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

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

Differentiation is one of the most powerful tools in A-Level mathematics, and among its many rules, the product rule stands out as a vital technique for handling products of functions. When two functions are multiplied together, the derivative is not simply the product of their individual derivatives. In this article, we will explain what the product rule is, when to use it, how to apply it step by step, and how to avoid common pitfalls.

求导是 A-Level 数学中最强大的工具之一，而在众多求导法则中，乘积法则（product rule）是处理函数乘积的重要技巧。当两个函数相乘时，它们的导数并不是简单地把各自导数相乘。本文将详细讲解乘积法则是什么、何时使用、如何一步步应用，以及如何避免常见错误。

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

The product rule gives the derivative of the product of two differentiable functions. Suppose we have two functions u(x) and v(x). If y = u(x) × v(x), then the derivative of y with respect to x is given by:

乘积法则给出了两个可导函数相乘后的导数。设有两个函数 u(x) 和 v(x)，若 y = u(x) × v(x)，则 y 对 x 的导数为：

dy/dx = u × (dv/dx) + v × (du/dx)

In words: the derivative of a product is the first function multiplied by the derivative of the second, plus the second function multiplied by the derivative of the first. Many students remember this as “uv = u dv + v du”, but be careful: the “dv” and “du” actually mean the derivatives, not differentials.

用语言表述：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。很多同学把它记成 “uv = u dv + v du”，但要注意，这里的 “dv” 和 “du” 实际上代表导数，而不是微分。

2. When Should You Use the Product Rule? | 何时应使用乘积法则？

You should use the product rule whenever the expression you need to differentiate is a genuine product of two or more simpler functions that cannot be easily expanded into a single, manageable expression. For example, y = x² sin x, y = eˣ cos x, and y = (x³ + 1) ln x all require the product rule.

当你需要求导的表达式是两个或多个简单函数真正的乘积，且不容易展开成一个简洁表达式时，就应该使用乘积法则。例如 y = x² sin x、y = eˣ cos x 和 y = (x³ + 1) ln x 都需要使用乘积法则。

However, if the product can be expanded easily into a sum of terms, you may not need the product rule. For instance, y = (x + 1)(x + 2) can be expanded to y = x² + 3x + 2 and differentiated term by term. The product rule would still give the same answer, but expanding is often faster.

然而，如果乘积可以轻松展开成多项之和，那么不一定需要乘积法则。例如，y = (x + 1)(x + 2) 可以展开为 y = x² + 3x + 2，然后逐项求导。使用乘积法则也能得到相同答案，但展开往往更快。

The key is to look at the structure of the function. If you see two different “blocks” multiplied together — a polynomial times a trig function, an exponential times a logarithm, etc. — the product rule is your go-to method.

关键是要观察函数的结构。如果你看到两个不同的”模块”相乘——例如多项式乘以三角函数、指数函数乘以对数函数等——乘积法则就是你的首选方法。

3. The Formula and Notation | 公式与记号

There are several common ways to write the product rule. If y = u(x)v(x), then:

乘积法则有几种常见的写法。如果 y = u(x)v(x)，则：

dy/dx = u × (dv/dx) + v × (du/dx)

Alternative notation using prime symbols: (uv)’ = u’v + uv’, where the prime denotes differentiation with respect to x. Some textbooks also write d/dx [u v] = u (dv/dx) + v (du/dx). All of these mean exactly the same thing.

另一种使用撇号的记号：(uv)’ = u’v + uv’，其中撇号表示对 x 求导。有些教科书也写成 d/dx [u v] = u (dv/dx) + v (du/dx)。所有这些写法含义完全相同。

A useful memory trick is to say “the derivative of a product is the first times the derivative of the second, plus the second times the derivative of the first.” It does not matter which factor you call u and which you call v, because addition is commutative — but for complex problems, choose u and v so that their derivatives are simple.

一个有用的记忆技巧是：”乘积的导数等于第一个乘以第二个的导数，再加上第二个乘以第一个的导数。” 把哪个因子称为 u、哪个称为 v 并不重要，因为加法满足交换律——但在复杂问题中，选择 u 和 v 时最好让它们的导数尽量简单。

4. Worked Example 1: Polynomial × Exponential | 例题1：多项式 × 指数函数

Differentiate y = x² eˣ with respect to x.

