设方程ez=1+xz+x2+y2确定隐函数z=z(x,y),求dz与z"xy。
设方程ez=1+xz+x2+y2确定隐函数z=z(x,y),求dz与z"xy。
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: \( { { x\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\)
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: \( { { x\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\)
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: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$
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: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$
由方程\({z^3} - 3xyz = {a^3}\)所确定的隐函数\(z= f(x,y)\)的偏导数\( { { \partial z} \over {\partial x}} = \) A: \( { { yz} \over { { z^2} - xy}}\) B: \(- { { yz} \over { { z^2} - xy}}\) C: \( { { yz} \over { { z^2} +xy}}\) D: \(- { { yz} \over { { z^2}+xy}}\)
由方程\({z^3} - 3xyz = {a^3}\)所确定的隐函数\(z= f(x,y)\)的偏导数\( { { \partial z} \over {\partial x}} = \) A: \( { { yz} \over { { z^2} - xy}}\) B: \(- { { yz} \over { { z^2} - xy}}\) C: \( { { yz} \over { { z^2} +xy}}\) D: \(- { { yz} \over { { z^2}+xy}}\)
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\(z = u{e^v}\),\(u = x + y\),\(v = xy\),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: \({e^{xy}}(y+ xy + {y^2})\)
(1)z=√(x^2+y^2)表示曲面为:(2)z=xy表示什么曲面。
(1)z=√(x^2+y^2)表示曲面为:(2)z=xy表示什么曲面。
z= x 2 y 2 +sin(xy),则z对y的偏导为()。
z= x 2 y 2 +sin(xy),则z对y的偏导为()。
z= x 2 y 2 +cos(xy),则z对y的偏导为()。
z= x 2 y 2 +cos(xy),则z对y的偏导为()。
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\( { { \partial z} \over {\partial x}}=\) 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\(z = u{e^v}\),\(u = {x^2} + {y^2}\),\(v = xy\),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: \({e^{xy}}({x}y+ {x^3} + 2y)\)