📚 Core Differentiation Rules for Edexcel A-Level Mathematics | Edexcel A-Level 数学核心微分法则
Differentiation is one of the most important skills in Edexcel A-Level Mathematics. It is used to find gradients of curves, calculate rates of change, and identify maximum or minimum points in optimisation problems. This guide brings together the key rules and techniques that candidates need to apply confidently in Pure Mathematics and in applied contexts.
微分是 Edexcel A-Level 数学中最重要的技能之一。它用于求曲线斜率、计算变化率，并在最优化问题中确定最大值或最小值点。本指南汇总了考生在纯数学和应用背景中需要熟练运用的核心法则与技巧。
1. The Derivative as a Limit | 导数的极限定义
The derivative measures the instantaneous rate of change of a function. For a curve y = f(x), the gradient at a point is found by taking the limit of the gradient of a chord as the two points get closer together.
导数度量函数在某一点的瞬时变化率。对于曲线 y = f(x)，某点处的斜率通过让弦上两点不断接近并取极限来求得。
In Edexcel notation, the derivative from first principles is written as:
在 Edexcel 记号中，从第一性原理出发的导数写作：
f'(x) = lim(h→0) [f(x+h) – f(x)] / h
The expression [f(x+h) – f(x)] / h represents the average rate of change over a small interval h. As h approaches zero, this gives the gradient of the tangent at x.
表达式 [f(x+h) – f(x)] / h 表示在小区间 h 上的平均变化率。当 h 趋近于零时，它给出 x 处切线的斜率。
Although first principles is mainly used to prove basic results, it underpins the rules below. In the exam it appears occasionally, so candidates should be able to set up the limit correctly.
虽然第一性原理主要用于证明基本结果，但它构成了以下所有法则的基础。考试中偶尔会考查此方法，因此考生应能够正确列出极限表达式。
2. Power Rule and Constant Multiple | 幂函数法则与常数倍法则
The power rule is the most frequently used differentiation rule. If n is any real constant, then the derivative of xⁿ is n xⁿ⁻¹.
幂函数法则是最常用的求导法则。如果 n 是任意实数常数，那么 xⁿ 的导数为 n xⁿ⁻¹。
d/dx (xⁿ) = n xⁿ⁻¹
For example, d/dx (x⁵) = 5x⁴ and d/dx (x¹/²) = ½ x⁻¹/². The rule also applies to negative powers, so d/dx (x⁻²) = -2x⁻³.
例如，d/dx (x⁵) = 5x⁴，d/dx (x¹/²) = ½ x⁻¹/²。该法则同样适用于负指数，因此 d/dx (x⁻²) = -2x⁻³。
If a function is multiplied by a constant, the constant can be taken outside the derivative:
如果函数乘以一个常数，该常数可以移到导数符号外：
d/dx [k f(x)] = k f'(x)
For example, d/dx (7x³) = 7 × 3x² = 21x². This combination of the power rule and constant multiple is essential for polynomial differentiation.
例如，d/dx (7x³) = 7 × 3x² = 21x²。这种幂函数法则与常数倍法则的组合是多项式求导的基础。
3. Sum and Difference Rules | 和差法则
Differentiation is linear, which means the derivative of a sum or difference is the sum or difference of the individual derivatives.
微分具有线性性质，这意味着和或差的导数等于各部分导数之和或之差。
d/dx [f(x) ± g(x)] = f'(x) ± g'(x)
This rule allows candidates to differentiate polynomial terms one by one. For example, if y = 4x³ – 2x² + 5x – 9, then dy/dx = 12x² – 4x + 5. The constant term -9 differentiates to zero.
该法则允许考生逐一求多项式中每一项的导数。例如，如果 y = 4x³ – 2x² + 5x – 9，则 dy/dx = 12x² – 4x + 5。常数项 -9 的导数为零。
It is important to keep the signs accurate, especially when a term has a negative coefficient. A common error is to lose a minus sign when differentiating an expression such as -3x².
保持符号准确非常重要，尤其是当某项系数为负时。常见错误是在求 -3x² 这类表达式的导数时丢失负号。
4. Chain Rule | 链式法则
The chain rule is used when one function is inside another. If y is a function of u, and u is a function of x, then the derivative of y with respect to x is the product of the two separate derivatives.
当一个函数嵌套在另一个函数内部时使用链式法则。如果 y 是 u 的函数，u 是 x 的函数，那么 y 对 x 的导数等于两个单独导数的乘积。
dy/dx = dy/du × du/dx
For 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 = 4(3x² + 5)³ × 6x = 24x(3x² + 5)³.
例如，求 y = (3x² + 5)⁴ 的导数。设 u = 3x² + 5，则 y = u⁴。于是 dy/du = 4u³，du/dx = 6x。相乘可得 dy/dx = 4(3x² + 5)³ × 6x = 24x(3x² + 5)³。
The chain rule also works for trigonometric and exponential functions. For example, d/dx [sin(2x)] = 2 cos(2x), and d/dx [e³ˣ] = 3e³ˣ.
链式法则也适用于三角函数和指数函数。例如，d/dx [sin(2x)] = 2 cos(2x)，d/dx [e³ˣ] = 3e³ˣ。
Forgetting to multiply by du/dx is one of the most common errors in A-Level differentiation. Always identify the inner function clearly.
忘记乘以内层函数的导数 du/dx 是 A-Level 微分中最常见的错误之一。务必先明确内层函数。
5. Product Rule | 积法则
The product rule is used when differentiating the product of two functions. If y = u v, where u and v are both functions of x, then the derivative is formed by taking u times the derivative of v plus v times the derivative of u.
