A-Level Maths: The Quotient Rule – Methods & Tips | A-Level 数学：商法则求导方法与技巧

📚 A-Level Maths: The Quotient Rule – Methods & Tips | A-Level 数学：商法则求导方法与技巧

In A-Level Mathematics, differentiation is one of the most important skills you will develop. The quotient rule, also known as the rule for differentiating a fraction, allows you to differentiate functions of the form y = u/v, where both u and v are differentiable functions of x. Mastery of this rule is essential for handling rational functions, trigonometric expressions, and many problems in pure mathematics.

在 A-Level 数学中，求导是你要掌握的最重要技能之一。商法则（也称分数求导法则）用于对形如 y = u/v 的函数求导，其中 u 和 v 都是关于 x 的可导函数。熟练运用商法则，对于处理有理函数、三角表达式以及纯数学中的诸多问题都至关重要。

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1. The Quotient Rule Formula | 商法则公式

If y = u/v, where u and v are differentiable functions of x, then the derivative of y with respect to x is given by:

若 y = u/v，其中 u 和 v 都是关于 x 的可导函数，则 y 对 x 的导数为：

dy/dx = (v · du/dx − u · dv/dx) / v²

Another common notation is: if f(x) = g(x)/h(x), then f'(x) = (h(x)g'(x) − g(x)h'(x)) / [h(x)]². This form makes the order of the terms clear: high d low minus low d high, all over low squared.

另一种常见记法是：若 f(x) = g(x)/h(x)，则 f'(x) = (h(x)g'(x) − g(x)h'(x)) / [h(x)]²。这种形式清楚地体现了顺序：低导高减高导低，除以低平方。

Since division by zero is undefined, we require v ≠ 0 (or h(x) ≠ 0) for the quotient rule to be valid.

由于除以零没有意义，商法则成立的前提是 v ≠ 0（或 h(x) ≠ 0）。

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2. Deriving the Quotient Rule | 从乘积法则推导商法则

The quotient rule is not a completely new rule; it can be derived from the product rule and the chain rule. Write y = u · v⁻¹. Then, applying the product rule:

商法则并不是一条全新的法则；它可以从乘积法则和链式法则推导出来。将 y 写成 y = u · v⁻¹，然后使用乘积法则：

dy/dx = du/dx · v⁻¹ + u · (−1)v⁻² · dv/dx = (v · du/dx − u · dv/dx) / v²

This derivation is a useful check: if you ever forget the quotient rule, you can recover it by rewriting the quotient as a product and using the product rule.

这一推导很有用：如果你忘记了商法则，可以先将商改写为乘积，再使用乘积法则来推导。

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3. When to Use the Quotient Rule | 何时使用商法则

The quotient rule is most useful when the denominator is not a single power of x and cannot be easily split into separate terms. For example, y = x / (x² + 1) is best differentiated using the quotient rule because the denominator is a sum of terms.

当分母不是 x 的单一幂次、且无法轻松拆分成多项时，商法则最为适用。例如，y = x / (x² + 1) 的分母是多项式，最好使用商法则求导。

However, for functions like y = (x² + 1)/x, you can rewrite as y = x + 1/x and differentiate term-by-term without the quotient rule. Always ask: can I simplify first?

但对于 y = (x² + 1)/x 这样的函数，可改写为 y = x + 1/x，逐项求导即可，无需使用商法则。因此，先问自己：能不能先化简？

• Use the quotient rule when both u and v are more complex than simple powers.

  当 u 和 v 都比简单幂函数更复杂时，使用商法则。

• Consider alternative methods when the quotient can be simplified algebraically or by using index laws.

  当商可以通过代数方法或指数法则化简时，考虑替代方法。

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4. Worked Example 1 | 例题 1：分步解析

Differentiate y = x² / (x + 1).

求 y = x² / (x + 1) 的导数。

Let u = x² and v = x + 1. Then u’ = 2x and v’ = 1. Applying the quotient rule:

令 u = x²，v = x + 1，则 u’ = 2x，v’ = 1。应用商法则：

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

Simplify the numerator: (2x² + 2x − x²) = x² + 2x = x(x + 2). Therefore:

化简分子：(2x² + 2x − x²) = x² + 2x = x(x + 2)。因此：

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

The final answer is factorised, which makes it easy to locate stationary points at x = 0 and x = -2.

最终答案已因式分解，便于找到驻点 x = 0 和 x = -2。

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5. Worked Example 2: Trigonometric Functions | 例题 2：含三角函数的商

Differentiate y = sin x / x.

求 y = sin x / x 的导数。

Let u = sin x and v = x. Then u’ = cos x and v’ = 1. Applying the quotient rule:

令 u = sin x，v = x，则 u’ = cos x，v’ = 1。应用商法则：

dy/dx = (x cos x − sin x) / x²

As a classic application, consider tan x = sin x / cos x. Using the quotient rule gives:

一个经典应用是 tan x = sin x / cos x。使用商法则可得：

d/dx (tan x) = (cos x·cos x − sin x·(−sin x)) / cos²x = (cos²x + sin²x) / cos²x = 1 / cos²x = sec²x

This is one of the most important trigonometric derivative results, and the quotient rule gives it directly.

这是最重要的三角导数公式之一，商法则可以很直接地推导出它。

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6. Worked Example 3: Exponential and Logarithmic Functions | 例题 3：含指数和对数的函数

Differentiate y = eˣ / x.

求 y = eˣ / x 的导数。

Let u = eˣ and v = x. Then u’ = eˣ and v’ = 1. Applying the quotient rule:

令 u = eˣ，v = x，则 u’ = eˣ，v’ = 1。应用商法则：

dy/dx = (x eˣ − eˣ) / x² = eˣ(x − 1) / x²

Now differentiate y = ln x / x².

再求 y = ln x / x² 的导数。

Let u = ln x and v = x². Then u’ = 1/x and v’ = 2x. Applying the quotient rule:

令 u = ln x，v = x²，则 u’ = 1/x，v’ = 2x。应用商法则：

dy/dx = [x²·(1/x) − (ln x)(2x)] / x⁴ = (x − 2x ln x) / x⁴ = (1 − 2 ln x) / x³Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

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