📚 A-Level AQA Maths: Calculus Basics | A-Level AQA 数学：微积分基础考点精讲
Calculus is a fundamental branch of mathematics that deals with rates of change and accumulation. In the AQA A-Level Mathematics syllabus, calculus basics form the essential foundation for both differentiation and integration. These concepts are not only central to pure mathematics but are also powerful tools for solving problems in physics, engineering, and beyond. This revision guide covers the key examination points, from first principles to area under a curve, with clear explanations and tips to help you master the core techniques.
微积分是数学的一个基础分支，研究变化率与累积量。在 AQA A-Level 数学大纲中，微积分基础构成了微分和积分的核心基石。这些概念不仅在纯数学中至关重要，也是解决物理、工程等学科问题的有力工具。本考点精讲涵盖从第一原理到曲线下方面积的关键内容，并提供清晰的解释与应试技巧，助你彻底掌握核心方法。
1. Gradient of a Curve and the Derivative | 曲线的斜率与导数
The derivative of a function at a point measures the gradient of the tangent to the curve at that point. For a curve y = f(x), the derivative is written as f'(x) or dy/dx. It represents the instantaneous rate of change of y with respect to x. On a graph, the derivative at a specific x-value gives the steepness and direction of the curve at that instant.
函数在某点的导数度量了曲线在该点处切线的斜率。对于曲线 y = f(x)，导数记作 f'(x) 或 dy/dx，它表示 y 关于 x 的瞬时变化率。在图像上，某特定 x 值处的导数给出了曲线在该瞬时的倾斜程度与方向。
The idea of a gradient is extended from straight lines to curves by considering the limiting position of a chord. As the two points on the curve get infinitesimally close, the chord approaches the tangent, and its gradient tends to the derivative.
梯度的概念通过考虑弦的极限位置从直线推广到曲线。当曲线上两点无限趋近时，弦逼近切线，其斜率便趋近于导数。
2. Differentiation from First Principles | 从第一原理求导
Differentiation from first principles uses the limit definition of the derivative: f'(x) = limₕ→₀ [f(x+h) – f(x)] / h. This process is fundamental to understanding how derivatives are constructed and why the rules work.
从第一原理求导使用导数的极限定义：f'(x) = limₕ→₀ [f(x+h) – f(x)] / h。这一过程对于理解导数如何构建以及规则为何成立至关重要。
Example: Find the derivative of f(x) = x² from first principles. First, f(x+h) = (x+h)² = x² + 2xh + h². Then f(x+h) – f(x) = 2xh + h². Dividing by h gives 2x + h. Taking the limit as h → 0 yields f'(x) = 2x. This confirms the power rule for n = 2.
示例：从第一原理求 f(x) = x² 的导数。首先，f(x+h) = (x+h)² = x² + 2xh + h²，于是 f(x+h) – f(x) = 2xh + h²。除以 h 得 2x + h，取 h → 0 的极限得 f'(x) = 2x，这验证了 n=2 时的幂函数求导法则。
Even when questions do not ask for a first-principles proof, understanding this limit definition helps avoid errors when applying rules, especially with signs and coefficients.
即使题目不要求给出第一原理的证明，理解这一定义也有助于在应用法则时避免错误，尤其在处理符号和系数时。
3. Basic Differentiation Rules | 基本求导法则
The AQA syllabus expects fluency with basic differentiation rules. The most important are the power rule, the constant multiple rule, and the sum/difference rule. These allow you to differentiate polynomials quickly.
AQA 大纲要求学生熟练运用基本求导法则。最重要的是幂函数法则、常数倍法则和和差法则，它们使你能够快速对多项式求导。
Key rules summarised:
关键法则总结：
Power Rule: d/dx (xⁿ) = n xⁿ⁻¹, for any real constant n.
幂函数法则：d/dx (xⁿ) = n xⁿ⁻¹，其中 n 为任意实常数。
Constant Multiple Rule: d/dx [k f(x)] = k f'(x).
常数倍法则：d/dx [k f(x)] = k f'(x)。
Sum Rule: d/dx [f(x) ± g(x)] = f'(x) ± g'(x).
和差法则：d/dx [f(x) ± g(x)] = f'(x) ± g'(x)。
| f(x) | f'(x) |
|---|---|
| x³ | 3x² |
| 5x⁴ | 20x³ |
| 2x² – 7x + 3 | 4x – 7 |
Always rewrite terms using negative or fractional indices before differentiating. For example, 1/x² becomes x⁻², and √x becomes x½. Then apply the power rule cleanly.
求导前务必将各项用负指数或分数指数重写。例如，1/x² 写成 x⁻²，√x 写成 x½，然后再清晰应用幂函数法则。
4. Derivatives of sin x and cos x | sin x 与 cos x 的导数
For the AQA AS and A-Level mathematics course, the derivatives of the trigonometric functions sin x and cos x must be known and used without reference to the first-principles proof.
在 AQA AS 和 A-Level 数学课程中，必须熟记并直接使用三角函数 sin x 和 cos x 的导数，无需引用第一原理证明。
The standard results are: d/dx (sin x) = cos x, and d/dx (cos x) = –sin x. Note the negative sign with the derivative of cos x — a very common mistake.
标准结果为：d/dx (sin x) = cos x，d/dx (cos x) = –sin x。注意 cos x 导数前的负号，这是一个极其常见的错误。
Example: Differentiate y = 4 sin x – 2 cos x. Using the sum rule and constant multiples, dy/dx = 4 cos x – 2 (–sin x) = 4 cos x + 2 sin x. Always double-check the sign when differentiating cosine.
