题目

2.设z=f(xy,(y)/(x))，其中f具有二阶连续偏导数，则(partial^2z)/(partial xpartial y)=（）.(A)f_(1)^prime-f_(2)^prime+xyf_(11)^primeprime-(y)/(x^2)f_(22)^primeprime (B)f_(1)^prime-(1)/(x)f_(2)^prime+xyf_(11)^primeprime+(y)/(x^3)f_(22)^primeprime(C)yf_(1)^prime-(1)/(x)f_(2)^prime+xyf_(11)^primeprime-(y)/(x^3)f_(22)^primeprime (D)f_(1)^prime-(1)/(x^2)f_(2)^prime+xyf_(11)^primeprime-(y)/(x^3)f_(22)^primeprime

2.设$z=f(xy,\frac{y}{x})$，其中f具有二阶连续偏导数，则$\frac{\partial^{2}z}{\partial x\partial y}=$（）. (A)$f_{1}^{\prime}-f_{2}^{\prime}+xyf_{11}^{\prime\prime}-\frac{y}{x^{2}}f_{22}^{\prime\prime}$ (B)$f_{1}^{\prime}-\frac{1}{x}f_{2}^{\prime}+xyf_{11}^{\prime\prime}+\frac{y}{x^{3}}f_{22}^{\prime\prime}$ (C)$yf_{1}^{\prime}-\frac{1}{x}f_{2}^{\prime}+xyf_{11}^{\prime\prime}-\frac{y}{x^{3}}f_{22}^{\prime\prime}$ (D)$f_{1}^{\prime}-\frac{1}{x^{2}}f_{2}^{\prime}+xyf_{11}^{\prime\prime}-\frac{y}{x^{3}}f_{22}^{\prime\prime}$

题目解答

答案

设 $ u = xy $，$ v = \frac{y}{x} $，则 $ z = f(u, v) $。由链式法则， \[ \frac{\partial z}{\partial x} = f_u \cdot y + f_v \cdot \left(-\frac{y}{x^2}\right) = y f_u - \frac{y}{x^2} f_v \] 再对 $ y $ 求导， \[ \frac{\partial^2 z}{\partial x \partial y} = f_u + y \left( f_{uu} \cdot x + f_{uv} \cdot \frac{1}{x} \right) - \frac{1}{x^2} f_v - \frac{y}{x^2} \left( f_{uv} \cdot x + f_{vv} \cdot \frac{1}{x} \right) \] 化简得 \[ \frac{\partial^2 z}{\partial x \partial y} = f_1' - \frac{1}{x^2} f_2' + xy f_{11}'' - \frac{y}{x^3} f_{22}'' \] 对应选项 **D**。答案：$\boxed{D}$

解析

本题考查多元复合函数的二阶偏导数计算，核心思路是链式法则的两次应用。关键在于：

中间变量替换：设$u=xy$，$v=\frac{y}{x}$，将$z=f(u,v)$转化为标准形式；
一阶偏导数计算：对$x$求导时，需分别对$u$和$v$的导数叠加；
二阶偏导数计算：对一阶结果再对$y$求导，注意乘积法则和二阶混合偏导数的对称性（利用$f$的二阶连续性）。

步骤1：设定中间变量

设$u=xy$，$v=\frac{y}{x}$，则$z=f(u,v)$。

步骤2：计算一阶偏导数$\frac{\partial z}{\partial x}$

根据链式法则：
$\frac{\partial z}{\partial x} = f_u \cdot \frac{\partial u}{\partial x} + f_v \cdot \frac{\partial v}{\partial x} = f_1' \cdot y + f_2' \cdot \left(-\frac{y}{x^2}\right) = y f_1' - \frac{y}{x^2} f_2'$

步骤3：计算二阶偏导数$\frac{\partial^2 z}{\partial x \partial y}$

对$\frac{\partial z}{\partial x}$关于$y$求导：

第一项$y f_1'$的导数：
- 直接导数：$\frac{\partial}{\partial y}(y f_1') = f_1' + y \cdot \frac{\partial f_1'}{\partial y}$
- 对$f_1'$应用链式法则：
  $\frac{\partial f_1'}{\partial y} = f_{11}'' \cdot \frac{\partial u}{\partial y} + f_{12}'' \cdot \frac{\partial v}{\partial y} = f_{11}'' \cdot x + f_{12}'' \cdot \frac{1}{x}$
- 结果：$f1' + y \left( x f{11}'' + \frac{1}{x} f_{12}'' \right) = f1' + x y f{11}'' + \frac{y}{x} f_{12}''
  $2. **第二项$-\frac{y}{x^2} f_2'$的导数**： - 直接导数：$\frac{\partial}{\partial y}\left(-\frac{y}{x^2} f_2'\right) = -\frac{1}{x^2} f_2' - \frac{y}{x^2} \cdot \frac{\partial f_2'}{\partial y}$ - 对$f_2'$应用链式法则：$
  \frac{\partial f2'}{\partial y} = f{21}'' \cdot \frac{\partial u}{\partial y} + f{22}'' \cdot \frac{\partial v}{\partial y} = f{21}'' \cdot x + f_{22}'' \cdot \frac{1}{x}
  $- 结果：$-\frac{1}{x^2} f_2' - \frac{y}{x^2} \left( x f_{21}'' + \frac{1}{x} f_{22}'' \right) = -\frac{1}{x^2} f_2' - \frac{y}{x} f_{21}'' - \frac{y}{x^3} f_{22}''$
合并所有项：
$\frac{\partial^2 z}{\partial x \partial y} = \left( f_1' + x y f_{11}'' + \frac{y}{x} f_{12}'' \right) + \left( -\frac{1}{x^2} f_2' - \frac{y}{x} f_{21}'' - \frac{y}{x^3} f_{22}'' \right)$
- 利用二阶混合偏导数对称性$f_{12}'' = f_{21}''$，抵消$\frac{y}{x} f_{12}'' - \frac{y}{x} f_{21}''$，最终化简为：
  $\frac{\partial^2 z}{\partial x \partial y} = f_1' - \frac{1}{x^2} f_2' + x y f_{11}'' - \frac{y}{x^3} f_{22}''$