# 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

## 二、解析

### 第 (1) 问

\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}

\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}

\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}

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

### 第 (2) 问

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

\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}

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

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

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

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

\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}

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

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

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

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

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