The Product and Quotient Rules | 乘积法则与商法则

📚 The Product and Quotient Rules | 乘积法则与商法则

When a function is formed by multiplying or dividing two or more simpler functions, we cannot simply differentiate each part separately. The product rule and quotient rule provide systematic methods for finding derivatives of such combined functions, essential tools for any A-Level mathematics student.

当一个函数由两个或多个更简单的函数相乘或相除构成时，我们不能简单地分别对每个部分求导。乘积法则和商法则为这类组合函数的求导提供了系统方法，是每位A-Level数学学生必备的核心工具。

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1. Review of Basic Differentiation | 基本求导回顾

Before tackling the product and quotient rules, recall the basic power rule: if y = xⁿ, then dy/dx = nxⁿ⁻¹. For a constant multiple, d/dx[cf(x)] = c·f'(x). The sum and difference rules state that d/dx[f(x) ± g(x)] = f'(x) ± g'(x), valid only for addition and subtraction.

在掌握乘积法则和商法则之前，先回顾基本幂法则：若 y = xⁿ，则 dy/dx = nxⁿ⁻¹。对于常数倍，d/dx[cf(x)] = c·f'(x)。和差法则表明 d/dx[f(x) ± g(x)] = f'(x) ± g'(x)，但这仅适用于加减运算。

d/dx(xⁿ) = nxⁿ⁻¹   |   d/dx[f(x) ± g(x)] = f'(x) ± g'(x)

Many students mistakenly apply the sum rule to products. For example, d/dx(x²·x³) is not 2x + 3x². In fact, x²·x³ = x⁵, whose derivative is 5x⁴. This simple example shows that multiplication of functions requires a dedicated rule.

许多学生错误地将和差法则应用于乘积。例如，d/dx(x²·x³) 不等于 2x + 3x²。事实上，x²·x³ = x⁵，其导数为 5x⁴。这个简单例子说明函数相乘需要专门的法则。

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2. The Product Rule — Statement | 乘积法则——表述

If y = u·v, where u and v are both functions of x, then the derivative is given by the product rule:

若 y = u·v，其中 u 和 v 都是 x 的函数，则其导数由乘积法则给出：

dy/dx = u·(dv/dx) + v·(du/dx)

In words: the derivative of a product is the first factor times the derivative of the second, plus the second factor times the derivative of the first. The order of the two terms does not matter, but each original function must appear exactly once as a multiplier.

用语言表述：乘积的导数等于第一个因式乘以第二个因式的导数，加上第二个因式乘以第一个因式的导数。两个项的先后顺序无关紧要，但每个原函数必须恰好作为乘数出现一次。

Using the notation f and g: if h(x) = f(x)·g(x), then h'(x) = f(x)·g'(x) + g(x)·f'(x). This compact form is often easier to remember and apply in exam questions.

使用 f 和 g 的记号：若 h(x) = f(x)·g(x)，则 h'(x) = f(x)·g'(x) + g(x)·f'(x)。这种简洁形式在考试中更容易记忆和应用。

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3. Worked Example — Product Rule | 乘积法则实例

Differentiate y = x²·sin x. Here u = x² and v = sin x. We compute du/dx = 2x and dv/dx = cos x. Applying the product rule:

求 y = x²·sin x 的导数。这里 u = x²，v = sin x。计算得 du/dx = 2x，dv/dx = cos x。应用乘积法则：

dy/dx = x²·cos x + 2x·sin x

The result can be factored as x(x·cos x + 2·sin x), though leaving it in expanded form is perfectly acceptable in most examinations.

结果可以提取公因式 x(x·cos x + 2·sin x)，不过在大多数考试中保留展开形式也是完全可以接受的。

Another example: y = eˣ·ln x. Then u = eˣ, v = ln x, giving du/dx = eˣ and dv/dx = 1/x. Therefore:

另一个例子：y = eˣ·ln x。则 u = eˣ，v = ln x，可得 du/dx = eˣ，dv/dx = 1/x。因此：

dy/dx = eˣ·(1/x) + eˣ·ln x = eˣ(1/x + ln x)

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4. The Quotient Rule — Statement | 商法则——表述

If y = u/v, where u and v are functions of x and v ≠ 0, then the derivative is given by the quotient rule:

若 y = u/v，其中 u 和 v 是 x 的函数且 v ≠ 0，则其导数由商法则给出：

dy/dx = (v·(du/dx) − u·(dv/dx)) / v²

A helpful mnemonic is “low d-high minus high d-low over low squared”: (low × d(high) − high × d(low)) / low². The order of subtraction is critical — reversing it introduces a sign error.

一个有用的记忆口诀是“低乘高导减高乘低导，除以低平方”：(低 × 高导 − 高 × 低导) / 低²。减法顺序至关重要——颠倒会产生符号错误。

Note that the numerator of the quotient rule resembles the product rule but with a minus sign. This minus sign is the most frequent source of errors. Always write v·du/dx first, then subtract u·dv/dx.

