📚 Mastering Differentiation for Edexcel A-Level Pure Maths | 掌握 Edexcel A-Level 纯数学中的微分
Differentiation allows us to move from a function that describes position, area or total cost to a function that describes the rate at which that quantity changes. In the Edexcel A-Level Pure Mathematics syllabus, differentiation appears in Pure Paper 1 and Pure Paper 2, and it underpins many problems in mechanics, kinematics, curve sketching and optimisation. This article revises the core techniques, the notation examiners expect, and the standard applications, with paired English and Chinese explanations.
微分使我们能够从描述位置、面积或总成本的函数，转向描述该量变化率的函数。在 Edexcel A-Level 纯数学大纲中，微分出现在 Pure Paper 1 和 Pure Paper 2 中，并为力学、运动学、曲线绘制和最优化中的许多问题奠定基础。本文复习核心技术、考官期望的记号以及标准应用，并给出中英文对照解释。
1. What Differentiation Measures | 微分衡量什么
A straight line has a constant gradient. A curve does not have a single gradient, but we can still ask how steep the curve is at a particular point. The derivative dy/dx is the gradient of the tangent to the curve at that point, and it is therefore a local, point-by-point measure of steepness.
直线具有恒定的斜率。曲线没有单一的斜率，但我们仍然可以考察曲线在某一点的陡峭程度。导数 dy/dx 就是曲线在该点的切线斜率，因此它是一种逐点的、局部的陡峭程度度量。
If a curve represents displacement, the derivative represents velocity. If a curve represents cost, the derivative represents marginal cost. In pure mathematics, differentiation is the main tool for analysing the local behaviour of functions: where they rise, where they fall, where they turn, and how sharply they curve.
如果曲线表示位移，则导数表示速度。如果曲线表示成本，则导数表示边际成本。在纯数学中，微分是分析函数局部行为的主要工具：函数在哪里上升、在哪里下降、在哪里转向，以及弯曲得有多剧烈。
A useful bridge between the two views is the word "rate". Whenever a quantity y depends on another quantity x, the derivative dy/dx tells you how many units y changes for each extra unit of x, at least approximately, for a small change in x.
连接这两种视角的一个有用桥梁是"变化率"这个词。只要一个量 y 依赖于另一个量 x，导数 dy/dx 就告诉你，在 x 发生微小变化时，每增加一个单位的 x，y 大约会改变多少个单位。
2. The First Principles Definition | 第一原理定义
d joining the points with x-coordinates x and x + h is (f(x + h) − f(x)) / h. This is just "rise over run" applied to two points on the curve.
导数由一个极限定义。对于 y = f(x)，连接横坐标为 x 与 x + h 的两点的弦的斜率为 (f(x + h) − f(x)) / h。这不过是把"纵向增量除以横向增量"应用到曲线上的两个点而已。
f'(x) = lim(h → 0) [ f(x + h) − f(x) ] / h
gives x² + 2xh + h², so the difference quotient simplifies to 2x + h and the limit is 2x.
当 h 趋近于 0 时，弦变为切线，这个极限给出曲线在 x 处的斜率。例如，若 f(x) = x²，展开 (x + h)² 得到 x² + 2xh + h²，因此差商化简为 2x + h，极限为 2x。
me method gives f(x) = x³ → 3x² and f(x) = 1/x → −1/x². In an exam, when the question says "differentiate from first principles" t step. Quoting the power rule is not an acceptable answer to that particular instruction.
同样的方法可得到 f(x) = x³ → 3x²，以及 f(x) = 1/x → −1/x²。在考试中，当题目要求"用第一原理求导"时，你必须写出差商、代数展开或化简，以及明确的取极限步骤。对于这一特定指令，直接引用幂法则是不被接受的作答。
3. Notation and the Gradient Function | 记号与导函数
e same thing, and Edexcel questions switch between them freely. If y = f(x), then dy/dx, f' bol d/dx is an operator: it tells you to differentiate whatever follows with respect to x.
有几种记号含义相同，而 Edexcel 的题目会自由地在它们之间切换。若 y = f(x)，那么 dy/dx、f'(x) 与 d/dx (y) 都表示导数。符号 d/dx 是一个算子：它告诉你要对后面跟着的表达式关于 x 求导。
n. To find the gradient at a specific point, substitute the x-value: for f(x) = x³ − 4x, f'(x) = 3x² − 4, so f'(2) = 8. The second derivative, written d²y/dx² or f''x), is obtained by differentiating again and measures how the gradient itself is changing.
导数本身也是一个函数，称为导函数（梯度函数）。要求某一点处的斜率，就代入该 x 值：对于 f(x) = x³ − 4x，f'(x) = 3x² − 4，所以 f'(2) = 8。二阶导数记作 d²y/dx² 或 f''(x)，通过再次求导得到，它衡量斜率本身如何变化。
Although dy/dx is not literally a fraction, it behaves like one in many manipulations, which is why the chain rule and related-rates arguments work so naturally. Treat this as a helpful notational convenience rather than a formal proof.
