📚 Mastering Differentiation for Edexcel A-Level Pure Maths | 精通 Edexcel A-Level 纯数学微分
Differentiation is one of the most heavily examined topics in Edexcel A-Level Mathematics, appearing in both Pure Mathematics Paper 1 and Paper 2. This revision guide consolidates the core rules, standard derivatives, and common exam-style applications, so you can move from routine practice to high-mark questions with confidence.
微分是 Edexcel A-Level 数学中考查频率最高的主题之一，出现在纯数学试卷 1 和试卷 2 中。本复习指南整合了核心法则、常见导数以及常见考试题型，帮助你从常规练习顺利过渡到高分题。
1. The Power Rule | 幂法则
The power rule states that if y = xⁿ, then dy/dx = n xⁿ⁻¹. This rule works for any real constant n, including negative and fractional powers.
幂法则指出，如果 y = xⁿ，那么 dy/dx = n xⁿ⁻¹。该法则适用于任何实常数 n，包括负指数和分数指数。
If y = xⁿ, then dy/dx = n xⁿ⁻¹
For example, if y = x⁵, then dy/dx = 5x⁴. If y = x⁻², then dy/dx = −2x⁻³. If y = √x, rewrite it as x^½ and differentiate to get dy/dx = 1/(2√x).
例如，若 y = x⁵，则 dy/dx = 5x⁴。若 y = x⁻²，则 dy/dx = −2x⁻³。若 y = √x，先改写为 x^½，求导得 dy/dx = 1/(2√x)。
2. Standard Derivatives of Common Functions | 常见函数的标准导数
You must memorise the derivatives of exponential, logarithmic, and trigonometric functions. These appear constantly in Edexcel questions, often combined with the chain rule.
你必须熟记指数函数、对数函数和三角函数的导数。它们频繁出现在 Edexcel 考题中，并且经常与链式法则结合使用。
| Function | Derivative |
|---|---|
| eˣ | eˣ |
| ln x | 1/x |
| sin x | cos x |
| cos x | −sin x |
| tan x | sec² x |
| cosec x | −cosec x cot x |
| sec x | sec x tan x |
| cot x | −cosec² x |
Notice that the derivative of ln x is only valid for x > 0. For ln kx, use the chain rule or simplify to ln k + ln x first.
注意 ln x 的导数仅在 x > 0 时成立。对于 ln kx，可先用链式法则，或先化为 ln k + ln x 再求导。
3. Sum, Difference and Constant Multiple Rules | 和差与常数倍法则
If y = a f(x) + b g(x), then dy/dx = a f ‘(x) + b g ‘(x). Differentiate term by term and keep constant multipliers unchanged.
若 y = a f(x) + b g(x)，则 dy/dx = a f ‘(x) + b g ‘(x)。逐项求导，常数倍保持不变。
dy/dx = a f ‘(x) + b g ‘(x)
For example, if y = 4x³ − 2 sin x + 5, then dy/dx = 12x² − 2 cos x. The constant 5 differentiates to 0.
例如，若 y = 4x³ − 2 sin x + 5，则 dy/dx = 12x² − 2 cos x。常数 5 的导数为 0。
4. The Chain Rule | 链式法则
The chain rule is used when one function is inside another. Let y = f(u) and u = g(x). Then dy/dx = dy/du × du/dx.
当一个函数嵌套在另一个函数中时，使用链式法则。设 y = f(u)，u = g(x)，则 dy/dx = dy/du × du/dx。
dy/dx = dy/du × du/dx
Example: y = (3x² + 5)⁴. Let u = 3x² + 5, so y = u⁴. Then dy/du = 4u³ and du/dx = 6x, giving dy/dx = 24x(3x² + 5)³.
示例：y = (3x² + 5)⁴。设 u = 3x² + 5，则 y = u⁴。于是 dy/du = 4u³，du/dx = 6x，因此 dy/dx = 24x(3x² + 5)³。
5. The Product Rule | 乘积法则
For y = u v, where u and v are functions of x, the derivative is dy/dx = u dv/dx + v du/dx.
对于 y = u v，其中 u 和 v 都是 x 的函数，导数为 dy/dx = u dv/dx + v du/dx。
dy/dx = u dv/dx + v du/dx
Example: y = x² eˣ. Take u = x² and v = eˣ. Then du/dx = 2x and dv/dx = eˣ, so dy/dx = x² eˣ + 2x eˣ = x eˣ(x + 2).
示例：y = x² eˣ。取 u = x²，v = eˣ。则 du/dx = 2x，dv/dx = eˣ，所以 dy/dx = x² eˣ + 2x eˣ = x eˣ(x + 2)。
6. The Quotient Rule | 商法则
For y = u/v, where u and v are functions of x, the derivative is dy/dx = (v du/dx − u dv/dx) / v².
