Example 9.33 求下列函数的二阶偏导数：

$$f(x, y, z) = xyz^2 + 3xy - z$$

Solution 在例中 9.26 我们求得各一阶偏导数为

$$f_x = \frac{\partial f}{\partial x} = yz^2 + 3y, \quad f_y = \frac{\partial f}{\partial y} = xz^2 + 3x, \quad f_z = \frac{\partial f}{\partial z} = 2xyz - 1$$

再次求导, 我们得到

$$\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial^2 f}{\partial x^2} = f_{xx} = 0, \quad \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial^2 f}{\partial y \partial x} = f_{xy} = z^2 + 3$$

$$\frac{\partial}{\partial z}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial^2 f}{\partial z \partial x} = f_{xz} = 2yz, \quad \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right) = \frac{\partial^2 f}{\partial x \partial y} = f_{yx} = z^2 + 3$$

$$\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right) = \frac{\partial^2 f}{\partial y^2} = f_{yy} = 0, \quad \frac{\partial}{\partial z}\left(\frac{\partial f}{\partial y}\right) = \frac{\partial^2 f}{\partial z \partial y} = f_{yz} = 2xz$$

$$\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial z}\right) = \frac{\partial^2 f}{\partial x \partial z} = f_{zx} = 2yz, \quad \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial z}\right) = \frac{\partial^2 f}{\partial y \partial z} = f_{zy} = 2xz$$

$$\frac{\partial}{\partial z}\left(\frac{\partial f}{\partial z}\right) = \frac{\partial^2 f}{\partial z^2} = f_{zz} = 2xy$$

注意, 正如所料,

$$f_{xy} = f_{yx}, \quad f_{xz} = f_{zx} \quad \text{and} \quad f_{yz} = f_{zy}$$

Example 9.34 $f(x, y)$ 是二元函数 $x$ and $y$ 我们希望将变量变换为 $s$ and $t$ , where

$$s = x^2 - y^2, \quad t = xy$$

Determine $f_{xx}$ and $f_{yy}$ 用……表示；以……的形式 $s$ , $t$ , $f_s$ , $f_t$ , $f_{ss}$ , $f_{tt}$ and $f_{st}$ . 证明

$$f_{xx} + f_{yy} = \sqrt{(s^2 + 4t^2)(4f_{ss} + f_{tt})}$$

Solution 利用链式法则,

$$f_x = \frac{\partial f}{\partial x} = \frac{\partial f}{\partial s} \frac{\partial s}{\partial x} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial x} = 2x \frac{\partial f}{\partial s} + y \frac{\partial f}{\partial t}$$

$$f_y = \frac{\partial f}{\partial y} = \frac{\partial f}{\partial s} \frac{\partial s}{\partial y} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial y} = -2y \frac{\partial f}{\partial s} + x \frac{\partial f}{\partial t}$$