Mastering Differentiation for Edexcel A-Level Maths | 攻克爱德思A-Level数学微分

📚 Mastering Differentiation for Edexcel A-Level Maths | 攻克爱德思A-Level数学微分

Differentiation is the branch of calculus that studies how functions change. In the Edexcel A-Level Mathematics specification, it appears across Pure Mathematics, mechanics and applied contexts, and it carries a large share of the marks in every paper. Mastery of differentiation rules, stationary points and optimisation is essential for high marks, because later topics such as integration, sequences of modelling questions and kinematics all depend on it.

微分是研究函数如何变化的微积分分支。在爱德思A-Level数学大纲中，它贯穿纯数学、力学和应用题，并且在每份试卷中都占据很大的分值比例。掌握微分法则、驻点和优化对于取得高分至关重要，因为之后的积分、建模题组和运动学等主题都依赖它。

This guide walks through the whole differentiation story in the order it is taught and examined: the limit definition, first principles, standard results, the three big rules, tangents and rates of change, stationary points, optimisation, and finally exam technique.

本文按照教学与考试的推进顺序，完整梳理微分的全部内容：极限定义、第一性原理、标准结果、三大求导法则、切线与变化率、驻点、优化，最后是应试技巧。

---

1. The Idea of a Derivative | 导数的思想

defined as the limit of the average rate of change over smaller and smaller intervals, so it measures the instantaneous rate of change rather than an average.

函数 f(x) 在 x = a 处的导数给出曲线 y = f(x) 在该点切线的斜率。它定义为当区间越来越小时平均变化率的极限，因此它衡量的是瞬时变化率，而不是平均变化率。

simply Δy / Δx. For a curve, we take the limit as Δx → 0. This limiting value is written f'(a), or dy/dx at x = a.

对于直线，斜率就是 Δy / Δx。对于曲线，我们取 Δx → 0 时的极限。这个极限值记作 f'(a)，或 x = a 处的 dy/dx。

The derivative is also a function in its own right. Differentiating f produces f', which maps each input x to the gradient at that point. You must be fluent with the three common notations: f'(x), dy/dx, and d/dx of f(x).

导数本身也是一个函数。对 f 求导得到 f'，它把每个输入 x 映射到该点的斜率。你必须熟练使用三种常见记号：f'(x)、dy/dx，以及 d/dx 作用于 f(x)。

A useful physical reading: if y is displacement and x is time, then dy/dx is velocity, and differentiating again gives acceleration. If y is the cost of producing x items, dy/dx is marginal cost. This interpretation is what makes differentiation a modelling tool, not just algebra.

一个有意义的物理解读：如果 y 是位移、x 是时间，那么 dy/dx 是速度，再求一次导就得到加速度。如果 y 是生产 x 件产品的成本，dy/dx 就是边际成本。正是这种解读让微分成为建模工具，而不仅仅是代数运算。

---

2. Differentiation from First Principles | 从第一性原理求导

The formal definition of the derivative is the result of the limiting process described above:

导数的正式定义就是上面所述极限过程的结果：

f'(x) = lim (h → 0) [f(x + h) − f(x)] / h

This process is called differentiation from first principles. It underpins every standard derivative result used in the exam, and it is a named skill in the Edexcel specification, so a first-principles question is always possible.

这个过程称为从第一性原理求导。它支撑着考试中使用的每一个标准导数结果，而且是爱德思大纲中明确列出的技能，因此第一性原理的题目随时可能出现。

x) = (x + h)² − x² = 2xh + h². Dividing by h gives 2x + h. As h → 0, the limit is 2x, so f'(x) = 2x.

示例：对于 f(x) = x²，f(x + h) − f(x) = (x + h)² − x² = 2xh + h²。除以 h 得到 2x + h。当 h → 0 时，极限为 2x，所以 f'(x) = 2x。

h)] = −1 / [x(x + h)]. As h → 0 this tends to −1/x², confirming the power rule for n = −1.

第二个标准例子使用负指数。对于 f(x) = 1/x，差商为 [1/(x + h) − 1/x] / h = [x − (x + h)] / [h x (x + h)] = −1 / [x(x + h)]。当 h → 0 时它趋于 −1/x²，这正好验证了 n = −1 时的幂法则。

e awarded at each stage, so never jump straight to the final answer without justification.

