人邮高数 第6章 第6-3-25题

[AI解答]

[AI解答]

已知 $ z = f(u, v) $，其中
$$
u = x^2 - y^2, \quad v = x^y.
$$
函数 $ f $ 具有二阶连续偏导数，记
$$
f_1 = \frac{\partial f}{\partial u}, \quad f_2 = \frac{\partial f}{\partial v},
$$
且 $ f_{11}, f_{12}, f_{21}, f_{22} $ 表示相应的二阶偏导，由连续性知 $ f_{12} = f_{21} $。

**第一步：求一阶偏导 $\frac{\partial z}{\partial x}$**
由链式法则：
$$
\frac{\partial z}{\partial x} = f_1 \cdot \frac{\partial u}{\partial x} + f_2 \cdot \frac{\partial v}{\partial x}.
$$
计算：
$$
\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial v}{\partial x} = y x^{y-1}.
$$
因此：
$$
\frac{\partial z}{\partial x} = 2x f_1 + y x^{y-1} f_2.
$$

**第二步：求混合偏导 $\frac{\partial^2 z}{\partial x \partial y}$**
对 $\frac{\partial z}{\partial x}$ 关于 $y$ 求偏导：
$$
\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y} \left( 2x f_1 + y x^{y-1} f_2 \right).
$$
分别对两项求导。

第一项 $2x f_1$ 对 $y$ 求导：
$$
\frac{\partial}{\partial y}(2x f_1) = 2x \left( f_{11} \frac{\partial u}{\partial y} + f_{12} \frac{\partial v}{\partial y} \right).
$$
其中：
$$
\frac{\partial u}{\partial y} = -2y, \quad \frac{\partial v}{\partial y} = x^y \ln x.
$$
所以：
$$
\frac{\partial}{\partial y}(2x f_1) = 2x \left( -2y f_{11} + x^y \ln x \, f_{12} \right) = -4xy f_{11} + 2x^{y+1} \ln x \, f_{12}.
$$

第二项 $y x^{y-1} f_2$ 对 $y$ 求导，使用乘积法则：
$$
\frac{\partial}{\partial y} \left( y x^{y-1} f_2 \right) = \frac{\partial}{\partial y} \left( y x^{y-1} \right) f_2 + y x^{y-1} \frac{\partial f_2}{\partial y}.
$$
先计算：
$$
\frac{\partial}{\partial y} \left( y x^{y-1} \right) = x^{y-1} + y \cdot x^{y-1} \ln x = x^{y-1} (1 + y \ln x).
$$
再计算：
$$
\frac{\partial f_2}{\partial y} = f_{21} \frac{\partial u}{\partial y} + f_{22} \frac{\partial v}{\partial y} = -2y f_{21} + x^y \ln x \, f_{22}.
$$
由于 $f_{12}=f_{21}$，所以第二项整体为：
$$
x^{y-1} (1 + y \ln x) f_2 + y x^{y-1} \left( -2y f_{12} + x^y \ln x \, f_{22} \right).
$$
整理为：
$$
x^{y-1} (1 + y \ln x) f_2 - 2y^2 x^{y-1} f_{12} + y x^{2y-1} \ln x \, f_{22}.
$$

第三步：合并两项结果
将第一项和第二项相加：
$$
\frac{\partial^2 z}{\partial x \partial y} = \left[ -4xy f_{11} + 2x^{y+1} \ln x \, f_{12} \right] + \left[ x^{y-1} (1 + y \ln x) f_2 - 2y^2 x^{y-1} f_{12} + y x^{2y-1} \ln x \, f_{22} \right].
$$
合并 $f_{12}$ 的系数：
$$
2x^{y+1} \ln x - 2y^2 x^{y-1} = 2x^{y-1} (x^2 \ln x - y^2).
$$
因此最终结果为：
$$
\boxed{\frac{\partial^{2} z}{\partial x \partial y} = -4xy f_{11} + 2x^{y-1}(x^2 \ln x - y^2) f_{12} + x^{y-1}(1 + y \ln x) f_2 + y x^{2y-1} \ln x \, f_{22}}.
$$

难度：★★★★☆