A-Level Maths: The Product Rule Explained | A-Level 数学：乘积法则求导详解

📚 A-Level Maths: The Product Rule Explained | A-Level 数学：乘积法则求导详解

The product rule is one of the most essential differentiation techniques in A-Level Mathematics. It allows us to differentiate functions that are formed by multiplying two or more differentiable functions together. Without it, expressions like x²·sin x or eˣ·ln x would be extremely difficult, if not impossible, to differentiate from first principles on a regular basis.

乘积法则是 A-Level 数学中最核心的求导技巧之一。它使我们能够对由两个或多个可导函数相乘形成的函数进行求导。如果没有它，像 x²·sin x 或 eˣ·ln x 这样的表达式，通常很难甚至无法通过导数的定义直接求导。

This guide will take you step by step through the product rule, its derivation, worked examples, common mistakes, and examination-style questions. By the end, you will have a complete understanding of when and how to apply it with confidence.

本指南将带你一步一步地学习乘积法则，包括它的推导过程、例题解析、常见错误以及考试风格的练习题。学完本指南，你将全面理解何时以及如何自信地运用这一法则。

1. What Is the Product Rule? | 什么是乘积法则？

The product rule states that if a function y can be written as the product of two functions u(x) and v(x), i.e. y = u × v, then its derivative is given by:

乘积法则指出：如果函数 y 可以写成两个函数 u(x) 和 v(x) 的乘积，即 y = u × v，那么它的导数为：

dy/dx = u·(dv/dx) + v·(du/dx)

In words: the derivative of a product is the first function times the derivative of the second, plus the second function times the derivative of the first. It is essential that you learn this formula exactly as written — in the order “u·v’ + v·u’” — because reversing the order is a common source of confusion.

用文字表述就是：两个函数乘积的导数等于第一个函数乘以第二个函数的导数，再加上第二个函数乘以第一个函数的导数。请务必原样记住这个公式 —— 顺序是 “u·v’ + v·u’” —— 因为把顺序反过来是常见的混淆来源。

In Leibniz notation, where u = u(x) and v = v(x), the rule is often written as:

在使用莱布尼茨记号时，即 u = u(x) 且 v = v(x)，该法则通常写作：

d(uv)/dx = u·(dv/dx) + v·(du/dx)

The product rule is particularly useful because the derivative of a product is NOT equal to the product of the derivatives. In other words, (uv)’ ≠ u’v’. Many students mistakenly try to apply this “simple multiplication rule” — but it is only valid when one of the factors is a constant.

乘积法则之所以特别有用，是因为乘积的导数并不等于导数的乘积。换句话说，(uv)’ ≠ u’v’。许多学生错误地尝试直接使用这种”简单相乘法则”——但只有当其中一个因子是常数时才成立。

2. Where Does the Product Rule Come From? | 乘积法则的来源

The product rule can be derived from the definition of the derivative. Let y = u(x)·v(x). Consider a small change Δx in x, leading to changes Δu and Δv in u and v respectively. The new value of y is:

乘积法则可以从导数的定义推导出来。设 y = u(x)·v(x)。考虑 x 的一个微小变化 Δx，导致 u 和 v 分别产生变化 Δu 和 Δv。y 的新值为：

y + Δy = (u + Δu)(v + Δv) = uv + u·Δv + v·Δu + Δu·Δv

Subtracting y = uv from both sides gives:

两边同时减去 y = uv，得到：

Δy = u·Δv + v·Δu + Δu·Δv

Now divide both sides by Δx:

现在两边同时除以 Δx：

Δy/Δx = u·(Δv/Δx) + v·(Δu/Δx) + (Δu/Δx)·Δv

As Δx → 0, the last term (Δu/Δx)·Δv tends to 0 because Δv → 0, and Δu/Δx tends to du/dx. Taking the limit, we obtain the product rule formula.

当 Δx → 0 时，最后一项 (Δu/Δx)·Δv 趋于 0，因为 Δv → 0，而 Δu/Δx 趋于 du/dx。取极限后，我们就得到了乘积法则公式。

This derivation helps explain why the product rule has two terms: the “cross terms” from expanding the product survive in the limit, while the “second-order” term Δu·Δv vanishes. Understanding this derivation is useful because it reinforces why the product rule is true rather than just a memorised formula.

