Differentiation and Integration for Edexcel A-Level Maths | Edexcel A-Level 数学：微分与积分复习指南

📚 Differentiation and Integration for Edexcel A-Level Maths | Edexcel A-Level 数学：微分与积分复习指南

Welcome to this focused revision guide for the Edexcel A-Level Mathematics specification. Differentiation and integration form the core of pure mathematics and appear in both Paper 1 and Paper 2. In this article, we review essential techniques, typical exam questions, and common pitfalls to help you build confidence before your assessment.

欢迎阅读本 Edexcel A-Level 数学专项复习指南。微分与积分是纯数学的核心内容，在卷一和卷二中都会出现。本文回顾关键技巧、常见题型和易错点，帮助你在考试前建立信心。

1. Differentiation Fundamentals | 微分基础

Before applying advanced rules, make sure you can quickly recall the standard derivatives. For the Edexcel specification, you need to differentiate powers, exponentials, logarithms, and trigonometric functions fluently.

在应用高级法则之前，请确保你能快速回想起标准导数公式。Edexcel 考试要求你熟练地求幂函数、指数函数、对数函数和三角函数的导数。

d/dx (xⁿ) = n xⁿ⁻¹, d/dx (eˣ) = eˣ, d/dx (ln x) = 1/x

d/dx (sin x) = cos x, d/dx (cos x) = -sin x, d/dx (tan x) = sec²x

A common mistake is forgetting to rewrite roots and reciprocals as powers before differentiating. For example, √x becomes x^(1/2) and 1/x² becomes x^(-2), which makes the power rule easier to apply.

常见错误是在求导之前忘记将根式和倒数改写为幂的形式。例如，√x 应写成 x^(1/2)，1/x² 应写成 x^(-2)，这样使用幂法则会更方便。

Function	Derivative
xⁿ	n xⁿ⁻¹
eˣ	eˣ
ln x	1/x
sin x	cos x
cos x	-sin x
tan x	sec²x

2. Chain, Product & Quotient Rules | 链式法则、乘积法则与商法则

These three rules are tested in almost every Edexcel Pure Mathematics paper. You must be able to identify which rule fits a given function and apply it without hesitation.

这三条法则几乎在每份 Edexcel 纯数学试卷中都会考查。你必须能够识别给定函数适用哪条法则，并毫不犹豫地加以应用。

Chain rule: dy/dx = dy/du × du/dx

Product rule: d/dx (uv) = u’v + uv’

Quotient rule: d/dx (u/v) = (u’v – uv’) / v²

For example, to differentiate y = (3x² + 1)⁵, use the chain rule with u = 3x² + 1, giving dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴.

例如，求 y = (3x² + 1)⁵ 的导数，使用链式法则，令 u = 3x² + 1，得到 dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴。

For y = x² sin x, the product rule gives dy/dx = 2x sin x + x² cos x. For y = x / (x² + 1), use the quotient rule to obtain dy/dx = (1 – x²) / (x² + 1)².

对于 y = x² sin x，乘积法则给出 dy/dx = 2x sin x + x² cos x。对于 y = x / (x² + 1)，使用商法则得到 dy/dx = (1 – x²) / (x² + 1)²。

3. Parametric Differentiation | 参数方程求导

When a curve is defined by x = f(t) and y = g(t), the derivative dy/dx is found by dividing the derivative of y with respect to t by the derivative of x with respect to t.

当曲线由参数方程 x = f(t) 和 y = g(t) 给出时，dy/dx 等于 y 对 t 的导数除以 x 对 t 的导数。

dy/dx = (dy/dt) / (dx/dt)

This technique is especially useful for finding tangents and normals to parametric curves. Always ensure dx/dt is not zero at the point of interest.

该方法特别适用于求参数曲线的切线和法线。务必确保在相关点处 dx/dt 不为零。

Example: Given x = t² + 1 and y = t³ – 2t, we have dx/dt = 2t and dy/dt = 3t² – 2, so dy/dx = (3t² – 2) / (2t).

