设函数 z=xy+xyz=xy+xy, 则 dz=
选项:
A:(y+1y)dx+(x−xy2)dy(y+1y)dx+(x−xy2)dy
B:(x−1y2)dx+(x+1y)dy(x−1y2)dx+(x+1y)dy
C:(y+1y)dx+(x+xy2)dy(y+1y)dx+(x−xy2)dy
D:(y+xy2)dx+(y+1y)dy(y+xy2)dx+(y+1y)dy.
发布时间:2024-05-19 16:28:30