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：

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

§2.1 $\frac{\partial z}{\partial x}$

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

$$\begin{array}{rl}\frac{\partial z}{\partial x}& ={\mathrm{e}}^{-y}(\frac{\partial w}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial w}{\partial v}\cdot \frac{\partial v}{\partial x})\\ \\ & ={\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial w}{\partial v})\end{array}$$

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

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

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

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：

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

又因为：

于是：

$$\begin{array}{rl}\frac{{\partial }^{2}z}{\partial x^2}& ={\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial }{\partial x}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial }{\partial x}\frac{\partial w}{\partial v})\\ \\ & ={\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial w}{\partial x}\frac{\partial }{\partial u}+\frac{1}{2}\frac{\partial w}{\partial x}\frac{\partial }{\partial v})\\ \\ & ={\mathrm{e}}^{-y}[\frac{1}{2}(\frac{\partial w}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial w}{\partial v}\cdot \frac{\partial v}{\partial x})\frac{\partial }{\partial u}+\\ & \frac{1}{2}(\frac{\partial w}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial w}{\partial v}\cdot \frac{\partial v}{\partial x})\frac{\partial }{\partial v}]\\ \\ & ={\mathrm{e}}^{-y}[\frac{1}{2}(\frac{\partial w}{\partial u}\cdot \frac{1}{2}+\frac{\partial w}{\partial v}\cdot \frac{1}{2})\frac{\partial }{\partial u}+\\ & \frac{1}{2}(\frac{\partial w}{\partial u}\cdot \frac{1}{2}+\frac{\partial w}{\partial v}\cdot \frac{1}{2})\frac{\partial }{\partial v}]\\ \\ & ={\mathrm{e}}^{-y}\cdot \frac{1}{4}[(\frac{\partial w}{\partial u}+\frac{\partial w}{\partial v})\frac{\partial }{\partial u}+(\frac{\partial w}{\partial u}+\frac{\partial w}{\partial v})\frac{\partial }{\partial v}]\\ \\ & ={\mathrm{e}}^{-y}(\frac{1}{4}\frac{{\partial }^{2}w}{\partial u^2}+\frac{1}{4}\frac{{\partial }^{2}w}{\partial u\partial v}+\frac{1}{4}\frac{{\partial }^{2}w}{\partial v\partial u}+\frac{1}{4}\frac{{\partial }^{2}w}{\partial v^2})\end{array}$$

§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：

$$\begin{array}{rl}\frac{\partial u}{\partial x}\frac{\partial }{\partial u}& =\frac{\partial }{\partial x}\\ \\ \frac{\partial v}{\partial x}\frac{\partial }{\partial v}& =\frac{\partial }{\partial x}\end{array}$$

所以，我们可以认为：

$$\begin{array}{rl}& \frac{\partial }{\partial x}=\frac{\partial u}{\partial x}\frac{\partial }{\partial u}+\frac{\partial v}{\partial x}\frac{\partial }{\partial v}\\ \\ \Rightarrow & \frac{\partial }{\partial x}\frac{\partial w}{\partial u}=\left(\frac{\partial u}{\partial x}\frac{\partial }{\partial u}+\frac{\partial v}{\partial x}\frac{\partial }{\partial v}\right)\frac{\partial w}{\partial u}\\ \\ \Rightarrow & \frac{\partial }{\partial x}\frac{\partial w}{\partial v}=\left(\frac{\partial u}{\partial x}\frac{\partial }{\partial u}+\frac{\partial v}{\partial x}\frac{\partial }{\partial v}\right)\frac{\partial w}{\partial v}\end{array}$$

