📚 Edexcel A-Level Pure Mathematics: Differentiation and Integration Masterclass | Edexcel A-Level 纯数学：微分与积分核心突破
This revision note covers the core techniques of differentiation and integration required for the Edexcel A-Level Mathematics specification. Understanding these methods is essential for success in Pure Mathematics papers and for applied contexts such as kinematics, optimisation and rates of change. Every rule below is examinable, and most appear every single year in some form.
本文复习笔记涵盖 Edexcel A-Level 数学大纲要求的微分与积分核心技巧。掌握这些方法对于纯数学试卷以及运动学、最优化、变化率等应用情境至关重要。下文每一条法则都在考纲之内，其中大多数每年都会以某种形式出现。
1. Specification Overview and Assessment Weighting | 大纲概览与考核权重
Edexcel A-Level Mathematics (9MA0) is split into three equally weighted papers: Pure Mathematics 1, Pure Mathematics 2, and Statistics and Mechanics. Calculus is the single largest strand of the two pure papers, and it also feeds directly into mechanics questions on variable acceleration and kinematics. A confident student should be able to differentiate any composite, product, quotient, implicit or parametric expression, and integrate using standard results, substitution, parts and the trapezium rule.
Edexcel A-Level 数学（9MA0）分为权重相等的三份试卷：纯数学 1、纯数学 2，以及统计与力学。微积分是两份纯数学试卷中占比最大的板块，同时直接支撑力学中变加速度与运动学题目。一名有把握的考生应当能够对复合函数、乘积、商、隐函数和参数方程求导，并能运用标准公式、换元法、分部积分法和梯形法则完成积分。
- Differentiate from first principles and apply all standard results.
- Use the chain, product and quotient rules fluently.
- Differentiate implicitly and parametrically.
- Find tangents, normals, stationary points and points of inflection.
- Integrate standard functions and evaluate definite integrals.
- Solve simple first-order differential equations by separation of variables.
- 从基本原理求导，并熟练运用全部标准导数公式。
- 流畅使用链式法则、乘法法则与除法法则。
- 进行隐函数求导与参数方程求导。
- 求切线、法线、驻点与拐点。
- 对标准函数积分并计算定积分。
- 用分离变量法求解简单的一阶微分方程。
2. Differentiation from First Principles and Standard Results | 从基本原理求导与标准导数
Differentiation from first principles defines the gradient function as the limit of a chord gradient as the chord shrinks to a tangent. Edexcel uses the small-increment notation δx, so the definition should be quoted exactly as it appears in the formula booklet.
从基本原理求导，是把弦的斜率在弦收缩为切线时的极限定义为导函数。Edexcel 使用小增量记号 δx，因此引用定义时应与公式册中的写法完全一致。
f'(x) = lim(δx → 0) [ f(x + δx) − f(x) ] ÷ δx
ubtract x³, divide by δx to get 3x² + 3xδx + (δx)², then let δx → 0 to obtain dy/dx = 3x².
示例：对于 y = x³，展开 (x + δx)³ = x³ + 3x²δx + 3x(δx)² + (δx)³，减去 x³，再除以 δx 得到 3x² + 3xδx + (δx)²，令 δx → 0，即得 dy/dx = 3x²。
| f(x) | f'(x) | f(x) | f'(x) |
| xⁿ | nxⁿ⁻¹ | sin x | cos x |
| eˣ | eˣ | cos x | −sin x |
| ln x | 1 ÷ x | tan x | sec² x |
| aᵏˣ | k aᵏˣ ln a | eᵏˣ | k eᵏˣ |
Remember that trigonometric calculus results only hold in radians, so always switch your calculator to radian mode before attempting these questions.
请注意，三角函数的微积分结论只在弧度制下成立，因此做这类题目前务必把计算器切换到弧度模式。
3. The Chain Rule | 链式法则
ed extensively in A-Level questions involving powers, exponentials, logarithms and trigonometric functions, and it is often embedded inside a product or quotient rule question.
