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The Product Rule | 乘积法则

2026-08-02 23:01 · 本文部分内容由AI辅助创作

📚 The Product Rule | 乘积法则

of two functions. If you have a function h(x) = f(x)·g(x), the derivative is not simply f'(x)·g'(x); instead, the product rule states that h'(x) = f'(x)g(x) + f(x)g' it allows you to tackle complex derivatives involving polynomials, exponentials, logarithms, and trigonometric functions.

乘积法则是微积分中基本的求导技巧，用于求两个函数乘积的导数。若 h(x) = f(x)·g(x)，其导数并非简单为 f'(x)·g'(x)；乘积法则指出 h'(x) = f'(x)g(x) + f(x)g'(x)。该法则在 IB 数学分析与方法(AA)以及应用与解释(AI)的 SL 和 HL 课程中均为核心内容。掌握它能让你处理涉及多项式、指数、对数和三角函数的复杂导数。

1. Introduction to the Product Rule | 乘积法则简介

x = u·(dv/dx) + v·(du/dx). This formula can also be expressed using prime notation as (uv)' = u'v + uv' the chain rule and the quotient rule (which itself is a special case of the product rule).

当我们对两个简单函数的乘积求导时，不能简单地对每个因子分别求导再相乘。乘积法则给出了正确的方法：若 y = u(x)·v(x)，则 dy/dx = u·(dv/dx) + v·(du/dx)。该公式也可以用撇号记法表示为 (uv)' = u'v + uv'。这个法则来源于导数的极限定义，并且是三大求导法则之一，另外两个是链式法则和商法则（商法则实际上是乘积法则的特例）。

2. Derivation from First Principles | 从第一原理推导

Let h(x) = f(x)g(x). The derivative from first principles is h' h + g(x) (f(x+h)-f(x)) / h]. As h→0, f(x+h)→f(x), and the difference quotients approach g'(x) and f'(x). Hence, h'(x) = f(x)g'(x) + g(x)f' y the product of derivatives does not appear; the interaction terms must be accounted for.

设 h(x) = f(x)g(x)。从第一原理出发，导数为 h'(x) = lim (h→0) [f(x+h)g(x+h) - f(x)g(x)] / h。我们在分子中加上并减去 f(x+h)g(x)：= lim (h→0) [f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)] / h。因式分解后得到 lim (h→0) [f(x+h)(g(x+h)-g(x))/h + g(x)(f(x+h)-f(x))/h]。当 h→0 时，f(x+h)→f(x)，两个差商分别趋于 g'(x) 和 f'(x)。因此，h'(x) = f(x)g'(x) + g(x)f'(x)。这个推导说明了为什么不是导数直接相乘——必须考虑交叉项。

3. The Formula: dy/dx = u dv/dx + v du/dx | 公式：dy/dx = u dv/dx + v du/dx

The product rule is often memorised as: " nction times derivative of the second, plus second function times derivative of the first." If we set u = f(x) and v = g(x), then:

d/dx (u·v) = u·(dv/dx) + v·(du/dx)

Or in prime notation: (uv)' = u v' + v u' rmula extends naturally to more than two functions by repeatedly applying the product rule.

乘积法则常被记作：“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”。若设 u = f(x)，v = g(x)，则：

d/dx (u·v) = u·(dv/dx) + v·(du/dx)

或使用撇号记法：(uv)' = u v' + v u'。这种对称形式很容易记忆。在应用时，务必先明确写出 u、v、du/dx、dv/dx，再代入公式。该公式通过反复使用乘积法则，可自然推广到多于两个函数的情形。

4. How to Choose u and v | 如何选择 u 和 v

For any product, you can assign either factor as u and the other as v; the result will be the same. However, a smart choice can simplify the algebra. Consider: if one factor is a simple power of x, set it as u to keep du/dx uncomplicated. If a factor becomes simpler when differentiated (e.g., ln x), you might set it as v so that its derivative is simple in the second term. There is no fixed rule, but practice builds intuition. In an IB exam, clearly labelling u and v helps avoid algebraic slips.

对任何乘积，你可以将任意一个因式设为 u，另一个设为 v；结果都是相同的。但巧妙的选择可以简化运算。例如，若其中一个因式是 x 的简单幂次，将其设为 u 可使 du/dx 保持简洁。若某个因式求导后变得更简单（如 ln x），可将其设为 v，这样第二项中其导数就很简单。并没有固定规则，但通过练习能培养直觉。在 IB 考试中，清楚标明 u 和 v 有助于避免代数错误。

5. Basic Examples | 基础例题

x³ + 5x² - 3x, derivative is 6x² + 10x - 3, matching the product rule result. This double-check reinforces the validity of the rule.

