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Edexcel A-Level Maths: Differentiation Rules, Tangents and Stationary Points | 爱德思 A-Level 数学：微分法则、切线与驻点

2026-08-29 17:45 · 本文部分内容由AI辅助创作

📚 Edexcel A-Level Maths: Differentiation Rules, Tangents and Stationary Points | 爱德思 A-Level 数学：微分法则、切线与驻点

Differentiation is one of the largest topics in Edexcel A-Level Pure Mathematics. It connects algebraic skills, the geometry of curves and real-world rates of change, so exam questions often combine several ideas at once. A single question may ask you to differentiate, find a tangent, locate a stationary point, justify its nature with the second derivative and then interpret the answer in a physical context.

微分是爱德思 A-Level 纯数学中最大的主题之一。它将代数技能、曲线几何和现实变化率联系起来，因此考试题经常同时综合多个知识点。一道题里可能要求你求导、求切线、找驻点、用二阶导数判断其性质，最后还要在物理情境中解释答案的意义。

1. What Differentiation Measures | 微分度量什么

Differentiation gives the instantaneous rate of change of a function. For a graph y = f(x), the derivative f'(x), also written dy/dx, is the gradient of the tangent at any given point. It tells you how fast the output changes for a very small change in the input, not an average over an interval.

微分给出函数的瞬时变化率。对于图像 y = f(x)，导数 f'(x)（也写作 dy/dx）是任意给定点处切线的斜率。它描述的是输入发生极小变化时输出的变化速度，而不是某一区间上的平均变化。

The sign of the derivative carries geometric meaning, and examiners expect you to state it clearly.

导数的符号具有几何意义，考官期望你能清楚说明这一点。

This idea underpins optimisation, curve sketching and many applied problems in Edexcel examinations. Whenever a question mentions "maximum", "minimum", "rate of change" or "tangent", differentiation is almost certainly the intended method.

这一思想是爱德思考试中最优化、曲线作图以及许多应用问题的基础。只要题目出现"最大值""最小值""变化率""切线"等字眼，几乎可以肯定要用微分方法。

2. First Principles: From Secant to Tangent | 第一性原理：从割线到切线

The formal definition of the derivative comes from the gradient of a chord between two nearby points on the curve y = f(x). If the two points have x-coordinates x and x + h, the gradient of the chord is the difference quotient.

导数的正式定义来自曲线 y = f(x) 上两个邻近点之间弦的斜率。如果两个点的横坐标为 x 与 x + h，则弦的斜率为差商。

[f(x + h) − f(x)] ÷ h

As h shrinks towards zero, the chord rotates and its gradient approaches the gradient of the tangent. Taking the limit gives the derivative from first principles.

当 h 趋近于零时，割线不断旋转，其斜率趋近切线的斜率。取极限便得到第一性原理下的导数。

f'(x) = limh→0 [f(x + h) − f(x)] ÷ h

h to obtain 2x + h, then let h → 0. The key mark is the limit statement, so never skip it.

典型考题要求用此方法证明 x² 的导数是 2x：展开 (x + h)² = x² + 2xh + h²，减去 x²，除以 h 得 2x + h，再令 h → 0。得分的关键是极限这一步，千万不要省略。

3. Core Rules: Powers, Constants and Sums | 核心法则：幂、常数与和差

Most Edexcel questions are solved with the standard rules rather than first principles. The power rule applies to any real power, including negative and fractional ones, and it is the workhorse of the whole topic.

爱德思大部分题目用标准法则而不是第一性原理求解。幂法则适用于任何实数指数，包括负指数和分数指数，是整个主题中最常用的工具。

y = xⁿ ⟹ dy/dx = nxⁿ⁻¹

Before differentiating, rewrite roots and reciprocals as powers: √x = x^½, 1/x = x⁻¹, 1/x² = x⁻². This single habit prevents most careless errors in the exam.

求导之前，先把根式和倒数改写为指数形式：√x = x^½，1/x = x⁻¹，1/x² = x⁻²。养成这个习惯可以避免考试中绝大多数粗心错误。

Function | 函数Derivative | 导数
xⁿnxⁿ⁻¹
constant k0
k f(x)k f'(x)
f(x) ± g(x)f'(x) ± g'(x)
eˣeˣ
ln x1/x

Differentiation is linear: you may differentiate a sum term by term, and constants multiply straight through. You cannot, however, differentiate a product or a quotient term by term, which is why the next rules matter.

