IGCSE CIE Maths: Fundamentals of Calculus - Key Points | IGCSE CIE 数学：微积分基础考点精讲

📚 IGCSE CIE Maths: Fundamentals of Calculus – Key Points | IGCSE CIE 数学：微积分基础考点精讲

Calculus is one of the most powerful tools in mathematics, and at IGCSE level, you will be introduced to the basics of differentiation. This topic lays the foundation for understanding rates of change, gradients of curves, and optimization problems. Mastering these concepts is essential for success in CIE IGCSE Mathematics (Extended) and for further studies in A-level Maths.

微积分是数学中最强大的工具之一，在 IGCSE 阶段，你将初步学习微分的基础知识。该主题为理解变化率、曲线梯度以及优化问题奠定了基础。掌握这些概念对于在 CIE IGCSE 数学（拓展）中取得好成绩以及进一步学习 A-level 数学至关重要。

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 f‘(x) or dy/dx, gives the gradient (slope) of the curve y = f(x) at any point. It measures how fast y is changing with respect to x.

微分是求函数导数的过程。导数通常写作 f‘(x) 或 dy/dx，它给出了曲线 y = f(x) 在任意一点的梯度（斜率）。导数衡量了 y 相对于 x 的变化速率。

For example, if y = x², the derivative is dy/dx = 2x. This means at x = 3, the gradient of the curve is 2 × 3 = 6.

例如，若 y = x²，其导数为 dy/dx = 2x。这表示当 x = 3 时，曲线的梯度为 2 × 3 = 6。

2. The Gradient Function and the Derivative | 梯度函数与导数

The derivative is called a gradient function because it provides the slope at any x-value. While the average rate of change between two points is given by the slope of the secant line, the derivative gives the exact gradient at a single point, which is the limiting value as the two points get infinitely close.

导数被称为梯度函数，因为它能给出任意 x 值处的斜率。两点之间的平均变化率由割线的斜率表示，而导数则给出了单点处的确切梯度，它是当两点无限接近时的极限值。

In IGCSE, you do not need to derive the limit definition formally, but you must understand that the derivative dy/dx is the gradient of the tangent to the curve at a point.

在 IGCSE 中，你无需严格推导极限定义，但必须明白导数 dy/dx 就是曲线在某点处切线的梯度。

3. Power Rule of Differentiation | 幂函数求导法则

The most important rule for IGCSE is the power rule. If y = xⁿ, where n is a real number constant, then the derivative is given by:

IGCSE 最重要的求导法则是幂法则。若 y = xⁿ（n 为实数常量），则导数为：

d/dx (xⁿ) = nx^n-1

This rule works for positive integers, negative integers, and fractions alike. Here are some examples:

该法则对正整数、负整数以及分数均适用。示例如下：

If y = x⁴, then dy/dx = 4x³.

若 y = x⁴，则 dy/dx = 4x³。
If y = x^-2, then dy/dx = -2x^-3 (which is -2/x³).

若 y = x^-2，则 dy/dx = -2x^-3（即 -2/x³）。
If y = x^½ (square root of x), dy/dx = ½ x^-½ = 1/(2√x).

若 y = x^½（即 √x），则 dy/dx = ½ x^-½ = 1/(2√x)。

Always rewrite roots and reciprocals as powers of x before differentiating. For instance, y = 1/x³ should be written as x^-3.

求导前，务必先将根式和倒数改写为 x 的幂形式。例如，y = 1/x³ 应先写成 x^-3。

4. Sum, Difference, and Constant Multiple Rules | 和、差与常数倍法则

Differentiation is a linear operation, meaning you can differentiate term by term. The derivative of a sum or difference is the sum or difference of the derivatives:

微分是线性运算，意味着你可以逐项求导。和或差的导数等于导数的和或差：

d/dx (u ± v) = du/dx ± dv/dx

If a function is multiplied by a constant, the constant tag along:

若函数乘以一个常数，该常数保留在导数中：

d/dx (c·u) = c·du/dx, where c is constant

Also, the derivative of a constant alone is zero: d/dx (5) = 0.

另外，常数项的导数为零：d/dx (5) = 0。

Example: For y = 3x⁴ – 2x² + 7x – 9, the derivative is dy/dx = 12x³ – 4x + 7.

示例：对 y = 3x⁴ – 2x² + 7x – 9，导数为 dy/dx = 12x³ – 4x + 7。

5. Finding the Gradient at a Given Point | 求曲线上一点的梯度

To find the slope of a curve at a specific x-value, first differentiate the function, then substitute the x-coordinate into the derivative. That value is the gradient of the tangent at that point.

要找到曲线在特定 x 值处的斜率，先对函数求导，然后将 x 坐标代入导函数。所得数值即该点切线的梯度。

Example: Find the gradient of y = x³ – 2x at x = -1.

示例：求 y = x³ – 2x 在 x = -1 处的梯度。

Solution: dy/dx = 3x² – 2. Substitute x = -1: gradient = 3(-1)² – 2 = 3 – 2 = 1.

