• 2022-06-19
    \(已知u=e^{xyz},则\frac{\partial^3u}{\partial x\partial y\partial z}=(\,)\)
    A: \[(x^2y^2z^2+zyz+1)e^{xyz}\]
    B: \[(x^2yz^2+3zyz+1)e^{xyz}\]
    C: \[(x^2y^2z^2+3zyz+1)e^{xyz}\]
    D: \[(x^2y^2z^2+3zyz)e^{xyz}\]
  • C

    举一反三

    内容

    • 0

      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

      设\(z = z\left( {x,y} \right)\)是由方程\(2{x^2} + {y^2} + {z^2} - 2z = 0\)确定的隐函数,则\( { { \partial z} \over {\partial x}}=\)( )。 A: \( { { 2x} \over {1 - z}}\) B: \( { { 2x} \over {z - 1}}\) C: \({z \over {1 - y}}\) D: \({z \over {y - 1}}\)

    • 2

      设\(w = f(x + y + z,xyz)\),其中\(f\)有连续偏导数,则\( { { {\partial}w} \over {\partial {x}}} =\) A: \({f'_1} + yz{f'_2}\) B: \(x{f'_1} + yz{f'_2}\) C: \(yz{f'_1} +x{f'_2}\) D: \({f'_1} +{f'_2}\)

    • 3

      设\(z = {u^2}{\rm{ + }}{v^2}\),\(u = x + y\),\(v = x - y\),则\( { { \partial z} \over {\partial x}}=\) A: \(4y\) B: \(4x\) C: \(2(x+y)\) D: \(2(x-y)\)

    • 4

      设方程\(z^2+y^2+z^2 = 4z\)确定函数\(z=z(x,y)\),则\( { { {\partial ^2}z} \over {\partial {x^2}}} =\) A: \( { { { { (2 - z)}^2} + {x^2}} \over { { {(2+ z)}^3}}}\) B: \( { { { { (2 - z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\) C: \( { { { { (2 - z)}^2} -{x^2}} \over { { {(2 - z)}^3}}}\) D: \( { { { { (2 + z)}^2} + {x^2}} \over { { {(2 - z)}^3}}}\)