考虑 y = (x² + 3x)(2x - 1)。取 u = x² + 3x，则 du/dx = 2x + 3；v = 2x - 1，则 dv/dx = 2。于是 dy/dx = (x² + 3x)·2 + (2x - 1)(2x + 3) = 2x² + 6x + (4x² + 6x - 2x - 3) = 6x² + 10x - 3。我们可以通过先展开来验证：(x²+3x)(2x-1) = 2x³ - x² + 6x² - 3x = 2x³ + 5x² - 3x，导数为 6x² + 10x - 3，与乘积法则结果一致。这一交叉检查强化了该法则的正确性。

/dx = cos x. So dy/dx = x² cos x + sin x (2x) = x² cos x + 2x sin x. Such mixed polynomial–trigonometric products appear frequently in IB questions.

另一个例子：y = x² sin x。令 u = x²，v = sin x；则 du/dx = 2x，dv/dx = cos x。因此 dy/dx = x² cos x + sin x (2x) = x² cos x + 2x sin x。这类多项式与三角函数的混合乘积在 IB 考题中经常出现。

6. Product of Three Functions | 三个函数的乘积

factors together, say f·g as the first function, and h as the second. The result is (fgh)' = f'gh + fg'h + fgh' of . Notice the symmetry: each term takes the derivative of one function while leaving the other two unchanged. For example, differentiate y = x·ex·ln x. Using the triple product rule: dy/dx = (1)·ex·ln x + x·ex·ln x + x·ex·(1/x) = ex ln x + x ex ln x + ex. Simplifying yields ex (ln x + x ln x + 1).

当对 y = f(x)g(x)h(x) 求导时，我们可以两次应用乘积法则。将两个因式合并，比如把 f·g 看作第一个函数，h 作为第二个。得到 (fgh)' = f'gh + fg'h + fgh'。注意其对称性：每一项只对其中一个函数求导，其余两个保持不变。例如，对 y = x·ex·ln x 求导，利用三重乘积法则：dy/dx = (1)·ex·ln x + x·ex·ln x + x·ex·(1/x) = ex ln x + x ex ln x + ex。化简后为 ex (ln x + x ln x + 1)。

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7. Combining with Chain Rule | 与链式法则结合

he product rule and chain rule simultaneously. For instance, differentiate y = sin(x²) · e3x erivative of sin(x²) requires the chain rule: derivative = cos(x²)·2x. The derivative of e3x is 3e3x. Applying the product rule: dy/dx = sin(x²)·3e3x + e3x·cos(x²)·2x = e3x[3 sin(x²) + 2x cos(x²)]. Always apply the chain rule to inner functions first, then use the product rule to combine the results.

许多 IB 问题要求同时使用乘积法则和链式法则。例如，对 y = sin(x²) · e3x 求导。其中 sin(x²) 的导数需要用链式法则：导数为 cos(x²)·2x。e3x 的导数是 3e3x。应用乘积法则：dy/dx = sin(x²)·3e3x + e3x·cos(x²)·2x = e3x[3 sin(x²) + 2x cos(x²)]。务必先对内部函数使用链式法则，然后再用乘积法则将结果组合起来。

Another example: y = (2x+1)5 · ln(4x). Let u = (2x+1)5, v = ln(4x). Then du/dx = 5(2x+1)4·2 = 10(2x+1)4 (chain rule), and dv/dx = 1/x (note derivative of ln(4x) = 1/x). Hence dy/dx = (2x+1)5·(1/x) + ln(4x)·10(2x+1)4 = (2x+1)4 [ (2x+1)/x + 10 ln(4x) ].

另一个例子：y = (2x+1)5 · ln(4x)。设 u = (2x+1)5，v = ln(4x)。则 du/dx = 5(2x+1)4·2 = 10(2x+1)4（链式法则），而 dv/dx = 1/x（注意 ln(4x) 的导数为 1/x）。因此 dy/dx = (2x+1)5·(1/x) + ln(4x)·10(2x+1)4 = (2x+1)4 [ (2x+1)/x + 10 ln(4x) ]。

8. Trigonometric Functions | 三角函数乘积

tive is (1/2)·2 cos 2x = cos 2x. This consistency illustrates how product rule integrates with trigonometric identities.

