📚 Mastering Exercise 4C.2: Product and Quotient Rules in IB Mathematics | 掌握练习 4C.2：IB 数学中的乘积与商法则
Exercise 4C.2 brings together the core differentiation tools that every IB Mathematics student must master: the product rule and the quotient rule. This article revisits the underlying formulas, walks through carefully chosen worked examples, and highlights the subtle pitfalls that often appear under exam pressure. Whether you are preparing for Analysis & Approaches or Applications & Interpretation, fluency with these rules is essential for tackling trigonometric, exponential, logarithmic, and algebraic functions confidently.
练习 4C.2 汇集了每位 IB 数学学生必须掌握的核心微分工具：乘积法则与商法则。本文重温基本公式，逐步解析精选例题，并指出考试压力下常出现的微妙陷阱。无论你正为分析与方法还是应用与解释课程备考，熟练运用这些法则对于自信地处理三角函数、指数函数、对数函数和代数函数都至关重要。
1. Introduction to Exercise 4C.2 | 练习 4C.2 简介
er expressions. The problems move beyond basic polynomial derivatives and require systematic application of the product rule (u·v)' = u'·v + u·v' and the quotient rule (u/v)' = (u'·v − u·v' re unlocks later topics such as optimisation, implicit differentiation, and related rates.
在 IB 数学课程中，练习 4C.2 旨在巩固由较简单表达式乘积或商构成的函数的微分。题目不再局限于基本多项式求导，而要求系统应用乘积法则 (u·v)' = u'·v + u·v' 和商法则 (u/v)' = (u'·v − u·v') / v²。掌握这部分内容将解锁后续主题，如最优化、隐函数微分和相关变化率。
2. The Product Rule – Formula and Derivation | 乘积法则 – 公式与推导
differentiable functions of x, then the derivative is given by the product rule. The formula arises from the limit definition of the derivative and the intuitive idea that the change in the product depends on the change in each factor. The formal statement is:
如果 y = u(x)·v(x)，其中 u 和 v 都是关于 x 的可微函数，那么其导数由乘积法则给出。该公式源于导数的极限定义，直观上反映出乘积的变化依赖于每个因子的变化。正式表述为：
dy/dx = u'·v + u·v'
dx). It is essential to remember the plus sign between the two terms; a common error is to multiply the derivatives directly or to use a minus sign.
用莱布尼茨记号常写作 d(uv)/dx = (du/dx)·v + u·(dv/dx)。务必牢记两个项之间的加号；常见错误是直接相乘导数或误用减号。
3. Applying the Product Rule – Worked Examples | 应用乘积法则 – 示例解析
Example 1: Differentiate f(x) = x²·sin x.
示例 1：求 f(x) = x²·sin x 的导数。
Let u = x² → u' = 2x; v = sin x → v' = cos x. Then f' , but it is perfectly acceptable in this form. The product rule keeps the structure clean.
令 u = x² → u' = 2x；v = sin x → v' = cos x。则 f'(x) = 2x·sin x + x²·cos x。注意结果无法简化为单项，但这种形式完全可接受。乘积法则保持了结构清晰。
Example 2: Find dy/dx when y = eˣ·ln x.
示例 2：求 y = eˣ·ln x 时的导数 dy/dx。
u = eˣ, u' = eˣ; v = ln x, v' = 1/x. Then dy/dx = eˣ·ln x + eˣ·(1/x) = eˣ /x). Presenting the final answer in factorised form is often preferred in IB mark schemes.
u = eˣ, u' = eˣ；v = ln x, v' = 1/x。则 dy/dx = eˣ·ln x + eˣ·(1/x) = eˣ(ln x + 1/x)。在 IB 评分方案中，通常偏爱以因式分解形式呈现最终答案。
4. The Quotient Rule – Formula and Derivation | 商法则 – 公式与推导
ent rule gives its derivative. The formula can be derived by applying the product rule to u·(v)⁻¹ and the chain rule, or directly from the limit definition.
当函数表示为比值 y = u(x)/v(x) 且 v(x) ≠ 0 时，商法则给出其导数。该公式可通过将乘积法则和链式法则应用于 u·(v)⁻¹ 导出，或直接由极限定义得到。
dy/dx = (u'·v − u·v') / v²
The numerator must always be "bottom times derivative of the top minus top times derivative of the bottom". The denominator is the square of the original bottom function. Reversing the subtraction order is a classic mistake that will lose marks in an examination.
分子必须始终是“底乘顶的导数减去顶乘底的导数”。分母是原有底函数的平方。颠倒减法顺序是考试中导致失分的典型错误。
5. Applying the Quotient Rule – Worked Examples | 应用商法则 – 示例解析
Example 1: Differentiate g(x) = (x² + 1)/(x − 1).