微分是线性的：可以逐项对和式求导，常数可以直接提出。但乘积与商不能逐项求导，这正是后续法则存在的原因。

4. Exponential, Logarithmic and Trigonometric Derivatives | 指数、对数与三角函数的导数

These standard results must be memorised, because Edexcel does not provide them in the formula booklet for every paper. They appear constantly in combination with the chain, product and quotient rules.

这些标准结果必须背熟，因为爱德思并非每份试卷的公式册中都会提供。它们会不断与链式、乘积、商法则结合出现。

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

Notice that tan x can also be written as sin x ÷ cos x, and its derivative as 1 ÷ cos²x. Angles must be in radians whenever you differentiate trigonometric functions in calculus; using degrees will lose marks.

注意 tan x 也可写成 sin x ÷ cos x，其导数为 1 ÷ cos²x。在微积分中对三角函数求导时，角度必须使用弧度制，用度数会失分。

For exponentials with a coefficient, the chain rule gives d/dx (eᵏˣ) = k eᵏˣ hms, d/dx [ln(kx)] = 1/x for k > 0, a useful result in integration later in the course.

对于带系数的指数函数，由链式法则得 d/dx (eᵏˣ) = k eᵏˣ。对于对数，当 k > 0 时 d/dx [ln(kx)] = 1/x，这一结果在后续积分学习中很有用。

5. The Chain Rule | 链式法则

The chain rule handles composite functions, that is, a function inside another function. If y depends on u and u depends on x, then the rates multiply.

链式法则处理复合函数，即"函数套函数"。若 y 依赖于 u，而 u 依赖于 x，则变化率相乘。

dy/dx = (dy/du) × (du/dx)

In the quicker form, differentiate the outer function, keep the inside unchanged, then multiply by the derivative of the inside. Practise identifying the inner function first, because that is where most mistakes begin.

更快的写法是：先对外层函数求导（内层保持不变），再乘以内层函数的导数。练习时先辨认内层函数，因为多数错误都始于这一步。

Example: y = (3x² + 1)⁵. Take u = 3x² + 1, so y = u⁵ and dy/du = 5u⁴, du/dx = 6x.

例：y = (3x² + 1)⁵。令 u = 3x² + 1，则 y = u⁵，dy/du = 5u⁴，du/dx = 6x。

dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴

lso covers exponentials, logarithms and trigonometry of composite arguments, such as y = e2x+3 or y = sin(4x).

链式法则同样适用于复合角度的指数、对数与三角函数，例如 y = e2x+3 或 y = sin(4x)。

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

When two functions are multiplied, you cannot simply multiply their derivatives. The product rule keeps both contributions and is often written with u and v.

当两个函数相乘时，不能简单地把导数相乘。乘积法则保留两部分贡献，通常用 u 与 v 表示。

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y = uv ⟹ dy/dx = u(dv/dx) + v(du/dx)

Example: y = x² e³ˣ. Let u = x², v = e³ˣ, so du/dx = 2x and dv/dx = 3e³ˣ.

例：y = x² e³ˣ。令 u = x²，v = e³ˣ，则 du/dx = 2x，dv/dx = 3e³ˣ。

dy/dx = 2x e³ˣ + 3x² e³ˣ = x e³ˣ (2 + 3x)

For a quotient, the numerator of the result is "bottom times derivative of top, minus top times derivative of bottom", all divided by the square of the bottom.

对于商，结果的分子是"分母乘分子的导数，减去分子乘分母的导数"，再除以分母的平方。

y = u ÷ v ⟹ dy/dx = [v(du/dx) − u(dv/dx)] ÷ v²

le: y = (2x + 1) ÷ (x − 3) gives dy/dx = [2(x − 3) − (2x + 1)] ÷ (x − 3)² = −7 ÷ (x − 3)².

例：y = (2x + 1) ÷ (x − 3)，则 dy/dx = [2(x − 3) − (2x + 1)] ÷ (x − 3)² = −7 ÷ (x − 3)²。

7. Higher Derivatives and Notation | 高阶导数与记法

Differentiating again gives the second derivative, written f''(x) or d²y/dx². It measures how the gradient itself is changing, which links directly to the curvature of the graph.

再求导一次得到二阶导数，记作 f''(x) 或 d²y/dx²。它度量斜率本身如何变化，与图像的弯曲方向直接相关。

Both notations appear in Edexcel papers. Use f' , especially in applied contexts where the variables carry meaning such as volume or time.

