📚 Exercise 23B.3: Product and Quotient Rules for Derivatives | 练习23B.3：导数的乘积法则与商法则
Exercise 23B.3 in the IB Mathematics Analysis and Approaches textbook is a focused set of problems designed to build fluency with two fundamental differentiation techniques: the product rule and the quotient rule. Mastering these rules is essential for tackling composite, rational, and trigonometric functions that appear throughout calculus sections and in IB exam questions.
IB 数学分析与方法教材中的练习 23B.3 是一组针对性练习，旨在帮助同学们熟练运用求导的两大基本法则——乘积法则与商法则。掌握这些法则对于处理微积分中广泛出现的复合函数、有理函数和三角函数至关重要，也是 IB 考试中的常见考点。
1. Understanding the Structure of Exercise 23B.3 | 认识练习23B.3的结构
This exercise typically includes a variety of functions that require the product rule, the quotient rule, or a combination of both. Some problems also involve trigonometric, exponential, and logarithmic functions to test your ability to choose and apply the correct rule in different contexts. Many items are arranged in increasing difficulty, starting with simple polynomial products and moving toward rational expressions with trigonometric numerators.
本练习通常包含需要乘积法则、商法则或两者结合使用的多种函数。部分题目还涉及三角函数、指数函数和对数函数，以检验你能否在不同情境下选择并正确使用相应的求导法则。题目难度通常逐步递增，从简单的多项式乘积开始，逐渐过渡到分子含三角函数的复杂有理表达式。
2. The Product Rule: Formal Statement and When to Use It | 乘积法则：公式及其适用场景
If you have a function defined as the product of two differentiable functions, h(x) = f(x) g(x), then its derivative is given by h'(x) = f'(x) g(x) + f(x) g'(x). The product rule should be applied whenever you are differentiating an expression that is the multiplication of two simpler functions, such as x² sin x, eˣ ln x, or (3x+1)√x.
若一个函数是可导函数 f(x) 与 g(x) 的乘积，即 h(x) = f(x) g(x)，那么它的导数公式为 h'(x) = f'(x) g(x) + f(x) g'(x)。每当需要对两个较简单函数的乘积表达式求导时，例如 x² sin x、eˣ ln x 或 (3x+1)√x，就应使用乘积法则。
h'(x) = f'(x) g(x) + f(x) g'(x)
The logic behind the rule is that a small change in the product is influenced by changes in both factors, and the derivative captures this combined effect. You can also remember it as “derivative of the first times the second, plus the first times the derivative of the second.”
该法则的原理在于：乘积的微小变化受到两个因子变化的共同影响，导数恰好捕捉了这一组合效应。你也可以这样记忆：”第一个函数的导数乘以第二个函数，加上第一个函数乘以第二个函数的导数”。
3. Worked Example 1: Applying the Product Rule | 例题1：乘积法则的应用
Differentiate y = (x³ + 2x) eˣ.
求函数 y = (x³ + 2x) eˣ 的导数。
Let f(x) = x³ + 2x and g(x) = eˣ. Then f'(x) = 3x² + 2, and g'(x) = eˣ. Using the product rule, y’ = (3x² + 2) eˣ + (x³ + 2x) eˣ = eˣ (3x² + 2 + x³ + 2x) = eˣ (x³ + 3x² + 2x + 2).
设 f(x) = x³ + 2x，g(x) = eˣ。则 f'(x) = 3x² + 2，g'(x) = eˣ。应用乘积法则，y’ = (3x² + 2) eˣ + (x³ + 2x) eˣ = eˣ (3x² + 2 + x³ + 2x) = eˣ (x³ + 3x² + 2x + 2)。
Notice that factoring out eˣ simplifies the final expression. This technique is common in IB exam questions, as it allows further analysis such as finding stationary points.
注意，提取公因子 eˣ 可以简化最终表达式。这种技巧在 IB 考试题中非常常见，因为它便于后续分析，例如寻找驻点。
4. Common Mistakes with the Product Rule | 乘积法则的常见错误
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Forgetting to differentiate one of the factors or simply multiplying the derivatives: (fg)’ is not f’ g’.
忘记对其中一个因子求导，或错误地将两个导数直接相乘：(fg)’ 并不等于 f’ g’。
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Misapplying the chain rule inside the product rule, especially when one factor is a composite function like sin(2x). Always differentiate the inner function.
在乘积法则内部误用链式法则，特别是当某个因子为复合函数（如 sin(2x)）时。务必对内层函数求导。
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Failing to simplify the final expression by factoring, which may cost marks in an “hence” or “find the turning point” question.
