• 2022-06-05
    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: \({e^{xy}}(y+ xy + {y^2})\)
  • A

    举一反三

    内容

    • 0

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

    • 1

      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: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$

    • 2

      分解因式()x()3()y()-()2()x()2()y()2()+()xy()3()正确的是A.()xy()(()x()+()y())()2()B.()xy()(()x()2()﹣()2()xy()+()y()2())()C.()xy()(()x()2()+2()xy()﹣()y()2())()D.()xy()(()x()﹣()y())()2

    • 3

      设E(X) =E(Y)= 1/3 , E(XY)= 0, D(X) =D(Y)=2/9 , , 则ρXY =____

    • 4

      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: \(- { { yz} \over { { z^2}+xy}}\)