📚 A-Level Edexcel Pure Maths: Differentiation Techniques & Applications | A-Level Edexcel 纯数：微分技巧与应用
Differentiation is one of the most heavily examined topics in Edexcel A-Level Pure Mathematics. This article covers the full range of differentiation techniques you need to master: first principles, standard derivatives, chain, product, quotient, implicit and parametric differentiation, stationary points, optimisation and connected rates of change. Each section presents the key idea in English followed by its Chinese equivalent, with worked examples and common exam pitfalls.
微分是 Edexcel A-Level 纯数学中考查频率最高的主题之一。本文涵盖了需要掌握的全部微分技巧：第一原理求导、基本导数公式、链式法则、乘积法则、商法则、隐函数微分、参数方程微分、驻点、优化问题以及相关变化率。每一节先用英文给出核心概念，再给出对应的中文解释，并配有例题和常见考试易错点。
1. Differentiation from First Principles | 从第一原理求导
For Edexcel A-Level, you must be able to use the limit definition of the derivative. The derivative of f(x) at a point x is defined as:
在 Edexcel A-Level 考试中，你需要掌握导数的极限定义。函数 f(x) 在 x 处的导数定义为：
f ‘(x) = lim h→0 (f(x+h) − f(x)) / h
To differentiate x² from first principles, expand (x+h)² = x² + 2xh + h², subtract f(x), divide by h and then let h tend to 0. The result is f ‘(x) = 2x.
要从第一原理对 x² 求导，先展开 (x+h)² = x² + 2xh + h²，减去 f(x)，除以 h，再令 h 趋于 0。结果为 f ‘(x) = 2x。
Exam questions often ask you to apply this definition to a simple polynomial, such as x³ or x² + 3x. Always show the full expansion and the limiting step to gain all method marks.
考试题常要求对简单多项式应用这一定义，例如 x³ 或 x² + 3x。务必写出完整的展开式和取极限的步骤，才能获得全部过程分。
2. Standard Derivatives and Basic Rules | 基本导数公式与基础法则
Memorising the standard derivatives is essential because they are the building blocks for all harder problems. The table below lists the most common functions and their derivatives.
熟记基本导数公式非常重要，因为它们是所有较难问题的基础。下表列出了最常见的函数及其导数。
| f(x) = xⁿ | f ‘(x) = n xⁿ⁻¹ |
| f(x) = sin x | f ‘(x) = cos x |
| f(x) = cos x | f ‘(x) = −sin x |
| f(x) = eˣ | f ‘(x) = eˣ |
| f(x) = ln x | f ‘(x) = 1/x |
| f(x) = aˣ | f ‘(x) = aˣ ln a |
The constant multiple rule and the sum rule allow you to differentiate expressions such as 5x⁴ − 3x² + 2x − 7 term by term. For example, f ‘(x) = 20x³ − 6x + 2.
常数倍法则和加法法则允许你对 5x⁴ − 3x² + 2x − 7 这样的表达式逐项求导。例如，f ‘(x) = 20x³ − 6x + 2。
Make sure you rewrite square roots and reciprocals as powers before differentiating. For instance, √x becomes x½, and 1/x³ becomes x⁻³, so you can apply the power rule.
求导前一定要把根式和倒数改写为幂的形式。例如，√x 写成 x½，1/x³ 写成 x⁻³，这样就可以应用幂法则。
3. Chain Rule | 链式法则
The chain rule is used to differentiate composite functions. If y = f(u) and u = g(x), then:
链式法则用于对复合函数求导。如果 y = f(u) 且 u = g(x)，那么：
dy/dx = dy/du × du/dx
This rule is sometimes written as: if y = [g(x)]ⁿ, then y ‘ = n [g(x)]ⁿ⁻¹ g ‘(x).
该法则有时写作：如果 y = [g(x)]ⁿ，则 y ‘ = n [g(x)]ⁿ⁻¹ g ‘(x)。
Worked example: Differentiate y = (3x² + 5)⁴. Let u = 3x² + 5, so y = u⁴. Then dy/du = 4u³ and du/dx = 6x. Multiplying gives dy/dx = 24x(3x² + 5)³.
例题：对 y = (3x² + 5)⁴ 求导。设 u = 3x² + 5，则 y = u⁴。于是 dy/du = 4u³，du/dx = 6x。相乘得到 dy/dx = 24x(3x² + 5)³。
A common mistake is to forget to multiply by the derivative of the inner function. In the example above, missing du/dx would give 4(3x² + 5)³, which is incomplete.
常见的错误是忘记乘以内层函数的导数。在上面的例子中，漏掉 du/dx 会得到 4(3x² + 5)³，这是不完整的。
4. Product Rule | 乘积法则
When differentiating a product of two functions, use the product rule. If y = u v, where u and v are functions of x, then:
当对两个函数的乘积求导时，使用乘积法则。如果 y = u v，其中 u 和 v 都是 x 的函数，那么：
dy/dx = u dv/dx + v du/dx
This is often remembered as ‘first times derivative of second plus second times derivative of first’.