示例 1：求 g(x) = (x² + 1)/(x − 1) 的导数。
Set u = x² + 1 → u' = 2x; v = x − 1 → v' = 1. Then g' − 2x − 1)/(x − 1)². Always simplify the numerator unless the question specifies otherwise.
设 u = x² + 1 → u' = 2x；v = x − 1 → v' = 1。则 g' (x − 1) − (x² + 1)·1] / (x − 1)² = [2x² − 2x − x² − 1] / (x − 1)² = (x² − 2x − 1)/(x − 1)²。除非题目另有要求，务必化简分子。
Example 2: Derive the derivative of tan x using the quotient rule.
示例 2：用商法则推导 tan x 的导数。
Write tan x = sin x / cos x. u = sin x, u' = cos x; v = cos x, v' in² x)/cos² x = 1/cos² x = sec² x. This standard result is frequently used in IB problems.
将 tan x 写作 sin x / cos x。u = sin x, u' = cos x；v = cos x, v' = −sin x。则 ) = [cos x·cos x − sin x·(−sin x)] / cos² x = (cos² x + sin² x)/cos² x = 1/cos² x = sec² x。这个标准结果在 IB 题目中频繁使用。
6. Extending to Higher Derivatives | 高阶导数的扩展
The product and quotient rules are not limited to first derivatives. When a question asks for the second derivative f''(x), you must differentiate f'(x) using the rules again. This process requires careful bookkeeping of terms.
乘积和商法则不限于一阶导数。当题目要求二阶导数 f''(x) 时，你必须再次使用法则对 f'(x) 求导。这一过程需要仔细整理各项。
Consider f(x) = x·eˣ. f'(x) = 1·eˣ + x·eˣ = eˣ(1 + x). To find f''(x), apply the product rule to eˣ(1 + x): u = eˣ, v = 1 + x, giving f''(x) = eˣ·(1 + x) + eˣ·1 = eˣ 2). For more complex products, a table or systematic grouping can prevent algebraic slips.
考虑 f(x) = x·eˣ。f'(x) = 1·eˣ + x·eˣ = eˣ(1 + x)。为求 f''(x)，对 eˣ(1 + x) 应用乘积法则：u = eˣ, v = 1 + x，得到 f''(x) = eˣ·(1 + x) + eˣ·1 = eˣ(x + 2)。对于更复杂的乘积，表格或系统分组可防止代数错误。
7. Common Mistakes and How to Avoid Them | 常见错误及避免方法
Even students who understand the formulas can lose marks through avoidable errors. The most frequent slip in the product rule is writing u'·v' instead of u'·v + u·v'. In the quotient rule, the subtraction order u'·v − u·v' is often reversed, producing an incorrect sign.
即使理解公式的学生也可能因可避免的错误而失分。乘积法则中最常见的手误是将 u'·v' 误写为 u'·v + u·v'。在商法则中，减法顺序 u'·v − u·v' 经常被颠倒，导致符号错误。
Another pitfall is forgetting to square the denominator in the quotient rule – writing just v instead of v². Also, when simplifying, students sometimes expand incorrectly or miss that a common factor can be cancelled. A disciplined approach: write u, v, u', v' clearly before substituting, and always double-check the signs.
另一陷阱是商法则中忘记将分母平方——只写 v 而非 v²。此外，化简时学生有时展开错误或未发现可以约去的公因子。规范做法是：在代入前清晰地写出 u、v、u'、v'，并始终检查符号。
8. Product and Quotient Rules Combined | 乘积与商法则的组合
x·cos x)/(x² + 1). You can treat the entire numerator as one function and the denominator as another, applying the quotient rule; within that, the derivative of the numerator itself requires the product rule.
有些函数需要在一个求导过程中同时使用两个法则。例如，求 f(x) = (x·cos x)/(x² + 1) 的导数。你可以将整个分子视为一个函数、分母视为另一个函数，应用商法则；在此过程中，分子本身的导数又需要乘积法则。
Step by step: Let N = x·cos x, N' = 1·cos x + x·(−sin x) = cos x − x sin x. Let D = x² + 1, D' = 2x. Then f'(x) = [N'·D − N·D' en acceptable to leave the derivative in this structured form unless instructed otherwise.
逐步进行：令 N = x·cos x，N' = 1·cos x + x·(−sin x) = cos x − x sin x。令 D = x² + 1，D' = 2x。则 f'(x) = [N'·D − N·D']/D² = [(cos x − x sin x)(x² + 1) − (x·cos x)·2x]/(x² + 1)²。尽管可以进一步化简，但在 IB 考试中，除非另有指示，通常可以接受这种结构化形式。
9. Trigonometric Functions: Product and Quotient Rules | 三角函数：乘积与商法则
Trigonometric expressions appear very frequently in Exercise 4C.2. Memorising the derivatives of sin x, cos x, and tan x, together with the chain rule for composite angles, allows you to apply the product and quotient rules smoothly to functions like x sin 2x or sec x tan x.
