2 Specific Skills: Differentiation and Integration Techniques | 专项技能2：微分与积分技巧

📚 2 Specific Skills: Differentiation and Integration Techniques | 专项技能2：微分与积分技巧

In A-Level Edexcel Mathematics, differentiation and integration are not just isolated topics; they are the two specific skills that underpin much of pure mathematics and applied modelling. Mastering these techniques will allow you to handle rates of change, curve analysis, area calculation, and the solution of differential equations with confidence. This article provides a structured revision of the core differentiation and integration methods required for success in the Edexcel examinations.

在A-Level Edexcel数学中，微分与积分并非孤立的课题，而是支撑纯数学与应用建模的两项专项技能。掌握这些技巧将使你能够自信地处理变化率、曲线分析、面积计算以及微分方程的求解。本文对Edexcel考试所需的核心微分与积分方法进行了结构化复习。

1. Introduction to Differentiation | 微分简介

Differentiation is the process of finding the derivative of a function, which measures the instantaneous rate of change at a point. The notation f'(x), dy/dx, or d/dx[f(x)] is used interchangeably. In Edexcel Pure Mathematics, you are expected to differentiate polynomial, trigonometric, exponential and logarithmic functions, and to apply derivatives to find gradients of curves, tangents and normals.

微分是求函数导数的过程，导数衡量函数在某一点的瞬时变化率。记号 f'(x)、dy/dx 或 d/dx[f(x)] 可以互换使用。在Edexcel纯数学中，你需要对多项式、三角函数、指数函数和对数函数求导，并应用导数求曲线的梯度、切线和法线。

2. Basic Derivative Formulas | 基本求导公式

The foundational rules for differentiation are essential to memorise. The power rule states that for y = xⁿ, dy/dx = n x^n–1. Exponential and logarithmic derivatives follow: d/dx (e^x) = e^x, and d/dx (ln x) = 1/x. Trigonometric derivatives include d/dx (sin x) = cos x, d/dx (cos x) = –sin x, and d/dx (tan x) = sec²x. Constants are straightforward: d/dx (k) = 0, and d/dx (k f(x)) = k f'(x).

基本的求导法则是必须熟记的。幂函数法则指出，对于 y = xⁿ，dy/dx = n x^n–1。指数和对数函数的导数分别为：d/dx (e^x) = e^x，d/dx (ln x) = 1/x。三角函数的导数包括：d/dx (sin x) = cos x，d/dx (cos x) = –sin x，d/dx (tan x) = sec²x。常数的导数为零：d/dx (k) = 0，且 d/dx (k f(x)) = k f'(x)。

These formulas form the building blocks for more advanced techniques such as the chain, product and quotient rules.

这些公式构成了链式法则、乘积法则和商法则等更高级技巧的基础。

3. The Chain Rule | 链式法则

The chain rule is used to differentiate composite functions. If y = f(u) and u = g(x), then dy/dx = dy/du × du/dx. In function notation, the derivative of f(g(x)) is f'(g(x)) g'(x). A typical Edexcel example is differentiating y = (3x² + 5)⁴. Set u = 3x² + 5, then y = u⁴, dy/du = 4u³, du/dx = 6x. Hence dy/dx = 4u³ × 6x = 24x(3x² + 5)³.

链式法则用于复合函数的求导。若 y = f(u) 且 u = g(x)，则 dy/dx = dy/du × du/dx。用函数记号表示，f(g(x)) 的导数为 f'(g(x)) g'(x)。Edexcel典型例题是对 y = (3x² + 5)⁴ 求导。令 u = 3x² + 5，则 y = u⁴，dy/du = 4u³，du/dx = 6x。因此 dy/dx = 4u³ × 6x = 24x(3x² + 5)³。

4. The Product Rule | 乘积法则

For the product of two functions u(x) and v(x), the product rule is d/dx (uv) = u dv/dx + v du/dx. It is vital to maintain the correct order: differentiate one function at a time and sum the results. For example, if y = x² sin x, let u = x², v = sin x. Then du/dx = 2x, dv/dx = cos x. So dy/dx = 2x sin x + x² cos x. When simplifying, look for common factors to present your answer neatly.

