∴
| du |
| dx |
| ∂f |
| ∂x |
| ∂f |
| ∂y |
| dy |
| dx |
| ∂f |
| ∂z |
| dz |
| dx |
又由exy-y=0，两边对x求导得：exy(y+x
| dy |
| dx |
| dy |
| dx |
∴
| dy |
| dx |
| yexy |
| 1−xexy |
| y2 |
| 1−xy |
由ez-xz=0，两边对x求导得：ez
| dz |
| dx |
| dz |
| dx |
∴
| dz |
| dx |
| z |
| ez−x |
| z |
| x(z−1) |
∴代入①得：
| du |
| dx |
| ∂f |
| ∂x |
| y2 |
| 1−xy |
| ∂f |
| ∂y |
| z |
| x(z−) |
| ∂f |
| ∂z |
| du |
| dx |
| du |
| dx |
| ∂f |
| ∂x |
| ∂f |
| ∂y |
| dy |
| dx |
| ∂f |
| ∂z |
| dz |
| dx |
| dy |
| dx |
| dy |
| dx |
| dy |
| dx |
| yexy |
| 1−xexy |
| y2 |
| 1−xy |
| dz |
| dx |
| dz |
| dx |
| dz |
| dx |
| z |
| ez−x |
| z |
| x(z−1) |
| du |
| dx |
| ∂f |
| ∂x |
| y2 |
| 1−xy |
| ∂f |
| ∂y |
| z |
| x(z−) |
| ∂f |
| ∂z |