2022考研数二第04题解析：二元偏导数、变上限积分求导

一、题目

A. $\frac { \partial F } { \partial x }$ $=$ $\frac { \partial F } { \partial y }$, $\frac { \partial ^ { 2 } F } { \partial x ^ { 2 } }$ $=$ $\frac { \partial ^ { 2 } F } { \partial y ^ { 2 } }$

B. $\frac { \partial F } { \partial x }$ $=$ $\frac { \partial F } { \partial y }$, $\frac { \partial ^ { 2 } F } { \partial x ^ { 2 } }$ $=$ $- \frac { \partial ^ { 2 } F } { \partial y ^ { 2 } }$

C. $\frac { \partial F } { \partial x }$ $=$ $- \frac { \partial F } { \partial y }$, $\frac { \partial ^ { 2 } F } { \partial x ^ { 2 } }$ $=$ $\frac { \partial ^ { 2 } F } { \partial y ^ { 2 } }$

D. $\frac { \partial F } { \partial x }$ $=$ $- \frac { \partial F } { \partial y }$, $\frac { \partial ^ { 2 } F } { \partial x ^ { 2 } }$ $=$ $- \frac { \partial ^ { 2 } F } { \partial y ^ { 2 } }$

难度评级：

二、解析

首先，对题目所给的式子进行变形：

$$
\begin{aligned}
F ( x , y ) \\ \\
& = \int _{ 0 } ^ { x – y } ( x – y – t ) f ( t ) \mathrm{ ~d } t \\ \\
& = ( x – y ) \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t – \int _{ 0 } ^ { x – y } t f ( t ) \mathrm{ ~d } t \\ \\
& = x \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t – y \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t – \int _{ 0 } ^ { x – y } t f ( t ) \mathrm{ ~d } t
\end{aligned}
$$

于是，对 “$\textcolor{springgreen}{x}$” 求偏导的时候，”$\textcolor{yellow}{y}$” 与 “$\textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t}$” 都要当作常数处理：

$$
\begin{aligned}
\frac { \partial F } { \partial x } \\ \\
& = \left[ \textcolor{springgreen}{x} \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} – \textcolor{yellow}{y \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} – \textcolor{yellow}{\int _{ 0 } ^ { x – y } t f ( t ) \mathrm{ ~d } t} \right]^{\prime}_{\textcolor{springgreen}{x}} \\ \\
& = \left[ \textcolor{springgreen}{x} \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} \right] ^{\prime}_{x} \\ \\
& = \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t}
\end{aligned}
$$

类似的，对 “$\textcolor{springgreen}{y}$” 求偏导的时候，”$\textcolor{yellow}{x}$” 与 “$\textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t}$” 都要当作常数处理：

$$
\begin{aligned}
\frac { \partial F } { \partial y } \\ \\
& = \left[ \textcolor{yellow}{x \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} – \textcolor{springgreen}{y} \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} – \textcolor{yellow}{\int _{ 0 } ^ { x – y } t f ( t ) \mathrm{ ~d } t} \right] ^{\prime}_{\textcolor{springgreen}{y}} \\ \\
& = \left[ – \textcolor{springgreen}{y} \textcolor{yellow}{\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t} \right] ^{\prime} _{\textcolor{springgreen}{y}} \\ \\
& = \textcolor{yellow}{- \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t}
\end{aligned}
$$

于是可知：

$$
\textcolor{magenta}{
\boldsymbol{
\frac { \partial F } { \partial x } = – \frac { \partial F } { \partial y }
}
}
$$

于是可以排除 [A] 和 [B] 两个选项。

接着，由 $\frac { \partial F } { \partial x }$ $=$ $\int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t$ 可得：

$$
\begin{aligned}
\frac { \partial ^ { 2 } F } { \partial x ^ { 2 } } \\ \\
& = \frac{\partial}{\partial x} \left( \frac { \partial F } { \partial x } \right) \\ \\
& = \left( \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t \right) ^{\prime} _{x} \\ \\
& = \textcolor{orangered}{x ^{\prime}} + f(x – y) \\ \\
& = f ( x – y )
\end{aligned}
$$

类似的，由 $\frac { \partial F } { \partial y }$ $=$ $- \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t$ 可得：

$$
\begin{aligned}
\frac { \partial ^ { 2 } F } { \partial y ^ { 2 } } \\ \\
& = \frac{\partial}{\partial y} \left( \frac { \partial F } { \partial y } \right) \\ \\
& = \left( – \int _{ 0 } ^ { x – y } f ( t ) \mathrm{ ~d } t \right) ^{\prime} _{y} \\ \\
& = \textcolor{orangered}{-(-y) ^{\prime}} + f(x – y) \\ \\
& = f ( x – y )
\end{aligned}
$$

于是可知：

$$
\textcolor{magenta}{
\boldsymbol{
\frac { \partial ^ { 2 } F } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } F } { \partial y ^ { 2 } }
}
}
$$

综上可知，本 题 应 选 C

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