考试中要写出这些步骤：写出 f(x + h)；减去 f(x)；展开并化简分子；除以 h；最后令 h → 0。每一步都有分值，所以不要在没有论证的情况下直接跳到最终答案。

---

3. Standard Derivatives and the Power Rule | 标准导数与幂法则

The power rule states that if f(x) = xⁿ, then f' and fractional powers, which is why it is the single most used result in the whole course.

幂法则指出，如果 f(x) = xⁿ，那么 f'(x) = n xⁿ⁻¹。它适用于任何实数常数 n，包括负指数和分数指数，这也是它成为整个课程中使用最多的结论的原因。

Key standard derivatives you must know by heart:

你必须牢记的关键标准导数：

Function f(x)	Derivative f'(x)
k (a constant)	0
xⁿ	n xⁿ⁻¹
eˣ	eˣ
eᵏˣ	k eᵏˣ
ln x	1 / x
sin x	cos x
cos x	−sin x
tan x	sec² x

rules complete the toolkit: a constant multiple can be pulled out, so d/dx [k f(x)] = k f'(x); and derivatives are added term by term, so d/dx [f(x) + g(x)] = f'(x) + g'(x). Together with the power rule these let you differentiate any polynomial on sight.

还有两条结构性法则补全工具包：常数因子可以提到外面，即 d/dx [k f(x)] = k f'(x)；导数可以逐项相加，即 d/dx [f(x) + g(x)] = f'(x) + g'(x)。结合幂法则，你可以一眼求出任何多项式的导数。

---

4. Polynomials, Negative Powers and Roots | 多项式、负指数与根式

dy/dx = 12x³ − 10x + 7. The constant term disappears because a constant has zero gradient.

早期考题大多是多项式，因此速度和准确性比技巧更重要。对 y = 3x⁴ − 5x² + 7x − 2 逐项求导得 dy/dx = 12x³ − 10x + 7。常数项消失，因为常数的斜率为零。

Roots and reciprocals must first be rewritten as powers. For y = √x = x^½, dy/dx = ½ x^−½ = 1 / (2√x). For y = 1/x³ = x⁻³, dy/dx = −3x⁻⁴ = −3 / x⁴.

根式和倒数必须先改写为幂的形式。对于 y = √x = x^½，dy/dx = ½ x^−½ = 1 / (2√x)。对于 y = 1/x³ = x⁻³，dy/dx = −3x⁻⁴ = −3 / x⁴。

1 and differentiate to get 8x + 4, then check with the chain rule: 2(2x + 1) × 2 = 8x + 4.

当括号内是整数次幂时，展开往往比使用链式法则更快。对于 y = (2x + 1)²，展开为 4x² + 4x + 1 后求导得 8x + 4；再用链式法则验证：2(2x + 1) × 2 = 8x + 4。

广告 / Sponsored

n. If the question uses root signs or fractions, converting back from x^½ or x⁻³ into 1/(2√x) or −3/x⁴ can be worth a mark on its own.

最终答案的表达形式应尽量与题目一致。如果题目使用根号或分式，把 x^½ 或 x⁻³ 还原成 1/(2√x) 或 −3/x⁴ 本身就可能单独得分。

---

5. Exponential and Logarithmic Functions | 指数函数与对数函数

The exponential function is its own derivative: if y = eˣ, then dy/dx = eˣ. More generally, y = eᵏˣ gives dy/dx = k eᵏˣ, so y = e^3x differentiates to 3e^3x and y = e^−2x differentiates to −2e^−2x.

指数函数的导数就是它本身：若 y = eˣ，则 dy/dx = eˣ。更一般地，y = eᵏˣ 的导数为 dy/dx = k eᵏˣ，所以 y = e^3x 求导得 3e^3x，y = e^−2x 求导得 −2e^−2x。

= 1 / x, valid only for x > 0. Combining with the chain rule, y = ln(f(x)) gives dy/dx = f'(x) / f(x), so y = ln(5x + 2) differentiates to 5 / (5x + 2).

对于自然对数，y = ln x 给出 dy/dx = 1 / x，仅在 x > 0 时成立。结合链式法则，y = ln(f(x)) 的导数为 dy/dx = f'(x) / f(x)，所以 y = ln(5x + 2) 求导得 5 / (5x + 2)。

These results are also reverse-engineered in integration questions: recognising that 2 / (2x + 1) is the derivative of ln(2x + 1) turns a difficult integral into a one-line answer. Learning derivatives forwards and backwards is therefore doubly useful.

