复合函数求偏导的两种理解方式

一、题目

已知 $u$ $=$ $\frac{x+y}{2}$, $v$ $=$ $\frac{x-y}{2}$, $w$ $=$ $z \mathrm{e}^{y}$, 取 $u$, $v$ 为新自变量，$w$ $=$ $w(u, v)$ 为新函数，请将下面的方程变换为以 $u$ 和 $v$ 为自变量的表示形式：

$$
\frac{\partial^{2} z}{\partial x^{2}} + \frac{\partial^{2 } z}{\partial x \partial y} + \frac{\partial z}{\partial x} = z
$$

难度评级：

二、解析

首先，我们要明白，在本题中，真正的自变量就是 $x$ 和 $y$, 之后再把 $x$ 和 $y$ 封装到 $u$ 和 $v$ 中，形成函数 $u(x, y)$ 和 $v(x, y)$, 之后，将 $u$ 和 $v$ 看作自变量，对 $\frac{\partial^{2} z}{\partial x^{2}}$ $+$ $\frac{\partial^{2 } z}{\partial x \partial y}$ $+$ $\frac{\partial z}{\partial x}$ $=$ $z$ 这个式子进行变形，如图 01 所示：

接着，由 $w$ $=$ $z \mathrm{e}^{y}$ 得：

$$
z = \mathrm{e}^{-y} w
$$

且：

$$
\begin{cases}
u = \frac{x+y}{2} \\
v = \frac{x-y}{2}
\end{cases}
$$

§2.1 $\frac{\partial z}{\partial x}$

对于 $\textcolor{pink}{ \frac{\partial z}{\partial x} }$, 我们有：

$$
\begin{aligned}
\textcolor{pink}{ \frac{\partial z}{\partial x} } & = \mathrm{e}^{-y} \left(\frac{\partial w}{\partial u} \cdot \textcolor{yellow}{ \frac{\partial u}{\partial x} } + \frac{\partial w}{\partial v} \cdot \textcolor{tan}{ \frac{\partial v}{\partial x} } \right) \\ \\
& = \mathrm{e}^{-y} \left(\textcolor{yellow}{ \frac{1}{2} } \frac{\partial w}{\partial u} + \textcolor{tan}{ \frac{1}{2} } \frac{\partial w}{\partial v} \right)
\end{aligned}
$$

§2.2 $\frac{\partial^{2} z}{\partial x^{2}}$

对于 $\textcolor{springgreen}{\frac{\partial^{2} z}{\partial x^{2}}}$, 我们有：

$$
\begin{aligned}
\textcolor{springgreen}{\frac{\partial^{2} z}{\partial x^{2}}} & = \frac{\partial}{\partial x} \left[ \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) \right] \\ \\
& = \mathrm{e}^{-y} \cdot \frac{\partial}{\partial x} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) \\ \\
& = \textcolor{yellow}{ \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\partial}{\partial x} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial}{\partial x} \frac{\partial w}{\partial v} \right) }
\end{aligned}
$$

接下来，我们要对 $w(u, v)$ 这个复合函数求偏导。

§2.2.1 解法一

其中一种对复合函数求偏导的方法就是直接使用下面的复合函数求导公式：

$$
\textcolor{yellow}{
\frac{\partial w(u, v)}{\partial x} = \frac{\partial w}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial w}{\partial v} \cdot \frac{\partial v}{\partial x}
}
$$

又因为：

于是：

$$
\begin{aligned}
\textcolor{springgreen}{\frac{\partial^{2} z}{\partial x^{2}}} & = \textcolor{yellow}{ \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\textcolor{springgreen}{ \partial }}{\partial x} \frac{\textcolor{red}{ \partial w }}{\partial u} + \frac{1}{2} \frac{\textcolor{springgreen}{ \partial }}{\partial x} \frac{\textcolor{red}{\partial w}}{\partial v} \right) } \\ \\
& = \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\textcolor{red}{ \partial w }}{\partial x} \frac{\textcolor{springgreen}{ \partial }}{\partial u} + \frac{1}{2} \frac{\textcolor{red}{ \partial w }}{\partial x} \frac{\textcolor{springgreen}{ \partial }}{\partial v} \right) \\ \\
& = \mathrm{e}^{-y} \left[\frac{1}{2} \left( \frac{\partial w}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial w}{\partial v} \cdot \frac{\partial v}{\partial x} \right) \frac{\textcolor{springgreen}{ \partial }}{\partial u} + \right. \\
& \left. \frac{1}{2} \left( \frac{\partial w}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial w}{\partial v} \cdot \frac{\partial v}{\partial x} \right) \frac{\textcolor{springgreen}{ \partial }}{\partial v} \right] \\ \\
& = \mathrm{e}^{-y} \left[\frac{1}{2} \left( \frac{\partial w}{\partial u} \cdot \frac{1}{2} + \frac{\partial w}{\partial v} \cdot \frac{1}{2} \right) \frac{\textcolor{springgreen}{ \partial }}{\partial u} + \right. \\
& \left. \frac{1}{2} \left( \frac{\partial w}{\partial u} \cdot \frac{1}{2} + \frac{\partial w}{\partial v} \cdot \frac{1}{2} \right) \frac{\textcolor{springgreen}{ \partial }}{\partial v} \right] \\ \\
& = \mathrm{e}^{-y} \cdot \frac{1}{4} \left[\left( \frac{\partial w}{\partial u} + \frac{\partial w}{\partial v} \right) \frac{\textcolor{springgreen}{ \partial }}{\partial u} + \left( \frac{\partial w}{\partial u} + \frac{\partial w}{\partial v} \right) \frac{\textcolor{springgreen}{ \partial }}{\partial v} \right] \\ \\
& = \mathrm{e}^{-y} \left(\frac{1}{4} \frac{\partial^{2} w}{\partial u^{2}} + \frac{1}{4} \frac{\partial^{2} w}{\partial u \partial v} + \frac{1}{4} \frac{\partial^{ 2} w}{\partial v \partial u} + \frac{1}{4} \frac{\partial^{2} w}{\partial v^{2}} \right)
\end{aligned}
$$