考虑 y = sin x cos x。令 u = sin x，v = cos x；du/dx = cos x，dv/dx = -sin x。则 dy/dx = sin x·(-sin x) + cos x·cos x = -sin² x + cos² x = cos² x - sin² x。利用恒等式 cos 2x = cos² x - sin² x，可知 dy/dx = cos 2x。这与(1/2) sin 2x 的导数一致，因为 y = sin x cos x = (1/2) sin 2x，其导数为 (1/2)·2 cos 2x = cos 2x。这一一致性展示了乘积法则如何与三角恒等式结合。

implified using 1+tan² x = sec² x, but the differentiation step itself is straightforward.

更复杂的例子：y = tan x sec x。已知 sec x = 1/cos x，tan x = sin x/cos x。乘积法则给出 dy/dx = tan x·sec x tan x + sec x·sec² x = sec x (tan² x + sec² x)。可利用 1+tan² x = sec² x 进一步化简，但求导步骤本身是直接的。

9. Exponential and Logarithmic Functions | 指数与对数函数乘积

The product rule is essential when differentiating expressions like y = x ex. Here u = x, v = ex; du/dx = 1, dv/dx = ex. So dy/dx = x·ex + ex·1 = ex re worth memorising for IB exams, as they appear in optimisation and integration problems.

乘积法则对求导如 y = x ex 这样的表达式至关重要。此处 u = x，v = ex；du/dx = 1，dv/dx = ex。因此 dy/dx = x·ex + ex·1 = ex (x+1)。另一个经典例子：y = x ln x，其中 u = x，v = ln x，得到 dy/dx = x·(1/x) + ln x·1 = 1 + ln x。这些结果值得为 IB 考试记住，因为它们出现在优化和积分问题中。

For y = ex sin x, we get dy/dx = ex cos x + sin x ex = ex (sin x + cos x). The pattern of a repeated product with exponential often leads to tidy factorisation, which exam markers expect.

对于 y = ex sin x，我们得到 dy/dx = ex cos x + sin x ex = ex (sin x + cos x)。指数与三角函数的乘积求导通常能整理出整齐的因式，这是阅卷老师期望看到的。

10. Applications in Kinematics and Optimization | 在运动学与优化中的应用

In kinematics, displacement may be given as s(t) = t e-t. Velocity v(t) = s'(t) requires the product rule: v(t) = 1·e-t + t·(-e-t) = e-t(1 - t). Acceleration a(t) = v'(t) = -e-t(1 - t) + e-t(-1) = e-t to find the maximum profit. The product rule is indispensable in such contextual problems.

在运动学中，位移可能表示为 s(t) = t e-t。速度 v(t) = s'(t) 需要用乘积法则：v(t) = 1·e-t + t·(-e-t) = e-t(1 - t)。加速度 a(t) = v'(t) = -e-t(1 - t) + e-t(-1) = e-t(t - 2)。这些计算有助于分析阻尼运动。在优化中，你可能将利润建模为 P(x) = x·ln(x)，并求导以找到最大利润点。乘积法则在这类应用问题中不可或缺。

11. Common Mistakes | 常见错误

dents initially assume the derivative of a product is the product of the derivatives: (fg)' = f'g'. This is incorrect. The table below lists frequent errors and their corrections:

许多学生起初会假定乘积的导数是导数的乘积：(fg)' = f'g'，这是错误的。下表列出了常见错误及其更正：

Common Mistake Correction / Explanation
(uv)' = u'v' Use (uv)' = u'v + uv'
Forgetting to differentiate one of the factors entirely Always compute both u' and v', then substitute.
Misidentifying u and v when functions are nested with chain rule Write the overall function as a product of two functions first, then apply chain rule within each derivative calculation.
Sign errors, especially with trigonometric derivatives Double-check dv/dx sign; remember derivative of cos x is -sin x.
Not simplifying the final answer Factorise where possible; IB exam often rewards simplest form.

Avoid these pitfalls by systematically listing u, v, u', v' each time, and always re-reading the question to confirm you have differentiated all parts correctly.

通过每次系统地列出 u、v、u'、v'，并反复审题确认所有部分求导正确，可以避免这些陷阱。

12. IB Exam Tips | IB考试技巧