虽然 dy/dx 严格来说并不是分数，但在许多运算中它的行为像分数，这正是链式法则与相关变化率论证如此自然的原因。请把它当作一种有用的记号便利，而不是形式化证明。
4. The Power Rule and Polynomials | 幂法则与多项式
The single most used result in the whole topic is the power rule: to differentiate a power of x, multiply by the index and reduce the index by one. The rule works for any real index, including negative and fractional ones.
整个主题中使用最多的结果就是幂法则：对 x 的幂求导，乘以指数，再把指数减一。该法则对任何实数指数都成立，包括负数指数和分数指数。
d/dx (xⁿ) = n xⁿ⁻¹
- d/dx (x⁵) = 5x⁴
- d/dx (x⁻²) = −2x⁻³
- d/dx (√x) = d/dx (x^½) = ½ x−½ = 1 / (2√x)
- d/dx (7) = 0, because a constant does not change
le, y = 4x³ − 5x² + 7x − 2 gives dy/dx = 12x² − 10x + 7, and the constant term disappears.
微分是线性的，因此你可以逐项对多项式求导并把结果相加。例如，y = 4x³ − 5x² + 7x − 2 给出 dy/dx = 12x² − 10x + 7，常数项消失。
Many marks are lost simply because the expression was not rewritten before differentiating. If the question gives (x + 3)/x, first write it as 1 + 3x⁻¹ and differentiate that to get −3x⁻². If it gives (2x + 1)², expand to 4x² + 4x + 1 first, unless the chain rule is intended.
许多分数仅仅因为求导前没有先改写表达式而丢失。如果题目给出 (x + 3)/x，先把它写成 1 + 3x⁻¹，再求导得到 −3x⁻²。如果给出 (2x + 1)²，先展开为 4x² + 4x + 1，除非题目本意是使用链式法则。
5. Tangents, Normals and Equations of Lines | 切线与法线
e tangent at the point where x = a, compute the y-coordinate from the curve, compute m = f'(a), then use the straight-line form y − y₁ = m(x − x₁).
微分把曲线问题转化为直线问题。要求 x = a 处的切线，先从曲线求出 y 坐标，再求出 m = f'(a)，然后使用直线形式 y − y₁ = m(x − x₁)。
a. A common exam requirement is to give the equation in the form ax + by + c = 0 with integer coefficients, or with exact fractions rather than rounded decimals.
法线是同一点处与切线垂直的直线。当 m 不为零时，其斜率为 −1/m，并用同样的公式求方程。常见的考试要求是给出形如 ax + by + c = 0 且系数为整数的方程，或使用精确分数而不是四舍五入的小数。
| Quantity | How to obtain it | 中文说明 |
| Point on curve | Substitute x = a into y = f(x) | 把 x = a 代入曲线 |
| Tangent gradient | m = f'(a) = dy/dx at x = a | 切线斜率 m = f'(a) |
| Normal gradient | −1 / m | 法线斜率 −1 / m |
Always check whether the given point actually lies on the curve. If it does not, the question is probably asking for a tangent through an external point, which requires a different setup with an unknown point of contact.
务必检查题目给出的点是否真的在曲线上。如果不在，题目多半是在求过曲线外一点的切线，需要用带未知切点坐标的另一种设法。
6. Stationary Points and the Second Derivative | 驻点与二阶导数
A stationary point is a point where the gradient is zero, so it satisfies f' axima, local minima, or points of inflection with zero gradient. To classify them, use the second derivative test.
驻点是斜率为零的点，因此满足 f'(x) = 0。驻点可以是极大值点、极小值点，或斜率为零的拐点。要对它们分类，可使用二阶导数判别法。
- If f'(a) = 0 and f''(a) < 0, the point is a local maximum.
- If f'(a) = 0 and f''(a) > 0, the point is a local minimum.
- If f'(a) = 0 and f''(a) = 0, the test is inconclusive; use a sign table of f'(x) on both sides of a.
²y/dx² = 6x − 12, at x = 1 it is −6 (local maximum) and at x = 3 it is +6 (local minimum).
例题：对于 y = x³ − 6x² + 9x + 1，dy/dx = 3x² − 12x + 9 = 3(x − 1)(x − 3)，因此 dy/dx = 0 在 x = 1 与 x = 3 处成立。对应的 y 值为 y(1) = 5 与 y(3) = 1。由于 d²y/dx² = 6x − 12，在 x = 1 处为 −6（极大值点），在 x = 3 处为 +6（极小值点）。
Examiners expect three things: the derivative set equal to zero, the correct coordinates as ordered pairs, and an explicit justification of the nature of each point. Saying "it is a maximum because the graph looks like that" earns no method marks.
