IGCSE Maths: Differentiation - Key Points Explained | IGCSE 数学：微分考点精讲

📚 IGCSE Maths: Differentiation – Key Points Explained | IGCSE 数学：微分考点精讲

Differentiation is a fundamental concept in calculus that measures how a function changes as its input changes. In IGCSE Mathematics, you learn to differentiate polynomial functions, find gradients of curves, determine stationary points, and apply differentiation to real-world problems like kinematics and optimisation.

微分是微积分的基本概念，用来度量函数值随自变量变化的快慢。在 IGCSE 数学中，你将学习多项式函数的求导、计算曲线梯度、确定驻点，以及将微分应用于运动学和优化等实际问题。

1. What is Differentiation? | 什么是微分？

Differentiation is the process of finding the derivative of a function. The derivative, often written as dy/dx or f'(x), gives the instantaneous rate of change of y with respect to x. Geometrically, it represents the gradient of the tangent line to the curve y = f(x) at any point.

微分是求函数导数的过程。导数经常写作 dy/dx 或 f'(x)，它表示 y 随 x 变化的瞬时变化率。从几何角度，导数代表曲线 y = f(x) 上任意一点切线的斜率。

2. The Power Rule | 幂函数求导法则

The most important rule in IGCSE differentiation is the power rule. For any real constant n, if y = xⁿ, then:

IGCSE 微分中最重要的法则就是幂函数求导法则。对于任意实常数 n，如果 y = xⁿ，则有：

dy/dx = n xⁿ⁻¹

This works for positive integers, negative integers, and fractional powers. For example, y = x⁵ gives dy/dx = 5x⁴. y = x⁻² gives dy/dx = −2x⁻³. y = √x = x^{1/2} gives dy/dx = (1/2)x^{−1/2}.

该法则适用于正整数、负整数以及分数指数。例如，y = x⁵ 的导数为 5x⁴。y = x⁻² 导数为 −2x⁻³。y = √x = x^{1/2} 的导数为 (1/2)x^{−1/2}。

3. Differentiating Constants and Linear Functions | 常数与线性函数的导数

The derivative of a constant term is zero: if y = k, where k is a constant, then dy/dx = 0. For a linear function y = kx, the derivative is simply k.

常数项的导数为零：若 y = k（k 为常数），则 dy/dx = 0。对于线性函数 y = kx，导数就是 k。

y = 5 ⇒ dy/dx = 0
y = 3x ⇒ dy/dx = 3

When a constant multiplies a function, you can keep the constant and differentiate the function. If y = k·f(x), then dy/dx = k·f'(x).

当常数乘以函数时，可以保留常数，对函数求导。若 y = k·f(x)，则 dy/dx = k·f'(x)。

4. Sum and Difference Rules | 和差法则

Differentiation can be performed term by term. If y = f(x) ± g(x), then the derivative is the sum or difference of the individual derivatives.

微分可以逐项进行。若 y = f(x) ± g(x)，那么导数就是各自导数的和或差。

dy/dx = f'(x) ± g'(x)

Example: y = x³ + 4x² − 7x + 2. Differentiating each term gives dy/dx = 3x² + 8x − 7.

例如：y = x³ + 4x² − 7x + 2。逐项求导得 dy/dx = 3x² + 8x − 7。

5. Finding the Gradient of a Curve | 求曲线的梯度

To find the gradient of a curve at a specific point (x₁, y₁), first differentiate the function, then substitute the x-coordinate into the derivative. For example, for y = x² + 3x, dy/dx = 2x + 3. At x = 2, the gradient is 2(2) + 3 = 7.

要计算曲线上某点 (x₁, y₁) 的梯度，先对函数求导，再将 x 坐标代入导函数。例如，y = x² + 3x，dy/dx = 2x + 3。在 x = 2 处，梯度为 2(2) + 3 = 7。

This gradient tells you how steep the curve is at that point. A positive gradient means the function is increasing; a negative gradient means it is decreasing.

该梯度说明了曲线在该点的倾斜程度。正梯度意味着函数递增，负梯度意味着函数递减。

6. Equation of Tangent and Normal | 切线与法线方程

Once you know the gradient m_t at a point (x₁, y₁) on a curve, the equation of the tangent line can be written using point-slope form:

当知道曲线在点 (x₁, y₁) 处的斜率 m_t 后，切线方程就可以用点斜式写出：

y − y₁ = m_t (x − x₁)

The normal line is perpendicular to the tangent. Its gradient m_n is the negative reciprocal of the tangent’s gradient: m_n = −1 / m_t, provided m_t ≠ 0.

法线与切线垂直。其斜率 m_n 是切线斜率的负倒数：m_n = −1 / m_t，前提是 m_t ≠ 0。

Example: For y = x² at x = 1, the point is (1,1), dy/dx = 2x, so m_t = 2. Tangent: y − 1 = 2(x − 1). Normal: m_n = −½, equation: y − 1 = −½(x − 1).

例：y = x² 在 x = 1 处，点为 (1,1)，dy/dx = 2x，所以 m_t = 2。切线：y − 1 = 2(x − 1)。法线：m_n = −½，方程：y − 1 = −½(x − 1)。

7. Stationary Points | 驻点

Stationary points occur where the derivative is zero, i.e. dy/dx = 0. At these points, the tangent to the curve is horizontal. Stationary points can be:

驻点发生在导数为零的位置，即 dy/dx = 0。在这些点上，曲线的切线是水平的。驻点可以是：

Local maximum – the curve reaches a peak.
局部极大值点 – 曲线到达一个峰顶。
Local minimum – the curve reaches a trough.
局部极小值点 – 曲线到达一个谷底。
Point of inflection – the curve flattens but does not change direction, or changes the type of curvature.
拐点 – 曲线变平但方向不变，或者弯曲方向改变。

To find stationary points, solve dy/dx = 0 and then find the corresponding y-values.

