IGCSE AQA Maths: Differentiation – Revision Guide | IGCSE AQA 数学：偏微分考点精讲

📚 IGCSE AQA Maths: Differentiation – Revision Guide | IGCSE AQA 数学：偏微分考点精讲

Although the term ‘partial differentiation’ belongs to multi-variable calculus studied at advanced levels, IGCSE AQA Maths focuses on ordinary differentiation, which lays the essential groundwork for all future calculus topics. This article breaks down every key concept you need to master for the exam, from the power rule to tangents, stationary points and kinematic applications.

虽然“偏微分”在高等数学中才涉及多元函数，但 IGCSE AQA 数学考察的是普通微分，是为所有微积分学习打基础的核心内容。本文全面拆解考试必会的每个关键概念，从幂函数求导法则到切线、驻点以及运动学应用，助你稳扎稳打。

1. What is Differentiation? | 什么是微分？

Differentiation is a mathematical technique that finds the gradient (rate of change) of a function at any given point. If y = f(x), the derivative dy/dx or f'(x) measures how y changes with respect to x.

微分是一种数学方法，用来求函数在任意给定点的梯度（变化率）。如果 y = f(x)，则导数 dy/dx 或 f'(x) 衡量 y 随 x 变化的快慢。

The derivative represents the slope of the tangent to the curve and is fundamental for understanding motion, optimisation and curve sketching. At IGCSE, we focus solely on functions of one variable, typically polynomials.

导数代表曲线切线的斜率，是理解运动、优化和曲线绘图的基础。在 IGCSE 阶段，我们只关注单变量函数，主要是多项式函数。

2. The Power Rule | 幂函数求导法则

The most important rule for IGCSE is the power rule: if f(x) = xⁿ, then f'(x) = n xⁿ⁻¹. This rule works for any real number n, but at IGCSE n is usually a positive integer or zero.

IGCSE 最重要的求导法则是幂法则：若 f(x) = xⁿ，则 f'(x) = n xⁿ⁻¹。该法则对任何实数 n 都成立，但在 IGCSE 阶段 n 常为正整数或零。

When n = 0, f(x) = 1, so f'(x) = 0 (constant rule).
当 n = 0 时，f(x) = 1，因此 f'(x) = 0（常数的导数为零）。
When n = 1, f(x) = x, so f'(x) = 1.
当 n = 1 时，f(x) = x，则 f'(x) = 1。
When n = 2, f(x) = x², so f'(x) = 2x.
当 n = 2 时，f(x) = x²，则 f'(x) = 2x。
When n = 3, f(x) = x³, so f'(x) = 3x².
当 n = 3 时，f(x) = x³，则 f'(x) = 3x²。

Constants multiplied by a function are preserved: if f(x) = a·g(x), then f'(x) = a·g'(x). For example, the derivative of 5x³ is 5 × 3x² = 15x².

常数乘函数时保留常数：若 f(x) = a·g(x)，则 f'(x) = a·g'(x)。例如，5x³ 的导数是 5 × 3x² = 15x²。

3. Differentiating Polynomials Term by Term | 多项式逐项微分

A polynomial is differentiated by applying the power rule to each term separately. The derivative of a sum is the sum of the derivatives: d/dx [u(x) + v(x)] = u'(x) + v'(x).

多项式微分时，对每一项分别使用幂法则。和的导数等于导数的和：d/dx [u(x) + v(x)] = u'(x) + v'(x)。

Example: For f(x) = 2x³ − 4x² + 6x − 8, we differentiate each term: derivative of 2x³ is 6x², of −4x² is −8x, of 6x is 6, and constant −8 gives 0. So f'(x) = 6x² − 8x + 6.

例题：对于 f(x) = 2x³ − 4x² + 6x − 8，逐项求导：2x³ 的导数是 6x²，−4x² 的导数是 −8x，6x 的导数是 6，常数 −8 的导数是 0。所以 f'(x) = 6x² − 8x + 6。

Always write the final derivative in standard descending powers of x, unless the question asks for a specific form.

除非题目有特殊要求，最后得到的导数表达式应按 x 的降幂书写。

4. Tangents and Normals | 切线与法线

To find the equation of a tangent to a curve y = f(x) at a point with x-coordinate a, first find the y-coordinate f(a), then compute the gradient m = f'(a). The tangent line equation is y − f(a) = m(x − a).

求曲线 y = f(x) 在 x = a 处的切线方程，首先求出 y 坐标 f(a)，再计算梯度 m = f'(a)。切线方程为 y − f(a) = m(x − a)。

The normal is perpendicular to the tangent, so its gradient is −1/m (provided m ≠ 0). The equation of the normal uses the same point but the new gradient: y − f(a) = (−1/m)(x − a).