示例：求 y = 4 sin x – 2 cos x 的导数。运用和法则与常数倍法则，dy/dx = 4 cos x – 2 (–sin x) = 4 cos x + 2 sin x。求余弦导数时务必双重检查符号。
These derivatives will also appear in stationary point problems and integration later, so memorise them correctly from the start.
这些导数也将出现在后续的稳定点问题和积分中，因此从一开始就要准确记忆。
5. Tangents and Normals | 切线与法线
Once the derivative is evaluated at a specific point, it gives the gradient of the tangent. The equation of the tangent at point (a, f(a)) is y – f(a) = f'(a)(x – a). This is simply the point–gradient form of a straight line.
导数在某特定点处的值即为切线的斜率。点 (a, f(a)) 处的切线方程为 y – f(a) = f'(a)(x – a)，这实际上就是直线的点斜式。
The normal is perpendicular to the tangent. Therefore its gradient is m_N = –1 / f'(a). The equation of the normal is y – f(a) = m_N (x – a).
法线垂直于切线，因此其斜率为 m_N = –1 / f'(a)。法线方程为 y – f(a) = m_N (x – a)。
Example: For the curve y = x³ – 3x at x = 2, first find the y-coordinate: f(2) = 2. Then f'(x) = 3x² – 3, so f'(2) = 9. Tangent: y – 2 = 9(x – 2). Normal: gradient = –1/9, equation: y – 2 = –1/9 (x – 2).
示例：曲线 y = x³ – 3x 在 x = 2 处，先求 y 坐标：f(2) = 2；接着 f'(x) = 3x² – 3，f'(2) = 9。切线：y – 2 = 9(x – 2)；法线：斜率 = –1/9，方程为 y – 2 = –1/9 (x – 2)。
Always check that the final tangent and normal equations are simplified and, if required, written in the form ax + by + c = 0.
最后务必检查所求的切线与法线方程是否已化简，若题目要求则写成 ax + by + c = 0 的形式。
6. Stationary Points and Turning Points | 稳定点与转折点
Stationary points occur where the gradient is zero, i.e., f'(x) = 0. These points can be local maxima, local minima, or points of inflection. Identifying their nature is a core skill.
导数为零的点即稳定点，即 f'(x) = 0 处。这些点可能是局部极大值、局部极小值或拐点。判断它们的性质是一项核心技能。
Two methods can determine the nature: the second derivative test and the first derivative sign test. The second derivative test: compute f”(x) at the stationary point x = a. If f”(a) > 0, the point is a local minimum; if f”(a) < 0, it is a local maximum. If f''(a) = 0, the test is inconclusive and the sign test must be used.
有两种方法可判断性质：二阶导数检验法与一阶导数符号法。二阶导数检验法：计算稳定点 x = a 处的 f”(x)。若 f”(a) > 0，该点为局部极小值；若 f”(a) < 0，则为局部极大值。若 f''(a) = 0，则该检验无法确定，必须使用符号检验法。
Example: f(x) = x³ – 3x. f'(x) = 3x² – 3 = 0 gives x = ±1. f”(x) = 6x. At x = 1, f”(1) = 6 > 0, so minimum. At x = –1, f”(–1) = –6 < 0, so maximum.
示例：f(x) = x³ – 3x，f'(x) = 3x² – 3 = 0 得 x = ±1；f”(x) = 6x。x = 1 时 f”(1) = 6 > 0，极小值；x = –1 时 f”(–1) = –6 < 0，极大值。
When f”(x) = 0, examine the sign of f'(x) either side of the stationary point. If the gradient changes from positive to negative, it is a maximum; from negative to positive, a minimum; no sign change indicates a point of inflection.
当 f”(x) = 0 时，需检查稳定点左右两侧 f'(x) 的符号。若斜率由正变负则为极大值，由负变正则为极小值；若无符号变化则为拐点。
7. Introduction to Integration | 积分引入
Integration is the reverse process of differentiation. If F'(x) = f(x), then F(x) is an antiderivative of f(x). The indefinite integral is written as ∫ f(x) dx = F(x) + C, where C is an arbitrary constant, called the constant of integration.
积分是微分的逆过程。如果 F'(x) = f(x)，则 F(x) 是 f(x) 的一个原函数。不定积分记作 ∫ f(x) dx = F(x) + C，其中 C 为任意常数，称为积分常数。
Because the derivative of a constant is zero, any antiderivative can have an added constant. Forgetting to include +C is one of the most frequent errors in AQA exams and will lose a mark.
由于常数的导数为零，任何原函数都可加上一个常数。忘记写 +C 是 AQA 考试中最常见的错误之一，会直接丢分。
To find an antiderivative for a basic function, think: what function, when differentiated, gives this? For example, ∫ 2x dx = x² + C because d/dx (x²) = 2x.
对于基本函数，找出原函数的思路是：什么函数求导后会得到被积函数？例如，∫ 2x dx = x² + C，因为 d/dx (x²) = 2x。
This idea directly leads to the integration rule for powers.
这一思路直接引出了幂函数的积分法则。
8. Indefinite Integrals of Polynomials and Trigonometric Functions | 多项式和三角函数的不定积分
For integer powers n ≠ –1, the integration rule is: ∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C. This is simply the reverse of the power rule for differentiation. Always add one to the exponent and divide by the new exponent.
对于整数次幂 n ≠ –1，积分法则为：∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C。这不过是幂函数求导法则的反向操作。始终将指数加一，并除以新的指数。
Trigonometric integrals are equally important: ∫ sin x dx = –cos x + C, and ∫ cos x dx = sin x + C. Notice the negative sign when integrating sin x — easy to miss.
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