基础知识
对于 $z=f(u(x, y), v(x, y))$ 偏导数为：
\begin{array}{l}
\frac{\partial z}{\partial x}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x}=f_{1}^{\prime} \cdot \frac{\partial u}{\partial x}+f_{2}^{\prime} \cdot \frac{\partial v}{\partial x} \\
\frac{\partial z}{\partial y}=\frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y}=f_{1}^{\prime} \cdot \frac{\partial u}{\partial y}+f_{2}^{\prime} \cdot \frac{\partial v}{\partial y}
\end{array}
$$
复合函数偏导数求导，就是分析从外函数到自变量要走过的路，把每条路的求导加起来就是了。一般来说我们使用广度优先的策略，一层一层展开求导。
【要点】
- 层层展开求导。
- 导数的复合路径图不变。
题型
复合函数偏导数
【例子】 设 $w = f(x+y+z, xyz)$ 求 $\frac{\partial w}{\partial x}$，$ \frac{\partial ^2w}{\partial x \partial z} $
【分析与解答】
首先分析复合路径：
f
x+y+z
x
y
z
xyz
x
y
z\begin{align}
\frac{\partial w}{\partial x} &= f_1' (x+y+z)' + f_2' (xyz)'\\
&= f_1' + f_2' yz
\end{align}
$$
\begin{align}
\frac{\partial ^2w}{\partial x \partial z}
&=(f_{11}^{''}1 + f_{12}^{''} xy) + [(f_{21}^{''} 1 + f_{22}^{11})yz + f_2' y]\\
&= f_11^{''} + (xy+yz)f_{12}^{''} + xy^2z f_{22}^{''}+yf_2'
\end{align}
$$
【例子】 设 $u=f(x,y,z), y = g (x,t), t = h (x,z)$ ，都有一阶连续偏导数，求 $\frac{\partial u}{\partial x}, \frac{\partial u}{\partial z}$
【分析与解答】
首先分析复合路径
u = f
x
y = g
x
t = h
x
z
t = h
x
z到 $x$ 一共有 4 条路径。一层一层展开
\begin{align}
\frac{\partial u}{\partial x} &= f_1' \cdot (x)' + f_2' \cdot (y)' + f_3' \cdot (z)'\\
&= f_1' + f_2'(g_1' + g_2' \cdot t') + 0 \\
&= f_1' + f_2' (g_1' + g_2' \cdot (h_1' )) \\
&= f_1' + f_2'g_1' + f_2'g_2'h_1'
\end{align}
$$
【例子】 $z = f(u, x, y)$，$u = xe^y$ ，求 $\frac{\partial ^2 z}{\partial x \partial y} $
【分析与解答】
分析复合路径
z = f
u
x
y
x
y先求 $\frac{\partial z}{\partial x} $
$z\to y$ 的分支就不考虑了，一定是 $0$。
\begin{align}
\frac{\partial z}{\partial x}
&= f_1'\cdot u' + f_2'\cdot x'\\
&= f_1'e^y + f_2'
\end{align}
$$
然后求 $\frac{\partial ^2 z}{\partial x \partial y} $
\begin{align}
\frac{\partial ^2 z}{\partial x \partial y}
&= [(f_{11}^{''}u' + f_{13}^{''}y')e^y + f_1' e^y] + (f_{21}^{''}u'+f_{23}^{''}y')\\
&= [(f_{11}^{''}xe^y + f_{13}^{''}y')e^y + f_1' e^y] + (f_{21}^{''}xe^y+f_{23}^{''}y')\\
&= xe^{2y}f_{11}^{''} + e^y f_{13}^{''} + e^y f_1' + xe^yf_{21}^{''} + f_{23}^{''}
\end{align}
$$
【例子】 用变换 $\left\{\begin{array}{l}u=x-2 y \\v=x+a y\end{array}\right.$ 可把 方程 $6 \frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial x \partial y}-\frac{\partial^{2} z}{\partial y^{2}}=0$ 化简为 $\frac{\partial^{2} z}{\partial u \partial v}=0$，求 $a$；$z$ 有二阶连续偏导数 。
【分析与解答】
这道题看似恐怖，其实只是像写 CURD 一样的体力活。下面来计算吧：
$6 \frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial x \partial y}-\frac{\partial^{2} z}{\partial y^{2}}=0$ 是 $z$ 关于 $x,y$ 的方程，替换后变成了 $z$ 关于 $u,v$ 的方程。
分析复合路径：
- z
- u
- x
- y
- v
- x
- y
- u
\begin{align}
\frac{\partial z}{\partial x}
&= {\color{red}\frac{\partial z}{\partial u} \cdot 1} + {\color{blue}\frac{\partial z}{\partial v} \cdot 1}
\end{align}
$$
\begin{align}
\frac{\partial ^2 z}{\partial x^2}
&= {\color{red}[\frac{\partial z}{\partial u}(\frac{\partial z}{\partial u} \cdot 1 + \frac{\partial z}{\partial v} \cdot 1) ]} + {\color{blue}[\frac{\partial z}{\partial v} (\frac{\partial z}{\partial u} \cdot 1 + \frac{\partial z}{\partial v} \cdot 1)]}\\
&= \frac{\partial ^2z }{\partial u^2} + 2 \frac{\partial ^2 z}{\partial uv} + \frac{\partial ^2 z}{\partial v^2} \tag{1}
\end{align}
$$
PS：二阶偏导连续所以可以合并混合导
\begin{align}
\frac{\partial z}{\partial y}
&= \frac{\partial z}{\partial u} \cdot (-2) + \frac{\partial z}{\partial v} \cdot a
\end{align}
$$
套娃开始（
\begin{align}
\frac{\partial^2 z}{\partial x \partial y}
&= -2 \frac{\partial z}{\partial u} [{\color{red}\frac{\partial z}{\partial u} \cdot 1} + {\color{blue}\frac{\partial z}{\partial v} \cdot 1}] + a \frac{\partial z}{\partial v} [{\color{red}\frac{\partial z}{\partial u} \cdot 1} + {\color{blue}\frac{\partial z}{\partial v} \cdot 1}]\\
&= -2 \frac{\partial ^2 z}{\partial u^2} + (a -2) \frac{\partial ^2 z}{\partial u \partial v} + a \frac{\partial ^2 z}{\partial v^2}\tag{2}
\end{align}
$$
\begin{align}
\frac{\partial^2 z}{\partial^2 y}
&= -2 \frac{\partial z}{\partial u} [\frac{\partial z}{\partial u} \cdot (-2) + \frac{\partial z}{\partial v} \cdot a] + a \frac{\partial z}{\partial v} [\frac{\partial z}{\partial u} \cdot (-2) + \frac{\partial z}{\partial v} \cdot a]\\
&= 4 \frac{\partial ^2 z}{\partial u^2} -4a \frac{\partial ^2 z }{\partial u \partial v} + a^2 \frac{\partial ^2 z}{\partial v ^2} \tag{3}
\end{align}
$$
将 $1,2,3$ 式代入，得到：
(10+5a) \frac{\partial ^2 z}{\partial u \partial v} +
(6 + a - a^2) \frac{\partial ^2 z}{\partial v^2}
$$
想要 $\frac{\partial ^2 z}{\partial u \partial v} = 0 $ ，只需 $6 + a - a^2 = 0 \and 10 + 5a \ne 0$。方程解为
a = 3
$$