法线与切线垂直，因此其梯度为 −1/m（假设 m ≠ 0）。法线方程使用同一点和新梯度：y − f(a) = (−1/m)(x − a)。

Example: For y = x² + 3x at x = 1, we have y = 1 + 3 = 4. Derivative dy/dx = 2x + 3, so m = 2(1) + 3 = 5. Tangent: y − 4 = 5(x − 1). Normal gradient = −1/5, so normal: y − 4 = (−1/5)(x − 1).

例题：曲线 y = x² + 3x 在 x = 1 处，y = 1 + 3 = 4。导数 dy/dx = 2x + 3，因此 m = 5。切线：y − 4 = 5(x − 1)。法线梯度 = −1/5，法线：y − 4 = (−1/5)(x − 1)。

5. Increasing and Decreasing Functions | 函数的单调性

A function f(x) is increasing on an interval if f'(x) > 0 for all x in that interval. It is decreasing if f'(x) < 0. When analysing a function, find the derivative and solve inequalities to determine where the function rises or falls.

如果在某区间内 f'(x) > 0，则函数递增；如果 f'(x) < 0，则函数递减。分析函数时，先求导，再解不等式，判断函数在何处上升或下降。

Example: f(x) = x² − 4x + 3. Then f'(x) = 2x − 4. f'(x) = 0 gives x = 2. For x < 2, f'(x) < 0, so function is decreasing. For x > 2, f'(x) > 0, so function is increasing.

例题：f(x) = x² − 4x + 3，则 f'(x) = 2x − 4。f'(x) = 0 得 x = 2。当 x < 2 时 f'(x) < 0，函数递减；当 x > 2 时 f'(x) > 0，函数递增。

6. Stationary Points | 驻点

Stationary points occur where the derivative equals zero: f'(x) = 0. These are points where the gradient of the tangent is zero, meaning the curve has a horizontal tangent. They can be maximum points, minimum points or points of inflection.

驻点出现在导数为零的地方：f'(x) = 0。这些点处切线斜率为零，即曲线有水平切线。驻点可能是极大值点、极小值点或拐点。

To locate stationary points, solve f'(x) = 0 for x, then substitute back into f(x) to get the y-coordinates. Always check the nature of each stationary point.

求驻点时，解方程 f'(x) = 0 得到 x 值，再代回 f(x) 得到 y 坐标。每个驻点的性质都需要进一步判断。

7. Classifying Stationary Points | 驻点分类

There are two common methods: second derivative test and first derivative sign analysis. In IGCSE, both are accepted. Using the second derivative: compute f”(x) and evaluate at the stationary point x = a.

常见方法有两种：二阶导数检验和一阶导数符号分析。IGCSE 两种均可使用。使用二阶导数时，先求 f”(x)，再代入 x = a 计算。

If f”(a) < 0, the point is a local maximum.
若 f”(a) < 0，该点为局部极大值点。
If f”(a) > 0, the point is a local minimum.
若 f”(a) > 0，该点为局部极小值点。
If f”(a) = 0, the test is inconclusive; use the first derivative test.
若 f”(a) = 0，该检验无法判定，需使用一阶导数检验。

The first derivative test checks the sign of f'(x) just to the left and right of x = a. If the sign changes from positive to negative, it’s a maximum; from negative to positive, a minimum; if no sign change, it’s a point of inflection.

一阶导数检验观察 f'(x) 在 x = a 左右两侧的符号。若符号由正变负，则为极大值；由负变正，则为极小值；符号不变则为拐点。

8. Second Derivative and Concavity | 二阶导数与凹凸性

The second derivative, f”(x), is the derivative of f'(x). It tells us about the rate of change of the gradient, and hence the curvature of the graph. In kinematics, if displacement is s(t), velocity is v(t) = s'(t), and acceleration is a(t) = v'(t) = s”(t).

二阶导数 f”(x) 是一阶导数的导数。它反映梯度的变化率，从而体现曲线的弯曲特征。在运动学中，若位移为 s(t)，则速度 v(t) = s'(t)，加速度 a(t) = v'(t) = s”(t)。

A positive second derivative indicates the curve is concave up (like a cup); a negative second derivative indicates concave down (like a frown). This is closely related to the classification of stationary points.

二阶导数为正表示曲线向上凹（如杯状）；为负表示向下凹（如皱眉状）。这与驻点分类密切相关。

9. Kinematics Applications | 运动学应用

IGCSE questions often give displacement s as a function of time t, e.g., s = t³ − 6t² + 9t. Velocity v is the first derivative, and acceleration a is the second derivative. Typical tasks: find times when a particle is at rest (v = 0), find maximum displacement, or find distance travelled.

