📚 AQA A-Level Maths Question 3: Differentiation, Stationary Points and Tangents | AQA A-Level 数学第3题：微分、驻点与切线
In AQA A-Level Mathematics, Paper 1 Question 3 frequently tests your ability to differentiate polynomial and simple rational functions, find stationary points, classify their nature, and apply these ideas to tangents and normals. This revision guide explains the core techniques, common mistakes and exam-ready strategies to help you secure full marks on this question type.
在 AQA A-Level 数学卷一第3题中，常考查你对多项式和简单有理函数求导、寻找驻点并判断其类型，以及将这些思想应用于切线和法线的能力。本复习指南将讲解核心技巧、常见错误和应试策略，帮助你在这类题目中拿到满分。
1. What AQA Question 3 Usually Tests | AQA 第3题通常考查什么
A typical AQA Paper 1 Question 3 focuses on calculus, especially differentiation. You may be asked to find dy/dx, locate stationary points and classify their nature, or find equations of tangents and normals. Some questions also ask about increasing or decreasing intervals and points of inflection.
典型的 AQA 试卷一第3题以微积分为主，尤其是求导。题目可能要求你求 dy/dx、找出驻点并判断其类型，或者求切线和法线的方程。有些题目还会考查递增或递减区间以及拐点。
The question is often worth 6 to 10 marks and combines algebraic manipulation with graphical understanding. It may also include a simple real-world context such as a curve modelling the profile of a ramp or the shape of a container.
这类题目通常分值为6到10分，综合考查代数运算和图形理解。有时还会给出简单的实际背景，例如用曲线模拟坡道轮廓或容器的形状。
- Key themes: differentiation rules, stationary points, tangents, normals, increasing and decreasing intervals, points of inflection.
- 核心主题：求导法则、驻点、切线、法线、递增与递减区间、拐点。
2. Core Differentiation Rules You Must Know | 必须掌握的核心求导法则
ional number n. This rule also applies to negative and fractional powers, so you can differentiate expressions like 1/x² and √x.
首先要熟记 x 的幂函数求导公式：若 y = xⁿ，则 dy/dx = nxⁿ⁻¹，其中 n 为任意有理数。该法则同样适用于负指数和分数指数，因此你可以对 1/x² 和 √x 这样的式子求导。
dy/dx (xⁿ) = nxⁿ⁻¹
ines with zero gradient. For example, y = 4x³ - 6x² + 2x - 9 gives dy/dx = 12x² - 12x + 2.
对于和与差，可以逐项求导。常数项求导为零，因为常数对应水平直线，梯度为零。例如，y = 4x³ - 6x² + 2x - 9 的导数为 dy/dx = 12x² - 12x + 2。
Negative and fractional powers follow the same rule, so rewrite 1/x as x⁻¹ and √x fferentiating. For instance, y = 1/x² can be written as y = x⁻², so dy/dx = -2x⁻³ = -2/x³.
负指数和分数指数遵循相同法则，因此在求导前应先把 1/x 写成 x⁻¹，把 √x 写成 x¹⁄²。例如，y = 1/x² 可以写成 y = x⁻²，因此 dy/dx = -2x⁻³ = -2/x³。
3. Finding Stationary Points Step by Step | 分步求驻点
ng equation for x. This equation may be linear, quadratic or require simple factorisation.
驻点出现在曲线梯度为零的位置，因此令 dy/dx = 0，然后解方程求出 x。该方程可能是线性方程、二次方程，或者需要简单因式分解。
nding y-coordinate. This gives the full coordinates of the stationary point. Do not stop at finding x alone.
求出每个 x 值后，将其代回原方程 y = f(x)，得到对应的 y 坐标。这样就能得到驻点的完整坐标。不要只求出 x 就停止。
x = 1 or x = 3. Substituting into y gives two stationary points: (1, y(1)) and (3, y(3)).
例如，若 dy/dx = 3x² - 12x + 9，令其等于零得到 3(x² - 4x + 3) = 0，因此 x = 1 或 x = 3。代回 y 中可得到两个驻点：(1, y(1)) 和 (3, y(3))。
Always write the stationary points as coordinates in the form (x, y), because AQA mark schemes usually require both coordinates for full marks.
请始终将驻点写成 (x, y) 的坐标形式，因为 AQA 的评分标准通常要求两个坐标都给全才能得满分。
4. Classifying Stationary Points | 判断驻点类型
To classify a stationary point as a maximum, minimum or point of inflection, use either the second derivative test or the sign-change method.
要判断驻点是极大点、极小点还是拐点，可以使用二阶导数检验法或一阶导数符号变化法。
dx² = 0, the test is inconclusive and you should examine the sign of dy/dx on either side.
二阶导数检验法：若驻点处 d²y/dx² > 0，则为局部极小点；若 d²y/dx² < 0，则为局部极大点；若 d²y/dx² = 0，该检验无法判断，需要观察两侧 dy/dx 的符号。
| d²y/dx² value | Nature of stationary point |
|---|---|
| d²y/dx² > 0 | Local minimum |
| d²y/dx² < 0 | Local maximum |
| d²y/dx² = 0 | Inconclusive; test sign of dy/dx |
中文对照：d²y/dx² 值对应驻点类型，大于0为局部极小，小于0为局部极大，等于0时需用 dy/dx 符号变化来判断。
The sign-change method looks at dy/dx just to the left and right of the stationary point. If the gradient changes from negative to positive, the point is a minimum; if from positive to negative, it is a maximum; if the sign does not change, it is a point of inflection.
符号变化法观察驻点左侧和右侧附近 dy/dx 的符号。如果梯度由负变正，该点为极小点；如果由正变负，该点为极大点；如果符号没有变化，则为拐点。
5. Tangents and Normals Linked to Question 3 | 第3题中涉及的切线与法线
Question 3 may ask you to find the equation of the tangent at a given point. The tangent has gradient equal to dy/dx evaluated at that point.
第3题也可能要求你求某一点处切线的方程。切线的斜率等于在该点处计算出的 dy/dx 值。
u can then rearrange the equation into the form y = mx + c or ax + by + c = 0 if required.
使用直线方程 y - y₁ = m(x - x₁)，其中 m 为斜率，(x₁, y₁) 为曲线上已知点的坐标。然后可以根据题目要求将方程整理成 y = mx + c 或 ax + by + c = 0 的形式。
The normal is perpendicular to the tangent, so its gradient is the negative reciprocal: mnormal = -1 / m_tangent. Then apply the same straight-line formula.
法线与切线垂直，因此其斜率为切线斜率的负倒数：mnormal = -1 / m_tangent。然后应用相同的直线方程公式。
dx = 3x² - 3, so at x = 2 the gradient is 9. If y(2) = 2, the tangent is y - 2 = 9(x - 2).
例如，若 y = x³ - 3x，要求 x = 2 处的切线，先求 dy/dx = 3x² - 3，则 x = 2 时斜率为 9。若 y(2) = 2，则切线方程为 y - 2 = 9(x - 2)。
6. Increasing and Decreasing Functions | 递增与递减函数
A function is increasing where dy/dx > 0 and decreasing where dy/dx < 0. AQA Question 3 may ask you to find the interval on which a curve is increasing or decreasing.
当 dy/dx > 0 时函数递增，当 dy/dx < 0 时函数递减。AQA 第3题可能会要求你求曲线递增或递减
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