于是：

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

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

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

$$\begin{array}{rl}\frac{{\partial }^{2}z}{\partial x\partial y}& =\frac{\partial }{\partial y}\left[{\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial w}{\partial v})\right]\\ \\ & =–{\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial w}{\partial v})+\\ & {\mathrm{e}}^{-y}\cdot \frac{\partial }{\partial y}(\frac{1}{2}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial w}{\partial v})\\ \\ & =–{\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial w}{\partial v})+\\ & {\mathrm{e}}^{-y}\cdot (\frac{1}{2}\frac{\partial }{\partial y}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial }{\partial y}\frac{\partial w}{\partial v})\\ \\ & =–{\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial w}{\partial v})+\\ & {\mathrm{e}}^{-y}\cdot [\frac{1}{2}\left(\frac{\partial u}{\partial y}\frac{\partial }{\partial u}+\frac{\partial v}{\partial y}\frac{\partial }{\partial v}\right)\frac{\partial w}{\partial u}+\\ & \frac{1}{2}\left(\frac{\partial u}{\partial y}\frac{\partial }{\partial u}+\frac{\partial v}{\partial y}\frac{\partial }{\partial v}\right)\frac{\partial w}{\partial v}]\\ \\ & =–{\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial w}{\partial v})+\\ & {\mathrm{e}}^{-y}\cdot [\frac{1}{2}\left(\frac{1}{2}\frac{\partial }{\partial u}+\frac{-1}{2}\frac{\partial }{\partial v}\right)\frac{\partial w}{\partial u}+\\ & \frac{1}{2}\left(\frac{1}{2}\frac{\partial }{\partial u}+\frac{-1}{2}\frac{\partial }{\partial v}\right)\frac{\partial w}{\partial v}]\\ \\ & =–{\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial w}{\partial v})+\\ & {\mathrm{e}}^{-y}(\frac{1}{4}\frac{{\partial }^{2}w}{\partial u^2}–\frac{1}{4}\frac{{\partial }^{2}w}{\partial u\partial v}+\frac{1}{4}\frac{{\partial }^{2}w}{\partial v\partial u}–\frac{1}{4}\frac{{\partial }^{2}w}{\partial v^2})\end{array}$$

综上，我们有：

$$\begin{array}{rl}\frac{\partial z}{\partial x}& ={\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial w}{\partial v})\\ \\ \frac{{\partial }^{2}z}{\partial x^2}& ={\mathrm{e}}^{-y}(\frac{1}{4}\frac{{\partial }^{2}w}{\partial u^2}+\frac{1}{4}\frac{{\partial }^{2}w}{\partial u\partial v}+\frac{1}{4}\frac{{\partial }^{2}w}{\partial v\partial u}+\frac{1}{4}\frac{{\partial }^{2}w}{\partial v^2})\\ \\ \frac{{\partial }^{2}z}{\partial x\partial y}& =–{\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial w}{\partial v})+\\ & {\mathrm{e}}^{-y}(\frac{1}{4}\frac{{\partial }^{2}w}{\partial u^2}–\frac{1}{4}\frac{{\partial }^{2}w}{\partial u\partial v}+\frac{1}{4}\frac{{\partial }^{2}w}{\partial v\partial u}–\frac{1}{4}\frac{{\partial }^{2}w}{\partial v^2})\end{array}$$

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：

$$\begin{array}{rl}& –{\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial w}{\partial v})+\\ & {\mathrm{e}}^{-y}(\frac{1}{4}\frac{{\partial }^{2}w}{\partial u^2}–\frac{1}{4}\frac{{\partial }^{2}w}{\partial u\partial v}+\frac{1}{4}\frac{{\partial }^{2}w}{\partial v\partial u}–\frac{1}{4}\frac{{\partial }^{2}w}{\partial v^2})+\\ & {\mathrm{e}}^{-y}(\frac{1}{4}\frac{{\partial }^{2}w}{\partial u^2}+\frac{1}{4}\frac{{\partial }^{2}w}{\partial u\partial v}+\frac{1}{4}\frac{{\partial }^{2}w}{\partial v\partial u}+\frac{1}{4}\frac{{\partial }^{2}w}{\partial v^2})+\\ & {\mathrm{e}}^{-y}(\frac{1}{2}\frac{\partial w}{\partial u}+\frac{1}{2}\frac{\partial w}{\partial v})=z\\ \\ \iff & \frac{1}{2}{\mathrm{e}}^{-y}(\frac{{\partial }^{2}w}{\partial u^2}+\frac{{\partial }^{2}w}{\partial v\partial u})=z\\ \\ \iff & \frac{{\mathit{\partial }}^{\mathbf{2}}\mathit{w}}{\mathit{\partial }{\mathit{u}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathit{\partial }}^{\mathbf{2}}\mathit{w}}{\mathit{\partial }\mathit{u}\mathit{\partial }\mathit{v}}\mathbf{=}\mathbf{2}\mathit{w}\end{array}$$

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