Edexcel A-Level Pure Maths: Differentiation – Rules, Stationary Points and Tangents | 爱德思 A-Level 纯数学：微分——法则、驻点与切线

📚 Edexcel A-Level Pure Maths: Differentiation – Rules, Stationary Points and Tangents | 爱德思 A-Level 纯数学：微分——法则、驻点与切线

Differentiation is one of the most heavily examined topics in Edexcel A-Level Mathematics. It underpins questions on gradients, tangents, normals, stationary points, optimisation and curve sketching. This revision article covers first principles, standard rules, higher derivatives, stationary points and the equations of tangents and normals. Work through every example and pay close attention to the Edexcel command words ‘show that’, ‘find’ and ‘justify’.

微分是爱德思 A-Level 数学中考查频率最高的主题之一。它是梯度、切线、法线、驻点、优化和曲线作图等问题的基础。本复习文章涵盖第一原理、标准求导法则、高阶导数、驻点以及切线和法线方程。请务必完成每一道例题，并注意 Edexcel 题干中 ‘show that’、’find’ 和 ‘justify’ 等指令词。

1. First Principles | 第一原理

Differentiation from first principles uses the limit of the gradient of a chord as the interval tends to zero. For a function f(x), the derivative f'(x) is defined by:

第一原理求导使用弦的斜率在区间趋近于零时的极限。对于函数 f(x)，导数 f'(x) 定义为：

f'(x) = limₕ→₀ [f(x+h) − f(x)] / h

Edexcel questions often ask you to prove the derivative of x² or x³ from first principles using this limit. You must show all steps, including expanding brackets and cancelling h before taking the limit. For f(x) = x², write:

Edexcel 题目常要求用该极限证明 x² 或 x³ 的导数。你必须展示所有步骤，包括展开括号、约去 h，然后再取极限。对于 f(x) = x²，写出：

f'(x) = limₕ→₀ [(x+h)² − x²] / h = limₕ→₀ [x²+2xh+h²−x²] / h = limₕ→₀ (2x+h) = 2x

This gives the standard result d/dx (x²) = 2x. Marks are awarded for correct expansion, simplification and final conclusion.

由此得到标准结果 d/dx (x²) = 2x。得分点在于正确展开、化简以及最终结论。

2. Standard Derivatives and Rules | 标准导数与求导法则

The standard results you must know from memory include the derivatives of powers, trigonometric functions, exponentials and logarithms. Linear combinations can always be differentiated term by term: d/dx [a f(x) + b g(x)] = a f'(x) + b g'(x).

你必须从记忆中掌握的标准结果包括幂函数、三角函数、指数函数和对数函数的导数。线性组合始终可以逐项求导：d/dx [a f(x) + b g(x)] = a f'(x) + b g'(x)。