关于 $y$ $=$ $x$ 对称的二元函数的二阶偏导数也关于 $y$ $=$ $x$ 对称

一、题目

难度评级：

二、解析

§2.1 式子 ⟬A⟭ 求解

首先对 $z(x, y)$ 中的 $x$ 求二阶偏导：

$$
\begin{aligned}
& z(x, y) = \ x ^{2} + y^{2} – 3 x^{4} y^{4} \\ \\
\Rightarrow & \ \frac{\partial z}{\partial x} = 2x + 0 – 3 y^{4} \cdot 4 \cdot x ^{3} \\ \\
\Rightarrow & \ \textcolor{yellow}{ \frac{\partial z}{\partial x} = 2x – 12 y^{4} x ^{3} } \\ \\
\Rightarrow & \ \textcolor{springgreen}{ \frac{\partial ^{2} z}{\partial x ^{2}} = 2 – 36 y^{4} x ^{2} }
\end{aligned}
$$

又由于对原式 $z(x, y)$ $=$ $x ^{2}$ $+$ $y^{2}$ $-$ $3 x^{4}$ $y^{4}$ 而言，交换变量 $\textcolor{orangered}{x}$ 和变量 $\textcolor{springgreen}{y}$ 的位置后，式子和原来相等，即：

$$
\begin{aligned}
z(\textcolor{orangered}{x}, \textcolor{springgreen}{y}) = & \ \textcolor{orangered}{x^{2}} + \textcolor{springgreen}{y^{2}} – 3 \textcolor{orangered}{x^{4}} \textcolor{springgreen}{y^{4}} \\
z(\textcolor{springgreen}{y}, \textcolor{orangered}{x}) = & \ \textcolor{springgreen}{y^{2}} + \textcolor{orangered}{x^{2}} – 3 \textcolor{springgreen}{y^{4}} \textcolor{orangered}{x^{4}} \\
= & \ \textcolor{orangered}{x^{2}} + \textcolor{springgreen}{y^{2}} – 3 \textcolor{orangered}{x^{4}} \textcolor{springgreen}{y^{4}}
\end{aligned}
$$

也就是说，函数 $z(x, y)$ 是关于直线 $y = x$ 对称的函数，根据“ 对 称 的 二 元 函 数 的 二 阶 偏 导 数 也 对 称 ”的原理，将 $\frac{\partial ^{2} z}{\partial \textcolor{orangered}{x} ^{2}}$ 中的变量 $\textcolor{orangered}{x}$ 替换为变量 $\textcolor{springgreen}{y}$ 就可以直接得到 $\frac{\partial ^{2} z}{\partial \textcolor{springgreen}{y} ^{2}}$ 的表达式：

$$
\begin{aligned}
& \frac{\partial ^{2} z}{\partial \textcolor{orangered}{x} ^{2}} = 2 – 36 \textcolor{springgreen}{y^{4}} \textcolor{orangered}{x^{2}} \\ \\
\Rightarrow & \ \frac{\partial ^{2} z}{\partial \textcolor{springgreen}{y} ^{2}} = 2 – 36 \textcolor{orangered}{x^{4}} \textcolor{springgreen}{y^{2}}
\end{aligned}
$$

§2.2 式子 ⟬B⟭ 求解

首先对 $z(x, y)$ 中的 $x$ 求二阶偏导：

$$
\begin{aligned}
& z(x, y) = \ \frac{x^{2} + y^{2}}{xy} \\ \\
\Rightarrow & \ z(x, y) = \ \frac{x}{y} + \frac{y}{x} \\ \\
\Rightarrow & \ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \left( \frac{x}{y} + \frac{y}{x} \right) \\ \\
\Rightarrow & \ \frac{\partial z}{\partial x} = \frac{1}{y} + y \cdot \left( \frac{1}{x} \right)_{x} ^{\prime} \\ \\
\Rightarrow & \ \textcolor{yellow}{ \frac{\partial z}{\partial x} = \frac{1}{y} – \frac{y}{x^{2}} } \\ \\
\Rightarrow & \ \frac{\partial ^{2} z}{\partial x^{2}} = -y \cdot \left( \frac{1}{x ^{2}} \right)_{x}^{\prime} \\ \\
\Rightarrow & \ \textcolor{springgreen}{ \frac{\partial ^{2} z}{\partial x^{2}} = \frac{2y}{x^{3}} }
\end{aligned}
$$

同理，由于函数 $z(x, y)$ $=$ $\frac{x^{2} + y^{2}}{xy}$ 也关于 $y=x$ 对称：

$$
\begin{aligned}
z(\textcolor{orangered}{x}, \textcolor{springgreen}{y}) = \ & \frac{\textcolor{orangered}{x^{2}} + \textcolor{springgreen}{y^{2}}}{\textcolor{orangered}{x} \textcolor{springgreen}{y}} \\ \\
z(\textcolor{springgreen}{y}, \textcolor{orangered}{x}) = \ & \frac{\textcolor{springgreen}{y^{2}} + \textcolor{orangered}{x^{2}}}{\textcolor{springgreen}{y} \textcolor{orangered}{x}} \\ \\
= \ & \frac{\textcolor{orangered}{x^{2}} + \textcolor{springgreen}{y^{2}}}{\textcolor{orangered}{x} \textcolor{springgreen}{y}}
\end{aligned}
$$

所以：

$$
\begin{aligned}
& \frac{\partial ^{2} z}{\partial \textcolor{orangered}{x}^{2}} = \frac{2 \textcolor{springgreen}{y}}{\textcolor{orangered}{x^{3}}} \\ \\
\Rightarrow & \frac{\partial ^{2} z}{\partial \textcolor{springgreen}{y}^{2}} = \frac{2 \textcolor{orangered}{x}}{\textcolor{springgreen}{y^{3}}}
\end{aligned}
$$

综上，对于 $z(x, y)$ $=$ $x ^{2}$ $+$ $y^{2}$ $-$ $3 x^{4} y^{4}$, 我们有：

$$
\textcolor{springgreen}{
\begin{cases}
\frac{\partial ^{2} z}{\partial x ^{2}} = 2 – 36 y^{4} x^{2} \\ \\
\frac{\partial ^{2} z}{\partial y^{2}} = 2 – 36 x^{4} y^{2}
\end{cases}
}
$$

对于 $z(x, y)$ $=$ $\frac{x^{2} + y^{2}}{xy}$, 我们有：

$$
\textcolor{springgreen}{
\begin{cases}
\frac{\partial ^{2} z}{\partial x^{2}} = \frac{2 y}{x^{3}} \\ \\
\frac{\partial ^{2} z}{\partial y^{2}} = \frac{2 x}{y^{3}}
\end{cases}
}
$$

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