求复合函数偏导数的两种方式：先求导再代换、先代换再求导

一、题目

难度评级：

二、解析

解法一：先求导再代换

首先，由题目可知：

$$
f(u, v) = f(x + y, \frac{y}{x}) \textcolor{lightgreen}{ \leadsto } \begin{cases}
u = x + y \\ \\
v = \frac{y}{x}
\end{cases}
$$

接着，在等式 $f \left( x + y, \frac{y}{x} \right)$ $=$ $f \left( u, v \right)$ $=$ $x^{2} – y^{2}$ 两边对 $x$ 求偏导数，得：

$$
\begin{align}
& \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} = \left( x^{2} – y^{2} \right)^{\prime} _{x} \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{gray}{ f ^{\prime} _{u} \cdot \frac{\partial u}{\partial x} + f ^{\prime} _{v} \cdot \frac{\partial v}{\partial x} = 2x } \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{gray}{ f ^{\prime} _{u} \cdot \left( x+y \right) ^{\prime} _{x} – f ^{\prime} _{v} \cdot \left( \frac{y}{x} \right) ^{\prime} _{x} = 2x } \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{gray}{ f ^{\prime} _{u} \cdot 1 – f ^{\prime} _{v} \cdot \frac{y}{x^{2}} = 2x } \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{lightgreen}{ f ^{\prime} _{u} – \frac{y}{x^{2}} f ^{\prime} _{v} = 2x } \notag \tag{1}
\end{align}
$$

在等式 $f \left( x + y, \frac{y}{x} \right)$ $=$ $f \left( u, v \right)$ $=$ $x^{2} – y^{2}$ 两边对 $y$ 求偏导数，得：

$$
\begin{align}
& \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} = \left( x^{2} – y^{2} \right)^{\prime} _{y} \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{gray}{ f ^{\prime} _{u} \cdot \frac{\partial u}{\partial y} + f ^{\prime} _{v} \cdot \frac{\partial v}{\partial y} = -2y } \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{gray}{ f ^{\prime} _{u} \cdot \left( x+y \right) ^{\prime} _{y} – f ^{\prime} _{v} \cdot \left( \frac{y}{x} \right) ^{\prime} _{y} = -2y } \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{gray}{ f ^{\prime} _{u} \cdot 1 – f ^{\prime} _{v} \cdot \frac{1}{x} = -2y } \notag \\ \notag \\
\textcolor{lightgreen}{ \leadsto } \ & \textcolor{lightgreen}{ f ^{\prime} _{u} – \frac{1}{x} f ^{\prime} _{v} = -2y } \notag \tag{2}
\end{align}
$$

由于 $u = x + y$, $v = \frac{y}{x}$，所以，当 $u = v = 1$ 时，$x = y = \frac{1}{2}$，代入上面得 $(1)$ 式和 $(2)$ 式中，得：

$$
\begin{cases}
f^{\prime}_{u}(1, 1) – 2f^{\prime}_{v}(1, 1) = 1 \\ \\
f^{\prime}_{u}(1, 1) + 2f^{\prime}_{v}(1, 1) = -1
\end{cases}
$$

进而可得：

$$
\begin{cases}
f^{\prime}_{u}(1,1) = 0 \\ \\
f^{\prime}_{v}(1,1) = \frac{-1}{2}
\end{cases}
$$

即：

$$
\textcolor{springgreen}{
\begin{aligned}
& \left. \frac{\partial f}{\partial u} \right|_{\substack{u=1 \\ v=1}} = 0 \\ \\
& \left. \frac{\partial f}{\partial v} \right|_{\substack{u=1 \\ v=1}} = \frac{-1}{2}
\end{aligned}
}
$$

解法二：先代换再求导

首先，由题目可知：

$$
f(u, v) = f(x + y, \frac{y}{x}) \textcolor{lightgreen}{ \leadsto } \begin{cases}
u = x + y \\ \\
v = \frac{y}{x}
\end{cases} \textcolor{lightgreen}{ \leadsto } \begin{cases}
x = \frac{u}{1+v} \\ \\
y = \frac{uv}{1+v}
\end{cases}
$$

于是：

$$
\begin{aligned}
\textcolor{lightgreen}{f \left( u, v \right)} \\ \\
& = x^{2} + y^{2} \\ \\
& = \left( \frac{u}{1+v} \right)^{2} – \left( \frac{uv}{1+v} \right)^{2} \\ \\
& = \textcolor{gray}{ \frac{u^{2} – u^{2} v^{2}}{(1+v)^{2}} } \\ \\
& \textcolor{gray}{ = \frac{u^{2} (1 – v^{2}) }{(1+v)^{2}} } \\ \\
& \textcolor{gray}{ = \frac{u^{2} (1+v)(1-v) }{(1+v)^{2}} } \\ \\
& = \textcolor{lightgreen}{ \frac{u^{2}(1-v)}{1+v} }
\end{aligned}
$$

因此，对 $\textcolor{lightgreen}{ f \left( u, v \right) }$ $\textcolor{lightgreen}{=}$ $\textcolor{lightgreen}{\frac{u^{2}(1-v)}{1+v}}$ 中的 $u$ 和 $v$ 分别求偏导，得：

$$
\begin{aligned}
& \frac{\partial f}{\partial u} \\ \\
& = \textcolor{gray}{ \frac{1-v}{1+v} \cdot \left( u^{2} \right) ^{\prime} _{u} } \\ \\
& = \textcolor{lightgreen}{ \frac{2u(1-v)}{1+v} } \\ \\ \\
& \frac{\partial f}{\partial v} \\ \\
& = \textcolor{gray}{ u^{2} \cdot \left( \frac{1-v}{1+v} \right) ^{\prime} _{v} } \\ \\
& = \textcolor{gray}{ u^{2} \cdot \frac{-(1+v) – (1-v)}{(1+v)^{2}} } \\ \\
& = \textcolor{gray}{ u^{2} \cdot \frac{-2}{(1+v)^{2}} } \\ \\
& = \textcolor{lightgreen}{ \frac{-2u^{2}}{(1+v)^{2}} }
\end{aligned}
$$

综上可得：

$$
\textcolor{springgreen}{
\begin{aligned}
& \left. \frac{\partial f}{\partial u} \right|_{\substack{u=1 \\ v=1}} = 0 \\ \\
& \left. \frac{\partial f}{\partial v} \right|_{\substack{u=1 \\ v=1}} = \frac{-1}{2}
\end{aligned}
}
$$

拓展资料

1. 复合函数求偏导的两种理解方式

---