求 y = x² eˣ 对 x 的导数。

Step 1: Identify u and v. Let u = x² and v = eˣ.

第一步：确定 u 和 v。令 u = x²，v = eˣ。

Step 2: Differentiate each part. du/dx = 2x and dv/dx = eˣ.

第二步：分别求导。du/dx = 2x，dv/dx = eˣ。

Step 3: Apply the formula dy/dx = u (dv/dx) + v (du/dx).

第三步：套用公式 dy/dx = u (dv/dx) + v (du/dx)。

dy/dx = x² × eˣ + eˣ × 2x = eˣ(x² + 2x)

We can leave the answer as eˣ(x² + 2x) or write it as x eˣ(x + 2). Both are correct. Notice that the exponential term eˣ is its own derivative, which makes it a very convenient factor in product-rule problems.

答案可以保留为 eˣ(x² + 2x)，也可以写成 x eˣ(x + 2)。两种写法都正确。注意指数函数 eˣ 的导数是它本身，这使得它在乘积法则问题中非常方便。

5. Worked Example 2: Trigonometric Functions | 例题2：三角函数

Differentiate y = x sin x.

求 y = x sin x 的导数。

Let u = x and v = sin x. Then du/dx = 1 and dv/dx = cos x. Applying the product rule:

令 u = x，v = sin x。则 du/dx = 1，dv/dx = cos x。应用乘积法则：

dy/dx = x × cos x + sin x × 1 = x cos x + sin x

Now consider y = sin x cos x. If we let u = sin x and v = cos x, then du/dx = cos x and dv/dx = −sin x. Therefore:

再考虑 y = sin x cos x。令 u = sin x，v = cos x，则 du/dx = cos x，dv/dx = −sin x。因此：

dy/dx = sin x × (−sin x) + cos x × cos x = cos² x − sin² x

This expression can be simplified using the double-angle identity cos² x − sin² x = cos 2x, so dy/dx = cos 2x. This confirms that the product rule works correctly even when a trigonometric identity could simplify the original function.

该表达式可以利用二倍角公式 cos² x − sin² x = cos 2x 简化，因此 dy/dx = cos 2x。这验证了即使三角恒等式可以简化原函数，乘积法则仍然可以正确工作。

6. Worked Example 3: Exponential × Logarithm | 例题3：指数函数 × 对数函数

Differentiate y = eˣ ln x.

求 y = eˣ ln x 的导数。

Let u = eˣ and v = ln x. Then du/dx = eˣ and dv/dx = 1/x. Applying the product rule:

令 u = eˣ，v = ln x。则 du/dx = eˣ，dv/dx = 1/x。应用乘积法则：

dy/dx = eˣ × (1/x) + ln x × eˣ = eˣ (ln x + 1/x)

This example shows the importance of remembering the derivative of ln x, which is 1/x. A very common related problem is y = ln x / x. At first glance this looks like a quotient, but you can rewrite it as y = x⁻¹ ln x, where x⁻¹ = 1/x. Then use the product rule:

这个例子表明记住 ln x 的导数为 1/x 非常重要。一个相关的常见问题是 y = ln x / x。乍一看这是商的形式，但可以改写为 y = x⁻¹ ln x，其中 x⁻¹ = 1/x。然后使用乘积法则：

dy/dx = x⁻¹ × (1/x) + ln x × (−x⁻²) = x⁻² − x⁻² ln x = x⁻²(1 − ln x)

Alternatively, you could use the quotient rule, but rewriting as a product is a perfectly valid strategy that many students find easier to remember.

或者，你也可以使用商法则，但改写成乘积形式是完全可行的策略，许多同学觉得这样更容易记。

7. Common Mistakes and How to Avoid Them | 常见错误及避免方法

One of the most frequent mistakes is forgetting the second term. Some students write dy/dx = u (dv/dx) and ignore the contribution from v (du/dx). Always check that your answer has two terms before simplifying.