未通过因式分解简化最终表达式，这在”由此求出……”或”求驻点”的后续问题中可能造成失分。
5. The Quotient Rule: Formula and Intuition | 商法则：公式与直观理解
For a function expressed as a quotient of two differentiable functions, h(x) = f(x) / g(x) with g(x) ≠ 0, the derivative is h'(x) = [f'(x) g(x) – f(x) g'(x)] / [g(x)]². The numerator resembles the product rule but with a subtraction, and the denominator is simply the square of the original denominator.
若函数为两个可导函数的商，即 h(x) = f(x) / g(x) 且 g(x) ≠ 0，则其导数为 h'(x) = [f'(x) g(x) – f(x) g'(x)] / [g(x)]²。分子类似于乘积法则但使用减法，分母则是原分母的平方。
h'(x) = [f'(x) g(x) − f(x) g'(x)] / [g(x)]²
A helpful memory device is “low d-high minus high d-low, square the bottom and away you go” (where “high” is the numerator f and “low” is the denominator g). The quotient rule is indispensable when the denominator is not easily rewritten with a negative exponent.
一个记忆口诀是：”分母的导数乘分子减去分子的导数乘分母，再除以分母的平方”。当分母不易表示为负指数形式时，商法则不可或缺。
6. Worked Example 2: Applying the Quotient Rule | 例题2：商法则的应用
Differentiate y = (x² – 1) / (x² + 1).
求 y = (x² – 1) / (x² + 1) 的导数。
Let f(x) = x² – 1, g(x) = x² + 1. Then f'(x) = 2x, g'(x) = 2x. Apply the quotient rule: y’ = [(2x)(x² + 1) – (x² – 1)(2x)] / (x² + 1)² = [2x³ + 2x – (2x³ – 2x)] / (x² + 1)² = (2x³ + 2x – 2x³ + 2x) / (x² + 1)² = 4x / (x² + 1)².
设 f(x) = x² – 1，g(x) = x² + 1。则 f'(x) = 2x，g'(x) = 2x。应用商法则：y’ = [(2x)(x² + 1) – (x² – 1)(2x)] / (x² + 1)² = [2x³ + 2x – (2x³ – 2x)] / (x² + 1)² = (2x³ + 2x – 2x³ + 2x) / (x² + 1)² = 4x / (x² + 1)²。
The careful expansion and cancellation of terms is typical of Exercise 23B.3 problems. Always simplify the numerator fully before stating the final answer.
仔细展开并消去同类项是练习 23B.3 中的典型操作。在写出最终答案前，务必彻底化简分子。
7. Combining Product and Quotient Rules | 乘积法则与商法则的联用
Some exercises require using both rules in the same question. For instance, to differentiate y = (x sin x) / (1 + x²), the numerator itself is a product, so you must first apply the product rule to find the derivative of x sin x, and then use the quotient rule for the whole expression.
有些题目需要同时使用两种法则。例如，要对 y = (x sin x) / (1 + x²) 求导，分子本身就是一个乘积，因此必须先对 x sin x 使用乘积法则求出分子的导数，再对整个分式使用商法则。
Step 1: Let u = x sin x, then u’ = (1) sin x + x cos x = sin x + x cos x. Step 2: Now y’ = [u’ (1 + x²) – u (2x)] / (1 + x²)² = [(sin x + x cos x)(1 + x²) – (x sin x)(2x)] / (1 + x²)². Simplify accordingly.
步骤1：设 u = x sin x，则 u’ = (1) sin x + x cos x = sin x + x cos x。步骤2：y’ = [u’ (1 + x²) – u (2x)] / (1 + x²)² = [(sin x + x cos x)(1 + x²) – 2x² sin x] / (1 + x²)²。随后进行化简。
8. Using the Product Rule as an Alternative to the Quotient Rule | 以乘积法则替代商法则
You can rewrite a quotient f(x)/g(x) as f(x)[g(x)]⁻¹ and use the product rule combined with the chain rule. This approach is particularly useful if you find the quotient rule difficult to memorise, but you must be comfortable with negative exponents and the chain rule.
你可以把商 f(x)/g(x) 改写为 f(x)[g(x)]⁻¹，然后联用乘积法则与链式法则求导。如果你觉得商法则不易记忆，这种方法尤为有用，但你必须熟练掌握负指数和链式法则。
For example, y = (x² + 1) / eˣ can be written as (x² + 1) e⁻ˣ. Then y’ = (2x) e⁻ˣ + (x² + 1)(-e⁻ˣ) = e⁻ˣ (2x – x² – 1). This matches the result from the quotient rule.