考官期望三点：令导数等于零、给出正确的坐标（有序数对），以及明确说明每个点的性质。仅说"从图像看它是极大值"是拿不到方法分的。
7. Increasing and Decreasing Functions | 函数的增减性
On an interval, a function is increasing where f'(x) > 0 and decreasing where f'(x) < 0. This follows directly from the tangent interpretation: positive gradient means the curve is going up as x increases, negative gradient means it is going down.
在某一区间上，函数在 f'(x) > 0 处递增，在 f'(x) < 0 处递减。这直接来自切线的解释：斜率为正意味着随着 x 增大曲线上升，斜率为负意味着它下降。
Typical question: find the values of x for which f is decreasing. Solve the inequality f' ality, and present the answer as an interval such as 1 < x < 3. If the question says "strictly increasing", the inequality is f'(x) > 0, whereas "increasing" may include points where f'(x) = 0; follow the wording carefully.
典型题目：求使 f 递减的 x 的取值范围。解不等式 f'(x) < 0，通常是二次不等式，并把答案写成如 1 < x < 3 的区间形式。若题目说"严格递增"，则不等式为 f'(x) > 0；而"递增"可能包含 f'(x) = 0 的点；请严格按照措辞作答。
A second common phrasing uses the derivative to prove a function has no turning points, for example by completing the square to show f'(x) > 0 for all x. This links differentiation back to algebra and is a favourite of Pure Paper 1.
第二种常见表述是用导数证明函数没有拐点，例如通过配方说明对一切 x 都有 f'(x) > 0。这把微分与代数联系起来，是 Pure Paper 1 常考的题型。
8. The Chain Rule for Composite Functions | 复合函数的链式法则
The chain rule handles a function of a function. If y is a function of u and u is a function of x, then the rates multiply. In Leibniz notation the rule looks like cancellation, which is exactly why the notation is so convenient.
链式法则处理"函数的函数"。如果 y 是 u 的函数，而 u 是 x 的函数，那么变化率相乘。在莱布尼茨记号下，该法则看起来像约分，这正是这套记号如此便利的原因。
dy/dx = (dy/du) × (du/dx)
tice you can do this in one line by differentiating the outside and multiplying by the derivative of the inside.
对于 y = (3x² + 1)⁵，取 u = 3x² + 1，则 dy/du = 5u⁴，du/dx = 6x。因此 dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴。熟练之后可以一行完成：先对外层求导，再乘以内层的导数。
The chain rule also drives connected rates of change. If a balloon' /dt = 4πr² × dr/dt. Read the question to decide which rate is given and which is required.
链式法则同样驱动相关变化率问题。若气球的体积在增大，则 dV/dt = (dV/dr) × (dr/dt)，由 V = (4/3)πr³ 得 dV/dt = 4πr² × dr/dt。读题时要判断已知的是哪一个变化率，要求的是哪一个。
9. 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 says: differentiate the first, keep the second; then keep the first, differentiate the second; and add the two results.
当两个函数相乘时，你不能简单地把它们的导数相乘。乘积法则为：先对第一个求导、保留第二个；再保留第一个、对第二个求导；然后把两项相加。
(uv)' = u'v + uv'
(u/v)' = (u'v − uv') / v²
Example with the product rule: y = x² e³ˣ, so u = x², v = e³ˣ, u' = 2x and v' = 3e³ˣ. Then dy/dx = 2x e³ˣ + 3x² e³ˣ = x e³ˣ x). Factorising is usually desirable because it makes the structure of the answer visible.
乘积法则示例：y = x² e³ˣ，取 u = x²，v = e³ˣ，u' = 2x，v' = 3e³ˣ。则 dy/dx = 2x e³ˣ + 3x² e³ˣ = x e³ˣ (2 + 3x)。通常应当提取公因式，因为这能让答案的结构一目了然。
Example with the quotient rule: y = (x² + 1)/(x − 1), so u = x² + 1, v = x − 1, u' = 2x, v' numerator: reversing the subtraction gives a sign error and loses all the accuracy marks.
商法则示例：y = (x² + 1)/(x − 1)，取 u = x² + 1，v = x − 1，u' = 2x，v' = 1，得 dy/dx = [2x(x − 1) − (x² + 1)] / (x − 1)² = (x² − 2x − 1) / (x − 1)²。注意分子的顺序：把减法两项颠倒会导致符号错误，从而失掉全部精确分。
| Rule | Formula | Typical trigger |
| Chain | dy/dx = (dy/du)(du/dx) | brackets raised to a power, ekx, ln(ax + b) |
| Product | style="border:1px solid #ccc;padding:
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