链式法则用于对 y = f(g(x)) 形式的复合函数求导。它指出 dy/dx = dy/du × du/dx，其中 u = g(x)。该法则广泛用于涉及幂函数、指数函数、对数函数和三角函数的 A-Level 考题，而且常常嵌套在乘法法则或除法法则的题目中。
dy/dx = dy/du × du/dx
5)⁵. The same structure gives the quick result d/dx (ax + b)ⁿ = an(ax + b)ⁿ⁻¹ and d/dx [ esin x ] = cos x × esin x.
示例：若 y = (2x³ − 5)⁶，令 u = 2x³ − 5，则 dy/du = 6u⁵，du/dx = 6x²，因此 dy/dx = 36x²(2x³ − 5)⁵。同样的结构可直接得到 d/dx (ax + b)ⁿ = an(ax + b)ⁿ⁻¹，以及 d/dx [ esin x ] = cos x × esin x。
Common error: students differentiate the outside function but forget to multiply by the derivative of the inside. Always write out your substitution u before differentiating, even when the question is short, because method marks are awarded for that shown step.
常见错误：考生只对外层函数求导，却忘记乘以内部函数的导数。即使题目很短，也要先把换元 u 写出来再求导，因为这部分解题步骤是有方法分的。
4. The Product and Quotient Rules | 乘法法则与除法法则
The product rule is used when differentiating a function that is the product of two simpler functions, u(x)v(x). The quotient rule applies when one function is divided by another. Both rules appear in the formula booklet, but you must still identify u and v clearly to earn full marks.
乘法法则用于对两个简单函数的乘积 u(x)v(x) 求导。除法法则适用于一个函数除以另一个函数的情形。两条法则都列在公式册中，但仍需清楚写出 u 与 v，才能拿满分。
Product rule: d/dx (uv) = u dv/dx + v du/dx
Quotient rule: d/dx (u ÷ v) = ( v du/dx − u dv/dx ) ÷ v²
Example: for y = x²eˣ, take u = x² and v = eˣ. Then u' = 2x and v' = eˣ, so dy/dx = x²eˣ + 2xeˣ = eˣ(x² + 2x). For y = sin x ÷ x, the quotient rule gives dy/dx = (x cos x − sin x) ÷ x².
示例：对于 y = x²eˣ，取 u = x²、v = eˣ，则 u' = 2x、v' = eˣ，因此 dy/dx = x²eˣ + 2xeˣ = eˣ(x² + 2x)。对于 y = sin x ÷ x，用除法法则得 dy/dx = (x cos x − sin x) ÷ x²。
Exam tip: if a quotient has a single-term denominator, rewrite it as a product with a negative power and use the chain rule instead. This is usually faster and produces fewer sign errors than the quotient rule.
考试技巧：若分母只有一项，可把它改写成负指数乘积，再用链式法则处理。这通常比除法法则更快，也更不容易出现符号错误。
5. Implicit Differentiation | 隐函数求导
When y cannot easily be written as an explicit function of x, differentiate both sides of the equation with respect to x and apply the chain rule to every term involving y. Each time you differentiate a function of y, multiply by dy/dx; terms mixing x and y require the product rule as well.
当 y 无法方便地写成 x 的显函数时，可对方程两边关于 x 求导，并对每个含 y 的项应用链式法则。每对 y 的函数求导一次，都要乘以 dy/dx；同时含 x 与 y 的项还需配合乘法法则。
d/dx (yⁿ) = n yⁿ⁻¹ dy/dx d/dx (xy) = y + x dy/dx
. At the point (3, 4) the gradient is −3/4, and the tangent there is y − 4 = −0.75(x − 3).
示例：对于圆 x² + y² = 25，两边求导得 2x + 2y dy/dx = 0，因此 dy/dx = −x ÷ y。在点 (3, 4) 处斜率为 −3/4，该点切线为 y − 4 = −0.75(x − 3)。
Implicit differentiation is frequently combined with finding tangents and normals, and with proving that a given point lies on a curve. Always substitute the coordinates at the end, never at the start.