两种记法都会出现在爱德思试卷中。函数写成 f(x) 时用 f'(x)，写成 y = ... 时用 dy/dx，尤其在变量有实际含义（如体积或时间）的应用情境中。

8. Tangents and Normals | 切线与法线

To find the equation of a tangent, you need a point and a gradient. The gradient comes from the derivative evaluated at the given x-coordinate; then substitute into the straight-line formula.

求切线方程需要一点一斜率。斜率由在给定横坐标处求导得到，然后代入直线方程公式。

y − y₁ = m(x − x₁)    where m = f'(x₁)

The normal is the line perpendicular to the tangent at the same point, so its gradient is the negative reciprocal of m.

法线是在同一点处与切线垂直的直线，因此其斜率为 m 的负倒数。

mnormal = −1 ÷ mtangent    (valid when m ≠ 0)

The tangent is y − 9 = 6(x − 3), that is y = 6x − 9. The normal is y − 9 = −(1/6)(x − 3).

例：对 y = x² 在点 (3, 9) 处，dy/dx = 2x，故 m = 6。切线为 y − 9 = 6(x − 3)，即 y = 6x − 9。法线为 y − 9 = −(1/6)(x − 3)。

Common exam wording: "find the equation of the tangent to the curve at the point where x = ..." question wants simplified form ax + by + c = 0 or y = mx + c, and give the form requested.

常见题干："求曲线在 x = ... 处的切线方程"。注意题目要求的是 ax + by + c = 0 还是 y = mx + c 形式，按要求作答。

9. Stationary Points and the Second Derivative Test | 驻点与二阶导数判别法

A stationary point occurs where the gradient is zero. Solving f' x-coordinates; substitute back into the original function to obtain the full coordinates.

驻点出现在斜率为零处。解 f'(x) = 0 得到横坐标，再代回原函数得到完整坐标。

f'(x) = 0 at a stationary point

There are three types to distinguish, and the second derivative settles the question quickly in most cases.

驻点分三类，多数情况下用二阶导数可迅速判断。

so stationary points occur at x = 0 and x = 2. The second derivative is d²y/dx² = 6x − 6.

例题：y = x³ − 3x² + 2。则 dy/dx = 3x² − 6x = 3x(x − 2)，驻点在 x = 0 与 x = 2。二阶导数为 d²y/dx² = 6x − 6。

At x = 0, f''(0) = −6 < 0, so (0, 2) is a maximum. At x = 2, f'' 0, so (2, −2) is a minimum. Always state the conclusion in words to secure the final mark.

在 x = 0 处，f''(0) = −6 < 0，故 (0, 2) 为极大值点。在 x = 2 处，f''(2) = 6 > 0，故 (2, −2) 为极小值点。一定要用文字写明结论以拿到最后的分。

10. Optimisation and Rates of Change in Context | 情境中的最优化与变化率

Applied questions ask you to build a formula from the given geometry or physics, differentiate it, solve for the stationary value and then justify that it is the required maximum or minimum. The justification is worth marks on its own.

应用题要求你根据给定的几何或物理条件建立公式，求导，解出驻点值，再说明它确实是所求的最大或最小值。这一说明本身就占分。

ides are folded up to form an open box. The volume is V = x(12 − 2x)² for 0 < x < 6.

例：边长为 12 cm 的正方形纸板，四角各剪去边长 x cm 的小正方形，再把四边折起做成无盖盒子。其体积为 V = x(12 − 2x)²，0 < x < 6。

, or x = 2. Then V = 2(8)² = 128 cm³, and d²V/dx² = −96 + 24x = −48 < 0 at x = 2, confirming a maximum.

展开并求导：dV/dx = (12 − 2x)² − 4x(12 − 2x) = (12 − 2x)(12 − 6x)。令其为零得 x = 6（超出有效区间）或 x = 2。于是 V = 2(8)² = 128 cm³，且 d²V/dx² = −96 + 24x，在 x = 2 处为 −48 < 0，确认为最大值。

he form dy/dt = (dy/dx) × (dx/dt), which is a favourite in Edexcel mechanics-linked pure questions.

在运动学中，若已知位移 s(t)，则 ds/dt 是速度，d²s/dt² 是加速度。相关变化率使用链式法则的 dy/dt = (dy/dx) × (dx/dt) 形式，这是爱德思与力学相关的纯数题的常见考法。

11. Exam Technique: Typical Errors and Mark Scheme Habits | 考试技巧：典型错误与评分习惯

Most lost marks in this topic come from a handful of repeated slips rather than from conceptual confusion. Knowing them in advance is worth several grades at the margin.

本主题失分大多来自几个反复出现的失误，而非概念不清。提前了解这些失误，在临界分数上往往价值数分。