The Chain Rule: Differentiating Composite Functions | 链式法则：求复合函数的导数

📚 The Chain Rule: Differentiating Composite Functions | 链式法则：求复合函数的导数

One of the most powerful tools in calculus is the chain rule. It allows us to differentiate composite functions — that is, functions nested inside other functions, such as sin(2x) or (3x² + 1)⁵. Without it, many derivatives would be extremely difficult, if not impossible, to compute from first principles.

链式法则是微积分中最强大的工具之一。它让我们能够对复合函数求导——也就是嵌套在其他函数之中的函数，例如 sin(2x) 或 (3x² + 1)⁵。如果没有它，许多导数将极难甚至无法通过基本定义计算出来。

1. Understanding Composite Functions | 理解复合函数

A composite function is formed when one function is applied to the result of another function. If we have two functions f and g, then the composite function f(g(x)) means “first apply g to x, then apply f to the result.” The function g is called the inner function, and f is called the outer function.

复合函数是将一个函数作用于另一个函数的结果而形成的。如果我们有两个函数 f 和 g，那么复合函数 f(g(x)) 的意思是“先将 g 作用于 x，再将 f 作用于得到的结果”。其中 g 称为内层函数，f 称为外层函数。

For example, in y = sin(2x), the inner function is 2x and the outer function is sin(u), where u = 2x. In y = (3x² + 1)⁵, the inner function is 3x² + 1 and the outer function is u⁵.

例如，在 y = sin(2x) 中，内层函数是 2x，外层函数是 sin(u)，其中 u = 2x。在 y = (3x² + 1)⁵ 中，内层函数是 3x² + 1，外层函数是 u⁵。

2. The Chain Rule Formula | 链式法则公式

The chain rule states that if y = f(g(x)), then the derivative dy/dx is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

链式法则指出：如果 y = f(g(x))，那么导数 dy/dx 等于外层函数在内层函数处的导数乘以内层函数的导数。

dy/dx = f'(g(x)) × g'(x)

An equivalent form, using the substitution u = g(x), is written as:

等价形式可以通过代换 u = g(x) 表示为：

dy/dx = (dy/du) × (du/dx)

This second form is intuitive: the rate of change of y with respect to x is the product of the rate of change of y with respect to u and the rate of change of u with respect to x.

第二种形式非常直观：y 关于 x 的变化率等于 y 关于 u 的变化率乘以 u 关于 x 的变化率。

3. Identifying Inner and Outer Functions | 识别内层与外层函数

The first step in applying the chain rule is always to correctly identify which part of the expression is the “inside” and which is the “outside.” A helpful question to ask is: “What is the last operation I would perform if I evaluated this function for a specific value of x?”

应用链式法则的第一步始终是正确识别表达式中的“内部”部分和“外部”部分。一个有帮助的问题是：“如果我针对一个具体的 x 值来求这个函数的值，我最后进行的运算是什么？”

For y = e^(3x), if we plug in x = 2, we first compute 3·2 = 6, then compute e⁶. The last operation is the exponential, so the outer function is eᵘ and the inner function is 3x.

对于 y = e^(3x)，如果我们代入 x = 2，首先计算 3·2 = 6，然后计算 e⁶。最后一步运算是指数运算，因此外层函数是 eᵘ，内层函数是 3x。

For y = √(x² + 4), plugging in x = 1 gives 1² + 4 = 5, then √5. The last operation is the square root, so the outer function is √u and the inner is x² + 4.

对于 y = √(x² + 4)，代入 x = 1 得到 1² + 4 = 5，然后取 √5。最后一步运算是平方根，因此外层函数是 √u，内层是 x² + 4。

4. Basic Worked Example | 基础例题

Let us differentiate y = (2x + 7)³ step by step. Here the inner function is u = 2x + 7 and the outer function is u³.

让我们逐步对 y = (2x + 7)³ 求导。这里内层函数是 u = 2x + 7，外层函数是 u³。

First, differentiate the outer function with respect to u: d/du (u³) = 3u². Then differentiate the inner function with respect to x: d/dx (2x + 7) = 2.

首先，对外层函数关于 u 求导：d/du (u³) = 3u²。然后对内层函数关于 x 求导：d/dx (2x + 7) = 2。

Multiplying these results together and substituting u = 2x + 7 back, we obtain:

将这两个结果相乘，并代回 u = 2x + 7，我们得到：

dy/dx = 3(2x + 7)² × 2 = 6(2x + 7)²

A common beginner mistake is to write only 3(2x + 7)² and forget the factor of 2. This is why the chain rule is essential: the inner function’s derivative must always be included.