最常见的错误之一是漏掉第二项。有些同学只写 dy/dx = u (dv/dx)，而忽略了 v (du/dx) 这一项。在化简之前，永远检查你的答案是否有两项。

Another common error is mixing up u and v after differentiating. If you set u = x² and v = eˣ, make sure you substitute u back into the first term and v back into the second term exactly as the formula requires. Switching them by accident still gives the correct result here because addition is commutative, but in some problems with signs it can cause trouble.

另一个常见错误是在求导后弄混 u 和 v。如果你设 u = x²，v = eˣ，请确保按照公式要求把 u 代回第一项、v 代回第二项。这里由于加法交换律，偶尔交换也不会出错，但在涉及符号的问题中可能会带来麻烦。

Students also often misapply the product rule to expressions that are not products, such as y = sin(2x) or y = e^(x²). These require the chain rule, not the product rule. Look carefully: sin(2x) is a single function with an inner function 2x, not a product of sin and 2x in the sense of the product rule.

同学们还经常把乘积法则误用于并非乘积的表达式，例如 y = sin(2x) 或 y = e^(x²)。这些需要链式法则，而不是乘积法则。仔细观察：sin(2x) 是一个复合函数，内层函数是 2x，并不是乘积法则意义下 sin 和 2x 的乘积。

Finally, be careful with constant factors. For y = (3x²)(4x + 1), you could use the product rule, but it is much easier to first simplify to y = 12x³ + 3x² and then differentiate. Simplifying first can save time and reduce errors.

最后，注意常数因子。对于 y = (3x²)(4x + 1)，虽然可以使用乘积法则，但先化简为 y = 12x³ + 3x² 再求导会简单得多。先化简可以节省时间并减少错误。

8. Proving the Product Rule from First Principles | 从定义出发证明乘积法则

To truly understand the product rule, it is helpful to see where it comes from. We start with the definition of the derivative:

要真正理解乘积法则，看看它从何而来会很有帮助。我们从导数的定义出发：

dy/dx = lim_{h→0} [y(x+h) − y(x)] / h

If y = u(x)v(x), then y(x+h) = u(x+h)v(x+h). Subtracting y(x) = u(x)v(x), we obtain:

如果 y = u(x)v(x)，则 y(x+h) = u(x+h)v(x+h)。减去 y(x) = u(x)v(x)，得到：

dy/dx = lim_{h→0} [u(x+h)v(x+h) − u(x)v(x)] / h

Now add and subtract u(x)v(x+h) in the numerator. This step does not change the value of the expression because we are adding zero, just in a disguised form.

现在在分子中加上并减去 u(x)v(x+h)。这一步不改变表达式的值，因为我们是在以隐蔽的方式加零。

= lim_{h→0} [u(x+h)v(x+h) − u(x)v(x+h) + u(x)v(x+h) − u(x)v(x)] / h

Group the terms:

将项分组：

= lim_{h→0} {[u(x+h) − u(x)]/h × v(x+h)} + lim_{h→0} {u(x) × [v(x+h) − v(x)]/h}

As h approaches 0, [u(x+h) − u(x)]/h tends to du/dx, and v(x+h) tends to v(x) because v is continuous. Similarly, [v(x+h) − v(x)]/h tends to dv/dx. Therefore:

当 h 趋于 0 时，[u(x+h) − u(x)]/h 趋向于 du/dx，且 v(x+h) 趋向于 v(x)（因为 v 连续）。类似地，[v(x+h) − v(x)]/h 趋向于 dv/dx。因此：

dy/dx = (du/dx)×v(x) + u(x)×(dv/dx)

This is exactly the product rule. Understanding this proof is not just about passing an exam — it reveals why the product rule has a plus sign and why each term contains one derivative at a time.

