📚 IGCSE Mathematics: Calculus Basics – Key Points | IGCSE 数学：微积分基础 考点精讲
Calculus is the branch of mathematics that deals with rates of change and the accumulation of quantities. In IGCSE Mathematics, the focus is on the fundamental ideas of differentiation and integration, with a strong emphasis on polynomial expressions. Mastering these core skills allows you to solve problems involving gradients of curves, tangents, turning points, and areas under graphs. This guide provides a clear, bilingual walkthrough of every essential concept and technique you will encounter in the exam.
微积分是研究变化率与累积量的数学分支。在 IGCSE 数学中，重点放在微分和积分的基本思想上，尤其是针对多项式表达式的运算。掌握这些核心技能后，你就能解决涉及曲线梯度、切线、转向点以及图形下方面积等问题。本指南将以清晰的中英双语形式，带你逐一梳理考试中会遇到的每一个重要概念和解题技巧。
1. What is Calculus? | 什么是微积分？
function, while integration allows us to find the total accumulation of a quantity, such as the area under a curve. In IGCSE, we work primarily with functions of the form y = axⁿ, where n is a rational number, and we learn systematic rules to differentiate and integrate them.
微积分主要分为两大分支：微分和积分。微分帮助我们找到函数的瞬时变化率，而积分则让我们能够求出一个量的累积总和，例如曲线下方的面积。在 IGCSE 阶段，我们主要处理 y = axⁿ（其中 n 为有理数）形式的函数，并学习系统化的法则来对它们进行微分和积分。
2. The Gradient of a Curve and the Derivative | 曲线的梯度与导数
stant, but the gradient of a curve changes from point to point. Differentiation gives us a formula for the gradient at any point on the curve. We write the derivative of y with respect to x as dy/dx or f' erivative dy/dx tells us the slope of the tangent to the curve at a specific x-coordinate.
直线的梯度是恒定的，但曲线上各点的梯度却在不断变化。微分能给我们提供曲线上任意一点处的梯度公式。我们将 y 关于 x 的导数记作 dy/dx 或 f'(x)。对于函数 y = f(x)，导数 dy/dx 可以给出在某个特定 x 坐标处，曲线切线的斜率。
3. Power Rule for Differentiation | 幂函数的求导法则
a n xⁿ⁻¹. This rule works for any real number n, but in IGCSE n is usually a positive or negative integer, or a simple fraction. Always remember to multiply by the power and then reduce the power by 1.
IGCSE 微积分中最重要的法则是幂函数求导法则：如果 y = xⁿ，那么 dy/dx = n xⁿ⁻¹。更一般地，如果 y = a xⁿ（a 为常数），那么 dy/dx = a n xⁿ⁻¹。该法则适用于任何实数 n，但在 IGCSE 中 n 通常为正整数、负整数或简单的分数。请务必牢记：乘以指数，然后将指数减 1。
Example / 示例: If y = 3x⁴, then dy/dx = 3 × 4 × x⁴⁻¹ = 12x³.
4. Sum and Constant Multiple Rules | 和与常数倍法则
y constant factor can be taken outside the differentiation operator. For example, if y = 5x³ + 2x² − 7x + 4, then dy/dx = 15x² + 4x − 7. Note that the derivative of a constant (like +4) is zero because a constant has no rate of change.
当函数由多个项相加而成时，可以逐项进行微分。和的导数等于导数的和。此外，任何常数因子都可以提到微分运算外面。例如，若 y = 5x³ + 2x² − 7x + 4，则 dy/dx = 15x² + 4x − 7。注意，常数项（如 +4）的导数为零，因为常数没有变化率。
5. Finding the Equation of a Tangent | 求切线方程
tangent. Next, use the straight-line formula y − y₁ = m(x − x₁) to obtain the equation. Sometimes the question gives only the x-coordinate, so you need to find y₁ by substituting into the original function.
若要找出曲线上点 P(x₁, y₁) 处的切线方程，首先进行微分求得 dy/dx，然后代入 x = x₁ 算出切线的斜率 m。接着，利用直线公式 y − y₁ = m(x − x₁) 得到方程。有时题目只给出 x 坐标，此时你需要代入原函数求出 y₁。
6. Stationary Points and Turning Points | 驻点与转向点
mum or a minimum. To find the coordinates of stationary points, set dy/dx = 0 and solve for x, then find the corresponding y-values from the original function.