初学者常犯的错误是只写出 3(2x + 7)² 而忘记乘以 2。这正是链式法则的关键：内层函数的导数必须始终包含在内。

5. Powers of Functions | 函数的幂

When a function is raised to a power, the chain rule applies in a general form. If y = [f(x)]ⁿ, then:

当一个函数被提升到某个幂次时，链式法则以一种通用形式适用。如果 y = [f(x)]ⁿ，那么：

dy/dx = n[f(x)]ⁿ⁻¹ × f'(x)

This is often called the “generalised power rule.” For example, if y = (x³ + 2x)⁴, then with f(x) = x³ + 2x and f'(x) = 3x² + 2, we get:

这通常被称为“广义幂法则”。例如，如果 y = (x³ + 2x)⁴，那么取 f(x) = x³ + 2x 且 f'(x) = 3x² + 2，我们得到：

dy/dx = 4(x³ + 2x)³ × (3x² + 2)

Notice that we do not expand the bracket. In calculus, leaving the answer in factored form is generally preferred because it is easier to analyse, simplify, or evaluate at specific points.

注意我们不需要展开括号。在微积分中，答案通常保留因式分解的形式，因为这样更容易分析、化简或在特定点取值。

6. Trigonometric Composite Functions | 三角复合函数

The chain rule is frequently used with trigonometric functions. The derivatives of sin u, cos u, and tan u with respect to x must each be multiplied by du/dx:

链式法则经常与三角函数一起使用。sin u、cos u 和 tan u 关于 x 的导数都必须乘以 du/dx：

d/dx [sin(u)] = cos(u) × du/dx

d/dx [cos(u)] = −sin(u) × du/dx

d/dx [tan(u)] = sec²(u) × du/dx

For y = cos(5x), let u = 5x, so du/dx = 5. Then dy/dx = −sin(5x) × 5 = −5sin(5x). For y = sin(x²), let u = x², so du/dx = 2x. Then dy/dx = cos(x²) × 2x = 2x·cos(x²).

对于 y = cos(5x)，令 u = 5x，所以 du/dx = 5。那么 dy/dx = −sin(5x) × 5 = −5sin(5x)。对于 y = sin(x²)，令 u = x²，所以 du/dx = 2x。那么 dy/dx = cos(x²) × 2x = 2x·cos(x²)。

7. Exponential and Logarithmic Functions | 指数函数与对数函数

Exponential and logarithmic composite functions appear frequently in A-Level mathematics. Recall the standard results:

指数和对数复合函数在 A-Level 数学中频繁出现。回忆标准结果：

d/dx [eᵘ] = eᵘ × du/dx

d/dx [ln(u)] = (1/u) × du/dx

For y = e^(3x²), let u = 3x², so du/dx = 6x. Thus dy/dx = e^(3x²) × 6x = 6x·e^(3x²). For y = ln(x³ + 1), let u = x³ + 1, so du/dx = 3x². Thus dy/dx = (1/(x³ + 1)) × 3x² = 3x²/(x³ + 1).

对于 y = e^(3x²)，令 u = 3x²，所以 du/dx = 6x。因此 dy/dx = e^(3x²) × 6x = 6x·e^(3x²)。对于 y = ln(x³ + 1)，令 u = x³ + 1，所以 du/dx = 3x²。因此 dy/dx = (1/(x³ + 1)) × 3x² = 3x²/(x³ + 1)。

8. Chain Rule Combined with Product Rule | 链式法则与乘积法则结合

In many questions, the chain rule must be used alongside the product rule or quotient rule. Consider y = x²·sin(3x). This is a product of x² and sin(3x). Using the product rule:

在许多题目中，链式法则必须与乘积法则或商法则结合使用。考虑 y = x²·sin(3x)。这是 x² 与 sin(3x) 的乘积。使用乘积法则：

dy/dx = 2x·sin(3x) + x²·3cos(3x)

Here, the derivative of sin(3x) is 3cos(3x), where the factor 3 comes from the chain rule. Notice how the product rule handles the multiplication while the chain rule handles the derivative of the composite part.

这里，sin(3x) 的导数是 3cos(3x)，其中因子 3 来自链式法则。注意乘积法则处理乘法结构，而链式法则处理复合部分的求导。

Another example: y = e^(2x)·cos(x²). The product rule gives dy/dx = 2e^(2x)·cos(x²) + e^(2x)·(−2x·sin(x²)). Simplifying: dy/dx = e^(2x)[2cos(x²) − 2x·sin(x²)].

