Mastering Differentiation for Edexcel A-Level Pure Mathematics | 精通 Edexcel A-Level 纯数学微分法

📚 Mastering Differentiation for Edexcel A-Level Pure Mathematics | 精通 Edexcel A-Level 纯数学微分法

Differentiation is one of the most important topics in the Edexcel A-Level Pure Mathematics specification. It appears not only as a standalone skill but also in applied questions on optimisation, curve sketching, tangents, normals and rates of change. This revision guide covers first principles, the core differentiation rules, key applications and the common mistakes that can cost marks in the exam.

微分是 Edexcel A-Level 纯数学大纲中最重要的主题之一。它不仅作为独立技能出现，还出现在优化、曲线作图、切线、法线和变化率等应用问题中。本复习指南涵盖第一性原理、核心微分法则、关键应用以及考试中可能导致失分的常见错误。

1. Differentiation from First Principles | 从第一性原理出发的微分

In Edexcel A-Level Pure Mathematics, differentiation is introduced through the formal limit definition. The derivative of a function f at a point x is given by f'(x) = lim (h → 0) [f(x + h) – f(x)] / h. This definition gives the gradient of the tangent to the curve y = f(x). For example, for f(x) = x², expanding f(x + h) = (x + h)² = x² + 2xh + h² gives [f(x + h) – f(x)] / h = 2x + h, so as h → 0, f'(x) = 2x. This process reinforces the idea that differentiation measures instantaneous rate of change.

在 Edexcel A-Level 纯数学中，微分是通过正式的极限定义引入的。函数 f 在点 x 处的导数定义为 f'(x) = lim (h → 0) [f(x + h) – f(x)] / h。这个定义给出了曲线 y = f(x) 的切线斜率。例如，对于 f(x) = x²，展开 f(x + h) = (x + h)² = x² + 2xh + h²，得到 [f(x + h) – f(x)] / h = 2x + h，因此当 h → 0 时，f'(x) = 2x。这一过程强化了微分衡量瞬时变化率的概念。

2. Basic Differentiation Rules | 基本微分法则

The power rule states that if y = xⁿ, then dy/dx = n xⁿ⁻¹ for any real constant n. Edexcel candidates must also know the derivative of a constant is zero, the constant multiple rule d/dx [a f(x)] = a f'(x), and the sum rule d/dx [f(x) ± g(x)] = f'(x) ± g'(x). Familiar results include d/dx (sin x) = cos x, d/dx (cos x) = -sin x, d/dx (eˣ) = eˣ, and d/dx (ln x) = 1/x for x > 0.

幂法则指出，如果 y = xⁿ，那么 dy/dx = n xⁿ⁻¹，其中 n 为任意实常数。Edexcel 考生还必须掌握常数的导数为零、常数倍法则 d/dx [a f(x)] = a f'(x) 以及和差法则 d/dx [f(x) ± g(x)] = f'(x) ± g'(x)。常见结果包括 d/dx (sin x) = cos x、d/dx (cos x) = -sin x、d/dx (eˣ) = eˣ，以及 d/dx (ln x) = 1/x（x > 0）。

3. Chain Rule | 链式法则

The chain rule is used to differentiate composite functions. If y = g(u) and u = h(x), then dy/dx = dy/du × du/dx. In function notation, d/dx [g(h(x))] = g'(h(x)) h'(x). For example, for y = (3x² + 5)⁴, let u = 3x² + 5, so y = u⁴. Then dy/du = 4u³ and du/dx = 6x, giving dy/dx = 4(3x² + 5)³ × 6x = 24x(3x² + 5)³. This rule appears frequently in differentiating trigonometric, exponential and logarithmic functions of linear expressions such as sin(2x + 1) or e⁵ˣ.

链式法则用于对复合函数求导。如果 y = g(u) 且 u = h(x)，则 dy/dx = dy/du × du/dx。用函数记号表示为 d/dx [g(h(x))] = g'(h(x)) h'(x)。例如，对于 y = (3x² + 5)⁴，令 u = 3x² + 5，则 y = u⁴。于是 dy/du = 4u³，du/dx = 6x，得到 dy/dx = 4(3x² + 5)³ × 6x = 24x(3x² + 5)³。该法则在求线性表达式的三角、指数和对数函数的导数时经常出现，例如 sin(2x + 1) 或 e⁵ˣ。

4. Product Rule | 乘积法则

When differentiating a product of two functions y = u(x) v(x), Edexcel uses the product rule: dy/dx = u dv/dx + v du/dx. A memorable form is “u v’ + v u’”. For instance, if y = x² sin x, set u = x² and v = sin x. Then u’ = 2x and v’ = cos x, so dy/dx = x² cos x + 2x sin x. It is important to factorise the result where possible, as this helps with finding stationary points later. Do not multiply out the two parts unless the question specifically asks for a simplified expanded form.

当对两个函数的乘积 y = u(x) v(x) 求导时，Edexcel 使用乘积法则：dy/dx = u dv/dx + v du/dx。一个易记形式是 “u v’ + v u’”。例如，若 y = x² sin x，设 u = x²，v = sin x。则 u’ = 2x，v’ = cos x，所以 dy/dx = x² cos x + 2x sin x。在可能的情况下对结果进行因式分解很重要，因为这将有助于后续求驻点。除非题目明确要求展开简化形式，否则不要将两部分乘开。

5. Quotient Rule | 商法则

The quotient rule differentiates y = u/v, where u and v are functions of x. The formula is dy/dx = (v du/dx – u dv/dx) / v². Edexcel candidates often remember this as “v du minus u dv over v squared”. For example, for y = x / (x² + 1), let u = x and v = x² + 1. Then u’ = 1, v’ = 2x, giving dy/dx = [(x² + 1) × 1 – x × 2x] / (x² + 1)² = (1 – x²) / (x² + 1)². Watch for sign errors: it is always the denominator derivative subtracted, not added.