隐函数求导常与求切线、法线以及证明某点位于曲线上结合考查。务必在最后一步才代入坐标，而不要在开头就代入。
6. Parametric Differentiation | 参数方程求导
nd by dividing the derivative of y by the derivative of x with respect to the parameter t.
当曲线由 x = f(t)、y = g(t) 定义时，斜率为 y 对参数 t 的导数除以 x 对 t 的导数。
dy/dx = (dy/dt) ÷ (dx/dt)
and dy/dt = 3t² − 1, so dy/dx = (3t² − 1) ÷ (2t). At t = 2 the gradient is 11 ÷ 4 = 2.75.
示例：若 x = t²、y = t³ − t，则 dx/dt = 2t、dy/dt = 3t² − 1，因此 dy/dx = (3t² − 1) ÷ (2t)。当 t = 2 时，斜率为 11 ÷ 4 = 2.75。
x) ] ÷ (dx/dt). This step is required when a question asks you to determine the nature of a stationary point on a parametric curve.
求二阶导数时需再次使用链式法则：d²y/dx² = [ d/dt (dy/dx) ] ÷ (dx/dt)。当题目要求判断参数曲线上驻点的性质时，这一步是必需的。
7. Tangents, Normals and Stationary Points | 切线、法线与驻点
procal: −1 ÷ (dy/dx). Stationary points occur where dy/dx = 0, and the second derivative test classifies them.
在某点处 dy/dx 的值即为该点切线的斜率。法线与切线垂直，故其斜率为负倒数：−1 ÷ (dy/dx)。驻点出现在 dy/dx = 0 处，可用二阶导数判别法判断其性质。
d²y/dx² > 0 → local minimum d²y/dx² < 0 → local maximum
nd y = −4. Since d²y/dx² = 6x − 6, the point (0, 0) is a maximum and (2, −4) is a minimum.
示例：对于 y = x³ − 3x²，dy/dx = 3x² − 6x = 3x(x − 2)，故驻点在 x = 0 与 x = 2 处，对应 y = 0 与 y = −4。由于 d²y/dx² = 6x − 6，点 (0, 0) 为极大值点，(2, −4) 为极小值点。
If the second derivative equals zero, do not automatically conclude that the point is an inflection; check the sign of dy/dx on either side of the point instead.
若二阶导数为零，不要直接断定该点是拐点；应改而检查该点两侧 dy/dx 的符号。
8. Optimisation and Connected Rates of Change | 最优化与相关变化率
m or minimum. Connected rates of change use the chain rule in the form dy/dt = dy/dx × dx/dt, and are extremely common in both pure and mechanics papers.
最优化问题要求你先建立某个量的表达式，再求导、令导数为零，并说明所得结果是最大值还是最小值。相关变化率问题使用 dy/dt = dy/dx × dx/dt 形式的链式法则，在纯数学与力学试卷中都极为常见。
dy/dt = dy/dx × dx/dt
Example: a circle' t = 4πr. When r = 5 cm, the area is increasing at 20π cm² s⁻¹, approximately 62.8 cm² s⁻¹.
示例：圆的半径以 2 cm s⁻¹ 增加。由 A = πr² 得 dA/dt = 2πr × dr/dt = 4πr。当 r = 5 cm 时，面积增长率为 20π cm² s⁻¹，约 62.8 cm² s⁻¹。
In optimisation questions, always state the domain of the variable, show that the second derivative is negative for a maximum, and give your final answer with correct units and sensible rounding.
在做最优化题目时，务必写出变量的取值范围，通过二阶导数为负说明是最大值，并在最终答案中给出正确单位与合理的小数位数。
9. Integration as the Reverse of Differentiation | 积分：微分的逆运算
Indefinite
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