示例：已知 x = t² + 1，y = t³ – 2t，我们有 dx/dt = 2t，dy/dt = 3t² – 2，所以 dy/dx = (3t² – 2) / (2t)。

4. Implicit Differentiation | 隐函数求导

For equations where y is not explicitly given as a function of x, differentiate both sides with respect to x. Every time you differentiate a term involving y, use the chain rule and multiply by dy/dx.

对于没有显式给出 y 作为 x 函数的方程，对方程两边同时关于 x 求导。每次对含有 y 的项求导时，都要使用链式法则并乘以 dy/dx。

d/dx (y²) = 2y × dy/dx, d/dx (xy) = y + x dy/dx

After differentiating, collect all terms containing dy/dx on one side and factor out dy/dx to solve for it.

求导之后，把所有含 dy/dx 的项移到一边，提取 dy/dx 后解出它。

For example, differentiating x² + y² = 25 gives 2x + 2y dy/dx = 0, so dy/dx = -x/y. This is a common Edexcel exam question type.

例如，对 x² + y² = 25 求导得到 2x + 2y dy/dx = 0，因此 dy/dx = -x/y。这是 Edexcel 常见的考题类型。

5. Applications of Derivatives | 导数的应用

Edexcel regularly tests the use of derivatives to find equations of tangents and normals, stationary points, increasing and decreasing intervals, and points of inflection.

Edexcel 经常考查利用导数求切线和法线方程、驻点、单调区间以及拐点。

Tangent equation: y – y₁ = m(x – x₁), where m = dy/dx at the point.

切线方程：y – y₁ = m(x – x₁)，其中 m 为该点处的 dy/dx。
Stationary points occur where dy/dx = 0. Use the second derivative test or sign change test to classify maxima and minima.

驻点出现在 dy/dx = 0 处。使用二阶导数检验或符号变化检验来区分极大值和极小值。
For optimisation problems, model the quantity to be maximised or minimised as a function of one variable, differentiate, and solve.

对于优化问题，将需要最大化或最小化的量表示为单变量函数，求导并解方程。

A typical question asks: find the equation of the normal to y = x³ – 3x at x = 2. First compute dy/dx = 3x² – 3, then at x = 2 the gradient is 9, so the normal gradient is -1/9.

典型题目：求 y = x³ – 3x 在 x = 2 处法线方程。先算 dy/dx = 3x² – 3，在 x = 2 处斜率为 9，因此法线斜率为 -1/9。

6. Integration Fundamentals | 积分基础

Integration is the reverse of differentiation. For the Edexcel A-Level, you must know the fundamental integrals and always include the constant of integration for indefinite integrals.

积分是微分的逆运算。在 Edexcel A-Level 中，你必须掌握基本积分公式，并且不定积分一定要加上积分常数。

∫xⁿ dx = xⁿ⁺¹/(n+1) + c, n ≠ -1

∫1/x dx = ln|x| + c, ∫eˣ dx = eˣ + c

∫cos x dx = sin x + c, ∫sin x dx = -cos x + c

Remember that you can often check your answer by differentiating it. Also, rewrite terms like 1/x² as x⁻² before integrating.

请记住，你可以通过求导来检验积分结果。另外，在积分前把 1/x² 等项改写为 x⁻² 会更容易计算。

7. Integration by Substitution | 换元积分法

The substitution method is the reverse of the chain rule. Choose u as an inner function, then express du and replace all x terms with u terms before integrating.

换元法是链式法则的反向应用。选择一个内层函数设为 u，然后表示 du，并在积分前把所有含 x 的项替换为含 u 的项。

∫f(g(x)) g'(x) dx = ∫f(u) du

For definite integrals, you must change the limits from x-values to u-values. After integration, substitute back to the original variable for indefinite integrals.