这通常记为“第一个函数乘以第二个的导数，再加上第二个函数乘以第一个的导数”。
Worked example: Differentiate y = x² sin x. Let u = x² and v = sin x. Then du/dx = 2x and dv/dx = cos x. Therefore, y ‘ = x² cos x + 2x sin x.
例题：对 y = x² sin x 求导。设 u = x²，v = sin x。则 du/dx = 2x，dv/dx = cos x。因此，y ‘ = x² cos x + 2x sin x。
When one factor is a constant, the product rule still works, but it is simpler to use the constant multiple rule. However, examiners may ask you to identify u and v explicitly for products such as x ln x.
当其中一个因式为常数时，乘积法则仍然适用，但使用常数倍法则更为简单。然而，考官可能会要求你对 x ln x 这类乘积明确指出 u 和 v。
5. Quotient Rule | 商法则
For a quotient of two functions, the quotient rule is required. If y = u / v, then:
对于两个函数的商，需要使用商法则。如果 y = u / v，那么：
dy/dx = (v du/dx − u dv/dx) / v²
The order matters: numerator first uses v du/dx, then subtract u dv/dx, all divided by v squared.
顺序很重要：分子先写 v du/dx，再减去 u dv/dx，整体除以 v 的平方。
Worked example: Differentiate y = (3x + 1) / (x² − 4). Let u = 3x + 1 and v = x² − 4. Then du/dx = 3 and dv/dx = 2x. The derivative is:
例题：对 y = (3x + 1) / (x² − 4) 求导。设 u = 3x + 1，v = x² − 4。则 du/dx = 3，dv/dx = 2x。导数为：
y ‘ = [(x² − 4)(3) − (3x + 1)(2x)] / (x² − 4)²
Simplify the numerator carefully: 3x² − 12 − (6x² + 2x) = −3x² − 2x − 12. Then y ‘ = (−3x² − 2x − 12) / (x² − 4)².
仔细化简分子：3x² − 12 − (6x² + 2x) = −3x² − 2x − 12。于是 y ‘ = (−3x² − 2x − 12) / (x² − 4)²。
Many candidates lose marks by mixing up the numerator order. A useful check is that the quotient rule is not symmetric, unlike the product rule.
很多考生因为弄错分子的顺序而失分。一个有用的检验方法是：商法则不像乘积法则那样具有对称性。
6. Implicit Differentiation | 隐函数微分
When an equation gives y implicitly as a function of x, differentiate both sides with respect to x, treating y as a function of x. Each time you differentiate y, multiply by dy/dx.
当方程以隐函数形式给出 y 与 x 的关系时，对两边关于 x 求导，并将 y 视为 x 的函数。每次对 y 求导时，都要乘以 dy/dx。
Worked example: Given x² + y² = 25, differentiate both sides to obtain 2x + 2y dy/dx = 0. Solving gives dy/dx = −x / y.
例题：已知 x² + y² = 25，对两边求导得到 2x + 2y dy/dx = 0。解得 dy/dx = −x / y。
Implicit differentiation is particularly useful for curves such as circles, ellipses and hyperbolas where y cannot easily be made the subject.
隐函数微分对于圆、椭圆和双曲线等难以将 y 表示为显函数的曲线尤其有用。
Be careful with terms like xy. Using the product rule, d/dx (xy) = x dy/dx + y. Also remember that d/dx (y²) = 2y dy/dx.
要注意 xy 这类项。使用乘积法则，d/dx (xy) = x dy/dx + y。还要记住 d/dx (y²) = 2y dy/dx。
7. Parametric Differentiation | 参数方程微分
For a curve defined parametrically by x = f(t) and y = g(t), the gradient is found using:
对于由参数方程 x = f(t) 和 y = g(t) 定义的曲线，其斜率由下式求得：
dy/dx = (dy/dt) / (dx/dt)
Worked example: A curve has x = t² + 1 and y = 2t + 3. Then dx/dt = 2t and dy/dt = 2. Hence dy/dx = 2 / (2t) = 1 / t.
例题：某曲线的参数方程为 x = t² + 1，y = 2t + 3。则 dx/dt = 2t，dy/dt = 2。因此 dy/dx = 2 / (2t) = 1 / t。
The second derivative for parametric curves is found by d²y/dx² = d/dx (dy/dx) = d/dt (dy/dx) ÷ dx/dt. This is a common follow-up question in Edexcel papers.
参数曲线的二阶导数用 d²y/dx² = d/dx (dy/dx) = d/dt (dy/dx) ÷ dx/dt 来求。这是 Edexcel 试卷中常见的后续问题。
Always check that dx/dt is not zero at the point of interest, otherwise the tangent may be vertical.
始终要检查在所关注点处 dx/dt 是否为零，否则切线可能是竖直的。
8. Second Derivatives and Concavity | 二阶导数与凹凸性
The second derivative, written d²y/dx² or f ”(x), is the derivative of the first derivative. It measures the rate of change of the gradient, which tells
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