e^(x+2y-z)=1+xy^(2/3)z 且z=f(x,y) 则dz(1,0)=?我做的和答案有出入,

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/23 17:42:50
e^(x+2y-z)=1+xy^(2/3)z且z=f(x,y)则dz(1,0)=?我做的和答案有出入,e^(x+2y-z)=1+xy^(2/3)z且z=f(x,y)则dz(1,0)=?我做的和答案有出

e^(x+2y-z)=1+xy^(2/3)z 且z=f(x,y) 则dz(1,0)=?我做的和答案有出入,
e^(x+2y-z)=1+xy^(2/3)z 且z=f(x,y) 则dz(1,0)=?我做的和答案有出入,

e^(x+2y-z)=1+xy^(2/3)z 且z=f(x,y) 则dz(1,0)=?我做的和答案有出入,
答:
两边求微分得:
(dx+2dy-dz)e^(x+2y-z)=y^(2/3)zdx+2xy^(-1/3)z/3dy+xy^(2/3)dz
dz=[(1-y^(2/3)z)dx+(2-2xy^(-1/3)z/3)dy]/[1+xy^(2/3)]
当x=1,y=0时,e^(1-z)=1,解得z=1
代入x=1,y=0,z=1得:
dz=dx+2dy