📚 Exercise 9B.1: Product and Quotient Rules in Differentiation | 练习9B.1：微分中的乘积法则与商法则
In the IB Mathematics course, mastering differentiation techniques is essential for tackling calculus problems. Exercise 9B.1 typically focuses on the product rule and quotient rule, two fundamental tools that allow us to differentiate functions involving products or quotients of simpler functions. This article provides a thorough walkthrough of these rules, complete with proofs, worked examples, common mistakes, and IB-style practice questions to help you build confidence and precision.
在IB数学课程中，掌握微分技巧是解决微积分问题的关键。练习9B.1通常聚焦于乘积法则和商法则，这两个基本工具使我们能够对涉及较简单函数乘积或商的复合函数进行求导。本文全面讲解这些法则，包括证明、例题精讲、常见错误和IB风格练习题，帮助你建立信心和精确度。
1. Review of Basic Derivatives | 基本导数复习
Before diving into the product and quotient rules, it is crucial to recall the derivatives of elementary functions: for any power xⁿ, the derivative is nxⁿ⁻¹; the derivative of sin x is cos x; the derivative of cos x is −sin x; and the derivative of eˣ is eˣ. Also remember that the derivative of a constant is zero, and the constant multiple rule allows us to pull constants out of the differentiation operator.
在深入乘积法则和商法则之前，回顾基本函数的导数至关重要：对于任意幂次xⁿ，导数为nxⁿ⁻¹；sin x的导数是cos x；cos x的导数是−sin x；eˣ的导数是eˣ。还要记住常数的导数为零，常数倍法则允许我们把常数提到微分运算符外面。
sin x, we need the derivatives of x² and sin x individually. Having them at your fingertips speeds up the process and reduces careless errors.
这些基本导数将在应用乘积法则和商法则时充当构建块。例如，要对f(x) = x² sin x求导，我们需要单独知道x²和sin x的导数。把它们熟记于心能加快解题速度，减少粗心错误。
2. Introducing the Product Rule | 引入乘积法则
v + uv′. This rule reflects the interaction between the two functions as they both change.
当一个函数表示为两个可微函数u(x)和v(x)的乘积时，我们不能简单地将各自的导数相乘。乘积法则指出，乘积u(x)v(x)的导数是u′(x)v(x) + u(x)v′(x)。记作(uv)′ = u′v + uv′。该法则反映了两者都在变化时相互影响的关系。
Many IB textbooks introduce this rule shortly after the chain rule, and Exercise 9B.1 often serves as the first dedicated practice. A common mnemonic is "derivative of the first times the second plus the first times derivative of the second".
许多IB教材在链式法则之后不久引入该法则，练习9B.1通常是第一次专门的练习。一个常见的口诀是：“前导乘后不导，加前不导乘后导”。
Product Rule: d/dx [u(x)·v(x)] = u′(x)·v(x) + u(x)·v′(x)
3. Proof of the Product Rule (IB Exploration) | 乘积法则证明（IB探究）
n be proved using the limit definition of the derivative. Consider f(x) = u(x)v(x). Then f'(x) = limh→0 [u(x+h)v(x+h) − u(x)v(x)] / h. By adding and subtracting u(x+h)v(x) in the numerator, we split the limit into two terms: limh→0 u(x+h)[v(x+h) − v(x)]/h + limh→0 we obtain u(x)v′(x) + v(x)u′(x). This rigorous derivation is a great example of mathematical thinking valued in IB Internal Assessments.
乘积法则可以利用导数的极限定义来证明。考虑f(x)=u(x)v(x)，则f'(x)=limh→0 [u(x+h)v(x+h)−u(x)v(x)]/h。通过在分子中加减u(x+h)v(x)，我们将极限拆分为两项：limh→0 u(x+h)[v(x+h)−v(x)]/h + limh→0 v(x)[u(x+h)−u(x)]/h。利用u的连续性，u(x+h)→u(x)，我们得到u(x)v′(x)+v(x)u′(x)。这一严谨推导是IB内部评估中重视数学思维的一个很好例子。
Understanding the proof not only deepens conceptual knowledge but also helps avoid the mistake of thinking that (uv)′ equals u′v′. The added cross terms are vital. For an IB student aiming for a level 7, being able to outline this proof is advantageous.
