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求z=xy/(x-y)的全微分,急!
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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
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
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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
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
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dz=Z'χdx+Z'ydy
=-y^2/(x-y)^2 dx+x^2/(x-y)^2 dy
可对啊?
=-y^2/(x-y)^2 dx+x^2/(x-y)^2 dy
可对啊?
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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
=y^2/(x-y)^2dx+x^2/(x-y)^2dy
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