Mastering the Chain Rule: Differentiation of Composite Functions | 链式法则精通：复合函数求导

📚 Mastering the Chain Rule: Differentiation of Composite Functions | 链式法则精通：复合函数求导

The chain rule is arguably the most powerful and frequently used differentiation technique in calculus. It allows us to differentiate composite functions—functions of functions—which appear everywhere in mathematics, physics, economics, and engineering. For IB Mathematics students, mastering the chain rule is not merely about memorising a formula; it is about developing a flexible and rigorous approach to breaking down complex functions into manageable parts.

链式法则可以说是微积分中最强大、最常用的求导技巧。它使我们能够对复合函数——即“函数的函数”——进行求导，而这些复合函数在数学、物理、经济学和工程学中无处不在。对于IB数学学生而言，掌握链式法则不仅仅是记住一个公式，更重要的是培养一种灵活而严谨的思路，将复杂函数拆解为可处理的部分。

1. The Core Formula | 核心公式

Suppose we have a composite function y = f(g(x)). The chain rule states that the derivative of y with respect to x is the derivative of the outer function f evaluated at g(x), multiplied by the derivative of the inner function g(x):

假设我们有一个复合函数 y = f(g(x))。链式法则指出，y 关于 x 的导数等于外函数 f 在 g(x) 处的导数，乘以内函数 g(x) 的导数：

dy/dx = f'(g(x)) × g'(x)

An alternative and often more intuitive notation, known as the Leibniz notation, is:

另一种更直观的记号，即莱布尼茨记号，可以写为：

dy/dx = dy/du × du/dx

where u = g(x). This formulation emphasises that the derivative can be “cancelled” conceptually like fractions, which helps students remember the structure.

其中 u = g(x)。这种形式强调导数在概念上可以像分数一样“约分”，这有助于学生记住其结构。

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

The first and most critical step in applying the chain rule is correctly identifying the inner function u and the outer function f. Consider the function y = (3x² + 2x – 5)⁷. Here, the inner function is u = 3x² + 2x – 5, and the outer function is f(u) = u⁷.

应用链式法则的第一步也是最关键的一步，是正确识别内层函数 u 和外层函数 f。考虑函数 y = (3x² + 2x – 5)⁷，内层函数是 u = 3x² + 2x – 5，外层函数是 f(u) = u⁷。

A practical rule of thumb: the inner function is usually found inside brackets, under a root sign, inside a trigonometric argument, or inside a logarithm/exponential expression. Ask yourself: “If I were to evaluate this function for a specific x, what would I compute first?” The answer is the inner function.

一个实用的经验法则：内层函数通常位于括号内、根号下、三角函数参数内，或对数/指数表达式内部。问问自己：“如果我要为某个具体的 x 计算这个函数，我首先计算什么？”答案就是内层函数。

3. Extended Chain Rule for Three or More Functions | 三层及以上复合函数的链式法则

When a function is composed of three or more nested functions, the chain rule naturally extends. For y = f(g(h(x))), we apply the rule twice:

当一个函数由三个或更多嵌套函数复合而成时，链式法则可以自然推广。对于 y = f(g(h(x)))，我们应用两次链式法则：

dy/dx = f'(g(h(x))) × g'(h(x)) × h'(x)

For example, differentiating y = sin(e^(2x)) requires three layers: the outer sine, the middle exponential, and the innermost linear term 2x.

例如，对 y = sin(e^(2x)) 求导需要三层：外层是正弦函数，中间是指数函数，最内层是线性项 2x。

dy/dx = cos(e^(2x)) × e^(2x) × 2

This “peeling the onion” approach ensures no layer is missed.

这种“剥洋葱”式的方法确保不遗漏任何一层。

4. Chain Rule with Power Functions | 链式法则在幂函数中的应用

Perhaps the most common application in IB exams is differentiating a power of a function. For y = [g(x)]ⁿ, the derivative is:

在IB考试中，最常见的应用或许是对某个函数的幂进行求导。对于 y = [g(x)]ⁿ，其导数为：

dy/dx = n[g(x)]ⁿ⁻¹ × g'(x)

Consider y = (x² + 1)¹⁰. Here n = 10, g(x) = x² + 1, and g'(x) = 2x. Thus:

考虑 y = (x² + 1)¹⁰，这里 n = 10，g(x) = x² + 1，g'(x) = 2x。因此：

dy/dx = 10(x² + 1)⁹ × 2x = 20x(x² + 1)⁹

Notice that the exponent decreases by one, and we must not forget to multiply by the derivative of the inner function.

注意指数减少1，且不能忘记乘以内层函数的导数。

5. Chain Rule with Trigonometric Functions | 链式法则在三角函数中的应用

For trigonometric functions, the chain rule appears when the argument of the trigonometric function is itself a function of x. The key results are:

对于三角函数，当三角函数的参数本身是 x 的函数时，需要用到链式法则。关键结果如下：

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 example, differentiate y = cos(5x + 2). Let u = 5x + 2, then du/dx = 5:

例如，对 y = cos(5x + 2) 求导。令 u = 5x + 2，则 du/dx = 5：

dy/dx = -sin(5x + 2) × 5 = -5 sin(5x + 2)

When trigonometric functions are raised to a power, such as y = sin²(x), we must combine the chain rule with the power rule. Writing y = (sin x)², we get:

当三角函数被提升到某个幂次时，例如 y = sin²(x)，我们需要将链式法则与幂法则结合。将 y = (sin x)² 写作，可得：

dy/dx = 2(sin x) × cos x = 2 sin x cos x = sin(2x)

The final simplification using the double-angle identity is often expected in exam answers.

