A-Level Edexcel Maths: Differentiation Techniques | A-Level Edexcel 数学：微分技巧

📚 A-Level Edexcel Maths: Differentiation Techniques | A-Level Edexcel 数学：微分技巧

This revision guide covers the differentiation techniques required for Edexcel A-Level Mathematics. Differentiation underpins many pure, mechanics and applied questions, so a confident command of the rules below will strengthen Paper 1 and Paper 2 performance.

本复习指南涵盖 Edexcel A-Level 数学中要求的微分技巧。微分是许多纯数学、力学和应用题的基础，因此熟练掌握以下法则将提升 Paper 1 和 Paper 2 的考试表现。

1. Core Derivatives and Notation | 核心导数与记号

In Edexcel A-Level Mathematics, you are expected to use standard derivative results fluently. The notation dy/dx is used for the derivative of y with respect to x, and f'(x) is used for the derivative of f(x). If a question gives displacement, velocity or acceleration, the same notation may appear as ds/dt, dv/dt or d²s/dt².

在 Edexcel A-Level 数学中，你需要熟练运用标准导数结果。dy/dx 表示 y 对 x 的导数，f'(x) 表示 f(x) 的导数。如果题目涉及位移、速度或加速度，同样的记号可能写成 ds/dt、dv/dt 或 d²s/dt²。

d/dx (xⁿ) = n xⁿ⁻¹, d/dx (eˣ) = eˣ, d/dx (ln x) = 1/x

Function	Derivative
xⁿ	n xⁿ⁻¹
eˣ	eˣ
ln x	1/x
sin x	cos x
cos x	-sin x
tan x	sec² x

You also need the constant multiple rule and the sum rule: if y = a u + b v, then dy/dx = a du/dx + b dv/dx. These allow you to differentiate polynomials term by term, such as y = 4x³ – 5x² + 7x – 3, giving dy/dx = 12x² – 10x + 7.

你还需要掌握常数倍法则和加法法则：如果 y = a u + b v，则 dy/dx = a du/dx + b dv/dx。这些法则允许你逐项微分多项式，例如 y = 4x³ – 5x² + 7x – 3，得到 dy/dx = 12x² – 10x + 7。

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2. Chain Rule | 链式法则

The chain rule is used for composite functions of the form y = f(g(x)). It states that if y = f(u) and u = g(x), then dy/dx = dy/du × du/dx. This rule is essential for powers, exponentials, logarithms and trigonometric functions with linear or non-linear inner functions.

链式法则用于 y = f(g(x)) 形式的复合函数。它表明如果 y = f(u) 且 u = g(x)，则 dy/dx = dy/du × du/dx。该法则对于幂函数、指数函数、对数函数以及具有线性或非线性内层函数的三角函数至关重要。

dy/dx = dy/du × du/dx

Example: if y = (3x² + 1)⁵, let u = 3x² + 1. Then du/dx = 6x and y = u⁵, so dy/du = 5u⁴. Multiplying gives dy/dx = 5u⁴ × 6x = 30x(3x² + 1)⁴.

示例：如果 y = (3x² + 1)⁵，设 u = 3x² + 1。则 du/dx = 6x，且 y = u⁵，所以 dy/du = 5u⁴。相乘得到 dy/dx = 5u⁴ × 6x = 30x(3x² + 1)⁴。

For e^{kx}, the derivative is k e^{kx}. For ln(ax + b), the derivative is a/(ax + b). These are direct applications of the chain rule and appear regularly in Edexcel pure mathematics papers.

对于 e^{kx}，其导数为 k e^{kx}。对于 ln(ax + b)，其导数为 a/(ax + b)。这些都是链式法则的直接应用，经常出现在 Edexcel 纯数学试卷中。

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3. Product Rule | 乘积法则

When differentiating a product of two functions, use the product rule. If y = u v, where u = f(x) and v = g(x), then dy/dx = u dv/dx + v du/dx. This is often written as dy/dx = u v’ + v u’.

当对两个函数的乘积进行微分时，使用乘积法则。如果 y = u v，其中 u = f(x)，v = g(x)，则 dy/dx = u dv/dx + v du/dx。这通常写作 dy/dx = u v’ + v u’。

d/dx (u v) = u dv/dx + v du/dx

Example: differentiate y = x² sin x. Let u = x² and v = sin x. Then du/dx = 2x and dv/dx = cos x. Therefore dy/dx = x² cos x + 2x sin x.