例如 y = (x² + 1) / eˣ 可写作 (x² + 1) e⁻ˣ。那么 y’ = (2x) e⁻ˣ + (x² + 1)(-e⁻ˣ) = e⁻ˣ (2x – x² – 1)。这与商法则得出的结果一致。
9. Typical Problem Types from Exercise 23B.3 | 练习23B.3的典型题型
Based on the exercise set, here are the core categories you will encounter:
根据该练习集，以下是你会遇到的主要题型：
| Type 题型 | Example 示例 | Rule 法则 |
|---|---|---|
| Polynomial × trig function | x² cos x | Product rule |
| Exponential × logarithmic | eˣ ln x | Product rule |
| Simple rational function | (2x+3)/(x²+1) | Quotient rule |
| Trigonometric quotient | sin x / (1 + cos x) | Quotient rule (+ identity simplification) |
| Product inside quotient | (x eˣ) / (x+1) | Product then quotient |
Working through these standard forms builds the pattern recognition needed for Part A and B questions in the IB exam.
通过练习这些标准形式，可以建立起 IB 考试中 A、B 部分题目所要求的模式识别能力。
10. Step-by-Step Solution to a Challenging Problem | 一道挑战题的分步解析
Problem: Differentiate y = (ln x · tan x) / √x.
题目：求 y = (ln x · tan x) / √x 的导数。
Step 1: Recognise that the numerator is a product, say u = ln x · tan x. We need u’ via the product rule: u’ = (1/x) · tan x + ln x · sec² x.
步骤1：识别分子为一个乘积，设 u = ln x · tan x。需要通过乘积法则求出 u’：u’ = (1/x) tan x + ln x · sec² x。
Step 2: Write the denominator as g(x) = x^(1/2), so g'(x) = (1/2) x^(-1/2). Apply the quotient rule: y’ = [u’ · √x – u · (1/(2√x))] / x.
步骤2：将分母写作 g(x) = x^(1/2)，则 g'(x) = (1/2) x^(-1/2)。应用商法则：y’ = [u’ · √x – u · (1/(2√x))] / x。
Step 3: Multiply numerator and denominator by 2√x to clear the fractional term in the numerator, giving y’ = [2x u’ – u] / (2x √x). Now substitute u and u’ back in.
步骤3：分子分母同乘以 2√x 以消除分子中的分式项，得到 y’ = [2x u’ – u] / (2x √x)。再将 u 和 u’ 代回。
y’ = [2x ((1/x) tan x + ln x sec² x) – (ln x tan x)] / (2x √x) = [2 tan x + 2x ln x sec² x – ln x tan x] / (2x √x).
y’ = [2x ((1/x) tan x + ln x sec² x) – ln x tan x] / (2x √x) = [2 tan x + 2x ln x sec² x – ln x tan x] / (2x √x)。
Such multi-step simplification is excellent practice for maintaining algebraic accuracy under time pressure.
这种多步化简是提高时间压力下代数运算准确性的绝佳练习。
11. Connections to IB Exam-Style Questions | 衔接IB考试题型
In the IB Mathematics Analysis and Approaches exam, you may be asked not just to find a derivative, but also to find equations of tangents, normals, or to determine increasing/decreasing intervals. A typical Exam Question might say: “For the curve y = (x² eˣ) / (x+1), find the equation of the tangent at x = 0.” You will need to evaluate the derivative at x=0 using the quotient/product rules.
在 IB 数学分析与方法的考试中，不仅会要求你求导，还可能要求你求出切线或法线方程，或者判断增减区间。一道典型的考题可能是：”对于曲线 y = (x² eˣ) / (x+1)，求 x=0 处的切线方程。”你需要利用商法则/乘积法则求出 x=0 处的导数值。
Always practice writing down f(x), g(x), f'(x), g'(x) explicitly before substituting into the formula. This structured approach reduces sign errors and makes partial marks possible even if the final simplification is incomplete.
请始终养成先将 f(x)、g(x)、f'(x)、g'(x) 明确写出后再代入公式的习惯。这种有条理的方法可以减少符号错误，且即使最终化简未完成，也能获得部分步骤分。
12. Summary and Key Takeaways | 总结与关键要点
Exercise 23B.3 is more than repetitive drill – it builds the foundational skill of recognising when a function is a product or quotient and selecting the appropriate rule. Remember to simplify by factoring when possible, watch for opportunities to rewrite quotients as products, and always double-check the subtraction sign in the quotient rule numerator.
练习 23B.3 不仅仅是重复性的训练——它可以培养你识别函数是乘积还是商、并选择适当法则的基础技能。记住：尽量通过因式分解进行化简，留意将商改写为乘积的机会，同时务必反复检查商法则分子中的减号。
Key points: (1) Product rule: (fg)’ = f’ g + f g’. (2) Quotient rule: (f/g)’ = (f’ g – f g’) / g². (3) Always simplify the numerator before finalising your answer. (4) For complex functions, break the problem into small steps rather than trying to apply both rules in one mental leap.
关键点：(1) 乘积法则：(fg)’ = f’ g + f g’。(2) 商法则：(f/g)’ = (f’ g – f g’) / g²。(3) 在确定最终答案之前务必化简分子。(4) 对于复杂函数，应将问题分解为小步骤，而不要试图在脑海中使用多个法则一步到位。
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