已知u=f(x+2y,y-3z,z+4x)，其中函数f有二阶连续偏导数，则(partial^2 u)/(partial y partial z)=（）A. (partial^2 u)/(partial y partial z)=6f_(12)''+2f_(13)''+3f_(22)''+f_(23)''B. (partial^2 u)/(partial y partial z)=-6f_(12)''-3f_(22)''C. (partial^2 u)/(partial y partial z)=-6f_(12)''+f_(13)''-3f_(22)''+f_(23)''D. (partial^2 u)/(partial y partial z)=-6f_(12)''+2f_(13)''-3f_(22)''+f_(23)''

本题考查复合函数的二阶混合偏导数的计算。解题思路是先根据复合函数求导法则求出$\frac{\partial u}{\partial y}$，再对$\frac{\partial u}{\partial y}$关于$z$求偏导数，从而得到$\frac{\partial^2 u}{\partial y \partial z}$。

设$s = x + 2y$，$t = y - 3z$，$v = z + 4x$，则$u = f(s,t,v)$。

步骤一：求$\frac{\partial u}{\partial y}$

根据复合函数求导法则$\frac{\partial u}{\partial y}=\frac{\partial f}{\partial s}\cdot\frac{\partial s}{\partial y}+\frac{\partial f}{\partial t}\cdot\frac{\partial t}{\partial y}+\frac{\partial f}{\partial v}\cdot\frac{\partial v}{\partial y}$。
分别计算各项偏导数：

• $\frac{\partial s}{\partial y} = 2$；
• $\frac{\partial t}{\partial y} = 1$；
• $\frac{\partial v}{\partial y} = 0$。
  所以$\frac{\partial u}{\partial y}=2f_{1}'+f_{2}'$，这里$f_{1}'$表示$f$对第一个中间变量$s$的偏导数，$f_{2}'$表示$f$对第二个中间变量$t$的偏导数。

步骤二：求$\frac{\partial^2 u}{\partial y \partial z}$

对$\frac{\partial u}{\partial y}=2f_{1}'+f_{2}'$关于$z$求偏导数，根据求导的加法法则$\frac{\partial^2 u}{\partial y \partial z}=\frac{\partial}{\partial z}(2f_{1}')+\frac{\partial}{\partial z}(f_{2}')$。

• 计算$\frac{\partial}{\partial z}(2f_{1}')$：
  根据复合函数求导法则$\frac{\partial}{\partial z}(2f_{1}') = 2\frac{\partial f_{1}'}{\partial s}\cdot\frac{\partial s}{\partial z}+2\frac{\partial f_{1}'}{\partial t}\cdot\frac{\partial t}{\partial z}+2\frac{\partial f_{1}'}{\partial v}\cdot\frac{\partial v}{\partial z}$。
  分别计算各项偏导数：

  • $\frac{\partial s}{\partial z} = 0$；
  • $\frac{\partial t}{\partial z} = -3$；
  • $\frac{\partial v}{\partial z} = 1$。
    所以$\frac{\partial}{\partial z}(2f_{1}') = 2\times(-3)f_{12}'' + 2\times1\times f_{13}''=-6f_{12}'' + 2f_{13}''$，这里$f_{12}''$表示$f_{1}'$对第二个中间变量$t$的偏导数，$f_{13}''$表示$f_{1}'$对第三个中间变量$v$的偏导数。

• 计算$\frac{\partial}{\partial z}(f_{2}')$：
  同理，根据复合函数求导法则$\frac{\partial}{\partial z}(f_{2}') = \frac{\partial f_{2}'}{\partial s}\cdot\frac{\partial s}{\partial z}+\frac{\partial f_{2}'}{\partial t}\cdot\frac{\partial t}{\partial z}+\frac{\partial f_{2}'}{\partial v}\cdot\frac{\partial v}{\partial z}$。
  分别计算各项偏导数：

  • $\frac{\partial s}{\partial z} = 0$；
  • $\frac{\partial t}{\partial z} = -3$；
  • $\frac{\partial v}{\partial z} = 1$。
    所以$\frac{\partial}{\partial z}(f_{2}') = (-3)f_{22}'' + 1\times f_{23}''=-3f_{22}'' + f_{23}''$，这里$f_{22}''$表示$f_{2}'$对第二个中间变量$t$的偏导数，$f_{23}''$表示$f_{2}'$对第三个中间变量$v$的偏导数。

将上述结果相加可得：
$\frac{\partial^2 u}{\partial y \partial z}=-6f_{12}'' + 2f_{13}''-3f_{22}'' + f_{23}''$