这一推导有助于解释为什么乘积法则有两项：展开乘积后产生的”交叉项”在极限中保留下来，而”二阶”项 Δu·Δv 则消失。理解这一推导很有价值，因为它能让你明白乘积法则为什么成立，而不只是记住一个公式。

3. How to Apply the Product Rule: A Step-by-Step Method | 如何应用乘积法则：分步方法

To apply the product rule effectively, follow these steps every time you meet a product of two functions.

为了有效应用乘积法则，每当你遇到两个函数的乘积时，都请遵循以下步骤。

Step 1: Identify u and v. Choose one factor as u and the other as v. In most cases, the order does not matter, as long as you are consistent.
Step 2: Differentiate each factor separately. Find du/dx and dv/dx using your standard differentiation rules (power rule, chain rule, trigonometric rules, etc.).
Step 3: Substitute into the formula. Write u·(dv/dx) + v·(du/dx).
Step 4: Simplify. Expand brackets, collect like terms, and factorise where possible.

第 1 步：确定 u 和 v。 选择一个因子作为 u，另一个作为 v。大多数情况下，顺序并不重要，只要你保持一致即可。
第 2 步：分别对每个因子求导。 使用标准求导法则（幂法则、链式法则、三角函数法则等）求出 du/dx 和 dv/dx。
第 3 步：代入公式。 写出 u·(dv/dx) + v·(du/dx)。
第 4 步：化简。 展开括号、合并同类项，并在可能的情况下进行因式分解。

A useful way to keep track of your workings is to write u, v, du/dx and dv/dx in a small table before substituting into the formula. This prevents careless mistakes in exams and makes your method clear to the examiner, which often earns you method marks.

一个有用的做法是：在代入公式之前，把 u、v、du/dx 和 dv/dx 写在一个小表格中。这可以防止考试中的粗心错误，同时让你的解题过程对阅卷老师清晰可见，从而更容易获得方法分。

4. Worked Example 1: Differentiating x²·sin x | 例题 1：对 x²·sin x 求导

Let us start with a classic example: y = x²·sin x.

让我们从一个经典例子开始：y = x²·sin x。

We identify:

我们来确定：

u = x², v = sin x

Differentiating each:

对每个因子求导：

du/dx = 2x, dv/dx = cos x

Substituting into the product rule:

代入乘积法则：

dy/dx = x²·cos x + sin x·2x

Simplifying:

化简：

dy/dx = x²·cos x + 2x·sin x

We can factor out an x if required:

如果需要，可以提出一个公因子 x：

dy/dx = x(x·cos x + 2·sin x)

This example demonstrates the standard procedure. Note that the final answer is often left as a sum of two terms; you do not need to factorise unless the question asks you to. However, factorising can help if you later need to solve dy/dx = 0.

这个例子展示了解题的标准流程。注意最终答案通常保留为两项之和；除非题目要求，否则你不必进行因式分解。但如果之后需要解 dy/dx = 0，因式分解会很有帮助。

5. Worked Example 2: Differentiating eˣ·ln x | 例题 2：对 eˣ·ln x 求导

Now let us try a function that mixes an exponential and a logarithm: y = eˣ·ln x, where x > 0.

现在我们尝试一个混合了指数函数和对数函数的函数：y = eˣ·ln x，其中 x > 0。

We identify:

我们来确定：

u = eˣ, v = ln x

Differentiating each:

对每个因子求导：

du/dx = eˣ, dv/dx = 1/x

Substituting into the product rule:

代入乘积法则：

dy/dx = eˣ·(1/x) + ln x·eˣ

This can be factored neatly:

这可以简洁地因式分解：

dy/dx = eˣ(1/x + ln x)

Notice that the derivative of eˣ is itself, which is a special property of the exponential function. This often simplifies answers when eˣ appears as a factor.

注意 eˣ 的导数就是它本身，这是指数函数的一个特殊性质。当 eˣ 作为一个因子出现时，这常常会简化最终答案。

6. Worked Example 3: Differentiating x³·cos x | 例题 3：对 x³·cos x 求导

Let us explore a further example: y = x³·cos x.