示例：对 y = x² sin x 求导。设 u = x²，v = sin x。则 du/dx = 2x，dv/dx = cos x。因此 dy/dx = x² cos x + 2x sin x。

Sometimes the result can be factorised. For y = x eˣ, dy/dx = x eˣ + eˣ = eˣ(x + 1). Factorising can help when locating stationary points or simplifying a final answer.

有时结果可以因式分解。对于 y = x eˣ，dy/dx = x eˣ + eˣ = eˣ(x + 1)。因式分解有助于确定驻点位置或简化最终答案。

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4. Quotient Rule | 商法则

For a quotient of two functions, use the quotient rule. If y = u/v, where u = f(x) and v = g(x), then dy/dx = (v du/dx – u dv/dx) / v². Pay close attention to the minus sign, which is a common source of error.

对于两个函数的商，使用商法则。如果 y = u/v，其中 u = f(x)，v = g(x)，则 dy/dx = (v du/dx – u dv/dx) / v²。请特别注意负号，这是常见的错误来源。

d/dx (u/v) = (v du/dx – u dv/dx) / v²

Example: differentiate y = x/(x + 1). Let u = x and v = x + 1. Then du/dx = 1 and dv/dx = 1. The quotient rule gives dy/dx = ((x + 1)(1) – x(1)) / (x + 1)² = 1/(x + 1)².

示例：对 y = x/(x + 1) 求导。设 u = x，v = x + 1。则 du/dx = 1，dv/dx = 1。商法则给出 dy/dx = ((x + 1)(1) – x(1)) / (x + 1)² = 1/(x + 1)²。

If you reverse the subtraction and write u dv/dx – v du/dx, the entire sign of the derivative changes. Always check that the denominator term v du/dx comes first in the numerator.

如果你把减法顺序颠倒，写成 u dv/dx – v du/dx，导数的符号会完全改变。务必确认分子中先写分母项 v du/dx。

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5. Tangents and Normals | 切线与法线

Differentiation is used to find the gradient of a curve at a given point. The tangent has gradient m = dy/dx at that point. The normal is perpendicular to the tangent, so its gradient is -1/m, provided m is not zero.

微分可用于求曲线上某一点处的梯度。切线在该点处的梯度为 m = dy/dx。法线垂直于切线，因此其梯度为 -1/m，前提是 m 不为零。

Gradient of normal = -1 / (gradient of tangent)

Example: find the tangent and normal to y = x³ – 2x at x = 1. First, dy/dx = 3x² – 2. At x = 1, m = 1 and y = 1 – 2 = -1. The tangent is y + 1 = 1(x – 1), giving y = x – 2. The normal has gradient -1, so y + 1 = -1(x – 1), giving y = -x.

示例：求 y = x³ – 2x 在 x = 1 处的切线和法线。首先，dy/dx = 3x² – 2。在 x = 1 处，m = 1，且 y = 1 – 2 = -1。切线为 y + 1 = 1(x – 1)，即 y = x – 2。法线梯度为 -1，所以 y + 1 = -1(x – 1)，即 y = -x。

When writing the equation of a tangent or normal, use y – y₁ = m(x – x₁). If the gradient is zero, the tangent is horizontal and the normal is the vertical line x = x₁.

写切线或法线方程时，使用 y – y₁ = m(x – x₁)。如果梯度为零，切线为水平线，法线为竖直线 x = x₁。

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6. Stationary Points and the Second Derivative | 驻点与二阶导数

Stationary points occur where dy/dx = 0. These can be maximum points, minimum points or points of inflection. To determine the nature of a stationary point, use the second derivative or inspect the sign of dy/dx on either side of the point.

驻点出现在 dy/dx = 0 处。这些可能是极大值点、极小值点或拐点。要确定驻点的类型，可以使用二阶导数，或者检查驻点两侧 dy/dx 的符号。

If d²y/dx² > 0, the point is a minimum; if d²y/dx² < 0, the point is a maximum.

Example: find the stationary points of y = x³ – 3x. Differentiate: dy/dx = 3x² – 3. Solve 3x² – 3 = 0 to get x = 1 and x = -1. The second derivative is d²y/dx² = 6x. At x = 1, d²y/dx² = 6 > 0, so there is a minimum. At x = -1, d²y/dx² = -6 < 0, so there is a maximum.

示例：求 y = x³ – 3x 的驻点。求导：dy/dx = 3x² – 3。解 3x² – 3 = 0，得到 x = 1 和 x = -1。二阶导数为 d²y/dx² = 6x。在 x = 1 处，d²y/dx² = 6 > 0，因此为极小值点；在 x = -1 处，d²y/dx² = -6 < 0，因此为极大值点。

If d²y/dx² = 0 at a stationary point,

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