这正是乘积法则。理解这个证明不仅仅是为了通过考试——它揭示了为什么乘积法则中有一个加号，以及为什么每一项中只出现一个导数。

9. The Product Rule with Three Factors | 三个因子的乘积法则

Sometimes you will meet a product of three functions, such as y = x² sin x eˣ. You could apply the product rule twice, first treating (sin x eˣ) as a single factor, but there is also an extension of the product rule for three factors:

有时你会遇到三个函数的乘积，例如 y = x² sin x eˣ。你可以先应用两次乘积法则，把 (sin x eˣ) 看作一个整体因子，但乘积法则也有针对三个因子的扩展形式：

If y = u v w, then dy/dx = (du/dx) v w + u (dv/dx) w + u v (dw/dx)

In words: differentiate one factor at a time, keep the other two unchanged, and add the three terms. For y = x² sin x eˣ, let u = x², v = sin x, w = eˣ. Then:

用语言表述：每次只对一个因子求导，其他两个保持不变，然后把三项相加。对于 y = x² sin x eˣ，令 u = x²，v = sin x，w = eˣ，则：

dy/dx = 2x sin x eˣ + x² cos x eˣ + x² sin x eˣ

This can be factored as x eˣ (2 sin x + x cos x). The three-factor product rule is essentially a pattern: for n factors, the derivative is the sum of n terms, each of which has exactly one differentiated factor.

该结果可以因式分解为 x eˣ (2 sin x + x cos x)。三因子乘积法则本质上是一种模式：对于 n 个因子，导数就是 n 项之和，每一项中恰好有一个因子被求导。

10. Practice Questions | 练习题目

Try these questions yourself before checking the answers. They cover the main types of product-rule problems you will encounter in A-Level exams.

请先自己尝试以下题目，再核对答案。这些题目覆盖了 A-Level 考试中会遇到的乘积法则主要题型。

Question 1: Differentiate y = (2x + 1)(x³ + 4).

题目1：求 y = (2x + 1)(x³ + 4) 的导数。
Question 2: Differentiate y = e⁻²ˣ cos x.

题目2：求 y = e⁻²ˣ cos x 的导数。
Question 3: Differentiate y = (x³ + 5x)². Hint: rewrite as a product (x³ + 5x)(x³ + 5x) and use the product rule, or expand.

题目3：求 y = (x³ + 5x)² 的导数。提示：将其改写为乘积 (x³ + 5x)(x³ + 5x) 并使用乘积法则，或先展开。

Question 题目	Answer 答案
1. y = (2x + 1)(x³ + 4)	dy/dx = (2x + 1)(3x²) + 2(x³ + 4) = 8x³ + 3x² + 8
2. y = e⁻²ˣ cos x	dy/dx = −2e⁻²ˣ cos x − e⁻²ˣ sin x = −e⁻²ˣ(2 cos x + sin x)
3. y = (x³ + 5x)²	dy/dx = 2(x³ + 5x)(3x² + 5) = 6x⁵ + 40x³ + 50x

Notice that Question 3 can also be solved by expanding first: y = x⁶ + 10x⁴ + 25x², so dy/dx = 6x⁵ + 40x³ + 50x. Both methods agree, which is a good check.

注意题目3也可以先展开再求导：y = x⁶ + 10x⁴ + 25x²，所以 dy/dx = 6x⁵ + 40x³ + 50x。两种方法结果一致，这可以作为很好的检验。

11. Summary | 总结

The product rule is a fundamental differentiation technique that every A-Level mathematics student must master. It states that for y = u v, the derivative is u(dv/dx) + v(du/dx). The rule applies whenever two or more functions are multiplied, and it extends naturally to three or more factors.

乘积法则是每位 A-Level 数学学生都必须掌握的基本求导技巧。它表明对于 y = u v，其导数为 u(dv/dx) + v(du/dx)。该法则适用于两个或多个函数相乘的情形，并且可以自然扩展到三个或更多因子。

To use it successfully, always identify u and v clearly, differentiate each one carefully, and remember to include both terms in the final answer. Expand or simplify first when possible, and always check whether the chain rule might be more appropriate for composite functions.

要成功运用该法则，务必清楚地确定 u 和 v，仔细对每个因子求导，并记得在最终答案中包含两项。如果可能，先展开或化简，并且始终判断对于复合函数使用链式法则是否更合适。

With regular practice, the product rule becomes second nature. Master it now, and you will find it much easier to deal with more advanced topics such as implicit differentiation, parametric equations, and differential equations.

通过经常练习，乘积法则会变得非常熟练。现在掌握好它，你将会发现处理隐函数求导、参数方程和微分方程等更高级的主题时轻松很多。

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