📚 Mastering Differentiation for Edexcel A-Level Pure Maths | 精通 Edexcel A-Level 纯数微分
Differentiation is one of the most heavily examined topics across Edexcel A-Level Mathematics. It underpins tangent equations, stationary-point analysis, optimisation, kinematic modelling and connected rates of change, and it reappears inside almost every later topic in Pure Mathematics, Mechanics and even Statistics. This tutorial consolidates the techniques, notation and exam language you need in order to solve Pure Mathematics questions confidently.
微分是 Edexcel A-Level 数学中考查频率最高的主题之一。它是切线方程、驻点分析、最优化、运动学建模以及相关变化率问题的基础，并且在纯数、力学乃至统计的后续内容中反复出现。本教程将帮助你巩固相关技巧、记号与考试表述，使你能够自信地解答纯数试题。
1. First Principles and the Derivative as a Limit | 第一性原理与作为极限的导数
At AS Level, Edexcel often asks you to differentiate a simple function from first principles using the limit definition of the gradient of a chord. The derivative is defined by
在 AS 阶段，Edexcel 经常要求考生用弦的斜率极限定义，从第一性原理出发求一个简单函数的导数。导数定义为
f′(x) = lim(h→0) [ f(x + h) − f(x) ] ÷ h
You must show the full expansion, simplify the numerator, cancel h, and then let h tend to 0. Do not simply write the standard result, because the mark scheme awards method marks for each visible stage of the algebra.
你必须展示完整展开、化简分子、约去 h，再令 h 趋向于 0，而不是直接写出标准结果，因为评分标准会为每一个可见的代数步骤给出方法分。
² = x² + 2xh + h², subtract x², divide by h, and obtain 2x + h. As h → 0, the limit is 2x.
例如，若 f(x) = x²，展开 (x + h)² = x² + 2xh + h²，减去 x²，除以 h，得到 2x + h。当 h → 0 时，极限为 2x。
fractions over a common denominator before cancelling. For f(x) = 1/x the limit is −1/x².
同样的结构适用于 f(x) = x³，此时分子变为 3x²h + 3xh² + h³；也适用于 f(x) = 1/x，但你需要先通分再约分。对 f(x) = 1/x，极限结果为 −1/x²。
Typical exam wording includes: differentiate from first principles, find the gradient of the chord and let the width tend to zero, or show that the derivative of a given function is a stated expression.
常见考法表述包括：用第一性原理求导、求弦的斜率并令其宽度趋向于零，或证明给定函数的导数为题中所给的表达式。
2. Standard Results and the Power Rule | 标准结果与幂函数法则
Edexcel assumes fluency with standard derivatives of powers, exponentials, logarithms and trigonometric functions. The power rule states that multiplying by the old power and reducing the power by one gives the derivative of xⁿ.
Edexcel 要求考生熟练掌握幂函数、指数函数、对数函数和三角函数的标准化导数。幂函数法则指出，将原幂次乘到前面，再把幂次减一，即可得到 xⁿ 的导数。
d(xⁿ)/dx = n xⁿ⁻¹
| Function | Derivative |
|---|---|
| xⁿ | n xⁿ⁻¹ |
| eˣ | eˣ |
| eᵏˣ | k eᵏˣ |
| ln x | 1 ÷ x |
| sin x | cos x |
| cos x | −sin x |
| tan x | sec²x |
Remember that constants differentiate to zero, that a constant multiplying a function stays as a multiplier, and that sums and differences are differentiated term by term.
记住：常数求导为零；与函数相乘的常数保持不变；和与差可以逐项求导。
These standard results must be memorised, because Edexcel does not print a formula booklet entry for them at every stage of the question. Faster recall means fewer lost method marks under time pressure.
这些标准结果必须背熟，因为 Edexcel 在题目的每一个环节并不会都给出公式表条目。回忆越快，在时间压力下丢掉的方法分就越少。
3. Polynomials, Constants and Negative or Fractional Powers | 多项式、常数与负幂或分数幂
Most AS questions begin with rewriting the expression in index form. Roots and reciprocals must be converted before you differentiate.
大多数 AS 题目都要求先把表达式改写成指数形式。根式与倒数必须在使用幂法则之前先转换。
- √x = x^½, and 1/x² = x⁻², and 1/√x = x−½.
- √x = x^½，1/x² = x⁻²，1/√x = x−½。
- Differentiate term by term, treating each constant as a coefficient.
- 逐项求导，把每个常数当作系数处理。
- Simplify final answers by removing negative indices only if the question asks for a particular form.
- 只有在题目要求特定形式时，才把负指数改写回分数形式。
differentiating term by term gives dy/dx = 12x³ − 10x + 7, and the constant −2 disappears.