这些结论在积分题中也会被反向使用：认出 2 / (2x + 1) 是 ln(2x + 1) 的导数，就能把一个困难的积分变成一行答案。因此正反两向掌握导数会有双重收益。

---

6. Derivatives of Trigonometric Functions | 三角函数的导数

he minus sign in the cosine result is the single most commonly dropped mark in this topic.

三个标准结果是：y = sin x 的导数为 dy/dx = cos x；y = cos x 的导数为 dy/dx = −sin x；y = tan x 的导数为 dy/dx = sec² x。余弦结果中的负号是本主题中最常被漏掉的一分。

ation, assumes radians, so check the mode of your calculator and remember that 180° = π radians.

这些导数只在 x 以弧度为单位时成立。任何涉及三角求导或小角近似的题目都默认使用弧度，所以要检查计算器模式，并牢记 180° = π 弧度。

s(kx) gives −k sin(kx), and y = sin(3x²) gives 6x cos(3x²). Combined results such as y = eˣ sin x need the product rule, which is covered next.

结合链式法则，这一族结果立刻扩展：y = sin(kx) 的导数为 k cos(kx)，y = cos(kx) 的导数为 −k sin(kx)，y = sin(3x²) 的导数为 6x cos(3x²)。像 y = eˣ sin x 这样组合型的函数则需要乘积法则，下面就会讲到。

---

7. The Chain Rule | 链式法则

Functions of a function, often called composite functions, are differentiated with the chain rule:

复合函数（即"函数的函数"）要用链式法则求导：

dy/dx = (dy/du) × (du/dx)

he derivative of the inside. For y = (3x² + 1)⁵, dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴.

用文字说：先对外层函数求导、内层保持不动，再乘以对内层函数的导数。对于 y = (3x² + 1)⁵，dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴。

Mixed examples show how universal the rule is. y = e^{sin x} gives dy/dx = cos x e^{sin x}; y = ln(x² + 4) gives dy/dx = 2x / (x² + 4); y = √2x + 7 gives dy/dx = 1 / √2x + 7, because ½(2x + 7)^−½ × 2 simplifies to (2x + 7)^−½.

混合例子展示了这条法则的普适性。y = e^{sin x} 给出 dy/dx = cos x e^{sin x}；y = ln(x² + 4) 给出 dy/dx = 2x / (x² + 4)；y = √2x + 7 给出 dy/dx = 1 / √2x + 7，因为 ½(2x + 7)^−½ × 2 可化简为 (2x + 7)^−½。

wo marks and produces a derivative that is wrong in every subsequent part of the question.

要避免的错误是忘记乘第二个因子。只写 5(3x² + 1)⁴ 而不乘 × 6x，至少会丢两分，而且会让后续每一小问的导数都错。

---

8. The Product and Quotient Rules | 乘积法则与商法则

When two functions are multiplied, write u for the first factor and v for the second, then use:

当两个函数相乘时，设第一个因子为 u、第二个因子为 v，然后使用：

d/dx (uv) = u'v + uv'

y/dx = sin x + x cos x. The two terms are never combined into one, so leave them as a sum.

例如，y = x² ln x 给出 dy/dx = 2x ln x + x² × (1/x) = 2x ln x + x。对于 y = x sin x，dy/dx = sin x + x cos x。两项不能合并成一项，保持和的形式即可。

For a quotient, keep the same u and v names and use:

对于商的形式，仍用相同的 u、v 记号，使用：

d/dx (u / v) = (u'v − uv') / v²

tice the structure: numerator first, then minus the swapped product, then the square of the denominator.

对于 y = sin x / x，dy/dx = (x cos x − sin x) / x²。对于 y = (2x + 1) / (x − 3)，dy/dx = (2(x − 3) − (2x + 1)) / (x − 3)² = −7 / (x − 3)²。注意结构：先写分子项，再减去交换后的乘积，最后除以分母的平方。

Choose the easier route when both exist. Algebraic division of (2x + 1)/(x − 3) into 2 + 7/(x − 3) turns the problem into the power rule, which is faster and less error-prone under time pressure.

两种方法都可行时选更简单的。把 (2x + 1)/(x − 3) 做多项式除法写成 2 + 7/(x − 3)，问题就变成幂法则，在时间紧张时更快、更不易出错。

---

9. Tangents, Normals and Rates of Change | 切线、法线与变化率

The derivative is the gradient of the tangent, so the tangent to y = f(x) at x = x₁

更多咨询请联系16621398022（同微信）