# 4.4 多元复合函数求导

## 一、链式法则

### 1.1 中间变量为一元函数

设 $z = f(u, v)$，$u = \varphi(t)$，$v = \psi(t)$，则复合函数 $z = f[\varphi(t), \psi(t)]$ 对 $t$ 的导数为：

$$\frac{dz}{dt} = \frac{\partial z}{\partial u}\frac{du}{dt} + \frac{\partial z}{\partial v}\frac{dv}{dt}$$

或

$$\frac{dz}{dt} = \frac{\partial f}{\partial u}\frac{d\varphi}{dt} + \frac{\partial f}{\partial v}\frac{d\psi}{dt}$$

**特点**：
- 左边是**全导数**（$\frac{dz}{dt}$），因为 $z$ 最终只是 $t$ 的函数
- 右边是偏导数与导数的乘积之和

### 1.2 中间变量为二元函数

设 $z = f(u, v)$，$u = \varphi(x, y)$，$v = \psi(x, y)$，则：

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

$$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$$

### 1.3 一般情形

设 $z = f(u_1, u_2, \ldots, u_n)$，$u_i = u_i(x_1, x_2, \ldots, x_m)$，则：

$$\frac{\partial z}{\partial x_j} = \sum_{i=1}^{n} \frac{\partial z}{\partial u_i}\frac{\partial u_i}{\partial x_j}$$

**口诀**：连线相乘，分线相加。

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## 二、链式法则的图示

### 2.1 情形1：$z = f(u, v)$，$u = u(t)$，$v = v(t)$

```
    z
   / \
  u   v
   \   /
     t
```

$$\frac{dz}{dt} = \frac{\partial z}{\partial u}\frac{du}{dt} + \frac{\partial z}{\partial v}\frac{dv}{dt}$$

### 2.2 情形2：$z = f(u, v)$，$u = u(x, y)$，$v = v(x, y)$

```
      z
     / \
    u   v
   / \ / \
  x   y   x   y
```

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

$$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y}$$

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## 三、例题

### 3.1 全导数

**例1**：设 $z = u^2 v$，$u = \cos t$，$v = \sin t$，求 $\frac{dz}{dt}$。

**解**：
$$\frac{\partial z}{\partial u} = 2uv, \quad \frac{\partial z}{\partial v} = u^2$$
$$\frac{du}{dt} = -\sin t, \quad \frac{dv}{dt} = \cos t$$

$$\frac{dz}{dt} = 2uv(-\sin t) + u^2(\cos t) = -2\cos t \sin^2 t + \cos^3 t$$

$$= \cos t(\cos^2 t - 2\sin^2 t) = \cos t(1 - 3\sin^2 t)$$

### 3.2 偏导数

**例2**：设 $z = e^u \sin v$，$u = xy$，$v = x + y$，求 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$。

**解**：
$$\frac{\partial z}{\partial u} = e^u \sin v, \quad \frac{\partial z}{\partial v} = e^u \cos v$$
$$\frac{\partial u}{\partial x} = y, \quad \frac{\partial v}{\partial x} = 1$$
$$\frac{\partial u}{\partial y} = x, \quad \frac{\partial v}{\partial y} = 1$$

$$\frac{\partial z}{\partial x} = e^u \sin v \cdot y + e^u \cos v \cdot 1 = e^{xy}[y\sin(x+y) + \cos(x+y)]$$

$$\frac{\partial z}{\partial y} = e^u \sin v \cdot x + e^u \cos v \cdot 1 = e^{xy}[x\sin(x+y) + \cos(x+y)]$$

### 3.3 含抽象函数

**例3**：设 $z = f(x^2 - y^2, xy)$，其中 $f$ 具有连续偏导数，求 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$。

**解**：
设 $u = x^2 - y^2$，$v = xy$

$$\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u}\cdot 2x + \frac{\partial f}{\partial v}\cdot y = 2x f_1 + y f_2$$

$$\frac{\partial z}{\partial y} = \frac{\partial f}{\partial u}\cdot (-2y) + \frac{\partial f}{\partial v}\cdot x = -2y f_1 + x f_2$$

其中 $f_1 = \frac{\partial f}{\partial u}$，$f_2 = \frac{\partial f}{\partial v}$。

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## 四、一阶全微分形式不变性

### 4.1 内容

设 $z = f(u, v)$，无论 $u, v$ 是自变量还是中间变量，全微分形式不变：

$$dz = \frac{\partial z}{\partial u}du + \frac{\partial z}{\partial v}dv$$

### 4.2 应用

利用全微分形式不变性可以方便地求复合函数的偏导数。

**例**：设 $z = f(x^2 + y^2)$，求 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$。

**解**：
设 $u = x^2 + y^2$，则 $z = f(u)$

$$dz = f'(u)du = f'(x^2 + y^2)d(x^2 + y^2) = f'(x^2 + y^2)(2x dx + 2y dy)$$

$$= 2x f'(x^2 + y^2)dx + 2y f'(x^2 + y^2)dy$$

所以：
$$\frac{\partial z}{\partial x} = 2x f'(x^2 + y^2)$$
$$\frac{\partial z}{\partial y} = 2y f'(x^2 + y^2)$$

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## 五、高阶偏导数

### 5.1 抽象函数的高阶偏导

设 $z = f(u, v)$，$u = u(x, y)$，$v = v(x, y)$，求 $\frac{\partial^2 z}{\partial x^2}$。

**步骤**：
1. 先求 $\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}$
2. 再对 $x$ 求偏导，注意 $\frac{\partial f}{\partial u}$ 和 $\frac{\partial f}{\partial v}$ 仍是 $u, v$ 的函数

### 5.2 例题

**例**：设 $z = f(x + y, xy)$，$f$ 具有二阶连续偏导数，求 $\frac{\partial^2 z}{\partial x \partial y}$。

**解**：
设 $u = x + y$，$v = xy$

$$\frac{\partial z}{\partial x} = f_1 + y f_2$$

$$\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y}(f_1 + y f_2)$$

$$= \frac{\partial f_1}{\partial y} + f_2 + y\frac{\partial f_2}{\partial y}$$

$$= (f_{11} \cdot 1 + f_{12} \cdot x) + f_2 + y(f_{21} \cdot 1 + f_{22} \cdot x)$$

$$= f_{11} + xf_{12} + f_2 + yf_{21} + xyf_{22}$$

由于 $f$ 具有二阶连续偏导数，$f_{12} = f_{21}$：

$$= f_{11} + (x + y)f_{12} + xyf_{22} + f_2$$

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## 六、考研重点

1. **链式法则**：掌握各种情形的公式
2. **全导数与偏导数的区别**：中间变量是一元还是多元
3. **抽象函数的偏导数**：正确识别中间变量
4. **高阶偏导数**：注意 $f_1, f_2$ 仍是复合函数
5. **全微分形式不变性**：简化计算的工具
6. **变量关系图**：理清变量之间的关系

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*下一节：4.5 隐函数求导*