对 y = 3x⁴ − 5x² + 7x − 2，逐项求导得 dy/dx = 12x³ − 10x + 7，其中常数 −2 求导后消失。
For f(x) = √x + 1/x, first write f(x) = x^½ + x⁻¹, so f′(x) = ½x−½ − x⁻², which may be rewritten as 1 ÷ (2√x) − 1 ÷ x².
对 f(x) = √x + 1/x，先写成 f(x) = x^½ + x⁻¹，于是 f′(x) = ½x−½ − x⁻²，也可改写为 1 ÷ (2√x) − 1 ÷ x²。
A common exam instruction is to expand the brackets first, for instance (2x + 1)(x − 3) must be multiplied out before differentiating. Differentiating the product without expanding is only acceptable once you have learned the product rule.
常见指令要求先展开括号，例如 (2x + 1)(x − 3) 必须先乘开再求导。在尚未学习乘积法则之前，直接对乘积求导是不被接受的。
4. Tangents, Normals and the Gradient at a Point | 切线、法线与某点处的斜率
The value of dy/dx at a given point equals the gradient of the tangent at that point. This single idea generates a large family of exam questions.
在给定点处 dy/dx 的数值等于该点处切线的斜率。仅凭这一个想法就能衍生出大量考题。
y − y₁ = m(x − x₁), where m = dy/dx evaluated at (x₁, y₁)
To find a tangent: differentiate, substitute the x-coordinate to find m, find the y-coordinate from the original equation, then substitute into the straight-line formula.
求切线的方法：先求导，代入横坐标得到 m，再由原方程求出纵坐标，最后代入直线方程公式。
ts gradient is the negative reciprocal −1 ÷ m. Given m = 3, the normal gradient is −1 ÷ 3.
法线与切线垂直，因此其斜率为负倒数 −1 ÷ m。若 m = 3，则法线斜率为 −1 ÷ 3。
Examiners also ask whether a given line is a tangent to a curve, which is answered by setting dy/dx equal to the gradient of the line and checking that the resulting point also lies on the curve.
考官还会问某条直线是否为曲线的切线，方法是令 dy/dx 等于该直线的斜率，并检验所得的点是否同时位于曲线上。
5. Second Derivatives and Stationary Points | 二阶导数与驻点
Differentiating twice gives the second derivative, written as d²y/dx² or f″(x). It measures how the gradient itself is changing.
对导函数再求导得到二阶导数，记作 d²y/dx² 或 f″(x)，它刻画的是斜率本身的变化。
. Turning points, maximum points, minimum points and points of inflection are all stationary points.
驻点出现在斜率为零的地方，因此需要解方程 dy/dx = 0。极大值点、极小值点与拐点都属于驻点。
= 2 are the stationary values. Substituting back gives the coordinates (0, 4) and (2, 0).
对 y = x³ − 3x² + 4，dy/dx = 3x² − 6x = 3x(x − 2)，因此 x = 0 与 x = 2 是驻点处横坐标。代回原式得坐标 (0, 4) 与 (2, 0)。
For y = x²eˣ uct rule from Section 8 before setting dy/dx to zero, and the resulting equation x(x + 2)eˣ = 0 has solutions x = 0 and x = −2 because eˣ is never zero.
对 y = x²eˣ，则须先使用第 8 节的乘积法则再令 dy/dx 为零，得到的方程 x(x + 2)eˣ = 0 的解为 x = 0 与 x = −2，因为 eˣ 永远不为零。
6. Classifying Stationary Points and Optimisation | 驻点判定与最优化
The second derivative test is the standard classification tool. If f″(x) is negative the point is a maximum; if f″(x) is positive the point is a minimum; if f″(x) equals zero the test is inconclusive and you must examine the sign of the gradient on either side.
二阶导数检验是标准的判定工具。若 f″(x) 为负，该点为极大值点；若 f″(x) 为正，该点为极小值点；若 f″(x) 等于零，则该检验失效，必须考察该点两侧斜率的符号。
Optimisation questions require a fully algebraic approach: form an expression in one variable, differentiate it, set the derivative to zero, solve, and then justify that the answer is a maximum or a minimum.
最优化题目要求完整使用代数方法：用一个变量建立表达式，对其求导，令导数为零，求解，然后说明结果是最大值还是最小值。
² = −2 < 0, this is a maximum area of 100 cm², achieved when the rectangle is a square.
若矩形周长为 40 cm，两边为 x 与 20 − x，则面积为 A = 20x − x²，由 dA/dx = 20 − 2x = 0 得 x = 10。因为 d²A/dx² = −2 < 0，所以这是最大面积 100 cm²，此时矩形为正方形。
Mark schemes normally allocate one mark simply for the justification sentence, so always write a concluding statement such as: since the second derivative is negative, this value gives the maximum volume.