对于定积分，必须把积分限从 x 值转换为 u 值。对于不定积分，积分后要回代到原变量。

Example: To evaluate ∫x√(x² + 1) dx, let u = x² + 1, so du = 2x dx and the integral becomes (1/2) ∫u^(1/2) du = (1/3)(x² + 1)^(3/2) + c.

示例：计算 ∫x√(x² + 1) dx，令 u = x² + 1，则 du = 2x dx，原积分变为 (1/2) ∫u^(1/2) du = (1/3)(x² + 1)^(3/2) + c。

8. Integration by Parts | 分部积分法

Integration by parts is used for products of functions, especially when one part becomes simpler upon differentiation. The formula comes from the product rule for derivatives.

分部积分法用于函数乘积的积分，特别是当其中一个函数求导后变得更简单时。该公式由乘积法则逆向推出。

∫u dv = uv – ∫v du

Use the LIATE order to choose u: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. In many A-Level questions, u is a polynomial or ln x, and dv is the remaining part.

使用 LIATE 顺序选择 u：对数函数、反三角函数、代数函数、三角函数、指数函数。在 A-Level 题目中，u 通常是多项式或 ln x，dv 是剩余部分。

Example: To integrate ∫x eˣ dx, let u = x and dv = eˣ dx. Then du = dx and v = eˣ, giving ∫x eˣ dx = x eˣ – ∫eˣ dx = x eˣ – eˣ + c.

示例：积分 ∫x eˣ dx，令 u = x，dv = eˣ dx，则 du = dx，v = eˣ，得到 ∫x eˣ dx = x eˣ – ∫eˣ dx = x eˣ – eˣ + c。

9. Definite Integrals and Area | 定积分与面积

A definite integral gives the signed area between a curve and the x-axis. If the curve crosses the x-axis within the interval, you must split the integral and take absolute values to find the total area.

定积分表示曲线与 x 轴之间带符号的面积。如果曲线在区间内穿过 x 轴，你必须拆分积分并取绝对值来求总面积。

Area = ∫ₐᵇ f(x) dx, if f(x) ≥ 0 on [a, b]

To find the area between two curves, integrate the top function minus the bottom function over the interval where they bound a region.

求两条曲线之间的面积时，在它们围成区域的区间上对“上方函数减下方函数”进行积分。

Example: Find the area under y = 4 – x² between x = -1 and x = 1. Since the curve is above the x-axis throughout, Area = ∫₋₁¹ (4 – x²) dx = [4x – x³/3]₋₁¹ = 22/3.

示例：求 y = 4 – x² 在 x = -1 到 x = 1 之间的面积。由于曲线始终在 x 轴上方，面积 = ∫₋₁¹ (4 – x²) dx = [4x – x³/3]₋₁¹ = 22/3。

10. Differential Equations | 微分方程

In Edexcel Pure Mathematics, you will be asked to solve first-order separable differential equations. The method is to separate the variables so that all y terms are on one side and all x terms on the other.

在 Edexcel 纯数学中，你需要求解一阶可分离变量微分方程。方法是分离变量，使所有含 y 的项在一侧，所有含 x 的项在另一侧。

dy/dx = g(x) h(y) → ∫1/h(y) dy = ∫g(x) dx

After integration, write the general solution and then use any given initial condition to find the particular solution.

积分后写出通解，然后利用给定的初始条件求出特解。

Example: Solve dy/dx = 2xy with y(0) = 3. Separating gives ∫1/y dy = ∫2x dx, so ln|y| = x² + c. Using y(0) = 3 gives c = ln 3, hence y = 3e^(x²).

示例：解 dy/dx = 2xy，y(0) = 3。分离变量得 ∫1/y dy = ∫2x dx，所以 ln|y| = x² + c。利用 y(0) = 3 得 c = ln 3，因此 y = 3e^(x²)。

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Differentiation and Integration for Edexcel A-Level Maths | Edexcel A-Level 数学：微分与积分复习指南