如果e^(x+y)+xyz=e^z,则dz=?(e^(x+y)+yz)dx/(e^z-xy)+(e^x+y+xz)dy/(e^z-xy),

Question

如果e^(x+y)+xyz=e^z,则dz=?
(e^(x+y)+yz)dx/(e^z-xy)+(e^x+y+xz)dy/(e^z-xy),

Answer 1

两边同时求微分得：
e^(x+y) ·(dx＋dy)＋yzdx＋xzdy＋xydz＝e^z· dz
移项得：
(e^z－xy)dz＝e^(x+y) ·(dx＋dy)＋yzdx＋xzdy
＝e^(x+y)dx＋e^(x+y)dy＋yzdx＋xzdy
＝[e^(x+y) ＋yz]dx ＋ [e^(x+y) ＋xz]dy
∴dz＝﹛[e^(x+y) ＋yz]dx＋[e^(x+y) ＋xz]dy﹜／(e^z－xy)
＝[e^(x+y) ＋yz]dx／(e^z－xy) ＋ [e^(x+y) ＋xz]dy／(e^z－xy)
其实楼上的就对,只是没化简到最后而已