设z=x2+y2-xy，则dz=______．

2022-06-16

答案：

(2x-y)dx+(2y-x)dy

设$z = u{e^v}$，$u = {x^2} + {y^2}$，$v = xy$，则$ { { \partial z} \over {\partial x}}=$ A: ${e^{xy}}({x^2}y + {y^3} + 2x)$ B: ${e^{xy}}({x}y^2 + {y^3} + 2x)$ C: ${e^{xy}}({x}y + {y^3} + 2x)$ D: ${e^{xy}}({x^2}y + {y^2} + 2x)$
设$z = u{e^v}$，$u = x + y$，$v = xy$，则$ { { \partial z} \over {\partial x}}=$ A: ${e^{xy}}(1 + xy + {y^2})$ B: ${e^{xy}}(1 + xy + {y^3})$ C: ${e^{xy}}(x+ xy + {y^2})$ D: ${e^{xy}}(y+ xy + {y^2})$
设$z = u{e^v}$，$u = {x^2} + {y^2}$，$v = xy$，则$ { { \partial z} \over {\partial y}}=$（）。 A: ${e^{xy}}({x}y^2 + {x^3} + 2y)$ B: ${e^{xy}}({x^2}y + {x^3} + 2y)$ C: ${e^{xy}}({x}y^2 + {x^3} + 2x)$ D: ${e^{xy}}({x}y+ {x^3} + 2y)$
9. 已知函数$z=z(x,y)$由${{z}^{3}}-3xyz={{a}^{3}}$确定，则$\frac{{{\partial }^{2}}z}{\partial x\partial y}=$（） A: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ B: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-xy)}{{{({{z}^{2}}-xy)}^{2}}}$ C: $\frac{z({{z}^{3}}-2xyz-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ D: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$
设$z = z\left( {x,y} \right)$是由方程${z^3}{\rm{ + }}3xyz - 3\sin xy = 1$确定的隐函数，则$ { { \partial z} \over {\partial y}}=$( ) A: $ { { y\left( {\cos xy - z} \right)} \over { { z^2} + xy}}$ B: $ { { y\left( {z - \cos xy} \right)} \over { { z^2} + xy}}$ C: $ { { x\left( {\cos xy - z} \right)} \over { { z^2} + xy}}$ D: $ { { x\left( {z - \cos xy} \right)} \over { { z^2} + xy}}$

0
设$z = \int_ { { x^2}}^y { { e^t}\sin t} dt$，则${z''_{xy}} = $______ 。
1
已知:()x()-()y()=()1(),()z()-()y()=()2(),则()xy()+()yz()+()zx()-()x()2()-()y()2()-()z()2()的值是
2
设方程ez=1+xz+x2+y2确定隐函数z=z(x，y)，求dz与z"xy。
3
z= x 2 y 2 +sin(xy),则z对y的偏导为（）。
4
z= x 2 y 2 +cos(xy),则z对y的偏导为（）。