# 乘积、商与复合函数的求导运算

> 爱德思国际GCSE进阶纯数学 · 4PM1
> 来源: https://www.owlsprep.com/zh/study/edexcel-igcse-further-pure-math-s9-differentiation-of-products-quotients-and/

本指南讲解爱德思IGCSE进阶纯数学中三个核心求导法则：乘积法则、商法则与链式法则。搭配贴合考试要求的例题，带你掌握将这些法则应用于多项式、三角函数与指数函数求导的方法。

**先修:** [x的幂函数、sin(kx)、cos(kx)与e^(kx)的基础求导方法](https://www.owlsprep.com/zh/study/edexcel-igcse-further-pure-math-s9-basic-differentiation-rules/); 复合函数的相关概念

## 学习目标

- 熟记乘积法则与链式法则，并正确使用考纲提供的商法则进行求导运算
- 运用链式法则对简单复合函数求导
- 结合三条法则，对多项式、sin(kx)、cos(kx)与e^(kx)的组合函数进行求导
- 遵循考试标准的符号规范，清晰展示求导题目的解题步骤

## 1. 乘积法则

**乘积法则** — 用于对两个可导函数u(x)和v(x)的乘积求导，该公式需要你在考试中自行熟记。

*记法:* \frac{d}{dx}(uv) = u'v + uv'

*例:* Differentiate $f(x) = x^2 \sin(3x)$

应用乘积法则时，首先将相乘的两个函数标记为u和v，计算它们的导数u'和v'，再代入乘积法则公式。最终答案要尽可能化简。

**例题:** 对$y = e^{2x} \cos(5x)$关于x求导。

1. 步骤1：将两个因子标记为u和v：$u = e^{2x}$，$v = \cos(5x)$
2. 步骤2：计算u和v的导数：

   $$u' = 2e^{2x}, v' = -5\sin(5x)$$
3. 步骤3：代入乘积法则公式：

   $$\frac{dy}{dx} = (2e^{2x})(\cos 5x) + (e^{2x})(-5\sin 5x)$$
4. 步骤4：提取公因式进行化简：

   $$\frac{dy}{dx} = e^{2x}(2\cos 5x - 5\sin 5x)$$

> **考试提示:** 解题时务必明确写出u、v、u'和v'，这样即使后续计算出现算术错误，也能拿到步骤分。

*计算器:* allowed

## 2. 商法则

**商法则** — 用于对两个可导函数u(x)（分子）和v(x)（分母）的商求导，该公式会在考试提供的公式表中给出。

*记法:* \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

*例:* Differentiate $f(x) = \frac{3x + 1}{x^2 - 2}$

使用商法则时，确保将u设为分子，v设为分母。要特别注意公式分子部分的减法运算，此处非常容易出现符号错误。

> **warning**
>
> 商法则分子中的减法顺序至关重要：先写u'v，再减去uv'。如果将两项顺序颠倒，得到的结果会是正确答案的相反数。

**例题:** 对$y = \frac{\sin(2x)}{4x + 1}$关于x求导，最终结果保留为单个分式。

1. 步骤1：将分子设为u，分母设为v：$u = \sin(2x)$，$v = 4x + 1$
2. 步骤2：计算导数：$u' = 2\cos(2x)$，$v' = 4$
3. 步骤3：代入商法则公式：

   $$\frac{dy}{dx} = \frac{(2\cos 2x)(4x + 1) - (\sin 2x)(4)}{(4x + 1)^2}$$
4. 步骤4：展开第一项，化简分子：

   $$\frac{dy}{dx} = \frac{2(4x + 1)\cos 2x - 4\sin 2x}{(4x + 1)^2}$$

*计算器:* allowed

## 3. 链式法则（复合函数）

**链式法则** — 用于对复合函数（即函数的函数，一个函数嵌套在另一个函数内部）表达式求导，该公式需要你在考试中自行熟记。

*记法:* If $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \times g'(x)$ (or $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ when substituting $u = g(x)$)

*例:* Differentiate $f(x) = (x^3 - 2x)^5$

对于IGCSE进阶纯数学难度的简单复合函数，你通常可以直接套用链式法则：将幂次作为系数乘到前面，幂次减1，再乘以内层函数的导数。该方法适用于幂函数、三角函数和指数形式的复合函数。

