx = -(y/x?)f1+f2,Dz/Dy = (1/x)f1+2*f2,
於是
D?z/DxDy = (D/Dx)(Dz/Dy)
= (D/Dx)[(1/x)f1+2*f2]
= [(-1/x?)*f1+(1/x)*[-(y/x?)f11+f12]+2*[(1/x)f21+2*f22]