下列选项中属于二元函数的是（）. A: $ y = \tan x $ B: $ y = {x^3} $ C: $ z = 2{x^2} + xy - 4{y^2} $ D: $ u = {e^v} $

2022-07-26

下列选项中属于二元函数的是（）.
A: $ y = \tan x $
B: $ y = {x^3} $
C: $ z = 2{x^2} + xy - 4{y^2} $
D: $ u = {e^v} $

答案：

C

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$z = u{e^v}$，$u = {x^2} + {y^2}$，$v = xy$，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: ${e^{xy}}({x}y+ {x^3} + 2y)$
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: ${e^{xy}}(y+ xy + {y^2})$
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: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$
4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$，并且$z(x,0)=x,z(0,y)={{y}^{2}}$，则$z(x,y)=$（） A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$ B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$ C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$ D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$

0
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下列选项中属于二元函数的是（ ）. A: \( y = \tan x \) B: \( y = {x^3} \) C: \( z = 2{x^2} + xy - 4{y^2} \) D: \( u = {e^v} \)