另一个例子：y = e^(2x)·cos(x²)。乘积法则给出 dy/dx = 2e^(2x)·cos(x²) + e^(2x)·(−2x·sin(x²))。化简得到：dy/dx = e^(2x)[2cos(x²) − 2x·sin(x²)]。

9. The Quotient Rule and the Chain Rule | 商法则与链式法则

When a composite function appears in the numerator or denominator of a quotient, the quotient rule and chain rule work together. Consider y = e^x / (x² + 1)³.

当复合函数出现在商的分子或分母中时，商法则与链式法则需要协同工作。考虑 y = e^x / (x² + 1)³。

Using the quotient rule with numerator u = e^x and denominator v = (x² + 1)³, we have u’ = e^x and v’ = 3(x² + 1)²·2x = 6x(x² + 1)². Hence:

使用商法则，设分子 u = e^x，分母 v = (x² + 1)³，我们有 u’ = e^x，v’ = 3(x² + 1)²·2x = 6x(x² + 1)²。因此：

dy/dx = [e^x·(x² + 1)³ − e^x·6x(x² + 1)²] / (x² + 1)⁶

This can be simplified by factoring out e^x(x² + 1)² from the numerator:

可以通过从分子中提取公因式 e^x(x² + 1)² 来化简：

dy/dx = e^x(x² + 1 − 6x) / (x² + 1)⁴

10. Second Derivatives | 二阶导数

Sometimes the question asks for the second derivative of a composite function. This requires differentiating the first derivative. For y = (2x + 1)⁴, the first derivative is dy/dx = 8(2x + 1)³.

有时题目要求复合函数的二阶导数。这需要对一阶导数再次求导。对于 y = (2x + 1)⁴，一阶导数是 dy/dx = 8(2x + 1)³。

To find the second derivative, treat 8(2x + 1)³ as a composite function too. Using the chain rule again:

为了求二阶导数，将 8(2x + 1)³ 也视为复合函数。再次使用链式法则：

d²y/dx² = 8 × 3(2x + 1)² × 2 = 48(2x + 1)²

This is a valuable technique in kinematics and optimisation problems, where acceleration is the second derivative of displacement with respect to time.

这是运动学与优化问题中非常有价值的技术，例如加速度是位移关于时间的二阶导数。

11. Common Mistakes to Avoid | 需要避免的常见错误

One common mistake is forgetting to multiply by the derivative of the inner function altogether. This produces an incomplete derivative and loses marks. Another mistake is identifying the wrong inner function, particularly with expressions like sin²x, which means (sin x)².

一个常见错误是完全忘记乘以内层函数的导数。这会得出不完整的导数并丢分。另一个错误是识别错误的内层函数，尤其是在 sin²x 这类表达式中——它表示 (sin x)²。

For sin²x, the inner function is sin x and the outer is u². The derivative is 2sin x · cos x.
For sin(x²), the inner function is x² and the outer is sin u. The derivative is 2x · cos(x²) — a completely different result.

对于 sin²x，内层函数是 sin x，外层是 u²。导数是 2sin x · cos x。
对于 sin(x²)，内层函数是 x²，外层是 sin u。导数是 2x · cos(x²)——结果完全不同。

Another common error is algebraic: forgetting to simplify. For instance, after applying the chain rule to y = (3x − 1)⁵, you should simplify to dy/dx = 15(3x − 1)⁴, not leave it as 5(3x − 1)⁴ × 3.

另一个常见错误是代数层面的：忘记化简。例如，对 y = (3x − 1)⁵ 应用链式法则后，应化简为 dy/dx = 15(3x − 1)⁴，而不是保留为 5(3x − 1)⁴ × 3。

12. Practice Questions and Summary | 练习题与总结

Try these problems on your own before checking the answers:

先自己尝试以下题目，再对照答案：

Question \| 题目	Answer \| 答案
1. y = (4x − 3)⁶	dy/dx = 24(4x − 3)⁵
2. y = e^(5x²)	dy/dx = 10x·e^(5x²)
3. y = ln(cos x)	dy/dx = −tan x
4. y = √(x² + 9)	dy/dx = x/√(x² + 9)
5. y = sin³(2x)	dy/dx = 6sin²(2x)cos(2x)

In summary, the chain rule is a fundamental technique that unlocks differentiation of composite functions. Master it by consistently following three steps: identify the outer and inner functions, differentiate the outer function with respect to the inner one, and multiply by the derivative of the inner function. Practising this pattern will build your confidence and accuracy across all calculus topics.

总的来说，链式法则是一项基础技术，它解锁了复合函数求导的能力。通过始终如一地遵循三个步骤来掌握它：识别外层和内层函数，对内层函数求外层函数的导数，再乘以内层函数的导数。练习这个模式将建立你的信心并提高在微积分所有主题中的准确性。

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The Chain Rule: Differentiating Composite Functions | 链式法则：求复合函数的导数