Question

求z=xy/(x-y)的全微分，急！

我是怎么都做不出和答案一样的了。。。为什么呢，那位亲帮我做下详细过程吧，谢谢

Answer 1

z对x的偏导为 (y(x-y)-xy)/(x-y)² =-y²/(x-y)²
z对y的偏导为 (x(x-y)+xy) = x²/(x-y)²
所以全微分为 -y²/(x-y)² dx + x²/(x-y)² dy

Answer 2

z=xy/(x-y)
dz=f'x(x, y)dx+f'y(x, y)dy

xy/(x-y)对x取偏导得
[y(x-y)-xy]/(x-y)²dx

xy/(x-y)对y取偏导得
[x(x-y)+xy]/(x-y)²dy

所以
dz=[y(x-y)-xy]/(x-y)²dx+[x(x-y)+xy]/(x-y)²dy
dz=-y²/(x-y)²dx+x²/(x-y)²dy

Answer 3

dz=Z'χdx+Z'ydy
=-y^2/(x-y)^2 dx+x^2/(x-y)^2 dy
可对啊？

Answer 4

dz=[y(x-y)-xy]/(x-y)^2dx+[x(x-y)-xy(-1)]/(x-y)^2dy
=y^2/(x-y)^2dx+x^2/(x-y)^2dy