注意商法则的分子与乘积法则相似但为减号。这个减号是最常见的错误来源。务必先写下 v·du/dx，再减去 u·dv/dx。

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5. Worked Example — Quotient Rule | 商法则实例

Differentiate y = x / (x² + 1). Let u = x, v = x² + 1. Then du/dx = 1, dv/dx = 2x. Applying the quotient rule:

求 y = x / (x² + 1) 的导数。令 u = x，v = x² + 1。则 du/dx = 1，dv/dx = 2x。应用商法则：

dy/dx = ((x² + 1)·1 − x·2x) / (x² + 1)² = (1 − x²) / (x² + 1)²

Simplify the numerator carefully: (x² + 1) − 2x² = 1 − x². The denominator (x² + 1)² is always positive, so the sign of the derivative depends only on 1 − x².

仔细化简分子：(x² + 1) − 2x² = 1 − x²。分母 (x² + 1)² 始终为正，因此导数的符号仅取决于 1 − x²。

Another example: y = sin x / x². Here u = sin x, v = x². Thus du/dx = cos x, dv/dx = 2x, and:

另一个例子：y = sin x / x²。这里 u = sin x，v = x²。因此 du/dx = cos x，dv/dx = 2x，于是：

dy/dx = (x²·cos x − 2x·sin x) / x⁴ = (x·cos x − 2·sin x) / x³

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6. Proving the Product Rule | 乘积法则的证明

For a rigorous understanding, consider the definition of the derivative. Let h(x) = f(x)·g(x). Using first principles:

为获得严谨理解，考虑导数的定义。令 h(x) = f(x)·g(x)。用第一性原理：

h'(x) = lim_δx→0 [f(x+δx)g(x+δx) − f(x)g(x)] / δx

Add and subtract f(x+δx)g(x) in the numerator, then split into two limits. This manipulation reveals the product rule naturally:

在分子中加上并减去 f(x+δx)g(x)，然后拆分为两个极限。这种操作自然地揭示了乘积法则：

h'(x) = f(x)g'(x) + g(x)f'(x)

This proof highlights why both terms appear: one comes from the change in g, the other from the change in f. Understanding this derivation helps students remember the formula and avoid sign errors.

这一证明突显了两个项出现的原因：一个来自 g 的变化，另一个来自 f 的变化。理解这一推导过程有助于学生记忆公式并避免符号错误。

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7. Proving the Quotient Rule | 商法则的证明

The quotient rule can be derived from the product rule by rewriting y = u·v⁻¹. First differentiate v⁻¹ using the chain rule: d/dx(v⁻¹) = −v⁻²·(dv/dx). Then apply the product rule:

商法则可以通过将 y = u·v⁻¹ 重写，从乘积法则推导出来。首先用链式法则求 v⁻¹ 的导数：d/dx(v⁻¹) = −v⁻²·(dv/dx)。然后应用乘积法则：

dy/dx = u·(−v⁻²·dv/dx) + v⁻¹·du/dx

Combining over a common denominator v² gives the standard quotient rule formula. This derivation is elegant and reinforces the connection between the two rules.

将两项合并到公分母 v² 上，即可得到标准的商法则公式。这一推导非常优雅，并加强了两个法则之间的联系。

(v·du/dx − u·dv/dx) / v²

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8. Product Rule for Three or More Functions | 三个或更多函数的乘积法则

For a product of three functions, y = u·v·w, the product rule extends naturally:

对于三个函数的乘积 y = u·v·w，乘积法则自然扩展为：

dy/dx = u·v·(dw/dx) + u·w·(dv/dx) + v·w·(du/dx)

Each term omits exactly one of the original functions and differentiates that omitted function. For n functions, there are n terms, each being the product of all but one function times the derivative of the omitted one.

每一项恰好省略一个原函数，并对该省略的函数求导。对于 n 个函数，共有 n 项，每一项是除一个函数外的所有函数与该被省略函数导数的乘积。

Example: y = x²·eˣ·cos x. Then dy/dx = x²·eˣ·(−sin x) + x²·cos x·eˣ + 2x·eˣ·cos x. Factorising gives eˣ multiplied by the remaining terms.

例如：y = x²·eˣ·cos x。则 dy/dx = x²·eˣ·(−sin x) + x²·cos x·eˣ + 2x·eˣ·cos x。提取公因式后得到 eˣ 与剩余项相乘。

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9. Combining Product and Quotient Rules | 乘积法则与商法则的结合

Some functions require both rules. For example, y = (x²·sin x) / (1 + x²). Let u = x²·sin x and v = 1 + x². First find du/dx using the product rule:

有些函数需要同时使用两个法则。例如，y = (x²·sin x) / (1 + x²)。令 u = x²·sin x，v = 1 + x²。先用乘积法则求 du/dx：

du/dx = x²·cos x + 2x·sin x

Then apply the quotient rule with dv/dx = 2x. This two-step approach breaks a complex problem into manageable pieces, reducing the chance of arithmetic errors.

