📚 GCSE CIE Maths: Calculus Essentials – Key Exam Points | GCSE CIE 数学：微积分基础 考点精讲
Calculus is a fundamental topic in the CIE GCSE Mathematics syllabus that introduces the concepts of change and accumulation. It empowers you to analyse curves, optimise functions, and solve real-world problems involving motion and area. This revision guide breaks down every essential point you need for exam success, from differentiation rules to integration and their applications.
微积分是 CIE GCSE 数学大纲中的一个基础主题，介绍了变化与累积的概念。它能帮助你分析曲线、优化函数，并解决涉及运动与面积的实际问题。本复习指南将考试所需的每一个要点逐一分解，从求导法则到积分及其应用，助你考试成功。
1. What is Differentiation? | 什么是微分？
Differentiation is the process of finding the derivative of a function. The derivative measures how a function’s output changes as its input changes – in other words, it gives the gradient (slope) of a curve at any point. For a function y = f(x), the derivative is denoted as f'(x) or dy/dx.
微分是求函数导数的过程。导数衡量函数输出随输入变化的速度——换句话说，它给出了曲线上任意点的梯度（斜率）。对于函数 y = f(x)，导数记作 f'(x) 或 dy/dx。
In GCSE, we mainly work with polynomial functions, and differentiation helps us find the instantaneous rate of change, such as velocity from a displacement-time graph.
在 GCSE 中，我们主要处理多项式函数，微分帮助我们找到瞬时变化率，例如从位移-时间图中求速度。
2. The Power Rule | 幂函数求导法则
The most important rule for differentiating simple powers of x is the power rule: if y = xⁿ, then dy/dx = n xⁿ⁻¹. Multiply by the original power and reduce the power by one.
对 x 的简单幂函数求导最重要的法则是幂法则：若 y = xⁿ，则 dy/dx = n xⁿ⁻¹。乘以原来的指数，再将指数减1。
Example: y = x³ → dy/dx = 3x².
示例：y = x³ → dy/dx = 3x²。
This rule also applies to terms with coefficients: if y = a xⁿ, then dy/dx = a n xⁿ⁻¹.
该法则同样适用于带有系数的项：若 y = a xⁿ，则 dy/dx = a n xⁿ⁻¹。
Remember that x to the power of zero is 1, so differentiating a constant gives zero.
记住 x 的零次方为 1，因此常数的导数为零。
3. Differentiating Polynomials | 多项式求导
To differentiate a polynomial, apply the power rule to each term individually and sum the results – this is the sum rule. For example, y = 4x⁵ – 2x² + 3x – 7. Differentiating term by term: dy/dx = 20x⁴ – 4x + 3. The constant -7 becomes 0.
对多项式求导时，将幂法则分别应用于每一项，然后求和——这就是和法则。例如，y = 4x⁵ – 2x² + 3x – 7。逐项求导：dy/dx = 20x⁴ – 4x + 3。常数 -7 的导数为 0。
Always check for negative and fractional powers, but in the core CIE GCSE syllabus, powers are usually positive integers. However, you may be required to rewrite terms like 1/x² as x⁻² before differentiating.
对于负指数和分数指数要格外留意，不过在 CIE GCSE 核心大纲中，指数通常为正整数。不过，你可能会需要先将 1/x² 等项改写为 x⁻² 再求导。
4. Finding the Gradient of a Curve | 求曲线的梯度
Once you have the derivative dy/dx, you can find the gradient at a specific point by substituting the x-coordinate. For y = x² – 3x + 2, the gradient at x = 2 is given by dy/dx = 2x – 3 → 2(2) – 3 = 1.
一旦求得导数 dy/dx，你就能通过代入 x 坐标求出某一点的梯度。对于 y = x² – 3x + 2，x = 2 处的梯度为 dy/dx = 2x – 3 → 2(2) – 3 = 1。
This tells you the slope of the tangent line to the curve at that point. A positive gradient means the function is increasing; negative means decreasing.
这告诉你该点处曲线切线的斜率。正梯度表示函数递增；负梯度表示递减。
5. Equation of Tangent and Normal | 切线与法线方程
The tangent to a curve at a point is a straight line that touches the curve and has the same gradient as the curve at that point. Its equation can be found using y – y₁ = m (x – x₁), where m is the gradient dy/dx at that point.
曲线上某一点的切线是一条与该点处曲线相切且梯度相同的直线。其方程可用点斜式 y – y₁ = m (x – x₁) 求得，其中 m 为该点处的梯度 dy/dx。
The normal is perpendicular to the tangent. If the tangent’s gradient is m, the normal’s gradient is -1/m (provided m ≠ 0).
法线与切线垂直。若切线斜率为 m，则法线斜率为 -1/m（前提是 m ≠ 0）。
Example: Find the tangent to y = x² at x = 1. dy/dx = 2x, so m = 2. Point (1,1). Tangent: y – 1 = 2(x – 1) → y = 2x – 1.
示例：求 y = x² 在 x = 1 处的切线。dy/dx = 2x，因此 m = 2。点 (1,1)。切线：y – 1 = 2(x – 1) → y = 2x – 1。
6. Stationary Points & Turning Points | 驻点与极值点
Stationary points occur where the gradient is zero, i.e., dy/dx = 0. These points can be maximum, minimum, or points of inflection. To find them, set the first derivative equal to zero and solve for x.
驻点出现在梯度为零的位置，即 dy/dx = 0。这些点可能是极大值点、极小值点或拐点。要求驻点，令一阶导数为零并解出 x。
For example, y = x² – 4x + 5. dy/dx = 2x – 4. Set 2x – 4 = 0 → x = 2. Substitute back to find y = (2)² – 4(2) + 5 = 1. The stationary point is (2,1).
例如，y = x² – 4x + 5。dy/dx = 2x – 4。令 2x – 4 = 0 → x = 2。代回得 y = (2)² – 4(2) + 5 = 1。驻点为 (2,1)。
7. Second Derivative & Nature of Turning Points | 二阶导数与极值点判定
The second derivative, denoted d²y/dx² or f”(x), is the derivative of the first derivative. It tells you about the curvature of the function. If d²y/dx² > 0 at a stationary point, the curve is concave up → minimum point. If < 0, concave down → maximum point.
二阶导数，记作 d²y/dx² 或 f”(x)，是一阶导数的导数。它反映了函数的凹凸性。若在驻点处 d²y/dx² > 0，则曲线凹向上 → 极小值点；若 < 0，凹向下 → 极大值点。
If the second derivative equals zero, the test is inconclusive, and you should examine the sign of the first derivative on either side. However, this is less common at GCSE level.
若二阶导数为零，则测试无法判定，应检查两侧一阶导数的符号。但这在 GCSE 阶段较少见。
For y = x² – 4x + 5: dy/dx = 2x – 4, d²y/dx² = 2 > 0, so (2,1) is a minimum.
对于 y = x² – 4x + 5：dy/dx = 2x – 4，d²y/dx² = 2 > 0，故 (2,1) 为极小值点。
8. Introduction to Integration | 积分基础
Integration is the reverse process of differentiation. If you know the derivative, integration helps you find the original function (antiderivative). It is also used to calculate the area under a curve.
积分是微分的逆运算。如果已知导数，积分能帮你
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