Double-Parameter Chain Rule Derivation | 双参数链式法则推导

📚 Double-Parameter Chain Rule Derivation | 双参数链式法则推导

In IB Mathematics Analysis and Approaches HL, the chain rule is usually introduced for a composite function of one variable. However, many real-world models and IB calculus extension topics use a function of two variables, where each of those variables depends on two new parameters. The double-parameter chain rule explains exactly how to differentiate such a composite function.

在 IB 数学分析与方法 HL 中，链式法则最先以单变量复合函数的形式出现。然而，许多实际模型以及 IB 微积分拓展专题都会涉及二元函数，并且每个自变量又依赖于两个新参数。双参数链式法则正是用来处理这类复合函数求导问题的。

1. Why We Need a Double-Parameter Chain Rule | 为什么需要双参数链式法则

Suppose z = f(x, y), while x = x(s, t) and y = y(s, t). Here s and t are independent parameters. If s changes while t is held fixed, x and y may both change. Therefore z can be influenced along two separate paths: one through x and one through y. To find ∂z/∂s, we must add the contributions from both paths.

设 z = f(x, y)，其中 x = x(s, t)，y = y(s, t)。这里 s 和 t 是独立参数。若在保持 t 不变的前提下改变 s，则 x 和 y 都可能改变。因此 z 会沿两条路径受到影响：一条经过 x，另一条经过 y。为了求 ∂z/∂s，我们必须把两条路径的贡献相加。

The single-variable chain rule cannot handle this situation because it assumes only one intermediate variable. When several intermediate variables exist, every relevant path from the outer function to the independent variable must be included.

单变量链式法则无法处理这种情况，因为它假定只有一个中间变量。当存在多个中间变量时，必须考虑从外层函数到自变量之间的每一条相关路径。

2. Notation and Assumptions | 记号与假设

We use the notation z = f(x, y), with x = x(s, t) and y = y(s, t). Throughout this article we assume that f is differentiable, and that x and y have continuous first-order partial derivatives with respect to s and t. In short, we assume the functions are C¹, meaning all first-order partial derivatives exist and are continuous.

我们使用记号 z = f(x, y)，其中 x = x(s, t)，y = y(s, t)。在本文中，我们假设 f 可微，并且 x、y 对 s 和 t 具有连续一阶偏导数。简言之，我们假设函数为 C¹ 光滑函数，即所有一阶偏导数存在且连续。

These assumptions guarantee that small errors in the differentiation process become negligible. They also allow us to use the total differential method with confidence.

这些假设保证了求导过程中产生的误差可以忽略不计，也让我们能够放心使用全微分方法。

3. The Increment Method for ∂z/∂s | 用增量法推导 ∂z/∂s

To find ∂z/∂s, hold t fixed and give s a small increment Δs. This creates increments Δx and Δy in x and y. Because x and y both depend on s, we have

为了求 ∂z/∂s，我们保持 t 不变，给 s 一个微小增量 Δs。这会使得 x 和 y 分别产生增量 Δx 和 Δy。由于 x 和 y 都依赖于 s，我们有

Δx = x(s + Δs, t) – x(s, t), Δy = y(s + Δs, t) – y(s, t)

Since f is differentiable, for sufficiently small Δx and Δy we may write

因为 f 可微，当 Δx 与 Δy 足够小时，我们可以写出

Δz = ∂f/∂x · Δx + ∂f/∂y · Δy + ε₁Δx + ε₂Δy

where ε₁ and ε₂ tend to 0 as (Δx, Δy) approaches (0, 0). Now divide both sides by Δs and let Δs → 0. Because t is fixed, Δx/Δs → ∂x/∂s and Δy/Δs → ∂y/∂s. The error terms disappear because Δx and Δy tend to 0. Therefore

其中当 (Δx, Δy) 趋向 (0, 0) 时，ε₁ 和 ε₂ 都趋向于 0。现在两边同时除以 Δs，并令 Δs → 0。由于 t 固定，Δx/Δs → ∂x/∂s，Δy/Δs → ∂y/∂s。误差项因 Δx、Δy 趋向 0 而消失。因此

∂z/∂s = ∂f/∂x · ∂x/∂s + ∂f/∂y · ∂y/∂s

This is the first chain-rule formula. It says that the total rate of change of z with respect to s is the sum of the rates along the two paths z → x → s and z → y → s.