要找出驻点，解方程 dy/dx = 0，然后求出对应的 y 值。

8. Classifying Stationary Points (First Derivative Test) | 驻点分类（一阶导数判定）

One way to determine the nature of a stationary point is to examine the sign of dy/dx just to the left and right of the point.

判断驻点性质的一种方法是检查该点左侧和右侧 dy/dx 的符号。

If dy/dx changes from positive to negative → maximum.
如果 dy/dx 由正变负 → 极大值点。
If dy/dx changes from negative to positive → minimum.
如果 dy/dx 由负变正 → 极小值点。
If dy/dx does not change sign → point of inflection.
如果 dy/dx 符号不变 → 拐点。

Create a simple table of signs for values around the stationary point to decide its type.

制作一个驻点附近取值的符号表，从而判定其类型。

9. The Second Derivative | 二阶导数及其判定

The second derivative, written as d²y/dx² or f”(x), is the derivative of the first derivative. It measures the rate of change of the gradient, and it helps to determine concavity and classify stationary points more efficiently.

二阶导数，记为 d²y/dx² 或 f”(x)，是一阶导数的导数。它度量梯度的变化率，有助于确定凹性并更有效地对驻点进行分类。

If d²y/dx² > 0 at a stationary point: minimum
If d²y/dx² < 0 at a stationary point: maximum
If d²y/dx² = 0: the test is inconclusive — use the first derivative test.

若驻点处 d²y/dx² > 0：极小值点
若驻点处 d²y/dx² < 0：极大值点
若 d²y/dx² = 0：无法判定——需用一阶导数检验。

For example, y = x³ − 3x, dy/dx = 3x² − 3, d²y/dx² = 6x. At x = 1, d²y/dx² = 6 > 0 → minimum. At x = −1, d²y/dx² = −6 < 0 → maximum.

例如，y = x³ − 3x，dy/dx = 3x² − 3，d²y/dx² = 6x。在 x = 1 处，d²y/dx² = 6 > 0 → 极小值点。在 x = −1 处，d²y/dx² = −6 < 0 → 极大值点。

10. Increasing and Decreasing Functions | 函数的增减性

A function is increasing on an interval where its derivative is positive, and decreasing where its derivative is negative.

函数在导数大于零的区间内递增，在导数小于零的区间内递减。

f'(x) > 0 ⇒ function is increasing.
f'(x) > 0 ⇒ 函数递增。
f'(x) < 0 ⇒ function is decreasing.
f'(x) < 0 ⇒ 函数递减。

This is useful for sketching graphs and for solving inequalities in optimisation contexts.

这对绘制图像草图以及解决优化情境中的不等式非常有用。

11. Kinematics: Displacement, Velocity, Acceleration | 运动学：位移、速度、加速度

If an object moves in a straight line and its displacement from a fixed point is given by s(t), where t is time, then:

若物体沿直线运动，其从固定点出发的位移为 s(t)，其中 t 为时间，则有：

Velocity: v = ds/dt
Acceleration: a = dv/dt = d²s/dt²

速度：v = ds/dt
加速度：a = dv/dt = d²s/dt²

For example, if s(t) = 2t³ − 9t² + 12t, then v = 6t² − 18t + 12. The object is at rest when v = 0. Solving 6t² − 18t + 12 = 0 gives t = 1 and t = 2. Acceleration a = 12t − 18 tells you how the velocity is changing.

例如，若 s(t) = 2t³ − 9t² + 12t，则 v = 6t² − 18t + 12。当 v = 0 时物体静止。解 6t² − 18t + 12 = 0 得 t = 1 和 t = 2。加速度 a = 12t − 18 说明速度的变化情况。

12. Optimisation Problems | 优化问题

Differentiation is widely used to find maximum or minimum values of a quantity. Typical IGCSE problems involve maximising area, volume, or profit, and minimising cost or surface area.

微分被广泛用于求某量的最大值或最小值。典型的 IGCSE 题目涉及最大面积、体积或利润，以及最小成本或表面积。

The general steps are:

一般步骤如下：

Express the quantity to be optimised as a function of one variable, using given constraints.
利用给定的约束条件，将要优化的量表示为单一变量的函数。
Differentiate the function and set the derivative equal to zero to find stationary points.
对函数求导，并令导数为零以找出驻点。
Use the second derivative test or first derivative sign analysis to confirm whether the stationary point gives a maximum or minimum.
利用二阶导数检验或一阶导数符号分析，确认驻点是极大值还是极小值。
Answer the question in the context of the problem (include units).
在问题情境下给出答案（包括单位）。

Example: A rectangular pen is built against a wall, so three sides use fencing. With 60 m of fence, maximise the area. Let width = x, then length = 60 − 2x. Area A = x(60 − 2x) = 60x − 2x². dA/dx = 60 − 4x = 0 → x = 15. d²A/dx² = −4 < 0, so maximum area occurs when x = 15 m, length = 30 m, area = 450 m².

例：一面靠墙围一个矩形围栏，三边使用篱笆。现有60米篱笆，求最大面积。设宽度为 x，则长度为 60 − 2x。面积 A = x(60 − 2x) = 60x − 2x²。dA/dx = 60 − 4x = 0 → x = 15。d²A/dx² = −4 < 0，因此当 x = 15 m 时面积最大，长度 30 m，面积为 450 m²。

Published by TutorHao | Maths Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

IGCSE Maths: Differentiation – Key Points Explained | IGCSE 数学：微分 考点精讲