当对两个函数的乘积求导时使用积法则。如果 y = u v，其中 u 和 v 都是 x 的函数，那么导数等于 u 乘以 v 的导数加上 v 乘以 u 的导数。
dy/dx = u dv/dx + v du/dx
For example, differentiate y = x² sin x. Let u = x² and v = sin x. Then du/dx = 2x and dv/dx = cos x. The derivative is dy/dx = x² cos x + 2x sin x.
例如，求 y = x² sin x 的导数。设 u = x²，v = sin x。则 du/dx = 2x，dv/dx = cos x。导数为 dy/dx = x² cos x + 2x sin x。
The order of the two terms does not matter for the product rule, but candidates should keep structure clear. In some exam questions, factorising the answer may be useful for finding stationary points.
积法则中两项的顺序并不重要，但考生应保持结构清晰。在一些考试题中，将答案因式分解可能有助于求驻点。
6. Quotient Rule | 商法则
The quotient rule is used when one function is divided by another. If y = u / v, where u and v are functions of x, then the derivative has a denominator of v² and a numerator generated by v times du/dx minus u times dv/dx.
当一个函数除以另一个函数时使用商法则。如果 y = u / v，其中 u 和 v 是 x 的函数，那么导数的分母为 v²，分子为 v 乘以 du/dx 减去 u 乘以 dv/dx。
dy/dx = (v du/dx – u dv/dx) / v²
For example, differentiate y = (2x + 1) / (x² + 3). Let u = 2x + 1 and v = x² + 3. Then du/dx = 2, dv/dx = 2x. Substituting into the formula gives dy/dx = [(x² + 3)(2) – (2x + 1)(2x)] / (x² + 3)².
例如，求 y = (2x + 1) / (x² + 3) 的导数。设 u = 2x + 1，v = x² + 3。则 du/dx = 2，dv/dx = 2x。代入公式可得 dy/dx = [(x² + 3)(2) – (2x + 1)(2x)] / (x² + 3)²。
After simplifying the numerator, this becomes dy/dx = (6 – 2x – 2x²) / (x² + 3)². Candidates should take care with the subtraction, as the order in the numerator is a frequent source of sign errors.
化简分子后得到 dy/dx = (6 – 2x – 2x²) / (x² + 3)²。考生应特别注意减法运算，因为分子中的顺序是符号错误的常见来源。
7. Second Derivatives and Concavity | 二阶导数与凹凸性
The second derivative is the derivative of the first derivative. In Edexcel questions it is written as d²y/dx² or f”(x). It measures how the gradient itself is changing.
二阶导数是导数的导数。在 Edexcel 考试题中它写作 d²y/dx² 或 f”(x)。它度量斜率本身的变化率。
d²y/dx² = d/dx (dy/dx)
For example, if y = x⁴ – 3x² + 2, then dy/dx = 4x³ – 6x and d²y/dx² = 12x² – 6. At x = 1, the second derivative is 6, which is positive, so the curve is concave up at that point.
例如，如果 y = x⁴ – 3x² + 2，则 dy/dx = 4x³ – 6x，d²y/dx² = 12x² – 6。在 x = 1 处，二阶导数为 6，是正值，因此曲线在该点向上凹。
If d²y/dx² > 0, the curve is concave up; if d²y/dx² < 0, the curve is concave down. A point where the concavity changes from one to the other is called a point of inflection.
如果 d²y/dx² > 0，曲线向上凹；如果 d²y/dx² < 0，曲线向下凹。凹凸性发生变化的点称为拐点。
8. Tangents and Normals | 切线与法线
The derivative at a point gives the gradient of the tangent to the curve at that point. This is enough to find the equation of both the tangent and the normal.
函数在某点的导数值给出曲线在该点切线的斜率。这足以求出切线和法线的方程。
If a curve passes through the point (a, b) and the gradient of the tangent is m, then:
如果曲线经过点 (a, b)，且切线斜率为 m，则：
Tangent: y – b = m(x – a)
Normal: y – b = (-1/m)(x – a)
For example, y = x³ at x = 2 has y = 8 and dy/dx = 3x² = 12. The tangent is y – 8 = 12(x – 2), and the normal is y – 8 = (-1/12)(x – 2).
例如，曲线 y = x³ 在 x = 2 处有 y = 8，dy/dx = 3x² = 12。切线为 y – 8 = 12(x – 2)，法线为 y – 8 = (-1/12)(x – 2)。
When writing the final answer, the equation can be left in point-gradient form or simplified to y = mx + c. The normal gradient must be the negative reciprocal of the tangent gradient.
书写最终答案时，方程可以保留点斜式，也可以化简为 y = mx + c。法线斜率必须是切线斜率的负倒数。
9. Stationary Points and Optimisation | 驻点与最优化
A stationary point occurs where dy/dx = 0. At such a point the tangent is horizontal, and it can be a local maximum, a local minimum, or a point of inflection.
当 dy/dx = 0 时出现驻点。在这种点处切线水平，它可能是局部最大值、局部最小值或拐点。
The second derivative is commonly used to classify stationary points. If d²y/dx² > 0, the point is a local minimum; if d²y/dx² < 0, the point is a local maximum.
二阶导数通常用于判断驻点类型。如果 d²y/dx² > 0，该点是局部最小值；如果 d²y/dx² < 0，该点是局部最大值。
For example, y = x³ – 3x gives dy/dx = 3x² – 3 = 3(x – 1)(x + 1). The stationary points are at x = 1 and x = -1. Since d²y/dx² = 6x, at x = 1 the second derivative is 6, so there is a minimum; at x = -1 it is -6, so there is a maximum.
例如，y = x³ – 3x 给出 dy/dx = 3x² – 3 = 3(x – 1)(x + 1)。驻点位于 x = 1 和 x = -1。由于 d²y/dx² = 6x，在 x = 1 处二阶导数为 6，因此
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