让我们再探讨一个例子：y = x³·cos x。

We identify:

我们来确定：

u = x³, v = cos x

Differentiating each:

对每个因子求导：

du/dx = 3x², dv/dx = −sin x

Substituting into the product rule:

代入乘积法则：

dy/dx = x³·(−sin x) + cos x·3x²

Simplifying:

化简：

dy/dx = −x³·sin x + 3x²·cos x

Again we may factor out x²:

同样，我们可以提出 x²：

dy/dx = x²(3·cos x − x·sin x)

A common error here is forgetting the minus sign when differentiating cos x. Always double-check the derivative of trigonometric functions before substituting.

这里一个常见错误是在对 cos x 求导时忘记负号。在代入之前，请务必仔细检查三角函数的导数。

7. Combining the Product Rule with the Chain Rule | 乘积法则与链式法则的结合

In real exam questions, the product rule rarely appears in isolation. Often, one or both of the factors are composite functions requiring the chain rule as well.

在真正的考试题目中，乘积法则很少单独出现。通常，其中一个或两个因子都是复合函数，还需要同时使用链式法则。

For example, consider y = x²·sin(3x). Here the second factor is not simply sin x but sin(3x), a composite function.

例如，考虑 y = x²·sin(3x)。这里第二个因子不是简单的 sin x，而是 sin(3x)，这是一个复合函数。

Let:

令：

u = x², v = sin(3x)

The chain rule gives dv/dx = cos(3x)·3 = 3cos(3x). Then the product rule gives:

链式法则给出 dv/dx = cos(3x)·3 = 3cos(3x)。然后乘积法则给出：

dy/dx = u·(dv/dx) + v·(du/dx) = x²·3cos(3x) + sin(3x)·2x

Simplifying:

化简：

dy/dx = 3x²·cos(3x) + 2x·sin(3x)

We can factor out x:

可以提出 x：

dy/dx = x(3x·cos(3x) + 2·sin(3x))

In general, whenever you see a function such as (2x + 1)⁴, e⁵ˣ, sin(x²), ln(3x), or similar, you must first differentiate that factor using the chain rule before applying the product rule.

总的来说，每当你看到 (2x + 1)⁴、e⁵ˣ、sin(x²)、ln(3x) 等形式的函数时，必须先使用链式法则对该因子求导，然后再应用乘积法则。

8. The Product Rule for Three or More Factors | 三个或多个因子的乘积法则

What happens when you need to differentiate a product of three functions, such as y = x²·eˣ·sin x? You can apply the product rule twice, grouping two factors together.

当你需要对三个函数的乘积求导时，例如 y = x²·eˣ·sin x，该怎么办？你可以将两个因子组合在一起，连续应用两次乘积法则。

Method 1: Group the first two factors. Let y = (x²·eˣ)·sin x. Treat u = x²·eˣ and v = sin x. Then:

方法 1：将前两个因子组合。令 y = (x²·eˣ)·sin x。把 u = x²·eˣ、v = sin x。那么：

du/dx = x²·eˣ + eˣ·2x = eˣ(x² + 2x)

Then:

于是：

dy/dx = u·cos x + sin x·(du/dx)

dy/dx = x²·eˣ·cos x + sin x·eˣ(x² + 2x)

Method 2: The general formula for three factors y = u·v·w is:

方法 2：三个因子 y = u·v·w 的通用公式为：

dy/dx = uv·(dw/dx) + uw·(dv/dx) + vw·(du/dx)

This formula is quick to use once you are comfortable with the product rule. For the example above:

一旦你熟悉了乘积法则，这个公式用起来会很快。对于上面的例子：

dy/dx = (x²·eˣ)(cos x) + (x²·sin x)(eˣ) + (eˣ·sin x)(2x)

Both methods give the same answer. Method 1 is often safer for exam students because it reuses a formula you already know well.

两种方法得到相同的结果。方法 1 对考生来说通常更稳妥，因为它重复使用了你已经熟练掌握的公式。

9. Common Mistakes and How to Avoid Them | 常见错误及如何避免

Even strong students make small errors when applying the product rule. Below are the most frequent pitfalls and practical strategies to avoid them.

即使是成绩优秀的学生，在应用乘积法则时也会犯一些小错误。以下是最高频的陷阱和实用的规避策略。

Mistake \| 错误	Why it happens \| 原因	How to avoid \| 如何避免
Writing dy/dx = u’·v’ instead of u·v’ + v·u’	Confusing the product rule with the power rule or assuming multiplication is preserved under differentiation.	Memorise the formula as “first × derivative of second + second × derivative of first”. Practise the phrase aloud.
Forgetting to differentiate one of the factors.	Rushing through the substitution.	Always write u, v, du/dx, dv/dx in a table before substituting.
Missing a minus sign when differentiating cos x or when applying the chain rule.	Trigonometric derivatives have negative terms.	Write the derivative of each factor on a separate line; double-check signs.
Simplifying incorrectly after substitution.	Algebraic expansion mistakes.	Simplify step by step and re-expand to verify your answer.

