2024年考研数二第20题解析：多元复合函数求偏导、一重定积分的计算

一、题目

难度评级：

二、解析

第 (1) 问

对于多元复合函数求偏导的问题，我们首先要明白两个关于形式表示上的问题：

[1]. 如果我们用 “$\textcolor{springgreen}{u}$” 表示函数 $f$ 的第一组参数 $\textcolor{springgreen}{2x + y}$, 用 “$\textcolor{orangered}{v}$” 表示函数 $f$ 的第二组参数 $\textcolor{orangered}{3x – y}$, 则题目让我们求解 $\frac{\partial^{2} f}{\partial \textcolor{springgreen}{u} \partial \textcolor{orangered}{v}}$, 其实就是求解 $f^{\prime \prime}_{\textcolor{springgreen}{1} \textcolor{orangered}{2}}$.

[2]. 在本题中，由 $g(x, y)$ $=$ $f(2 x+y, 3 x-y)$ 可知：
$$
\begin{aligned}
\frac{\partial g}{\partial x} & \text{ 就是 } \frac{\partial f}{\partial x} \\
\frac{\partial g}{\partial y} & \text{ 就是 } \frac{\partial f}{\partial y} \\
\frac{\partial^{2} g}{\partial x^{2}} & \text{ 就是 } \frac{\partial^{2} f}{\partial x^{2}} \\
\frac{\partial^{2} g}{\partial y^{2}} & \text{ 就是 } \frac{\partial^{2} f}{\partial y^{2}} \\
\frac{\partial^{2} g}{\partial x \partial y} & \text{ 就是 } \frac{\partial^{2} f}{\partial x \partial y}
\end{aligned}
$$

已知 $g(x, y)$ $=$ $f(2 x+y, 3 x-y)$, 则：

$$
\begin{aligned}
\frac{\partial g}{\partial x} & = 2 f_{1}^{\prime}+3 f_{2}^{\prime} \\ \\
\frac{\partial g}{\partial y} & = f_{1}^{\prime}-f_{2}^{\prime}
\end{aligned}
$$

进而：

$$
\begin{aligned}
\textcolor{springgreen}{\frac{\partial^{2} g}{\partial x^{2}}} & = 2\left(2 f_{11}{ }^{\prime \prime}+3 f_{12}{ }^{\prime \prime}\right)+3\left(2 f_{21}{ }^{\prime \prime} + 3 f_{22}{ }^{\prime \prime}\right) \\ \\
& = \textcolor{pink}{4 f_{11}{ }^{\prime \prime}+12 f_{12}{ }^{\prime \prime} + 9 f_{22}{ }^{\prime \prime}} \\ \\ \\
\textcolor{springgreen}{\frac{\partial^{2} g}{\partial x \partial y}} & = 2\left(f_{11}{ }^{\prime \prime}-f_{12}{ }^{\prime \prime}\right)+3\left(f_{21}{ }^{\prime \prime}-f_{22}{ }^{\prime \prime}\right) \\
& = \textcolor{blue}{2 f_{11}{ }^{\prime \prime} + f_{12}{ }^{\prime \prime} – 3 f_{22}{ }^{\prime \prime}} \\ \\ \\
\textcolor{springgreen}{\frac{\partial^{2} g}{\partial y^{2}}} & = f_{11}{ }^{\prime \prime}-f_{12}{ }^{\prime \prime}-\left(f_{21}{ }^{\prime \prime}-f_{22}{ }^{\prime \prime}\right) \\
& = \textcolor{red}{f_{11}{ }^{\prime \prime}-2 f_{12}{ }^{\prime \prime}+f_{22}{ }^{\prime \prime}}
\end{aligned}
$$

于是，由 $\frac{\partial^{2} g}{\partial x^{2}}$ $+$ $\frac{\partial^{2} g}{\partial x \partial y}$ $-$ $6 \frac{\partial^{2} g}{\partial y^{2}}$ $=$ $1$, 得：

$$
\begin{aligned}
& \textcolor{pink}{4 f_{11}{ }^{\prime \prime}+12 f_{12}{ }^{\prime \prime}+9 f_{22}{ }^{\prime \prime}} \\
& + \textcolor{blue}{2 f_{11}{ }^{\prime \prime}+ f_{12}{ }^{\prime \prime} – 3 f_{22}{ }^{\prime \prime}} \\
& – 6\left( \textcolor{red}{f_{11}{ }^{\prime \prime}-2 f_{12}{ }^{\prime \prime}+f_{22}{ }^{\prime \prime}}\right) = 1
\end{aligned}
$$

即：

$$
f_{12}^{\prime \prime}=\frac{1}{25} \Rightarrow \frac{\partial^{2} f}{\partial u \partial v}=\frac{1}{25}
$$

第 (2) 问

由第 $(1)$ 问可知：

$$
\frac{\partial^{2} f}{\partial u \partial v}=\frac{1}{25}
$$

于是，积分得：

$$
\begin{aligned}
\int \frac{\partial^{2} f}{\partial u \partial v} \mathrm{~d} v \\ \\
& = \frac{\partial f}{\partial u} \\ \\
& = f_{u}^{\prime} \\ \\
& = \int \frac{1}{25} \mathrm{~d} v \\ \\
& = \frac{1}{25} v + C(u)
\end{aligned}
$$

即：

$$
\textcolor{springgreen}{
f_{u}^{\prime} (u, v) = \frac{1}{25} v + C(u)
} \tag{1}
$$

又由题知：

$$
\frac{\partial f(u, 0)}{\partial u} = f_{u}^{\prime}(u, 0) = u e^{-u}
$$

于是，将该题目已知条件代入上面的 $\textcolor{springgreen}{(1)}$ 式，可得：

$$
u e^{-u} = \frac{1}{25} \times 0 + C(u) \Rightarrow C(u) = u e^{-u}
$$

于是：

$$
f_{u}^{\prime}(u, v)=\frac{1}{25} v+u e^{-u}
$$

继续积分，得：

$$
\begin{aligned}
f(u, v) \\ \\
& = \int f_{u}^{\prime}(u, v) \mathrm{~d} u \\ \\
& = \int\left(\frac{1}{25} v+u e^{-u}\right) a \\ \\
& = \frac{1}{25} v \cdot u \textcolor{orangered}{+ \int u e^{-u} \mathrm{~d} u} \\ \\
& = \frac{1}{25} u v \textcolor{orangered}{- u e^{-u}-e^{-u}} + C(v)
\end{aligned}
$$

即：

$$
\textcolor{springgreen}{
f(u, v) = \frac{1}{25} u v – u e^{-u}-e^{-u} + C(v)
} \tag{2}
$$

又由题知：

$$
f(0, v)=\frac{1}{50} v^{2}-1
$$

于是，将该题目已知条件代入上面的 $\textcolor{springgreen}{(2)}$ 式，可得：

$$
0-0-1+c(v)=\frac{1}{50} v^{2}-1 \Rightarrow
$$

$$
C(v)=\frac{1}{50} v^{2}
$$

综上可得：

$$
\textcolor{springgreen}{
\boldsymbol{
f(u, v)=\frac{1}{25} u v-e^{-u}(u+1)+\frac{1}{50} v^{2}
}
}
$$

---