然后用商法则，其中 dv/dx = 2x。这种两步法将复杂问题分解为可管理的部分，减少计算错误的机会。

dy/dx = [(1 + x²)(x²·cos x + 2x·sin x) − (x²·sin x)(2x)] / (1 + x²)²

This expression can be simplified further, but leaving it in this factored form often suffices for exam credit. Always check whether simplification is requested.

该表达式可以进一步化简，但保留这种因式分解形式通常足以获得考试分数。务必检查题目是否要求化简。

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10. Choosing Between Product and Quotient Rules | 在乘积法则与商法则之间选择

Sometimes a quotient can be rewritten as a product. For instance, y = 1/x can be written as x⁻¹ and differentiated using the power rule. Similarly, y = (x + 1)/x = 1 + 1/x = 1 + x⁻¹ avoids the quotient rule entirely.

有时商可以改写为乘积。例如，y = 1/x 可以写成 x⁻¹，直接使用幂法则求导。类似地，y = (x + 1)/x = 1 + 1/x = 1 + x⁻¹，完全避免了商法则。

Consider y = x/(x + 1). Simplifying to 1 − 1/(x + 1) and using the chain rule is often faster than applying the quotient rule directly. However, for functions like sin x/x or eˣ/(x² + 1), the quotient rule is usually the most straightforward method.

考虑 y = x/(x + 1)。化简为 1 − 1/(x + 1) 并使用链式法则通常比直接应用商法则更快。然而，对于像 sin x/x 或 eˣ/(x² + 1) 这样的函数，商法则通常是最直接的方法。

Key advice: simplify the function first if possible. A simple algebraic manipulation can turn a quotient-rule problem into a simple power-rule problem, saving time and reducing errors.

关键建议：如果可能，先化简函数。简单的代数变换可以将一个商法则问题转化为简单的幂法则问题，节省时间并减少错误。

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11. Common Mistakes and How to Avoid Them | 常见错误及避免方法

The most common mistake is confusing the sign in the quotient rule. Always remember: numerator is v·du/dx − u·dv/dx, not the reverse. A quick check: for y = 1/x, the quotient rule with u = 1, v = x gives (x·0 − 1·1)/x² = −1/x², which matches the power rule result.

最常见的错误是混淆商法则中的符号。始终记住：分子是 v·du/dx − u·dv/dx，而不是反过来。快速检验：对于 y = 1/x，用 u = 1，v = x 的商法则得到 (x·0 − 1·1)/x² = −1/x²，与幂法则结果一致。

Another error is forgetting to square the denominator in the quotient rule. The v² is essential. A useful habit is to write down the formula before substituting, ensuring all components are accounted for.

另一个错误是忘记在商法则中将分母平方。v² 是必不可少的。一个有用的习惯是在代入前先写出公式，确保所有组成部分都被考虑到。

Forgetting to apply the chain rule within the product rule is also common. For example, differentiating y = x²·sin(3x) correctly requires dv/dx = 3·cos(3x), not cos(3x) alone. Always check for inner functions.

在乘积法则中忘记应用链式法则也很常见。例如，对 y = x²·sin(3x) 求导，正确需要 dv/dx = 3·cos(3x)，而不是单独的 cos(3x)。始终检查是否存在内层函数。

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12. Practice Problems | 练习题

Try differentiating the following functions using the appropriate rules. Solutions are outlined briefly below.

尝试使用适当法则对以下函数求导。简要解题思路如下。

• y = (x³ + 1)·eˣ — Product rule: dy/dx = (x³ + 1)eˣ + 3x²eˣ.
• y = ln x / x — Quotient rule: dy/dx = (x·(1/x) − ln x)/x² = (1 − ln x)/x².
• y = x²·cos x·e⁻ˣ — Triple product rule: three terms; simplify by factoring x·e⁻ˣ.
• y = (x² + 1)/(x² − 1) — Quotient rule: dy/dx = [(x²−1)(2x) − (x²+1)(2x)]/(x²−1)² = −4x/(x²−1)².
• y = sin x·cos x — Product rule: dy/dx = sin x·(−sin x) + cos x·cos x = cos²x − sin²x = cos 2x.

At least one solution demonstrates a useful simplification: y = sin x·cos x can also be written as (1/2)sin 2x, whose derivative cos 2x is immediate. Recognising such identities can be a powerful shortcut.

至少有一个解展现了有用的化简：y = sin x·cos x 也可以写成 (1/2)sin 2x，其导数为 cos 2x，立即可得。识别这类恒等式是一种强大的捷径。

Regular practice with these rules builds fluency. Always verify your result by testing a simple value of x or by checking units/dimensions where applicable.

定期练习这些法则可以提升熟练度。始终通过测试一个简单的 x 值或在适用时检查量纲来验证结果。

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