When using the chain rule inside a product factor, a particularly helpful trick is to substitute an intermediate variable. For example, if v = (2x + 1)⁵, set w = 2x + 1 and write v = w⁵; then dv/dx = 5w⁴·2 = 10(2x + 1)⁴. This avoids writing the chain rule incorrectly under time pressure.

在乘积因子中使用链式法则时，一个特别有用的技巧是引入中间变量。例如，如果 v = (2x + 1)⁵，设 w = 2x + 1，写出 v = w⁵；那么 dv/dx = 5w⁴·2 = 10(2x + 1)⁴。这样可以避免在时间压力下写错链式法则。

10. Exam-Style Questions and Solutions | 考试风格题目与解答

Let us apply everything we have learned to three exam-style questions. These are typical of the type you may encounter in an A-Level paper.

让我们将所学内容应用到三道考试风格的题目中。这些题目代表了你在 A-Level 试卷中可能遇到的典型题型。

Question 1: Differentiate y = x⁵·eˣ.

题目 1： 求 y = x⁵·eˣ 的导数。

Solution: Using u = x⁵ and v = eˣ, we get du/dx = 5x⁴ and dv/dx = eˣ.

解答： 令 u = x⁵、v = eˣ，得到 du/dx = 5x⁴、dv/dx = eˣ。

dy/dx = x⁵·eˣ + eˣ·5x⁴ = eˣ(x⁵ + 5x⁴) = x⁴eˣ(x + 5)

Question 2: Find the gradient of the tangent to the curve y = x·ln x at the point where x = e.

题目 2： 求曲线 y = x·ln x 在 x = e 处的切线斜率。

Solution: Let u = x and v = ln x. Then du/dx = 1 and dv/dx = 1/x.

解答： 令 u = x、v = ln x。则 du/dx = 1、dv/dx = 1/x。

dy/dx = x·(1/x) + ln x·1 = 1 + ln x

At x = e, we have ln e = 1, so:

当 x = e 时，ln e = 1，因此：

gradient = 1 + 1 = 2

Question 3: A curve has equation y = x·sin x. Find the equation of the tangent at x = π.

题目 3： 曲线方程为 y = x·sin x。求曲线在 x = π 处的切线方程。

Solution: Using the product rule, dy/dx = x·cos x + sin x·1 = x·cos x + sin x.

解答： 使用乘积法则，dy/dx = x·cos x + sin x·1 = x·cos x + sin x。

At x = π, sin π = 0 and cos π = −1, so the gradient is:

当 x = π 时，sin π = 0、cos π = −1，因此斜率为：

m = π·(−1) + 0 = −π

The y-coordinate at x = π is y = π·sin π = 0. Using y − y₁ = m(x − x₁):

x = π 处的 y 坐标为 y = π·sin π = 0。利用点斜式 y − y₁ = m(x − x₁)：

y − 0 = −π(x − π) → y = −πx + π²

These questions show how the product rule is tested both as a mechanical skill and in contextual settings involving tangents and stationary points.

这些题目展示了乘积法则如何既作为纯技巧被考查，又如何在涉及切线和驻点的情境中被考查。

11. Using the Product Rule to Find Stationary Points | 用乘积法则求驻点

A common follow-up task is finding stationary points. To do this, you differentiate the function using the product rule, set the derivative equal to zero, and solve for x.

一个常见的后续任务是求驻点。为此，你需要用乘积法则对函数求导，令导数等于零，然后解出 x。

Consider the function y = x²·e⁻ˣ. Using u = x² and v = e⁻ˣ, we have du/dx = 2x and, by the chain rule, dv/dx = −e⁻ˣ.