理解证明不仅能加深概念知识，还有助于避免误认为(uv)′等于u′v′。额外的交叉项至关重要。对于目标为7分的IB学生来说，能够概述该证明是有利的。
4. Worked Examples: Product Rule | 例题精讲：乘积法则
Example 1: the product rule, f′(x) = (2x)(3x − 2) + (x² + 1)(3) = 6x² − 4x + 3x² + 3 = 9x² − 4x + 3.
示例1：对f(x) = (x² + 1)(3x − 2)求导。设u(x) = x² + 1, v(x) = 3x − 2。则u′(x) = 2x，v′(x) = 3。应用乘积法则，f′(x) = (2x)(3x − 2) + (x² + 1)(3) = 6x² − 4x + 3x² + 3 = 9x² − 4x + 3。
Example 2: x. With u = x³, v = sin x, we have u′ = 3x², v′ = cos x. So g′(x) = 3x² sin x + x³ cos x.
示例2：对g(x) = x³ sin x求导。令u = x³，v = sin x，则u′ = 3x²，v′ = cos x。因此g′(x) = 3x² sin x + x³ cos x。
Notice that in both cases, the product rule produced the correct result neatly. An alternative method—expanding the brackets first and then differentiating—is sometimes possible for polynomials, but for trigonometric or exponential factors, the product rule is indispensable.
注意，在这两种情况下，乘积法则都简洁地给出了正确结果。另一种方法——先展开括号然后求导——对于多项式有时是可行的，但对于含有三角函数或指数因子的函数，乘积法则必不可少。
5. The Quotient Rule | 商法则
t rule applies. The derivative is given by (u′v − uv′)/v². This rule is often remembered as "low d high minus high d low, over the square of what's below". It is a direct consequence of the product rule and chain rule.
当一个函数是两个可微函数的商，即f(x) = u(x)/v(x)且v(x) ≠ 0时，应用商法则。其导数由(u′v − uv′)/v²给出。该法则常被记忆为“分母乘分子导数减去分子乘分母导数，除以分母的平方”。它是乘积法则和链式法则的直接结果。
Quotient Rule: d/dx [u(x)/v(x)] = (u′(x)v(x) − u(x)v′(x)) / [v(x)]²
The order in the numerator matters: it must be u′v minus uv′. Swapping the subtraction will give the wrong sign. In IB exams, forgetting the minus or misplacing terms is a frequent source of lost marks.
分子中的顺序很重要：必须是u′v减去uv′。交换减法顺序将导致符号错误。在IB考试中，忘记负号或错置项是常见的失分原因。
6. Deriving the Quotient Rule | 推导商法则
. Then use the product rule with the chain rule on v⁻¹. Differentiating gives u′ · v⁻¹ + u · (−1)v⁻² · v′. Rearranging terms with a common denominator v² yields (u′v − uv′)/v². This derivation shows why the minus sign appears and how the square in the denominator arises naturally.
要推导商法则，可以将商重写为乘积：u/v = u · v⁻¹。然后对v⁻¹使用乘积法则和链式法则。求导得到u′ · v⁻¹ + u · (−1)v⁻² · v′。通分整理后得到(u′v − uv′)/v²。这一推导说明了负号为何出现以及分母为何自然地为平方项。
IB learners who recognize this connection can better retain the formula and are less likely to confuse the quotient rule with the product rule. It also reinforces the concept that the quotient rule can be avoided by using negative exponents and the product rule in some cases, though it is often safer to apply the standard quotient rule directly.
