题目

18. (10.0分)设函数z=f(x+y,(x)/(y))，其中f具有二阶连续偏导数，求(partial z)/(partial x)，(partial ^2z)/(partial x^2)；

18. (10.0分) 设函数$z=f(x+y,\frac{x}{y})$，其中f具有二阶连续偏导数，求$\frac{\partial z}{\partial x}$，$\frac{\partial ^2z}{\partial x^2}$；

题目解答

答案

设 $ z = f(u, v) $，其中 $ u = x + y $，$ v = \frac{x}{y} $。

  1. 求 $\frac{\partial z}{\partial x}$：
    由链式法则，
    $\frac{\partial z}{\partial x} = f_u \frac{\partial u}{\partial x} + f_v \frac{\partial v}{\partial x} = f_u \cdot 1 + f_v \cdot \frac{1}{y} = f_1' + \frac{1}{y} f_2'$
    其中 $ f_1' = f_u $，$ f_2' = f_v $。

  2. 求 $\frac{\partial^2 z}{\partial x^2}$：
    对 $\frac{\partial z}{\partial x}$ 再次关于 $x$ 求偏导，
    $\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} \left( f_1' + \frac{1}{y} f_2' \right)$
    分别对两项求导：

    • 对 $ f_1' $ 求导：
      $\frac{\partial}{\partial x} (f_1') = f_{11}'' \frac{\partial u}{\partial x} + f_{12}'' \frac{\partial v}{\partial x} = f_{11}'' \cdot 1 + f_{12}'' \cdot \frac{1}{y} = f_{11}'' + \frac{1}{y} f_{12}''$
    • 对 $ \frac{1}{y} f_2' $ 求导：
      $\frac{\partial}{\partial x} \left( \frac{1}{y} f_2' \right) = \frac{1}{y} \left( f_{21}'' \frac{\partial u}{\partial x} + f_{22}'' \frac{\partial v}{\partial x} \right) = \frac{1}{y} \left( f_{21}'' \cdot 1 + f_{22}'' \cdot \frac{1}{y} \right) = \frac{1}{y} f_{21}'' + \frac{1}{y^2} f_{22}''$
      将两部分相加，
      $\frac{\partial^2 z}{\partial x^2} = f_{11}'' + \frac{1}{y} f_{12}'' + \frac{1}{y} f_{21}'' + \frac{1}{y^2} f_{22}'' = f_{11}'' + \frac{2}{y} f_{12}'' + \frac{1}{y^2} f_{22}''$
      （注意 $ f_{12}'' = f_{21}'' $，由连续性）。

最终结果：
$\frac{\partial z}{\partial x} = f_1' + \frac{1}{y} f_2'$
$\frac{\partial^2 z}{\partial x^2} = f_{11}'' + \frac{2}{y} f_{12}'' + \frac{1}{y^2} f_{22}''$

解析

本题考查复合函数求偏导数的知识，解题思路是利用复合函数求导的链式法则逐步求解。

设 $z = f(u, v)$，其中 $u = x + y$，$v = \frac{x}{y}$。

求 $\frac{\partial z}{\partial x}$

根据复合函数求导的链式法则，若 $z = f(u, v)$，$u = u(x,y)$，$v = v(x,y)$，则$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\cdot\frac{\partial v}{\partial x}$。

  • 先求$\frac{\partial u}{\partial x}$和$\frac{\partial v}{\partial x}$：
    • 对$u = x + y$关于$x$求偏导数，因为$y$看作常数，根据求导公式$(x^n)^\prime=nx^{n - 1}$，可得$\frac{\partial u}{\partial x}=\frac{\partial (x + y)}{\partial x}=1$。
    • 对$v = \frac{x}{y}$关于$x$求偏导数，因为$y$看作常数，根据求导公式$(x^n)^\prime=nx^{n - 1}$，可得$\frac{\partial v}{\partial x}=\frac{\partial (\frac{x}{y})}{\partial x}=\frac{1}{y}$。
  • 再根据链式法则计算$\frac{\partial z}{\partial x}$：
    令$f_1^\prime=\frac{\partial f}{\partial u}$，$f_2^\prime=\frac{\partial f}{\partial v}$，则$\frac{\partial z}{\partial x}=f_1^\prime\cdot\frac{\partial u}{\partial x}+f_2^\prime\cdot\frac{\partial v}{\partial x}=f_1^\prime\cdot 1 + f_2^\prime\cdot\frac{1}{y}=f_1^\prime+\frac{1}{y}f_2^\prime$。

