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：

$$\begin{array}{r}F(x,y)\\ \\ & ={\int }_0^{x–y}(x–y–t)f(t)\mathrm{d}t\\ \\ & =(x–y){\int }_0^{x–y}f(t)\mathrm{d}t–{\int }_0^{x–y}tf(t)\mathrm{d}t\\ \\ & =x{\int }_0^{x–y}f(t)\mathrm{d}t–y{\int }_0^{x–y}f(t)\mathrm{d}t–{\int }_0^{x–y}tf(t)\mathrm{d}t\end{array}$$

于是，对 “$x$” 求偏导的时候，”$y$” 与 “${\int }_0^{x–y}f(t)\mathrm{d}t$” 都要当作常数处理：

$$\begin{array}{r}\frac{\partial F}{\partial x}\\ \\ & ={\left[x{\int }_0^{x–y}f(t)\mathrm{d}t–y{\int }_0^{x–y}f(t)\mathrm{d}t–{\int }_0^{x–y}tf(t)\mathrm{d}t\right]}_x^{\prime }\\ \\ & ={\left[x{\int }_0^{x–y}f(t)\mathrm{d}t\right]}_x^{\prime }\\ \\ & ={\int }_0^{x–y}f(t)\mathrm{d}t\end{array}$$

类似的，对 “$y$” 求偏导的时候，”$x$” 与 “${\int }_0^{x–y}f(t)\mathrm{d}t$” 都要当作常数处理：

$$\begin{array}{r}\frac{\partial F}{\partial y}\\ \\ & ={\left[x{\int }_0^{x–y}f(t)\mathrm{d}t–y{\int }_0^{x–y}f(t)\mathrm{d}t–{\int }_0^{x–y}tf(t)\mathrm{d}t\right]}_y^{\prime }\\ \\ & ={\left[–y{\int }_0^{x–y}f(t)\mathrm{d}t\right]}_y^{\prime }\\ \\ & =-{\int }_0^{x–y}f(t)\mathrm{d}t\end{array}$$

于是可知：

$$\frac{\mathit{\partial }\mathit{F}}{\mathit{\partial }\mathit{x}}\mathbf{=}–\frac{\mathit{\partial }\mathit{F}}{\mathit{\partial }\mathit{y}}$$

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：

$$\begin{array}{r}\frac{{\partial }^{2}F}{\partial x^2}\\ \\ & =\frac{\partial }{\partial x}\left(\frac{\partial F}{\partial x}\right)\\ \\ & ={({\int }_0^{x–y}f(t)\mathrm{d}t)}_x^{\prime }\\ \\ & =x^{\prime }+f(x–y)\\ \\ & =f(x–y)\end{array}$$

类似的，由 $\frac{\partial F}{\partial y}$ $=$ $-{\int }_0^{x–y}f(t)\mathrm{d}t$ 可得：

$$\begin{array}{r}\frac{{\partial }^{2}F}{\partial y^2}\\ \\ & =\frac{\partial }{\partial y}\left(\frac{\partial F}{\partial y}\right)\\ \\ & ={(–{\int }_0^{x–y}f(t)\mathrm{d}t)}_y^{\prime }\\ \\ & =-(-y{)}^{\prime }+f(x–y)\\ \\ & =f(x–y)\end{array}$$

于是可知：

$$\frac{{\mathit{\partial }}^{\mathbf{2}}\mathit{F}}{\mathit{\partial }{\mathit{x}}^{\mathbf{2}}}\mathbf{=}\frac{{\mathit{\partial }}^{\mathbf{2}}\mathit{F}}{\mathit{\partial }{\mathit{y}}^{\mathbf{2}}}$$

综上可知，本 题 应 选 C

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