考虑函数 y = x²·e⁻ˣ。令 u = x²、v = e⁻ˣ，我们有 du/dx = 2x，并且根据链式法则，dv/dx = −e⁻ˣ。

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

Set dy/dx = 0. Since e⁻ˣ is never zero for any real x, we only need to solve 2x − x² = 0:

令 dy/dx = 0。由于 e⁻ˣ 对任意实数 x 均不为零，我们只需解 2x − x² = 0：

2x − x² = 0 → x(2 − x) = 0 → x = 0 或 x = 2

The stationary points are therefore at x = 0 and x = 2. To find their nature (maximum or minimum), you can use the second derivative test or a sign table.

因此驻点位于 x = 0 和 x = 2 处。要判断它们的性质（极大值还是极小值），你可以使用二阶导数测试或符号表。

The key insight here is that whenever you set a derivative equal to zero, you can often divide both sides by an exponential factor because it is never zero. Similarly, factors like x² are only zero at specific points, so you must include those values in your solution set.

这里的关键技巧是：当你令导数为零时，通常可以两边同时除以一个指数因子，因为它永远不会为零。类似地，像 x² 这样的因子只在特定点为零，所以你必须将这些值包含在解集中。

12. Practice Problems | 练习题

To consolidate your understanding, attempt the following problems. Full solutions are provided so you can check your work.

为了巩固你的理解，请尝试以下练习。每道题都提供了完整解答，以便你核对。

Problem 1: Differentiate y = x⁴·cos x.

练习 1： 求 y = x⁴·cos x 的导数。

Answer: dy/dx = −x⁴·sin x + 4x³·cos x = x³(4·cos x − x·sin x).

答案： dy/dx = −x⁴·sin x + 4x³·cos x = x³(4·cos x − x·sin x)。

Problem 2: Differentiate y = (2x + 1)³·ln x.

练习 2： 求 y = (2x + 1)³·ln x 的导数。

Answer: Let u = (2x + 1)³ and v = ln x. Then du/dx = 6(2x + 1)² and dv/dx = 1/x. Therefore:

答案： 令 u = (2x + 1)³、v = ln x。则 du/dx = 6(2x + 1)²、dv/dx = 1/x。因此：

dy/dx = (2x + 1)³·(1/x) + ln x·6(2x + 1)²

Problem 3: Find the derivative of y = e²ˣ·sin x.

练习 3： 求 y = e²ˣ·sin x 的导数。

Answer: Let u = e²ˣ and v = sin x. Then du/dx = 2e²ˣ and dv/dx = cos x.

答案： 令 u = e²ˣ、v = sin x。则 du/dx = 2e²ˣ、dv/dx = cos x。

dy/dx = e²ˣ·cos x + sin x·2e²ˣ = e²ˣ(cos x + 2·sin x)

Problem 4: The curve y = x·e⁻ˣ reaches a stationary point. Find its coordinates.

练习 4： 曲线 y = x·e⁻ˣ 有一个驻点。求该点的坐标。

Answer: dy/dx = x·(−e⁻ˣ) + e⁻ˣ·1 = e⁻ˣ(1 − x). Setting this to zero gives x = 1. Then y = 1·e⁻¹ = 1/e. The stationary point is (1, 1/e).

答案： dy/dx = x·(−e⁻ˣ) + e⁻ˣ·1 = e⁻ˣ(1 − x)。令其为零得到 x = 1。则 y = 1·e⁻¹ = 1/e。驻点为 (1, 1/e)。

Practising these problems repeatedly will build the fluency you need for examinations. The more comfortable you are with the table method — writing u, v, du/dx, dv/dx — the fewer mistakes you will make under time pressure.

反复练习这些题目将培养你在考试中所需的熟练度。你对”列表法”（写出 u、v、du/dx、dv/dx）越熟悉，在时间压力下犯的错误就越少。

In summary, the product rule is a straightforward yet powerful differentiation tool that every A-Level mathematics student must master. Always recall the formula u·v’ + v·u’, remember that the derivative of a product is not the product of derivatives, and keep your working organised. With regular practice and careful attention to signs and chain-rule steps, you will find product rule questions among the most predictable marks on your exam paper.

总而言之，乘积法则是一个简洁而强大的求导工具，每位 A-Level 数学学生都必须掌握。始终记住公式 u·v’ + v·u’，牢记乘积的导数并不是导数的乘积，并保持解题过程条理清晰。通过定期练习和对符号及链式法则步骤的细心关注，你会发现乘积法则类题目是试卷中最容易稳拿分的题目之一。

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A-Level Maths: The Product Rule Explained | A-Level 数学：乘积法则求导详解