最后使用二倍角公式进行化简，这通常是考试答案中期望的步骤。

6. Chain Rule with Exponential and Logarithmic Functions | 链式法则在指数与对数函数中的应用

The chain rule transforms the standard derivatives of exponential and logarithmic functions into their generalised forms:

链式法则将指数函数和对数函数的标准导数推广为一般形式：

d/dx [eᵘ] = eᵘ · du/dx
d/dx [aᵘ] = aᵘ · ln(a) · du/dx
d/dx [ln(u)] = 1/u · du/dx

For instance, differentiate y = e^(3x²). Here u = 3x², so du/dx = 6x:

例如，对 y = e^(3x²) 求导。这里 u = 3x²，所以 du/dx = 6x：

dy/dx = e^(3x²) × 6x = 6x e^(3x²)

For logarithmic functions, consider y = ln(2x – 1). Then:

对于对数函数，考虑 y = ln(2x – 1)，则：

dy/dx = 1/(2x – 1) × 2 = 2/(2x – 1)

A common IB question involves differentiating y = ln(ln x), which requires the chain rule applied twice:

一个常见的IB考题涉及对 y = ln(ln x) 求导，需要将链式法则应用两次：

dy/dx = 1/(ln x) × 1/x = 1/(x ln x)

7. Chain Rule in Implicit Differentiation | 链式法则在隐函数求导中的应用

In implicit differentiation, we differentiate both sides of an equation with respect to x, treating y as a function of x. Whenever we differentiate a term involving y, we apply the chain rule:

在隐函数求导中，我们对等式两边关于 x 求导，将 y 视为 x 的函数。每当对含有 y 的项求导时，我们都需要应用链式法则：

d/dx [f(y)] = f'(y) × dy/dx

For example, consider the unit circle equation x² + y² = 1. Differentiating both sides with respect to x:

例如，考虑单位圆方程 x² + y² = 1。两边关于 x 求导：

2x + 2y × dy/dx = 0

Solving for dy/dx gives:

解出 dy/dx 可得：

dy/dx = -x/y

This result is only valid when y ≠ 0, which corresponds to the horizontal tangent points on the circle.

该结果仅在 y ≠ 0 时成立，对应圆上的水平切点。

8. Chain Rule with Related Rates and Parametric Equations | 链式法则在相关变化率与参数方程中的应用

Related rates problems are a classic application of the chain rule. When two quantities both depend on time t, we relate their rates of change through implicit differentiation with respect to t.

相关变化率问题是链式法则的经典应用。当两个量都依赖于时间 t 时，我们通过关于 t 的隐函数求导来关联它们的变化率。

Consider a circle whose radius r is increasing at a rate of 2 cm/s. The area is A = πr². Differentiating with respect to time t:

考虑一个半径 r 以 2 cm/s 的速度增大的圆，其面积 A = πr²。关于时间 t 求导：

dA/dt = 2πr × dr/dt = 2πr × 2 = 4πr

When r = 5 cm, the area increases at dA/dt = 20π cm²/s.

当 r = 5 cm 时，面积以 dA/dt = 20π cm²/s 的速率增大。

For parametric equations x = f(t), y = g(t), the chain rule gives the beautiful formula:

对于参数方程 x = f(t)，y = g(t)，链式法则给出了一个优美的公式：

dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0

9. Common Mistakes and How to Avoid Them | 常见错误及规避方法

Despite its importance, the chain rule is often misapplied. Below are the most frequent errors and the strategies to avoid them.

尽管链式法则非常重要，但学生常常会误用。以下是最常见的错误及相应的规避策略。

Common Mistake \| 常见错误	Correct Approach \| 正确方法
Forgetting to multiply by the inner derivative	Always check: did you differentiate the “inside”?
Forgetting to multiply by the inner derivative（忘记乘以内层导数）	Always check: did you differentiate the “inside”？（始终检查：你是否对“内部”求导了？）
Confusing the order of composition	Recall that f(g(x)) ≠ g(f(x)) in general; identify the outermost function first.
Confusing the order of composition（混淆复合顺序）	Recall that f(g(x)) ≠ g(f(x)) in general; identify the outermost function first.（记住通常 f(g(x)) ≠ g(f(x))；先识别最外层函数。）
Applying chain rule where product rule is needed	If your function is a product of two unrelated functions, use the product rule instead.
Applying chain rule where product rule is needed（在需要用乘积法则时误用链式法则）	If your function is a product of two unrelated functions, use the product rule instead.（如果函数是两个不相关函数的乘积，应改用乘积法则。）

Strategy: Before differentiating, annotate the function by labelling each “layer” from outside to inside.