**例题:** 求$y = (\cos(3x))^4$关于x的导数。

1. 步骤1：识别内层函数$u = \cos(3x)$，因此$y = u^4$
2. 步骤2：分别计算y关于u的导数，以及u关于x的导数：

   $$\frac{dy}{du} = 4u^3, \frac{du}{dx} = -3\sin(3x)$$
3. 步骤3：应用链式法则$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$：

   $$\frac{dy}{dx} = 4u^3 \times (-3\sin 3x)$$
4. 步骤4：代回$u = \cos(3x)$并化简：

   $$\frac{dy}{dx} = -12\cos^3(3x) \sin(3x)$$

> **考试提示:** 对于函数的简单幂次形式，你可以使用速算公式$\frac{d}{dx}[f(x)]^n = n[f(x)]^{n-1} \times f'(x)$来节省考试时间，前提是你展示了足够的解题步骤。

*计算器:* allowed

## 4. 多种求导法则的综合运用

绝大多数考试题会要求你在同一道题中使用不止一种求导法则。例如，某乘积的其中一项为复合函数，或者某分式的分子为乘积形式。请始终将题目逐步拆解，先确定第一步要使用的法则。

**例题:** 求$y = (x^2 + 1)^3 e^{4x}$关于x的导数。

1. 步骤1：识别该式为两个函数的乘积：$u = (x^2 + 1)^3$，$v = e^{4x}$
2. 步骤2：直接计算v的导数：$v' = 4e^{4x}$
3. 步骤3：由于u是复合函数，使用链式法则计算u的导数：

   $$u' = 3(x^2 + 1)^2 \times 2x = 6x(x^2 + 1)^2$$
4. 步骤4：将u、v、u'、v'代入乘积法则公式：

   $$\frac{dy}{dx} = [6x(x^2 + 1)^2]e^{4x} + [(x^2 + 1)^3]4e^{4x}$$
5. 步骤5：提取公因式进行化简：

   $$\frac{dy}{dx} = 2e^{4x}(x^2 + 1)^2(2x^2 + 3x + 2)$$

**概念自测**

1. When differentiating $y = \frac{\sin(2x)}{x e^{3x}}$, which rules do you need to use, in order?

   - 1. Product rule, then quotient rule
   - 1. Quotient rule, then product rule for the denominator derivative
   - Only quotient rule
   - Only chain rule

   *答案:* 1. Quotient rule, then product rule for the denominator derivative

   *解析:* The expression is a quotient, so you first apply the quotient rule. To differentiate the denominator $v = x e^{3x}$, you need to use the product rule.

*计算器:* allowed

## 常见错误

- **错误做法:** Swapping the order of terms in the quotient rule numerator, writing $uv' - u'v$ instead of $u'v - uv'$
  - 原因: Sign error leads to the negative of the correct answer, losing accuracy marks
  - 正确做法: Always write u'v first, then subtract uv', and double-check the order before simplifying
- **错误做法:** Forgetting to multiply by the derivative of the inner function when applying the chain rule, e.g. writing $\frac{d}{dx}[(2x + 1)^3] = 3(2x + 1)^2$ instead of multiplying by 2
  - 原因: Misses a key step of the chain rule, leading to an incorrect derivative
  - 正确做法: After adjusting the power of the composite function, always multiply by the derivative of the inner function before simplifying
- **错误做法:** Failing to label u, v, u', v' when applying product or quotient rules
  - 原因: Makes it harder to spot errors, and you may lose method marks if your final answer is wrong
  - 正确做法: Explicitly write down u, v, u', v' as your first step for every product or quotient rule question
- **错误做法:** Overcomplicating simple problems by using the quotient rule when the product rule is easier, e.g. differentiating $\frac{x}{e^{2x}}$ as quotient instead of $x \times e^{-2x}$ with product rule
  - 原因: Increases the chance of sign or arithmetic errors
  - 正确做法: For simple denominators that are powers or exponentials, rewrite as a negative power and use the product + chain rules if you find it easier

## 速查表

| Rule | Formula | Recall Required? | Use Case |
| --- | --- | --- | --- |
| Product Rule | $\frac{d}{dx}(uv) = u'v + uv'$ | Yes | Product of two functions |
| Quotient Rule | $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ | No (given on formula sheet) | Fraction of two functions |
| Chain Rule | $\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)$ | Yes | Composite (function of a function) |

## 下一步

Now that you have mastered the three core differentiation rules for Edexcel IGCSE FPM, you are ready to apply these to practical problems including finding gradients, tangents, normals, and stationary points. These applications form a large portion of the calculus section of your exam, and often carry 5-8 mark questions that require you to combine differentiation with algebraic manipulation. You should also practice past paper questions that combine these rules with trigonometric and exponential functions to build speed and accuracy before your exam.

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