求 $\frac{\partial^2 z}{\partial x^2}$

对$\frac{\partial z}{\partial x}=f_1^\prime+\frac{1}{y}f_2^\prime$再次关于$x$求偏导数，根据求导的加法法则$(u+v)^\prime=u^\prime+v^\prime$，可得$\frac{\partial^2 z}{\partial x^2}=\frac{\partial}{\partial x}(f_1^\prime)+\frac{\partial}{\partial x}(\frac{1}{y}f_2^\prime)$。

  • 求$\frac{\partial}{\partial x}(f_1^\prime)$：
    同样根据链式法则，$f_1^\prime$仍是关于$u$和$v$的函数，所以$\frac{\partial}{\partial x}(f_1^\prime)=\frac{\partial f_1^\prime}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial f_1^\prime}{\partial v}\cdot\frac{\partial v}{\partial x}$。
    令$f_{11}^{\prime\prime}=\frac{\partial^2 f}{\partial u^2}$，$f_{12}^{\prime\prime}=\frac{\partial^2 f}{\partial v\partial u}$，则$\frac{\partial}{\partial x}(f_1^\prime)=f_{11}^{\prime\prime}\cdot 1 + f_{12}^{\prime\prime}\cdot\frac{1}{y}=f_{11}^{\prime\prime}+\frac{1}{y}f_{12}^{\prime\prime}$。
  • 求$\frac{\partial}{\partial x}(\frac{1}{y}f_2^\prime)$：
    因为$\frac{1}{y}$是常数，根据求导的常数乘法法则$(cf)^\prime=cf^\prime$，可得$\frac{\partial}{\partial x}(\frac{1}{y}f_2^\prime)=\frac{1}{y}\cdot\frac{\partial}{\partial x}(f_2^\prime)$。
    再根据链式法则，$\frac{\partial}{\partial x}(f_2^\prime)=\frac{\partial f_2^\prime}{\partial u}\cdot\frac{\partial u}{\partial x}+\frac{\partial f_2^\prime}{\partial v}\cdot\frac{\partial v}{\partial x}$。
    令$f_{21}^{\prime\prime}=\frac{\partial^2 f}{\partial u\partial v}$，$f_{22}^{\prime\prime}=\frac{\partial^2 f}{\partial v^2}$，则$\frac{\partial}{\partial x}(f_2^\prime)=f_{21}^{\prime\prime}\cdot 1 + f_{22}^{\prime\prime}\cdot\frac{1}{y}$，所以$\frac{\partial}{\partial x}(\frac{1}{y}f_2^\prime)=\frac{1}{y}(f_{21}^{\prime\prime}\cdot 1 + f_{22}^{\prime\prime}\cdot\frac{1}{y})=\frac{1}{y}f_{21}^{\prime\prime}+\frac{1}{y^2}f_{22}^{\prime\prime}$。
  • 因为$f$具有二阶连续偏导数，所以$f_{12}^{\prime\prime}=f_{21}^{\prime\prime}$，将$\frac{\partial}{\partial x}(f_1^\prime)$和$\frac{\partial}{\partial x}(\frac{1}{y}f_2^\prime)$相加可得：
    $\frac{\partial^2 z}{\partial x^2}=f_{11}^{\prime\prime}+\frac{1}{y}f_{12}^{\prime\prime}+\frac{1}{y}f_{21}^{\prime\prime}+\frac{1}{y^2}f_{22}^{\prime\prime}=f_{11}^{\prime\prime}+\frac{2}{y}f_{12}^{\prime\prime}+\frac{1}{y^2}f_{22}^{\prime\prime}$。