认识到这一联系的IB学习者能更好地记住公式，不太可能将商法则与乘积法则混淆。这也强化了一个概念：在某些情况下，可以通过使用负指数和乘积法则来避免直接使用商法则，不过直接套用标准商法则通常更稳妥。
7. Worked Examples: Quotient Rule | 例题精讲：商法则
Example 3: x)(x − 1) − (x² + 1)(1)]/(x − 1)² = [2x² − 2x − x² − 1]/(x − 1)² = (x² − 2x − 1)/(x − 1)².
示例3：对h(x) = (x² + 1)/(x − 1)求导。确定u = x² + 1，v = x − 1。则u′ = 2x，v′ = 1。应用商法则： 2x)(x − 1) − (x² + 1)(1)]/(x − 1)² = [2x² − 2x − x² − 1]/(x − 1)² = (x² − 2x − 1)/(x − 1)²。
Example 4: Differentiate k(x) = eˣ / cos x. Here u = eˣ, v = cos x. So u′ = eˣ, v′ = −sin x. Using the quotient rule: k′(x) = [eˣ · cos x − eˣ · (−sin x)]/(cos x)² = eˣ(cos x + sin x)/cos² x. This can also be written as eˣ sec x (1 + tan x) after simplification.
示例4：对k(x) = eˣ / cos x求导。此处u = eˣ，v = cos x，于是u′ = eˣ，v′ = −sin x。使用商法则：k′(x) = [eˣ · cos x − eˣ · (−sin x)]/(cos x)² = eˣ(cos x + sin x)/cos² x。简化后也可写为 eˣ sec x (1 + tan x)。
In these examples, it is advisable to simplify the numerator early to see if any further factoring or cancellation is possible. IB exam questions frequently ask for the derivative in its simplest form, so tidying up is essential.
这些示例中，建议尽早简化分子，看看是否能进一步因式分解或约分。IB考题经常要求以最简形式给出导数，因此整理化简至关重要。
8. Common Pitfalls | 常见陷阱
′ which is completely wrong. Another pitfall is forgetting to square the denominator in the quotient rule, or misplacing the minus sign as uv′ − u′v. Even stronger students sometimes make algebraic slips when simplifying the resulting expression.
最常见的错误之一是误将乘积法则用于商，或反之。例如，有些学生会写出(u/v)′ = u′/v′，这是完全错误的。另一个陷阱是在商法则中忘记将分母平方，或将减号位置错写为uv′ − u′v。即使是能力较强的学生，在简化所得表达式时有时也会犯代数错误。
To avoid these mistakes, always take a moment to identify u and v explicitly before differentiating. Write down u, v, u′, v′ and substitute carefully into the formula. Box or underline the final simplified answer, and if time permits, check by using an alternative method or a rough graph of the original function versus its derivative.
为避免这些错误，在求导前务必花点时间明确写出u和v。写下u、v、u′、v′，然后仔细代入公式。将最终简化答案框出或划线，如果时间允许，可以用其他方法检查，或者粗略画出原函数及其导数的图像进行比对。
9. IB Exam Tips | IB考试提示
In IB Mathematics: Analysis and Approaches, you can expect to see product and quotient rule problems in Paper 1 (non-calculator) and Paper 2 (calculator). Questions may ask for the derivative only, or they may embed differentiation in a larger problem, such as finding the equation of a tangent, determining stationary points, or solving optimisation tasks. Often, you will need to combine the product/quotient rule with the chain rule.
在IB数学分析与方法的考试中，你会在试卷一（无计算器）和试卷二（有计算器）中遇到乘积法则和商法则的题目。问题可能只要求求导，也可能将微分嵌入更大的问题中，例如求切线方程、确定驻点或解决优化任务。通常，你需要将乘积/商法则与链式法则结合起来使用。
Show clear steps: label u and v, state the derivatives, and write the rule before substituting. Marks are awarded for method, even if the final answer is slightly off. Additionally, be comfortable with higher derivatives: the second derivative of a product may require applying the product rule twice.
