二阶偏导数计算

设 $z = f(u, v)$，其中 $u = x^2 - y^2$ 且 $v = xy$。我们需要计算 $\frac{\partial^2 z}{\partial x \partial y}$。

首先，计算一阶偏导数：

$$\frac{\partial z}{\partial x} = f_u \cdot \frac{\partial u}{\partial x} + f_v \cdot \frac{\partial v}{\partial x} = f_u \cdot 2x + f_v \cdot y$$

然后，对上述结果再对 $y$ 求导：

$$\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y} \left( f_u \cdot 2x + f_v \cdot y \right)$$

利用链式法则和二阶连续可导性，可以展开为：

$$\frac{\partial^2 z}{\partial x \partial y} = \left( f_{uu} \cdot \frac{\partial u}{\partial y} + f_{uv} \cdot \frac{\partial v}{\partial y} \right) \cdot 2x + \left( f_{vu} \cdot \frac{\partial u}{\partial y} + f_{vv} \cdot \frac{\partial v}{\partial y} \right) \cdot y + f_v$$

代入 $\frac{\partial u}{\partial y} = -2y$ 和 $\frac{\partial v}{\partial y} = x$：

$$\frac{\partial^2 z}{\partial x \partial y} = (-2y f_{uu} + x f_{uv}) \cdot 2x + (-2y f_{uv} + x f_{vv}) \cdot y + f_v$$

简化后得到：

$$\frac{\partial^2 z}{\partial x \partial y} = -4xy f_{uu} + (2x^2 - 2y^2) f_{uv} + xy f_{vv} + f_v$$

最终结果为：

$$\frac{\partial^2 z}{\partial x \partial y} = -4xy\, f_{uu} + (2x^2 - 2y^2)\, f_{uv} + xy\, f_{vv} + f_v$$

下面给出更为完整的推导，并用明确的偏导数符号表示最终结果。

解答步骤：

1. 定义：
   已知 $z = f(u,v)$ ，其中

   $$u = x^2 - y^2,\quad v = xy.$$

2. 一阶偏导数：
   首先对 $x$ 求偏导：

   $$\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}.$$

   计算 $u,v$ 对 $x$ 的偏导：

   $$\frac{\partial u}{\partial x} = 2x,\quad \frac{\partial v}{\partial x} = y.$$

   故

   $$\frac{\partial z}{\partial x} = f_u(2x) + f_v(y).$$

   这里为简写起见，用 $f_u = \frac{\partial f}{\partial u}$, $f_v = \frac{\partial f}{\partial v}$ 表示。

3. 二阶混合偏导数：
   现在对上式再对 $y$ 求偏导：

   $$\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y}(f_u(2x) + f_v(y)).$$

   当对 $y$ 求导时， $f_u$ 和 $f_v$ 都是以 $u,v$ 为自变量的函数，需要用链式法则。
   首先再次求出 $u,v$ 对 $y$ 的偏导：

   $$\frac{\partial u}{\partial y} = -2y,\quad \frac{\partial v}{\partial y} = x.$$

   利用链式法则对 $f_u$ 和 $f_v$ 求对 $y$ 的偏导：

   $$\frac{\partial f_u}{\partial y} = \frac{\partial f_u}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f_u}{\partial v}\frac{\partial v}{\partial y} = f_{uu}(-2y) + f_{uv}(x),$$ $$\frac{\partial f_v}{\partial y} = \frac{\partial f_v}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f_v}{\partial v}\frac{\partial v}{\partial y} = f_{vu}(-2y) + f_{vv}(x).$$

   将上面结果代入：

   $$\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial y}(2x f_u + y f_v) = 2x \frac{\partial f_u}{\partial y} + \frac{\partial (y f_v)}{\partial y}.$$

   对 $y f_v$ 求导要用积的法则：

   $$\frac{\partial (y f_v)}{\partial y} = f_v + y \frac{\partial f_v}{\partial y}.$$

   综合起来：

   $$\frac{\partial^2 z}{\partial x \partial y} = 2x(f_{uu}(-2y) + f_{uv}(x)) + \left( f_v + y(f_{vu}(-2y) + f_{vv}(x)) \right).$$

   注意 $f_{uv} = f_{vu}$，所以合并同类项时可以直接将 $f_{uv}$ 和 $f_{vu}$ 视为同一个二阶混合偏导数。

4. 整理表达式：
   展开并合并项：

   $$\frac{\partial^2 z}{\partial x \partial y} = 2x(-2y f_{uu} + x f_{uv}) + f_v + y(-2y f_{uv} + x f_{vv}).$$

   将各项展开：

   $$= -4xy f_{uu} + 2x^2 f_{uv} + f_v - 2y^2 f_{uv} + xy f_{vv}.$$

   最后将 $f_{uv}$ 项合并：

   $$\frac{\partial^2 z}{\partial x \partial y} = -4xy \frac{\partial^2 f}{\partial u^2} + (2x^2 - 2y^2) \frac{\partial^2 f}{\partial u \partial v} + xy \frac{\partial^2 f}{\partial v^2} + \frac{\partial f}{\partial v}.$$

最终答案：

$$\frac{\partial^2 z}{\partial x \partial y} = -4xy\,\frac{\partial^2 f}{\partial u^2} + (2x^2 - 2y^2)\,\frac{\partial^2 f}{\partial u \partial v} + xy\,\frac{\partial^2 f}{\partial v^2} + \frac{\partial f}{\partial v}.$$