5. The Quotient Rule | 5. 商法则

📚 5. The Quotient Rule | 5. 商法则

quickly without having to rearrange or expand. From rational polynomials to trigonometric ratios like tan x, the quotient rule appears throughout Edexcel A-Level Pure Mathematics, especially in Papers 1 and 2 (Pure 3/4). Mastering it means you can handle rates of change, curve sketching, and optimisation problems where a division of functions is involved.

商法则是处理分式型函数求导的关键工具。它使我们能够快速对形如 f(x) = u(x)/v(x) 的函数求导，无需通分或展开。从有理多项式到三角比（如 tan x），商法则贯穿爱德思 A-Level 纯数课程，尤其在试卷 1 和 2（纯数 3/4）中频繁出现。掌握它意味着你能处理涉及函数相除的变化率、曲线作图以及优化问题。

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1. What Is the Quotient Rule? | 什么是商法则？

/dx) / v². In short notation: (u/v)′ = (u′v − uv′)/v². The numerator memorably starts with 'v du' minus 'u dv', and the denominator is simply the square of the original denominator.

如果 y = u(x)/v(x)，其中 u 和 v 都是 x 的可导函数且 v(x) ≠ 0，则导数由 dy/dx = (v·du/dx − u·dv/dx) / v² 给出。简写为：(u/v)′ = (u′v − uv′)/v²。分子的记忆口诀是 “下乘上导减上乘下导”，而分母就是原分母的平方。

You can think of the rule as applying the product rule to u·v⁻¹ and then using the chain rule on v⁻¹. This helps you understand why the 'minus' appears and why the v² is in the denominator. The formula is in the Edexcel formula booklet, but you must know how to apply it correctly under exam conditions.

你可以把该法则想象成对 u·v⁻¹ 应用乘积法则，然后对 v⁻¹ 使用链式法则。这有助于理解为什么会出现 “减号” 以及为何分母是 v²。虽然公式在爱德思公式表中有提供，但你必须在考试中能正确应用它。

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2. Derivation via the Product Rule | 用乘积法则推导商法则

bstituting back: dy/dx = u′/v − u·v′/v². Combine over the common denominator v²: dy/dx = (u′v − uv′)/v². This neat derivation shows that the quotient rule is essentially an extension of the product and chain rules.

考虑 y = u/v。改写为 y = u·v⁻¹。由乘积法则：dy/dx = u′·v⁻¹ + u·(d/dx)(v⁻¹)。对 v⁻¹ 使用链式法则得到 (d/dx)(v⁻¹) = −v⁻²·v′。回代得：dy/dx = u′/v − u·v′/v²。通分以 v² 为公分母：dy/dx = (u′v − uv′)/v²。这个简洁的推导表明商法则本质上是乘积法则和链式法则的延伸。

Many students find it easier to reconstruct the quotient rule on the spot by thinking of u·v⁻¹ rather than memorising the entire formula. However, in timed exams, using the memorised form saves valuable minutes. Understanding the proof also reinforces why the negative sign attaches to uv′, a common source of errors.

很多学生发现通过 u·v⁻¹ 当场重构商法则比死记硬背整个公式更容易。但在限时考试中，直接使用记忆公式能节省宝贵时间。理解推导过程也能加深对 “减号为何附在 uv′ 上” 的认识，这是常见错误来源。

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3. First Examples with Polynomials | 多项式入门示例

3)(2)] / (2x − 1)². Expand the numerator: 4x² − 2x − 2x² − 6 = 2x² − 2x − 6. Factor out 2: dy/dx = [2(x² − x − 3)] / (2x − 1)². The answer is usually left in this factorised form.

示例 1：求 y = (x² + 3) / (2x − 1) 的导数。设 u = x² + 3，v = 2x − 1。则 u′ = 2x，v′ = 2。代入商法则：dy/dx = [(2x)(2x − 1) − (x² + 3)(2)] / (2x − 1)²。展开分子：4x² − 2x − 2x² − 6 = 2x² − 2x − 6。提取公因数 2：dy/dx = [2(x² − x − 3)] / (2x − 1)²。答案通常保留因式分解形式。

ly: (12x³ − 5)(x² + 2) = 12x⁵ + 24x³ − 5x² − 10; (3x⁴ − 5x)(2x) = 6x⁵ − 10x². Subtract to get numerator: 12x⁵ + 24x³ − 5x² − 10 − 6x⁵ + 10x² = 6x⁵ + 24x³ + 5x² − 10. So dy/dx = (6x⁵ + 24x³ + 5x² − 10) / (x² + 2)².