商法则用于对 y = u/v 求导，其中 u 和 v 都是 x 的函数。公式为 dy/dx = (v du/dx – u dv/dx) / v²。Edexcel 考生通常将其记为“v du 减 u dv 除以 v 的平方”。例如，对于 y = x / (x² + 1)，设 u = x，v = x² + 1。则 u’ = 1，v’ = 2x，得到 dy/dx = [(x² + 1) × 1 – x × 2x] / (x² + 1)² = (1 – x²) / (x² + 1)²。注意符号错误：始终是减去分母的导数，而不是相加。

6. Tangents and Normals | 切线与法线

Differentiation allows you to find the equation of a tangent or normal to a curve at a given point. The gradient of the tangent at x = a is m = dy/dx evaluated at x = a. Since a normal is perpendicular to the tangent, its gradient is -1/m, provided m ≠ 0. If m = 0, the tangent is horizontal and the normal is vertical. The equation of a straight line is y – y₁ = m (x – x₁), where (x₁, y₁) lies on the curve. Always calculate y₁ by substituting into the original equation of the curve, not the derivative.

微分使你能够求出曲线在给定点处的切线和法线方程。在 x = a 处切线的斜率为 m = dy/dx 在 x = a 处的值。由于法线与切线垂直，其斜率为 -1/m（前提是 m ≠ 0）。如果 m = 0，则切线为水平线，法线为竖直线。直线方程为 y – y₁ = m (x – x₁)，其中 (x₁, y₁) 位于曲线上。始终通过代入原曲线方程来计算 y₁，而不是代入导数方程。

7. Stationary Points and the Second Derivative Test | 驻点与二阶导数判别法

Stationary points occur where dy/dx = 0. To determine their nature, Edexcel candidates can use the second derivative test: compute d²y/dx² by differentiating dy/dx again. If d²y/dx² > 0 at the stationary point, the point is a local minimum; if d²y/dx² < 0, it is a local maximum; if d²y/dx² = 0, the test is inconclusive and you should examine the sign of dy/dx on either side. For example, for y = x³ - 3x, dy/dx = 3x² - 3. Setting 3x² - 3 = 0 gives x = ±1. At x = 1, d²y/dx² = 6 > 0, so a local minimum; at x = -1, d²y/dx² = -6 < 0, so a local maximum.

驻点出现在 dy/dx = 0 的位置。为了判断其性质，Edexcel 考生可以使用二阶导数判别法：通过对 dy/dx 再次求导来计算 d²y/dx²。如果在驻点处 d²y/dx² > 0，则该点为局部极小值；如果 d²y/dx² < 0，则为局部极大值；如果 d²y/dx² = 0，则该判别法无法确定，你需要考察 dy/dx 在两侧的符号。例如，对于 y = x³ - 3x，dy/dx = 3x² - 3。令 3x² - 3 = 0 得 x = ±1。在 x = 1 处，d²y/dx² = 6 > 0，因此为局部极小值；在 x = -1 处，d²y/dx² = -6 < 0，因此为局部极大值。

8. Increasing and Decreasing Functions | 增函数与减函数

A function is increasing on an interval if dy/dx > 0 throughout that interval, and decreasing if dy/dx < 0. Edexcel questions often ask you to show that a function is increasing for all real x, or to find the interval on which it is decreasing. This requires solving inequalities involving dy/dx. For example, for f(x) = x³ - 3x + 2, f'(x) = 3x² - 3 = 3(x - 1)(x + 1). So f'(x) > 0 when x < -1 or x > 1, and f'(x) < 0 when -1 < x < 1. Hence f is increasing on (-∞, -1) and (1, ∞), and decreasing on (-1, 1). Use a sign diagram or a sketch of f'(x) to justify your answer.

如果在一个区间上 dy/dx > 0，则函数在该区间上递增；如果 dy/dx < 0，则函数递减。Edexcel 题目经常要求你证明函数对所有实数 x 都递增，或找出其递减区间。这需要求解涉及 dy/dx 的不等式。例如，对于 f(x) = x³ - 3x + 2，f'(x) = 3x² - 3 = 3(x - 1)(x + 1)。因此当 x < -1 或 x > 1 时 f'(x) > 0，当 -1 < x < 1 时 f'(x) < 0。所以 f 在 (-∞, -1) 和 (1, ∞) 上递增，在 (-1, 1) 上递减。使用符号图或 f'(x) 的草图来证明你的答案。

9. Modelling with Differentiation | 用微分进行建模

Differentiation is frequently tested in applied contexts such as optimisation. In these problems, a quantity such as area, volume, profit or cost is expressed as a function of one variable. The maximum or minimum value is found by setting the first derivative equal to zero and verifying the nature of the stationary point. For example, a company’s profit P thousand pounds from producing x hundred units might be modelled by P = 60x – 3x² – 100. Then dP/dx = 60 – 6x, so dP/dx = 0 gives x = 10. Since d²P/dx² = -6 < 0, profit is maximised at x = 10, giving P = 200 thousand pounds. Remember to check that any stationary point lies within the feasible domain of the model.

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）