对于 y = u/v，其中 u 和 v 都是 x 的函数，导数为 dy/dx = (v du/dx − u dv/dx) / v²。
dy/dx = (v du/dx − u dv/dx) / v²
Example: y = x / (x² + 1). Let u = x and v = x² + 1. Then du/dx = 1 and dv/dx = 2x, so dy/dx = (x² + 1 − 2x²) / (x² + 1)² = (1 − x²) / (x² + 1)².
示例：y = x / (x² + 1)。设 u = x，v = x² + 1。则 du/dx = 1，dv/dx = 2x，所以 dy/dx = (x² + 1 − 2x²) / (x² + 1)² = (1 − x²) / (x² + 1)²。
7. Implicit Differentiation | 隐函数微分
When y is not written explicitly as a function of x, differentiate both sides with respect to x and apply the chain rule to any y term.
当 y 没有写成 x 的显式函数时，对等式两边关于 x 求导，并对任何含 y 的项应用链式法则。
For x² + y² = 25, 2x + 2y dy/dx = 0, so dy/dx = −x/y
Example: x² + y² = 25. Differentiate both sides: 2x + 2y dy/dx = 0. Rearranging gives dy/dx = −x/y.
示例：x² + y² = 25。两边求导得 2x + 2y dy/dx = 0。整理后得到 dy/dx = −x/y。
8. Parametric Differentiation | 参数方程微分
If x = f(t) and y = g(t), then dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0.
若 x = f(t)，y = g(t)，则 dy/dx = (dy/dt) / (dx/dt)，前提是 dx/dt ≠ 0。
dy/dx = (dy/dt) / (dx/dt)
Example: x = t², y = 2t. Then dx/dt = 2t and dy/dt = 2, so dy/dx = 2 / 2t = 1/t.
示例：x = t²，y = 2t。则 dx/dt = 2t，dy/dt = 2，所以 dy/dx = 2 / 2t = 1/t。
9. Second Derivatives and Rates of Change | 二阶导数与变化率
The second derivative d²y/dx² is the derivative of dy/dx. It measures the rate of change of the gradient and is essential for classifying stationary points.
二阶导数 d²y/dx² 是 dy/dx 的导数。它衡量斜率的变化率，对于判断驻点类型至关重要。
d²y/dx² = d/dx(dy/dx)
For connected rates of change, use the chain rule. If A = πr² and dr/dt is given, then dA/dt = dA/dr × dr/dt = 2πr dr/dt.
对于相关变化率，使用链式法则。若 A = πr² 且已知 dr/dt，则 dA/dt = dA/dr × dr/dt = 2πr dr/dt。
10. Stationary Points and Curve Sketching | 驻点与曲线草图
Stationary points occur where dy/dx = 0. To determine their nature, evaluate the second derivative or check the sign of dy/dx either side of the point.
驻点出现在 dy/dx = 0 处。要判断其性质，可计算二阶导数，或检查该点两侧 dy/dx 的符号。
If d²y/dx² > 0, the point is a minimum; if d²y/dx² < 0, it is a maximum.
Example: y = x³ − 3x. dy/dx = 3x² − 3 = 0 gives x = ±1. Since d²y/dx² = 6x, at x = 1 we have d²y/dx² = 6 > 0, so (1, −2) is a minimum. At x = −1, d²y/dx² = −6 < 0, so (−1, 2) is a maximum.
示例：y = x³ − 3x。dy/dx = 3x² − 3 = 0 得 x = ±1。因为 d²y/dx² = 6x，在 x = 1 处 d²y/dx² = 6 > 0，所以 (1, −2) 是极小值点；在 x = −1 处 d²y/dx² = −6 < 0，所以 (−1, 2) 是极大值点。
11. Tangents and Normals | 切线与法线
The tangent to a curve at a point has gradient m = dy/dx. The normal is perpendicular to the tangent, with gradient −1/m, provided m ≠ 0.
曲线在某点的切线斜率 m = dy/dx。法线垂直于切线，斜率为 −1/m，前提是 m ≠ 0。
Tangent gradient m, normal gradient −1/m
Example: For y = x² at (1, 1), dy/dx = 2x so m = 2. The tangent is y − 1 = 2(x − 1), and the normal is y − 1 = −½(x − 1).
示例：对于 y = x² 在点 (1, 1) 处，dy/dx = 2x，所以 m = 2。切线方程为 y − 1 = 2(x − 1)，法线方程为 y − 1 = −½(x − 1)。
12. Optimisation Problems | 最优化问题
Optimisation questions ask you to find maximum or minimum values in a practical context. Express the quantity to be optimised in terms of one variable, differentiate, set the derivative to zero, and verify the nature of the stationary point.
最优化问题要求你求出实际情境中的最大值或最小值。先将需要优化的量表示为单一变量的函数，求导，令导数为零，并验证驻点性质。
Set dy/dx = 0 and test for maximum or minimum.
Example: A rectangle has perimeter 20 cm. Let one side be x and the other 10 − x. The area A = x(10 − x). Then dA/d
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