§2.2.2 解法二

如果考试的时候没有记住复合函数求导公式，则可以尝试下面这种计算思路，所得的结果和上面用复合函数求导公式计算出来的结果是一样的：

由于：

$$
\begin{aligned}
\frac{\textcolor{orange}{ \partial u }}{\partial x} \frac{\partial}{\textcolor{orange}{ \partial u }} & = \frac{\partial}{\partial x} \\ \\
\frac{\textcolor{yellow}{ \partial v }}{\partial x} \frac{\partial}{\textcolor{yellow}{ \partial v }} & = \frac{\partial}{\partial x}
\end{aligned}
$$

所以，我们可以认为：

$$
\begin{aligned}
& \frac{\partial}{\partial x} = \textcolor{yellow}{ \frac{\partial u}{\partial x} \frac{\partial}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial}{\partial v} } \\ \\
\Rightarrow & \ \textcolor{pink}{ \frac{\partial}{\partial x} } \frac{\partial w}{\partial u} = \left( \textcolor{pink}{ \frac{\partial u}{\partial x} \frac{\partial}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial}{\partial v} } \right) \frac{\partial w}{\partial u} \\ \\
\Rightarrow & \ \textcolor{pink}{ \frac{\partial}{\partial x} } \frac{\partial w}{\partial v} = \left( \textcolor{pink}{ \frac{\partial u}{\partial x} \frac{\partial}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial}{\partial v} } \right) \frac{\partial w}{\partial v}
\end{aligned}
$$

于是：

$$
\begin{aligned}
\textcolor{springgreen}{\frac{\partial^{2} z}{\partial x^{2}}} & = \mathrm{e}^{-y} \left(\frac{1}{2} \textcolor{pink}{ \frac{\partial}{\partial x} } \frac{\partial w}{\partial u} + \frac{1}{2} \textcolor{pink}{ \frac{\partial}{\partial x} } \frac{\partial w}{\partial v} \right) \\ \\
& = \mathrm{e}^{-y} \left[ \frac{1}{2} \left( \textcolor{pink}{ \frac{\partial u}{\partial x} \frac{\partial}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial}{\partial v} } \right) \frac{\partial w}{\partial u} + \right. \\
& \left. \frac{1}{2} \left( \textcolor{pink}{ \frac{\partial u}{\partial x} \frac{\partial}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial}{\partial v} } \right) \frac{\partial w}{\partial v} \right] \\ \\
& = \mathrm{e}^{-y} \left[ \frac{1}{2} \left( \textcolor{pink}{ \frac{1}{2} \frac{\partial}{\partial u} + \frac{1}{2} \frac{\partial}{\partial v} } \right) \frac{\partial w}{\partial u} + \right. \\
& \left. \frac{1}{2} \left( \textcolor{pink}{ \frac{1}{2} \frac{\partial}{\partial u} + \frac{1}{2} \frac{\partial}{\partial v} } \right) \frac{\partial w}{\partial v} \right] \\ \\
& = \mathrm{e}^{-y} \left(\frac{1}{4} \frac{\partial^{2} w}{\partial u^{2}} + \frac{1}{4} \frac{\partial^{2} w}{\partial u \partial v} + \frac{1}{4} \frac{\partial^{ 2} w}{\partial v \partial u} + \frac{1}{4} \frac{\partial^{2} w}{\partial v^{2}} \right)
\end{aligned}
$$