策略：求导之前，先给函数的每一“层”从外到内做标记。

10. Chain Rule Combined with Other Differentiation Rules | 链式法则与其它求导法则的联合使用

In many IB questions, the chain rule must be used alongside the product rule and the quotient rule. Consider differentiating the function:

在许多IB题目中，链式法则需要与乘积法则、商法则联合使用。考虑对以下函数求导：

y = x² · e^(3x)

This is a product of x² and e^(3x). Applying the product rule:

这是 x² 与 e^(3x) 的乘积。应用乘积法则：

dy/dx = 2x · e^(3x) + x² · 3e^(3x) = x e^(3x)(2 + 3x)

For a quotient such as y = sin(2x)/(x² + 1), we combine the quotient rule with the chain rule for the numerator:

对于商式 y = sin(2x)/(x² + 1)，我们将商法则与对分子的链式法则结合：

dy/dx = [2 cos(2x)(x² + 1) – sin(2x)(2x)] / (x² + 1)²

The systematic approach is to first identify the overall structure of the function, apply the corresponding rule (product/quotient), and then apply the chain rule whenever differentiating a nested component.

系统的方法是：首先识别函数的整体结构，应用相应的法则（乘积/商法则），然后在求导嵌套分量时应用链式法则。

11. Applications in IB Exam Questions | 链式法则在IB考题中的应用

IB Mathematics Analysis & Approaches (AA) and Applications & Interpretation (AI) exams frequently test the chain rule in both Paper 1 (no calculator) and Paper 2 (calculator allowed). Common question types include:

IB数学分析与方法（AA）和应用与解释（AI）考试在Paper 1（不允许使用计算器）和Paper 2（允许使用计算器）中都经常考查链式法则。常见题型包括：

Find the derivative of a given composite function and simplify.
Find the equation of the tangent or normal line to a curve at a given point.
Use second derivatives to determine the nature of stationary points.
Optimisation problems where the function includes composite expressions.

求给定复合函数的导数并化简。
求曲线在给定点处的切线或法线方程。
使用二阶导数判断驻点的性质。
包含复合表达式的最优化问题。

Consider the past-paper style question: “Find the coordinates of the stationary points of y = (x² – 2x)³ and determine their nature.” The first derivative is:

考虑一道真题风格的题目：“求 y = (x² – 2x)³ 的驻点坐标并判断其性质。”一阶导数为：

dy/dx = 3(x² – 2x)² × (2x – 2) = 6(x² – 2x)²(x – 1)

Setting dy/dx = 0 gives x = 0, x = 2, and x = 1. The second derivative test helps classify these points.

令 dy/dx = 0，可得 x = 0，x = 2 和 x = 1。二阶导数检验可用于分类这些点。

12. Practice Problems and Solutions | 练习题与解答

The following problems will help consolidate your understanding of the chain rule. Attempt each one before consulting the solution.

以下练习有助于巩固你对链式法则的理解。请先尝试解题，再查看解答。

Problem 1: Differentiate y = (4x – 1)⁵.

Solution: Let u = 4x – 1, du/dx = 4, so dy/dx = 5(4x – 1)⁴ × 4 = 20(4x – 1)⁴.

问题1：求 y = (4x – 1)⁵ 的导数。

解答：令 u = 4x – 1，du/dx = 4，因此 dy/dx = 5(4x – 1)⁴ × 4 = 20(4x – 1)⁴。

Problem 2: Differentiate y = tan(3x² – 1).

Solution: dy/dx = sec²(3x² – 1) × 6x = 6x sec²(3x² – 1).

问题2：求 y = tan(3x² – 1) 的导数。

解答：dy/dx = sec²(3x² – 1) × 6x = 6x sec²(3x² – 1)。

Problem 3: Given y = e^(cos x), find dy/dx.

Solution: dy/dx = e^(cos x) × (-sin x) = -sin x · e^(cos x).

问题3：已知 y = e^(cos x)，求 dy/dx。

解答：dy/dx = e^(cos x) × (-sin x) = -sin x · e^(cos x)。

Problem 4: Find the equation of the tangent line to y = ln(2x + 3) at x = 1.

Solution: dy/dx = 2/(2x + 3). At x = 1, dy/dx = 2/5, and y = ln(5). The tangent line is y – ln(5) = (2/5)(x – 1).

问题4：求曲线 y = ln(2x + 3) 在 x = 1 处的切线方程。

解答：dy/dx = 2/(2x + 3)。在 x = 1 处，dy/dx = 2/5，y = ln(5)。切线方程为 y – ln(5) = (2/5)(x – 1)。

Problem 5: Differentiate y = √(sin x + 2).

Solution: Write y = (sin x + 2)^(1/2). Then dy/dx = ½(sin x + 2)^(-1/2) × cos x = cos x / (2√(sin x + 2)).

问题5：求 y = √(sin x + 2) 的导数。

解答：将 y = (sin x + 2)^(1/2) 改写，则 dy/dx = ½(sin x + 2)^(-1/2) × cos x = cos x / (2√(sin x + 2))。

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Mastering the Chain Rule: Differentiation of Composite Functions | 链式法则精通：复合函数求导