当曲线梯度为零，即 dy/dx = 0 时，就会出现驻点。在这些点上，切线是水平的。转向点可能是极大值点或极小值点。要找出驻点的坐标，先令 dy/dx = 0 并解出 x，再通过原函数求出对应的 y 值。
7. Determining Nature of Turning Points | 判断转向点的性质
To decide whether a stationary point is a maximum or a minimum, you can use the second derivative test: find d²y/dx² by differentiating dy/dx. Substitute the x-coordinate of the stationary point. If d²y/dx² > 0, the point is a minimum; if d²y/dx² < 0, it is a maximum. Alternatively, you can examine the sign of dy/dx just before and after the point.
要判断驻点是极大值点还是极小值点，可以使用二阶导数检验法：对 dy/dx 再求导得到 d²y/dx²。代入驻点的 x 坐标。若 d²y/dx² > 0，则该点为极小值点；若 d²y/dx² < 0，则为极大值点。另一种方法是，检查该点前后 dy/dx 的符号变化。
8. Introduction to Integration | 积分入门
ral of f(x) with respect to x is written as ∫ f(x) dx = F(x) + C, where C is the constant of integration. In IGCSE, integration is mainly applied to polynomial functions and used to calculate areas under curves.
积分是微分的逆过程。当已知一个函数的导数 f(x) 时，可以用积分来求原函数 F(x)。f(x) 关于 x 的不定积分写作 ∫ f(x) dx = F(x) + C，其中 C 是积分常数。在 IGCSE 中，积分主要应用于多项式函数，并用于计算曲线下方的面积。
9. Power Rule for Integration | 幂函数的积分法则
his rule fails when n = −1, but such cases are not tested at IGCSE level. Always add the constant C when evaluating an indefinite integral.
幂函数的积分法则是：∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C，其中 n ≠ −1。对于常数倍，∫ a xⁿ dx = a (xⁿ⁺¹)/(n+1) + C。请记住，当 n = −1 时该法则失效，但这种情况不在 IGCSE 考试范围内。计算不定积分时，务必加上常数 C。
10. Definite Integrals and Area Under a Curve | 定积分与曲线下面积
nd x = b. We compute it as [F(x)]ₐᵇ = F(b) − F(a), where F(x) is any antiderivative of f(x). If the curve lies below the x-axis, the integral gives a negative value, so for total area we must split the interval and take absolute values where needed.
定积分带有上下限，其结果是一个数值，表示曲线 y = f(x)、x 轴以及直线 x = a 和 x = b 所围成的面积。计算方法为 [F(x)]ₐᵇ = F(b) − F(a)，其中 F(x) 是 f(x) 的任意一个原函数。如果曲线位于 x 轴下方，积分值会为负，因此求总面积时需分段，并在必要的地方取绝对值。
Area = ∫ₐᵇ f(x) dx = F(b) − F(a)
11. Applications and Exam Tips | 应用与考试技巧
Common IGCSE problems involve combining differentiation and integration: you may be given a gradient function dy/dx and a point on the curve, and asked to find the equation of the curve by integrating. Always use the given point to determine the constant C. For area questions, sketch the graph if possible to check which regions are above or below the x-axis. Show clear steps: differentiate/integrate term by term, substitute carefully, and express the final answer in simplest form.
常见的 IGCSE 题目会将微分与积分结合起来：比如给出梯度函数 dy/dx 以及曲线上的一点，要求通过积分求出曲线方程。此时一定要利用给定的点来确定常数 C。对于面积问题，尽可能画出草图，以便检查哪些区域在 x 轴上方或下方。解题时要展示清晰步骤：逐项微分或积分、仔细代入，并将最终答案化为最简形式。
12. Summary of Key Formulas | 核心公式总结
The following table summarises the essential differentiation and integration rules for IGCSE. Keep them on a revision card and practise applying them until they become second nature.
下表总结了 IGCSE 中关键的微分和积分法则。将它们记在复习卡片上，并反复练习运用，直到完全得心应手。
| Operation / 运算 | Rule / 法则 |
|---|---|
| Differentiation: y = a xⁿ | dy/dx = a n xⁿ⁻¹ |
| Second derivative | d²y/dx² = a n(n−1) xⁿ⁻² |
| Indefinite integration: ∫ a xⁿ dx | a xⁿ⁺¹/(n+1) + C, n ≠ −1 |
| Definite integral: ∫ₐᵇ f(x) dx | F(b) − F(a), where F'(x) = f(x) |
With a solid grasp of these fundamentals, you will be able to tackle any calculus question in the IGCSE exam with confidence.
牢牢掌握这些基础知识后，你就能满怀信心地应对 IGCSE 考试中的任何微积分题目了。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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