解答：dy/dx = 3x² – 2。代入 x = -1：梯度 = 3(-1)² – 2 = 3 – 2 = 1。

6. Equation of a Tangent | 切线方程

Once you have the gradient m at a point (x₁, y₁), the equation of the tangent can be found using the point-slope form:

一旦得到某点 (x₁, y₁) 的梯度 m，即可用点斜式求出切线方程：

y – y₁ = m(x – x₁)

Example: Find the tangent to y = x² – 3x at x = 2.

示例：求 y = x² – 3x 在 x = 2 处的切线方程。

Step 1: Find y₁: y = (2)² – 3(2) = 4 – 6 = -2 → point (2, -2).

步骤 1：求 y₁：y = 2² – 3×2 = 4 – 6 = -2 → 点 (2, -2)。

Step 2: dy/dx = 2x – 3. At x = 2, m = 2(2) – 3 = 1.

步骤 2：dy/dx = 2x – 3。当 x = 2，m = 2×2 – 3 = 1。

Step 3: Tangent equation: y – (-2) = 1(x – 2) → y + 2 = x – 2 → y = x – 4.

步骤 3：切线方程：y – (-2) = 1(x – 2) → y + 2 = x – 2 → y = x – 4。

7. The Second Derivative | 二阶导数

The second derivative is the derivative of the first derivative. It is written as f”(x) or d²y/dx². It describes the rate of change of the gradient and helps determine the nature of stationary points.

二阶导数是一阶导数的导数，记作 f”(x) 或 d²y/dx²。它描述了梯度的变化率，并用于判断驻点的类型。

Example: If y = 2x³ – 9x² + 12x, then dy/dx = 6x² – 18x + 12, and the second derivative d²y/dx² = 12x – 18.

示例：若 y = 2x³ – 9x² + 12x，则 dy/dx = 6x² – 18x + 12，二阶导数 d²y/dx² = 12x – 18。

8. Stationary Points: Maximum and Minimum | 驻点：最大值与最小值

Stationary points occur where the first derivative is zero (dy/dx = 0). At these points, the tangent is horizontal. To determine whether a stationary point is a maximum, minimum, or point of inflection, use the second derivative test.

驻点出现在一阶导数为零的位置（dy/dx = 0）。在这些点，切线是水平的。要判断驻点是极大值点、极小值点还是拐点，可使用二阶导数检验法。

Second derivative test:

二阶导数检验法：

If d²y/dx² > 0 at the stationary point → minimum point (curve is concave up).

若在驻点处 d²y/dx² > 0 → 最小值点（曲线上凹）。
If d²y/dx² < 0 at the stationary point → maximum point (curve is concave down).

若在驻点处 d²y/dx² < 0 → 最大值点（曲线下凹）。
If d²y/dx² = 0, the test is inconclusive; the point could be a point of inflection (beyond IGCSE scope).

若 d²y/dx² = 0，无法判定；可能是拐点（超出 IGCSE 范围）。

Worked example: Find and classify the stationary points of y = x³ – 3x².

完整示例：求 y = x³ – 3x² 的驻点并分类。

Step 1: dy/dx = 3x² – 6x. Set equal to 0: 3x(x – 2) = 0 → x = 0 or x = 2.

步骤 1：dy/dx = 3x² – 6x。令其为零：3x(x – 2) = 0 → x = 0 或 x = 2。

Step 2: Find y-coordinates: at x = 0, y = 0; at x = 2, y = 8 – 12 = -4. Points: (0, 0) and (2, -4).

步骤 2：求 y 坐标：x = 0 时，y = 0；x = 2 时，y = 8 – 12 = -4。点：(0, 0) 和 (2, -4)。

Step 3: Second derivative d²y/dx² = 6x – 6. At x = 0: d²y/dx² = -6 (negative) → maximum at (0, 0). At x = 2: d²y/dx² = 6 (positive) → minimum at (2, -4).

步骤 3：二阶导数 d²y/dx² = 6x – 6。在 x = 0：d²y/dx² = -6（负）→ 极大值点 (0, 0)。在 x = 2：d²y/dx² = 6（正）→ 极小值点 (2, -4)。

9. Application in Kinematics | 运动学应用

In kinematics, if displacement s is given as a function of time t, the derivative represents velocity, and the second derivative represents acceleration:

在运动学中，若位移 s 表示为时间 t 的函数，则一阶导数代表速度，二阶导数代表加速度：

v = ds/dt, a = dv/dt = d²s/dt²

Example: A particle moves so that s = t³ – 6t² + 9t metres after t seconds. Find the velocity and acceleration expressions, and the times when the particle is at rest.

示例：一质点运动满足 s = t³ – 6t² + 9t 米（t 秒后）。求速度和加速度表达式，以及质点静止的时刻。

Solution: v = ds/dt = 3t² – 12t + 9. a = dv/dt = 6t – 12.

解答：v = ds/dt = 3t² – 12t + 9。a = dv/dt = 6t – 12。

At rest means v = 0: 3t² – 12t + 9 = 0 → divide by 3: t² – 4t + 3 = 0 → (t – 1)(t – 3) = 0, so t = 1 s or t = 3 s.

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IGCSE CIE Maths: Fundamentals of Calculus – Key Points | IGCSE CIE 数学：微积分基础 考点精讲