这就是第一条链式法则公式。它表明 z 关于 s 的总变化率等于沿 z → x → s 与 z → y → s 两条路径的变化率之和。

4. The Increment Method for ∂z/∂t | 用增量法推导 ∂z/∂t

The same reasoning applies when s is held fixed and t changes. Give t an increment Δt, producing increments Δx and Δy through

同样的推理适用于保持 s 不变而改变 t 的情形。给 t 一个增量 Δt，通过下式产生 Δx 和 Δy

Δx = x(s, t + Δt) – x(s, t), Δy = y(s, t + Δt) – y(s, t)

Using differentiability of f, we obtain the same linear approximation

利用 f 的可微性，我们得到相同的线性近似

Δz = ∂f/∂x · Δx + ∂f/∂y · Δy + ε₁Δx + ε₂Δy

Dividing by Δt and passing to the limit gives the second chain-rule formula:

两边同时除以 Δt 并取极限，得到第二条链式法则公式：

∂z/∂t = ∂f/∂x · ∂x/∂t + ∂f/∂y · ∂y/∂t

Notice the symmetry: each formula contains two terms, one for each intermediate variable. The only difference is whether the denominator is ∂s or ∂t.

注意这里的对称性：每条公式都包含两项，分别对应每个中间变量。唯一的区别在于分母是 ∂s 还是 ∂t。

5. Total Differential Derivation | 全微分推导

A second method uses total differentials. Since z = f(x, y), its total differential is

第二种方法使用全微分。因为 z = f(x, y)，其全微分为

dz = ∂f/∂x · dx + ∂f/∂y · dy

Because x and y depend on s and t, we also have

因为 x 和 y 依赖于 s 和 t，我们还有

dx = ∂x/∂s · ds + ∂x/∂t · dt, dy = ∂y/∂s · ds + ∂y/∂t · dt

Substitute dx and dy into dz and collect the coefficients of ds and dt:

将 dx 和 dy 代入 dz，并整理 ds 与 dt 的系数：

dz = (∂f/∂x · ∂x/∂s + ∂f/∂y · ∂y/∂s) ds + (∂f/∂x · ∂x/∂t + ∂f/∂y · ∂y/∂t) dt

The coefficient of ds is ∂z/∂s, and the coefficient of dt is ∂z/∂t. This gives the same two formulas, but in a very compact way. The total differential method is often easier to remember and to apply when many variables are present.

ds 的系数就是 ∂z/∂s，dt 的系数就是 ∂z/∂t。这样得到的两条公式与增量法结果完全一致。全微分方法通常更容易记忆，也更容易在变量较多时使用。

6. Jacobian Matrix Form | 雅可比矩阵形式

The two chain-rule formulas can be written as one matrix product. Let the derivative of f with respect to (x, y) be the row vector

两条链式法则公式可以合并为一个矩阵乘积。设 f 关于 (x, y) 的导数为行向量

[ ∂f/∂x, ∂f/∂y ]

and let the Jacobian matrix of the transformation (x, y) with respect to (s, t) be

并设变换 (x, y) 关于 (s, t) 的雅可比矩阵为

[ ∂x/∂s, ∂x/∂t; ∂y/∂s, ∂y/∂t ]

Then the derivative of z with respect to (s, t) is

那么 z 关于 (s, t) 的导数为

[ ∂z/∂s, ∂z/∂t ] = [ ∂f/∂x, ∂f/∂y ] × [ ∂x/∂s, ∂x/∂t; ∂y/∂s, ∂y/∂t ]

This is the chain rule in matrix form. It shows that differentiating composite functions is exactly matrix multiplication of the corresponding derivative matrices. For IB purposes, the two separate formulas are usually sufficient, but the matrix form helps explain why the rule is true.

这就是矩阵形式的链式法则。它表明复合函数求导本质上就是相应导数矩阵做乘法。对于 IB 考试而言，通常掌握两个独立公式即可，但矩阵形式有助于解释链式法则的本质。

7. Worked Example 1: Polynomial and Trigonometric Substitution | 例 1：多项式与三角替换

Let z = x²y, with x = s cos t and y = s sin t. First compute the partial derivatives of x and y:

设 z = x²y，其中 x = s cos t，y = s sin t。首先计算 x 和 y 的偏导数：

∂x/∂s = cos t, ∂x/∂t = -s sin t
∂y/∂s = sin t, ∂y/∂t = s cos t

For f(x, y) = x²y, we have ∂f/∂x = 2xy and ∂f/∂y = x². Therefore

对于 f(x, y) = x²y，有 ∂f/∂x = 2xy，∂f/∂y = x²。因此

∂z/∂s = 2xy cos t + x² sin t

Substitute x = s cos t and y = s sin t:

代入 x = s cos t 和 y = s sin t：

∂z/∂s = 2(s cos t)(s sin t) cos t + (s cos t)² sin t = 3s² sin t cos²t

Similarly, for t:

类似地，对于 t：

∂z

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Double-Parameter Chain Rule Derivation | 双参数链式法则推导