§2.3 $\frac{\partial^{2} z}{\partial x \partial y}$

对于 $\textcolor{orange}{\frac{\partial^{2} z}{\partial x \partial y}}$, 我们有：

$$
\begin{aligned}
\textcolor{orange}{\frac{\partial^{2} z}{\partial x \partial y}} & = \frac{\partial}{\partial y} \left[ \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) \right] \\ \\
& = – \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) + \\
& \quad \quad \mathrm{e}^{-y} \cdot \textcolor{magenta}{ \frac{\partial}{\partial y} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) } \\ \\
& = – \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) + \\
& \quad \quad \mathrm{e}^{-y} \cdot \textcolor{magenta}{ \left(\frac{1}{2} \textcolor{yellow}{ \frac{\partial}{\partial y} } \frac{\partial w}{\partial u} + \frac{1}{2} \textcolor{yellow}{ \frac{\partial}{\partial y} } \frac{\partial w}{\partial v} \right) } \\ \\
& = – \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) + \\
& \quad \quad \mathrm{e}^{-y} \cdot \left[\frac{1}{2} \left( \textcolor{yellow}{ \frac{\partial u}{\partial y} \frac{\partial}{\partial u} + \frac{\partial v}{\partial y} \frac{\partial}{\partial v} } \right) \frac{\partial w}{\partial u} + \right. \\
& \quad \quad \left. \frac{1}{2} \left( \textcolor{yellow}{ \frac{\partial u}{\partial y} \frac{\partial}{\partial u} + \frac{\partial v}{\partial y} \frac{\partial}{\partial v} } \right) \frac{\partial w}{\partial v} \right] \\ \\
& = – \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) + \\
& \quad \quad \mathrm{e}^{-y} \cdot \left[\frac{1}{2} \left( \textcolor{yellow}{ \frac{1}{2} \frac{\partial}{\partial u} + \frac{-1}{2} \frac{\partial}{\partial v} } \right) \frac{\partial w}{\partial u} + \right. \\
& \quad \quad \left. \frac{1}{2} \left( \textcolor{yellow}{ \frac{1}{2} \frac{\partial}{\partial u} + \frac{-1}{2} \frac{\partial}{\partial v} } \right) \frac{\partial w}{\partial v} \right] \\ \\
& = – \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) + \\
& \quad \quad \mathrm{e}^{-y} \left(\frac{1}{4} \frac {\partial^{2} w}{\partial u^{2}} – \frac{1}{4} \frac{\partial^{2} w}{\partial u \partial v} + \frac{1}{4} \frac{\partial^{2} w} {\partial v \partial u} – \frac{1}{4} \frac{\partial^{2} w}{\partial v^{2} } \right)
\end{aligned}
$$

综上，我们有：

$$
\begin{aligned}
\textcolor{pink}{ \frac{\partial z}{\partial x} } & = \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) \\ \\
\textcolor{springgreen}{\frac{\partial^{2} z}{\partial x^{2}}} & = \mathrm{e}^{-y} \left(\frac{1}{4} \frac{\partial^{2} w}{\partial u^{2}} + \frac{1}{4} \frac{\partial^{2} w}{\partial u \partial v} + \frac{1}{4} \frac{\partial^{ 2} w}{\partial v \partial u} + \frac{1}{4} \frac{\partial^{2} w}{\partial v^{2}} \right) \\ \\
\textcolor{orange}{\frac{\partial^{2} z}{\partial x \partial y}} & = – \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) + \\
& \quad \quad \ \mathrm{e}^{-y} \left(\frac{1}{4} \frac {\partial^{2} w}{\partial u^{2}} – \frac{1}{4} \frac{\partial^{2} w}{\partial u \partial v} + \frac{1}{4} \frac{\partial^{2} w} {\partial v \partial u} – \frac{1}{4} \frac{\partial^{2} w}{\partial v^{2} } \right)
\end{aligned}
$$

将上面的式子代入 $\frac{\partial^{2} z}{\partial x^{2}}$ $+$ $\frac{\partial^{2} z}{\partial x \partial y}$ $+$ $\frac{\partial z}{\partial x}$ $=$ $z$ 得：

$$
\begin{aligned}
& \textcolor{magenta}{ – \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) } + \\
& \quad \ \mathrm{e}^{-y} \left(\frac{1}{4} \frac{\partial^{2} w}{\partial u^{2}} \textcolor{orangered}{ – \frac{1}{4} \frac{\partial^{2} w}{\partial u \partial v} } + \frac{1}{4} \frac{\partial^{2} w} {\partial v \partial u} \textcolor{orange}{ – \frac{1}{4} \frac{\partial^{2} w}{\partial v^{2} } } \right) + \\
& \quad \ \mathrm{e}^{-y} \left(\frac{1}{4} \frac{\partial^{2} w}{\partial u^{2}} \textcolor{orangered}{ + \frac{1}{4} \frac{\partial^{2} w}{\partial u \partial v} } + \frac{1}{4} \frac{\partial^{ 2} w}{\partial v \partial u} \textcolor{orange}{ + \frac{1}{4} \frac{\partial^{2} w}{\partial v^{2}} } \right) + \\
& \quad \ \textcolor{magenta}{ \mathrm{e}^{-y} \left(\frac{1}{2} \frac{\partial w}{\partial u} + \frac{1}{2} \frac{\partial w}{\partial v} \right) } = z \\ \\
\Leftrightarrow & \ \frac{1}{2} \mathrm{e}^{-y} \left( \frac{\partial^{2} w}{\partial u^{2}} + \frac{\partial^{2} w} {\partial v \partial u} \right) = z \\ \\
\Leftrightarrow & \ \textcolor{springgreen}{\boldsymbol{ \frac{\partial^{2} w}{\partial u^{2}} + \frac{\partial^{2} w}{\partial u \partial v} = 2w }}
\end{aligned}
$$

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