评分标准通常会为这一句论证单独给一分，因此务必写出结论句，例如：由于二阶导数为负，该值给出最大体积。
7. The Chain Rule | 链式法则
The chain rule differentiates a function of a function. In Leibniz notation it is written as follows.
链式法则用于对复合函数求导。用莱布尼茨记号可表示如下。
dy/dx = dy/du × du/dx
In words: differentiate the outer function with the inner function left unchanged, then multiply by the derivative of the inner function.
用文字表述：先把内层函数原样保留，对外层函数求导，再乘以对内层函数的导数。
Typical results include y = (3x + 2)⁵ giving dy/dx = 15(3x + 2)⁴, y = √x² + 1 giving dy/dx = x ÷ √x² + 1, and y = sin(2x) giving dy/dx = 2cos(2x).
典型结果包括：y = (3x + 2)⁵ 的 dy/dx = 15(3x + 2)⁴；y = √x² + 1 的 dy/dx = x ÷ √x² + 1；y = sin(2x) 的 dy/dx = 2cos(2x)。
The most common error is forgetting the inner derivative, which loses the accuracy mark even when the outer derivative is correct. Examiners describe this as not differentiating the inside.
最常见的错误是漏掉内层导数，即使外层求导正确，也会丢掉准确分。考官通常把这种错误描述为没有对内层求导。
8. The Product Rule | 乘积法则
When two functions of x are multiplied, neither the chain rule nor the power rule alone is enough. Use the product rule instead.
当两个关于 x 的函数相乘时，单用链式法则或幂法则都不够，必须使用乘积法则。
d(uv)/dx = u dv/dx + v du/dx
Label one factor u and the other v, differentiate each separately, then substitute into the formula and factorise the answer where possible.
把其中一个因式记为 u，另一个记为 v，分别求导，再代入公式，并在可能的情况下对结果提取公因式。
For y = x²eˣ, take u = x² and v = eˣ, so dy/dx = x²eˣ + 2xeˣ = xeˣ(x + 2), which is the factorised form examiners prefer.
对 y = x²eˣ，取 u = x²，v = eˣ，则 dy/dx = x²eˣ + 2xeˣ = xeˣ(x + 2)，这种因式分解形式是考官更喜欢的答案形式。
g u = x and v = ln x. Choosing the polynomial factor as u usually keeps the algebra short.
对 y = x ln x，令 u = x，v = ln x，得 dy/dx = ln x + 1。通常把多项式因式选作 u 可以让代数过程更简洁。
9. The Quotient Rule | 商法则
When one function is divided by another, use the quotient rule. Writing the numerator as u and the denominator as v gives the standard form.
当一个函数除以另一个函数时，使用商法则。把分子记为 u、分母记为 v，即得标准形式。
dy/dx = ( v du/dx − u dv/dx ) ÷ v²
The order of the numerator matters: the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
分子的顺序很重要：分母乘以分子的导数，减去分子乘以分母的导数，再整体除以分母的平方。
2), u = 3x + 1 and v = x − 2, so dy/dx = [3(x − 2) − (3x + 1)] ÷ (x − 2)² = −7 ÷ (x − 2)².
对 y = (3x + 1) ÷ (x − 2)，取 u = 3x + 1，v = x − 2，则 dy/dx = [3(x − 2) − (3x + 1)] ÷ (x − 2)² = −7 ÷ (x − 2)²。
Alternatively, if the denominator is a single term such as x², split the fraction into separate powers of x and use the power rule, which is usually quicker and safer under exam conditions.
若分母为单项式如 x²，也可以把分式拆成若干 x 的幂再使用幂法则，这在考试中通常更快也更稳妥。
10. Exponential, Logarithmic and Trigonometric Functions | 指数、对数与三角函数
The exponential and logarithmic derivatives are used constantly in Edexcel papers, and they combine directly with the chain rule.
指数函数与对数函数的导数在 Edexcel 试卷中反复出现，并且可与链式法则直接结合使用。
- d(eˣ)/dx = eˣ and d(eᵏˣ)/dx = k eᵏˣ, for example d(e³ˣ)/dx = 3e³ˣ.
- d(eˣ)/dx = eˣ，d(eᵏˣ)/dx = k eᵏˣ，例如 d(e³ˣ)/dx = 3e³ˣ。
- d(ln x)/dx = 1 ÷ x and d(ln(ax + b))/dx = a ÷ (ax + b).
- d(ln x)/dx = 1 ÷ x，d(ln(ax + b))/dx = a ÷ (ax + b)。
- d(sin ax)/dx = a cos ax and d(cos ax)/dx = −a sin ax.
- d(sin ax)/dx = a cos ax，d(cos ax)/dx = −a sin ax。
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