IGCSE 考题常给出位移 s 关于时间 t 的函数，如 s = t³ − 6t² + 9t。速度 v 为位移的一阶导数，加速度 a 为二阶导数。典型问题包括：求质点静止的时刻（v = 0）、最大位移或运动距离。

Example: For s = t³ − 6t² + 9t, v = 3t² − 12t + 9. Set v = 0 gives 3(t² − 4t + 3) = 0 → t = 1 or t = 3. Acceleration a = 6t − 12. At t = 1, a = −6 (negative, particle slowing down). At t = 3, a = 6 (positive, speeding up).

例题：对于 s = t³ − 6t² + 9t，v = 3t² − 12t + 9。令 v = 0 得 3(t² − 4t + 3) = 0 → t = 1 或 t = 3。加速度 a = 6t − 12。在 t = 1 时，a = −6（负值，减速）；t = 3 时 a = 6（正值，加速）。

10. Optimisation Problems | 优化问题

Optimisation involves finding maximum or minimum values of a quantity using differentiation. You model a real-life situation with a function, differentiate, find stationary points and classify them. Constraints are used to reduce the function to one variable.

优化问题是通过微分求某个量的最大值或最小值。你需要将实际情况建模为函数，求导，找到驻点并分类。约束条件用于将函数化简为一个变量。

A classic IGCSE example: an open box made from a square sheet of side 20 cm by cutting squares of side x from each corner. Volume V = x(20 − 2x)², simplify to V = 4x(10 − x)². Differentiate, find dV/dx = 0, solve for x, and determine the maximum volume.

经典 IGCSE 例题：用边长为 20 cm 的正方形纸板，每个角切去边长为 x 的小正方形做成无盖盒。体积 V = x(20 − 2x)²，化简为 V = 4x(10 − x)²。求导 dV/dx = 0，解出 x，确定最大体积。

11. Exam Tips and Common Mistakes | 应试技巧与常见错误

Always show clear steps: write ‘dy/dx =’ before differentiating. When finding stationary points, state ‘for stationary points, dy/dx = 0’ to earn method marks. Do not forget to substitute back to find the y-coordinate.

务必写出清晰步骤：在微分前写出“dy/dx =”。求驻点时，写明“驻点处 dy/dx = 0”以获取方法分。别忘了代回求出 y 坐标。

Common mistakes include differentiating incorrectly (forgetting the power rule or sign), misclassifying stationary points, and confusing tangents with normals. Also, ensure you answer the question fully – if asked for coordinates, give (x, y), not just x.

常见错误有：求导出错（忘记幂法则或符号）、驻点分类错误、混淆切线和法线。另外，若问题要求坐标，务必给出 (x, y)，不要只写 x。

12. Summary and Key Formulae | 总结与核心公式

Differentiation allows us to analyse gradients, rates of change and motion. The core rules are: d/dx (xⁿ) = n xⁿ⁻¹, constant multiple rule, sum rule. Tangent gradient = f'(a), normal gradient = −1/f'(a). Stationary points found at f'(x) = 0, classified by second derivative or first derivative test. Practice plenty of applied questions involving kinematics and optimisation.

微分让我们能够分析梯度、变化率和运动。核心法则为：d/dx (xⁿ) = n xⁿ⁻¹，常数乘法法则，和法则。切线梯度 = f'(a)，法线梯度 = −1/f'(a)。驻点在 f'(x) = 0 处取得，通过二阶导数或一阶导数检验分类。大量练习运动学和优化类应用题是成功的关键。

Key formula summary:

核心公式一览：

Power rule: d/dx (xⁿ) = n xⁿ⁻¹
幂法则：d/dx (xⁿ) = n xⁿ⁻¹
Derivative of sum: d/dx (u+v) = du/dx + dv/dx
和的导数：d/dx (u+v) = du/dx + dv/dx
Tangent equation: y − y₁ = m(x − x₁) with m = f'(x₁)
切线方程：y − y₁ = m(x − x₁)，其中 m = f'(x₁)
Normal gradient: m_normal = −1/m
法线梯度：m_法 = −1/m
Stationary points: f'(x) = 0; classification: f”(x) > 0 → min; f”(x) < 0 → max
驻点：f'(x) = 0；分类：f”(x) > 0 → 极小；f”(x) < 0 → 极大

Keep this guide handy as you revise, and work through past paper questions to build confidence. Differentiation appears in extended IGCSE AQA Papers 2H and 3H, contributing significantly to your final grade.

复习时随身携带这份指南，并通过历年真题建立信心。微分出现在 IGCSE AQA 高等卷 Paper 2H 和 3H 中，对最终成绩贡献很大。

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IGCSE AQA Maths: Differentiation – Revision Guide | IGCSE AQA 数学：偏微分 考点精讲