• 2022-07-25 问题

    设方程\(\sin z - xyz = 0\)确定函数\(z=z(x,y)\),则\( { { \partial z} \over {\partial x}}=\) A: \( { { yz} \over {\cos z + xy}}\) B: \( { { yz} \over {xy-cos z }}\) C: \( { { yz} \over {\cos z - xy}}\) D: \(- { { yz} \over { xy+cos z }}\)

    设方程\(\sin z - xyz = 0\)确定函数\(z=z(x,y)\),则\( { { \partial z} \over {\partial x}}=\) A: \( { { yz} \over {\cos z + xy}}\) B: \( { { yz} \over {xy-cos z }}\) C: \( { { yz} \over {\cos z - xy}}\) D: \(- { { yz} \over { xy+cos z }}\)

  • 2021-04-14 问题

    设u=cos(xy),则du=( ).? cos(xy)(ydx+xdy)|-sin(xy)(ydx+xdy)|sin(xy)(ydx+xdy)|-cos(xy)(ydx+xdy)

    设u=cos(xy),则du=( ).? cos(xy)(ydx+xdy)|-sin(xy)(ydx+xdy)|sin(xy)(ydx+xdy)|-cos(xy)(ydx+xdy)

  • 2022-06-16 问题

    设\(z = xy{e^{\sin xy}}\),则\({z'_y} = \)( )。 A: \(x{e^{\sin xy}}\left( {1 + xy\cos xy} \right)\) B: \(y{e^{\sin xy}}\left( {1 + xy\cos xy} \right)\) C: \(x{e^{\sin xy}}\left( {1 + y\cos xy} \right)\) D: \(x{e^{\sin xy}}\left( {1 - xy\cos xy} \right)\)

    设\(z = xy{e^{\sin xy}}\),则\({z'_y} = \)( )。 A: \(x{e^{\sin xy}}\left( {1 + xy\cos xy} \right)\) B: \(y{e^{\sin xy}}\left( {1 + xy\cos xy} \right)\) C: \(x{e^{\sin xy}}\left( {1 + y\cos xy} \right)\) D: \(x{e^{\sin xy}}\left( {1 - xy\cos xy} \right)\)

  • 2022-07-25 问题

    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: \( { { x\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\)

  • 2021-04-14 问题

    设u=cos(xy),则du=().

    设u=cos(xy),则du=().

  • 2022-06-05 问题

    设\(z = {e^u}\sin v,\;u = xy,\;v = x + y\),则\( { { \partial z} \over {\partial y}}=\)( ) A: \(x{e^{xy}}\sin \left( {x + y} \right) + {e^{xy}}\cos \left( {x + y} \right)\) B: \(x{e^{xy}}\sin \left( {x + y} \right) \) C: \( {e^{xy}}\cos \left( {x + y} \right)\) D: \(x{e^{xy}}\sin \left( {x + y} \right) - {e^{xy}}\cos \left( {x + y} \right)\)

    设\(z = {e^u}\sin v,\;u = xy,\;v = x + y\),则\( { { \partial z} \over {\partial y}}=\)( ) A: \(x{e^{xy}}\sin \left( {x + y} \right) + {e^{xy}}\cos \left( {x + y} \right)\) B: \(x{e^{xy}}\sin \left( {x + y} \right) \) C: \( {e^{xy}}\cos \left( {x + y} \right)\) D: \(x{e^{xy}}\sin \left( {x + y} \right) - {e^{xy}}\cos \left( {x + y} \right)\)

  • 2022-06-06 问题

    y=cos³xy=cos²x这类周期怎么求有公式么y=(1+cosx)1/2为什么周期是π

    y=cos³xy=cos²x这类周期怎么求有公式么y=(1+cosx)1/2为什么周期是π

  • 2021-04-14 问题

    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()。

  • 2022-07-25 问题

    设方程\(\sin z - xyz = 0\)确定函数\(z=z(x,y)\),则\( { { \partial z} \over {\partial y}}=\)( )。 A: \( { { xz} \over {xy+cos z }}\) B: \(- { { xz} \over {xy+cos z }}\) C: \(- { { xz} \over {\cos z - xy}}\) D: \( { { xz} \over {\cos z - xy}}\)

    设方程\(\sin z - xyz = 0\)确定函数\(z=z(x,y)\),则\( { { \partial z} \over {\partial y}}=\)( )。 A: \( { { xz} \over {xy+cos z }}\) B: \(- { { xz} \over {xy+cos z }}\) C: \(- { { xz} \over {\cos z - xy}}\) D: \( { { xz} \over {\cos z - xy}}\)

  • 2022-06-12 问题

    设随机变量服θ从[-π,π]上的均匀分布,令X=sinθ,Y=cosθ,则 (1) E(X)=1. (2)E(XY)=2. (3)ρXY=3.

    设随机变量服θ从[-π,π]上的均匀分布,令X=sinθ,Y=cosθ,则 (1) E(X)=1. (2)E(XY)=2. (3)ρXY=3.

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