对于两个函数的乘积 u(x) 和 v(x)，乘积法则为 d/dx (uv) = u dv/dx + v du/dx。关键在于保持正确的顺序：每次对一个函数求导，然后求和。例如，若 y = x² sin x，令 u = x²，v = sin x，则 du/dx = 2x，dv/dx = cos x。于是 dy/dx = 2x sin x + x² cos x。在进行化简时，寻找公因子可使答案更整洁。

5. The Quotient Rule | 商法则

The quotient rule is employed when differentiating a fraction. If y = u / v, then dy/dx = (v du/dx – u dv/dx) / v². A crucial tip is to square the denominator and be careful with signs. Consider y = (x² + 1) / sin x. Here u = x² + 1, du/dx = 2x; v = sin x, dv/dx = cos x. Substituting into the formula yields dy/dx = [2x sin x – (x² + 1) cos x] / sin²x. No further simplification is needed unless common factors appear.

商法则用于对分式进行求导。若 y = u / v，则 dy/dx = (v du/dx – u dv/dx) / v²。关键提示：将分母平方，并注意符号。考虑 y = (x² + 1) / sin x。此时 u = x² + 1，du/dx = 2x；v = sin x，dv/dx = cos x。代入公式得 dy/dx = [2x sin x – (x² + 1) cos x] / sin²x。除非出现公因子，否则无需进一步化简。

6. Implicit Differentiation | 隐函数微分

Implicit differentiation is used when y is not explicitly given as a function of x. Differentiate each term with respect to x, treating y as a function of x and applying the chain rule. For instance, for the circle equation x² + y² = 25, differentiate to get 2x + 2y dy/dx = 0. Solving gives dy/dx = –x/y. Implicit differentiation often leads to expressions involving both x and y, which is perfectly acceptable in Edexcel exam answers.

当 y 未显式表示成 x 的函数时，使用隐函数微分。对每一项关于 x 求导，将 y 视为 x 的函数并运用链式法则。例如，对于圆的方程 x² + y² = 25，求导得 2x + 2y dy/dx = 0。解得 dy/dx = –x/y。隐函数微分常得到同时包含 x 和 y 的表达式，这在 Edexcel 考试答案中是完全可接受的。

7. Parametric Differentiation | 参数方程微分

When a curve is defined parametrically by x = f(t) and y = g(t), the derivative dy/dx is given by (dy/dt) / (dx/dt), provided dx/dt ≠ 0. For example, given x = t² + 2t and y = t³ – 3t, first compute dx/dt = 2t + 2 and dy/dt = 3t² – 3. Then dy/dx = (3t² – 3) / (2t + 2). Factorising gives 3(t² – 1) / 2(t + 1) = 3(t – 1)(t + 1) / 2(t + 1) = 3(t – 1)/2, after cancelling (t + 1) provided t ≠ –1.

当曲线由参数方程 x = f(t) 和 y = g(t) 定义时，导数 dy/dx 等于 (dy/dt) / (dx/dt)，前提是 dx/dt ≠ 0。例如，给定 x = t² + 2t 和 y = t³ – 3t，首先计算 dx/dt = 2t + 2 与 dy/dt = 3t² – 3。则 dy/dx = (3t² – 3) / (2t + 2)。因式分解得 3(t² – 1) / 2(t + 1) = 3(t – 1)(t + 1) / 2(t + 1)。在 t ≠ –1 时约去 (t + 1) 得到 3(t – 1)/2。

8. Applications of Derivatives | 导数的应用

Derivatives have numerous practical applications. To find the equation of a tangent at x = a, calculate the gradient m = f'(a) and use y – f(a) = m(x – a). For a normal, the gradient is –1/m. Stationary points occur where f'(x) = 0. Their nature is determined by the second derivative: if f”(x) > 0, it is a local minimum; if f”(x) < 0, a local maximum. In optimisation problems, identify the quantity to be maximised or minimised, express it as a function of one variable, differentiate, and set f'(x) = 0. Always verify that your solution makes physical sense.

导数有许多实际应用。要求 x = a 处的切线方程，计算梯度 m = f'(a) 并使用 y – f(a) =

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）