展示清晰的步骤：标注u和v，写出导数，代入公式前先写出法则。即使最终答案稍有偏差，方法也得分。此外，要熟练应对高阶导数：乘积的二阶导数可能需要两次应用乘积法则。
10. Typical Exercise 9B.1 Questions | 典型练习9B.1题目
Exercise 9B.1 usually contains a set of progressively difficult problems. They begin with straightforward polynomial products and quotients, then move to trigonometric, exponential, and logarithmic functions. Here are a few representative questions:
练习9B.1通常包含一组难度递增的题目。它们从简单的多项式乘积和商开始，然后过渡到三角函数、指数和对数函数。以下是一些代表性题目：
- Find f′(x) for f(x) = (2x + 1)(x² − 3x). / 求 f(x) = (2x + 1)(x² − 3x) 的 f′(x)。
- If y = (sin x)/(x²), find dy/dx. / 若 y = (sin x)/(x²)，求 dy/dx。
- Let g(x) = eˣ ln x. Determine g′(x). / 设 g(x) = eˣ ln x，求 g′(x)。
- Find the derivative of h(x) = (x² − 4)/(x² + 4). / 求 h(x) = (x² − 4)/(x² + 4) 的导数。
- For the curve y = x² cos x, find the gradient at x = π. / 对于曲线 y = x² cos x，求在 x = π 处的梯度。
Notice that the last problem bridges differentiation and evaluation, a common IB approach. Students are expected to substitute the given x-value into the derived expression. Accuracy in both differentiation and substitution is key.
注意，最后一题将微分与求值相结合，这是IB常见的考法。要求学生将给定的x值代入导出的表达式。微分和代入的准确性都很关键。
11. Challenge Problem | 挑战题
hat combines product and quotient rules with the chain rule. Differentiate F(x) = (x² + 1)√3x − 1 elf is a product involving a chain rule term. Let the numerator be N(x) = (x² + 1)(3x − 1)1/2 the product rule and chain rule. Finally, apply the quotient rule to F(x) = N(x)/(x − 2).
尝试下面这道更具挑战性的题目，它综合了乘积法则、商法则和链式法则。对 F(x) = (x² + 1)√3x − 1 / (x − 2) 求导。识别整体结构：这是一个商，其分子本身是一个涉及链式法则项的乘积。设分子为 N(x) = (x² + 1)(3x − 1)1/2。那么 N′(x) 需要乘积法则和链式法则。最后，对 F(x) = N(x)/(x − 2) 应用商法则。
This type of question appears at the end of Exercise 9B.1 in many IB textbooks and serves as excellent preparation for extended response questions. Break it down systematically and do not rush the algebra.
这类题目出现在许多IB教材练习9B.1的末尾，是备考长答题的绝佳训练。务必系统地拆解问题，不要仓促进行代数运算。
12. Summary & Key Formulae | 总结与关键公式
Mastering differentiation of products and quotients is a cornerstone of IB calculus. The two rules can be summarised neatly:
掌握乘积和商的微分是IB微积分的基石。这两条法则可以简洁总结如下：
| Rule / 法则 | Formula / 公式 |
|---|---|
| Product Rule | d/dx [u·v] = u′v + uv′ |
| Quotient Rule | d/dx [u/v] = (u′v − uv′)/v² |
Always remember that these rules are derived from first principles and are consistent with the limit definition. Regular practice with Exercise 9B.1 and similar sets will make the application automatic. When solving problems, write down u, v, u′, v′ and then substitute carefully. Look out for opportunities to simplify before or after differentiating. With discipline and repetition, you will find these rules manageable and even enjoyable.
始终记住，这些法则源于基本原理，并与极限定义一致。通过练习9B.1及类似练习题的规律训练，将使应用变得自动化。解题时，写下u、v、u′、v′，然后仔细代入。留意在微分前或微分后进行简化的机会。凭借自律和反复练习，你会觉得这些法则轻松易懂，甚至趣味盎然。
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