函数z=(x+y) ∕ (x-y)的全微分dz=
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解决时间 2021-02-09 06:45
- 提问者网友：箛茗
- 2021-02-08 07:54
函数z=(x+y) ∕ (x-y)的全微分dz=
最佳答案
- 五星知识达人网友：不如潦草
- 2021-02-08 08:47
z=u/v: u=x+y v=x-y
Z'x= Z'uU'x+Z'vV'x
=1/v-u/v^2=-2y/(x+y)^2
Z'y=Z'uU'y+Z'vV'y
=1/v+u/v^2=2x/(x+y)^2
dz=Z'xdx+Z'ydy
=-2/(x+y)^2 * [xdx-ydy]
Z'x= Z'uU'x+Z'vV'x
=1/v-u/v^2=-2y/(x+y)^2
Z'y=Z'uU'y+Z'vV'y
=1/v+u/v^2=2x/(x+y)^2
dz=Z'xdx+Z'ydy
=-2/(x+y)^2 * [xdx-ydy]
全部回答
- 1楼网友：一叶十三刺
- 2021-02-08 10:20
z=e∧(-x+y)
z=e^(-x)*e^y
dz/dx=e^y *e^(-x) *(-1)
=-e^(-x)*e^y*
dz/dy=e^(-x)*e^y
∴dz=-e^(-x)*e^y*dx +e^(-x)*e^y*dy
=e^(-x+y) *(-dx+dy)
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