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高阶偏导数

## 高阶偏导数
设函数 $z=f(x, y)$ 在区域 $D$ 内具有偏导数

$$
\frac{\partial z}{\partial x}=f_x(x, y), \quad \frac{\partial z}{\partial y}=f_y(x, y),
$$

则在 $D$ 内 $f_x(x, y)$ 和 $f_y(x, y)$ 都是 $x, y$ 的函数. 如果这两个函数的偏导数存在，
则称它们是函数 $z=f(x, y)$ 的二阶偏导数. 按照对变量求导次序的不同，共有下 列四个二阶偏导数：

$$
\begin{aligned}
& \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial^2 z}{\partial x^2}=f_{x x}(x, y), \quad \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial^2 z}{\partial x \partial y}=f_{x y}(x, y) \\
& \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial^2 z}{\partial y \partial x}=f_{y x}(x, y), \quad \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial^2 z}{\partial y^2}=f_{y y}(x, y)
\end{aligned}
$$

其中第二、第三两个偏导称为混合偏导数.
类似地，可以定义三阶、四阶、拟及 $n$ 阶偏导数.
我们把二阶及二阶以上的偏导数统称为高阶偏导数.

`例`设 $z=4 x^3+3 x^2 y-3 x y^2-x+y$ ，求

$$
\frac{\partial^2 z}{\partial x^2}, \frac{\partial^2 z}{\partial y \partial x}, \frac{\partial^2 z}{\partial x \partial y}, \frac{\partial^2 z}{\partial y^2}, \frac{\partial^3 z}{\partial x^3}
$$

解 $\frac{\partial z}{\partial x}=12 x^2+6 x y-3 y^2-1, \quad \frac{\partial z}{\partial y}=3 x^2-6 x y+1$;

$$
\begin{aligned}
& \frac{\partial^2 z}{\partial x^2}=24 x+6 y, \frac{\partial^3 z}{\partial x^3}=24 \quad \frac{\partial^2 z}{\partial y^2}=-6 x, \\
& \frac{\partial^2 z}{\partial x \partial y}=6 x-6 y, \frac{\partial^2 z}{\partial y \partial x}=6 x-6 y .
\end{aligned}
$$

`例` 求 $z=x \ln (x+y)$ 的二阶偏导数.
解
$$
\begin{aligned}
& \frac{\partial z}{\partial x}=\ln (x+y)+\frac{x}{x+y} ， \frac{\partial z}{\partial y}=\frac{x}{x+y}, \\
& \frac{\partial^2 z}{\partial x^2}=\frac{1}{x+y}+\frac{x+y-x}{(x+y)^2}=\frac{x+2 y}{(x+y)^2} ， \frac{\partial^2 z}{\partial y^2}=\frac{-x}{(x+y)^2} ， \\
& \frac{\partial^2 z}{\partial x \partial y}=\frac{1}{x+y}+\frac{-x}{(x+y)^2}=\frac{y}{(x+y)^2} ， \\
& \frac{\partial^2 z}{\partial y \partial x}=\frac{(x+y)-x}{(x+y)^2}=\frac{y}{(x+y)^2} .
\end{aligned}
$$

`例` 求函数 $z=x^y$ 的二阶偏导数.
解

$$
\begin{aligned}
& \frac{\partial z}{\partial x}=y x^{y-1} ， \frac{\partial z}{\partial y}=x^y \ln x \\
& \frac{\partial^2 z}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)=y(y-1) x^{y-2}, \frac{\partial^2 z}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)=x^y \ln ^2 x ， \\
& \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=x^{y-1}+y x^{y-1} \ln x=x^{y-1}(1+y \ln x) ， \\
& \frac{\partial^2 z}{\partial y \partial x}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)=y x^{y-1} \ln x+x^y \cdot \frac{1}{x}=x^{y-1}(1+y \ln x)
\end{aligned}
$$

`例` 验证函数 $u(x, y)=\ln \sqrt{x^2+y^2}$ 满足方程

$$
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0 \text {. }
$$

证 因为 $\ln \sqrt{x^2+y^2}=\frac{1}{2} \ln \left(x^2+y^2\right)$ ，
所以

$$
\fra

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