三角函数表达式在练习 4C.2 中出现极其频繁。熟记 sin x、cos x 和 tan x 的导数，并结合复合角的链式法则，你就能流畅地将乘积与商法则应用于诸如 x sin 2x 或 sec x tan x 的函数。
Example: Differentiate f(x) = sin x·cos 2x. u = sin x, u' = cos x. v = cos 2x; using the chain rule, v' = −2 sin 2x. Thus f' at may simplify using double-angle identities is a valuable skill for longer IB questions.
示例：求 f(x) = sin x·cos 2x 的导数。u = sin x，u' = cos x。v = cos 2x；利用链式法则，v' = −2 sin 2x。因此 f'(x) = cos x·cos 2x + sin x·(−2 sin 2x) = cos x cos 2x − 2 sin x sin 2x。能利用倍角公式简化模式的识别能力对于较长的 IB 题目是一项宝贵技能。
10. Exponential and Logarithmic Functions | 指数与对数函数
Functions involving eˣ ally, but their products and quotients demand careful rule application. Recall that d/dx(eˣ) = eˣ and d/dx(ln x) = 1/x, so these often combine with polynomial or trigonometric factors.
涉及 eˣ 和 ln x 的函数单独求导很简单，但它们的乘积和商则需要仔细应用法则。回顾 d/dx(eˣ) = eˣ 以及 d/dx(ln x) = 1/x，因此它们经常与多项式或三角函数因子组合出现。
For f(x) = ln x / x, use the quotient rule: u = ln x, u' = 1/x; v = x, v' = 1. Then f' elps find the maximum of ln x / x, a common optimisation target. For products such as x² eˣ, the product rule yields f'(x) = 2x eˣ + x² eˣ = x eˣ (2 + x), which is easy to analyse for critical points.
对于 f(x) = ln x / x，使用商法则：u = ln x，u' = 1/x；v = x，v' = 1。则 f'(x) = [(1/x)·x − ln x·1]/x² = (1 − ln x)/x²。这个导数有助于求 ln x / x 的最大值，这是一个常见的最优化目标。对于 x² eˣ 这类乘积，乘积法则给出 f'(x) = 2x eˣ + x² eˣ = x eˣ (2 + x)，便于分析临界点。
11. Applications: Tangent Lines and Equations | 应用：切线方程
e equation of the tangent line to a curve at a given point. After finding the derivative f' t the x-coordinate to get the gradient m. The tangent equation is then y − y₁ = m(x − x₁).
练习 4C.2 系列中的典型题目要求找出曲线在给定点处的切线方程。在使用乘积或商法则求出导数 f'(x) 后，代入 x 坐标求得斜率 m。随后切线方程为 y − y₁ = m(x − x₁)。
Example: Find the tangent to f(x) = x·√x + 1 at x = 3. First write √x + 1 as (x + 1)½. Product rule: u = x, u' = 1; v = (x + 1)½, v' = ½ (x + 1)−½. So f'(x) = 1·(x + 1)½ + x·½ (x + 1)−½. At x = 3, f(3) = 3·√4 = 6; f'(3) = √4 + 3·½·(4)−½ = 2 + (3/2)·(1/2) = 2 + 3/4 = 11/4. Tangent: y − 6 = (11/4)(x − 3).
示例：求 f(x) = x·√x + 1 在 x = 3 处的切线。首先将 √x + 1 写作 (x + 1)½。乘积法则：u = x，u' = 1；v = (x + 1)½，v' = ½ (x + 1)−½。因此 f'(x) = 1·(x + 1)½ + x·½ (x + 1)−½。在 x = 3 处，f(3) = 3·√4 = 6；f'(3) = √4 + 3·½·(4)−½ = 2 + (3/2)·(1/2) = 2 + 3/4 = 11/4。切线为：y − 6 = (11/4)(x − 3)。
12. IB Exam-Style Questions from Exercise 4C.2 | 基于练习 4C.2 的 IB 考题
assessment, Exercise 4C.2 includes multi-part questions. One example: Let f(x) = (x² + 3)eˣ. (a) Find f'(x) and f'' onary points and their nature. (c) Find the equation of the tangent to the curve at x = 0.
为贴近 IB 评估风格，练习 4C.2 包含多问设置题。一例如下：设 f(x) = (x² + 3)eˣ。(a) 求 f'(x) 和 f''(x)。(b) 确定所有
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