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,
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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,](/uploads/image/z/18688505-41-5.jpg?t=e%5E%28x%2B2y-z%29%3D1%2Bxy%5E%282%2F3%29z+%E4%B8%94z%3Df%28x%2Cy%29+%E5%88%99dz%281%2C0%29%3D%3F%E6%88%91%E5%81%9A%E7%9A%84%E5%92%8C%E7%AD%94%E6%A1%88%E6%9C%89%E5%87%BA%E5%85%A5%2C)
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
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
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