示例 2：y = (3x⁴ − 5x) / (x² + 2)。此处 u = 3x⁴ − 5x，u′ = 12x³ − 5；v = x² + 2，v′ = 2x。则 dy/dx = [(12x³ − 5)(x² + 2) − (3x⁴ − 5x)(2x)] / (x² + 2)²。仔细展开：(12x³ − 5)(x² + 2) = 12x⁵ + 24x³ − 5x² − 10；(3x⁴ − 5x)(2x) = 6x⁵ − 10x²。相减得分子：12x⁵ + 24x³ − 5x² − 10 − 6x⁵ + 10x² = 6x⁵ + 24x³ + 5x² − 10。因此 dy/dx = (6x⁵ + 24x³ + 5x² − 10) / (x² + 2)²。

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4. Trigonometric Quotients | 三角函数的商

x·(−sin x)] / cos² x = (cos² x + sin² x) / cos² x = 1 / cos² x = sec² x. This neatly yields the known result sec² x.

tan x 的导数就是一个经典应用。将 tan x = sin x / cos x 写入。设 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。这就简洁地得到了熟知的结果 sec² x。

or cot x = cos x / sin x, you get −csc² x, and for csc x = 1/sin x, you get −csc x cot x. All these standard derivatives flow directly from the quotient rule.

类似地，对 sec x = 1/cos x 求导，设 u = 1，v = cos x。则 u′ = 0，v′ = −sin x，得到 (0·cos x − 1·(−sin x)) / cos² x = sin x / cos² x = sec x tan x。对于 cot x = cos x / sin x，会得到 −csc² x；对于 csc x = 1/sin x，会得到 −csc x cot x。所有这些标准导数都能从商法则直接得出。

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5. Exponential and Logarithmic Quotients | 指数与对数函数之商

Consider y = eˣ / (x + 1). Set u = eˣ, v = x + 1, with u′ = eˣ, v′ = 1. Then dy/dx = [eˣ (x + 1) − eˣ·1] / (x + 1)² = eˣ ile the algebra simplifies. When the denominator is more complicated, expand methodically.

考虑 y = eˣ / (x + 1)。设 u = eˣ，v = x + 1，u′ = eˣ，v′ = 1。则 dy/dx = [eˣ (x + 1) − eˣ·1] / (x + 1)² = eˣ x / (x + 1)²。注意指数部分保持不变，而代数部分被化简。当分母更复杂时，要有条理地展开。

x ln x) / x⁴ = (1 − 2 ln x) / x³. This shows how logarithmic growth affects the derivative's sign and critical points.

对于 y = ln x / x²，取 u = ln x，v = x²。则 u′ = 1/x，v′ = 2x。由商法则：dy/dx = [(1/x)·x² − ln x·2x] / x⁴ = (x − 2x ln x) / x⁴ = (1 − 2 ln x) / x³。这展示了对数增长怎样影响导数的符号和临界点。

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6. Combining with the Chain Rule | 结合链式法则

r v′ before using the quotient rule. For example, y = sin(3x) / (x² + 1). Here u = sin(3x) ⇒ u′ = 3 cos(3x). v = x² + 1 ⇒ v′ = 2x. Then dy/dx = [3 cos(3x)·(x² + 1) − sin(3x)·2x] / (x² + 1)².

当分子或分母本身就是复合函数时，先应用链式法则求出 u′ 或 v′，再使用商法则。例如 y = sin(3x) / (x² + 1)。这里 u = sin(3x) ⇒ u′ = 3 cos(3x)。v = x² + 1 ⇒ v′ = 2x。则 dy/dx = [3 cos(3x)·(x² + 1) − sin(3x)·2x] / (x² + 1)²。

Sometimes the numerator requires the chain rule as well. For y = (e^x²) / cos x, u = e^x² gives u′ = 2x e^x², and v = cos x gives v′ = −sin x. The quotient rule then yields dy/dx = [2x e^x² cos x − e^x²(−sin x)] / cos² x = e^x² [2x cos x + sin x] / cos² x. Note the sign change due to minus minus.

有时分子也需要链式法则。对于 y = (e^x²) / cos x，u = e^x² 给出 u′ = 2x e^x²，v = cos x 给出 v′ = −sin x。商法则给出 dy/dx = [2x e^x² cos x − e^x²(−sin x)] / cos² x = e^x² [2x cos x + sin x] / cos² x。注意由于 “减负” 导致的符号变化。

These layered examples reflect typical Edexcel exam questions where you must fluently combine rules. Always decide the order: differentiate the numerator and denominator using any needed rules, then plug into the quotient formula.

这些分层示例反映了典型的爱德思考题，要求你流畅地组合使用求导法则。始终确定顺序：先用所需法则分别求出分子和分母的导数，再代入商法则公式。

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7. Second Derivatives Using the Quotient Rule | 用商法则求二阶导数

ent rule again to the first derivative. For instance, from y = (x² + 1) / (x − 2), we found dy/dx = (x² − 4x − 1) / (x − 2)². To get d²y/dx², treat this as a new quotient with numerator N = x² − 4x − 1 and denominator D = (x − 2)², then differentiate N and D appropriately.

