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

## 二、解析

\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}

\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}

\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}

\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}

\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}

\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}

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

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