如果一个函数本身是分式，它的一阶导数往往也是分式。为求二阶导数，需对一阶导数再次使用商法则。例如，从 y = (x² + 1) / (x − 2) 得到 dy/dx = (x² − 4x − 1) / (x − 2)²。为求 d²y/dx²，将其视为新的分式，分子 N = x² − 4x − 1，分母 D = (x − 2)²，然后分别对 N 和 D 求导。

or (x − 2): = (x − 2)[(2x − 4)(x − 2) − 2(x² − 4x − 1)] / (x − 2)⁴ = [2(x − 2)² − 2x² + 8x + 2] / (x − 2)³. After expanding, the numerator simplifies to a constant: eventually you get d²y/dx² = 10 / (x − 2)³. This process illustrates how repeated use of the quotient rule remains manageable with careful algebra.

计算 N′ = 2x − 4，D′ = 2(x − 2)·1 = 2(x − 2)。则 d²y/dx² = [N′ D − N D′] / D²。代入并化简：[(2x − 4)(x − 2)² − (x² − 4x − 1)·2(x − 2)] / (x − 2)⁴。提取公因式 (x − 2)： − 2)[(2x − 4)(x − 2) − 2(x² − 4x − 1)] / (x − 2)⁴ = [2(x − 2)² − 2x² + 8x + 2] / (x − 2)³。展开后分子简化为一个常数，最终得到 d²y/dx² = 10 / (x − 2)³。这个过程说明只要代数操作细心，重复使用商法则依然是可控的。

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8. Common Mistakes and How to Avoid Them | 常见错误及如何避免

Mistake 1: Subtracting in the wrong order. The formula is (u′v − uv′), not (uv′ − u′v). Write 'lo dHi minus hi dLo' as a mnemonic, where 'lo' is the denominator and 'hi' is the numerator. Always double-check the subtraction sequence.

错误 1：减法顺序错误。公式是 (u′v − uv′)，而非 (uv′ − u′v)。记忆口诀 “下乘上导减上乘下导”，其中 “下” 表示分母，“上” 表示分子。始终检查减法次序。

Mistake 2: Forgetting to square the denominator. The denominator v² is not v. This leads to a completely wrong derivative. Highlight v² in your working from the very first step.

错误 2：忘记将分母平方。分母是 v² 而非 v。这会导致完全错误的导数。从第一步起就在计算中突出标记 v²。

Mistake 3: Not simplifying numerator first. Sometimes the expression can be simplified algebraically before differentiation, saving work. For example, (x² − 4)/(x − 2) simplifies to x + 2 for x ≠ 2, which differentiates to 1. Always check for cancellations.

错误 3：没有先化简分子。有时可以在求导前进行代数化简，从而节省工作量。例如，(x² − 4)/(x − 2) 在 x ≠ 2 时可化简为 x + 2，其导数为 1。务必检查能否约分。

res the chain rule: v′ = 5(x² + 1)⁴·2x. Forgetting the inner derivative is a classic slip.

错误 4：对分母或分子的复合函数求导错误。如果分母是 (x² + 1)⁵，其导数需要链式法则：v′ = 5(x² + 1)⁴·2x。忘记乘以内层导数是一个经典失误。

Mistake 5: Losing minus signs when simplifying. When expanding −u·v′, maintain brackets to avoid sign errors. A single missing sign can cascade through an entire solution.

错误 5：化简时丢失负号。展开 −u·v′ 时要保留括号，避免符号错误。一个遗漏的符号可能使整个解答崩溃。

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9. Exam-Style Worked Questions | 考试题型演练

Question 1: Differentiate f(x) = (3x² − 1) / eˣ. Let u = 3x² − 1, u′ = 6x; v = eˣ, v′ = eˣ. Then f′(x) = [6x·eˣ − (3x² − 1)eˣ] / e²ˣ = eˣ(6x − 3x² + 1) / e²ˣ = (6x − 3x² + 1) / eˣ. It is often cleaner to leave eˣ in the denominator rather than using negative exponents unless instructed otherwise.

问题 1：求 f(x) = (3x² − 1) / eˣ 的导数。设 u = 3x² − 1，u′ = 6x；v = eˣ，v′ = eˣ。则 f′(x) = [6x·eˣ − (3x² − 1)eˣ] / e²ˣ = eˣ(6x − 3x² + 1) / e²ˣ = (6x − 3x² + 1) / eˣ。除非特别要求，通常将 eˣ 留在分母比使用负指数更简洁。

quation of the tangent to y = x/(x² + 1) at x = 1. First, differentiate: u = x, v = x² + 1 ⇒ x = 1, y = 1/(1+1) = 1/2, and dy/dx = (1 − 1) / (2)² = 0. Tangent is horizontal: y = 1/2.

问题 2：

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