📚 Differentiation for CCEA IGCSE Mathematics | 微分考点精讲
Differentiation is a central topic in the CCEA IGCSE Mathematics higher-tier syllabus. It provides the tools to analyse how a curve changes, to find the gradient at any point, and to tackle real-world problems involving rates of change, tangents, and optimisation. A confident grasp of differentiation will significantly boost your exam performance.
微分是 CCEA IGCSE 数学高等卷的核心主题。它为你提供了分析曲线变化、求任意点斜率以及解决变化率、切线和最优化等实际问题的工具。扎实掌握微分将显著提升你在考试中的表现。
1. What is Differentiation? | 什么是微分？
Differentiation is the process used to find the gradient function of a curve. For a straight line, the gradient is constant and easily found. For a curve, the steepness varies from point to point – differentiation gives a new function, called the derivative, that tells us the gradient at any given x‑coordinate.
微分是求曲线斜率函数的过程。对于一条直线，斜率是恒定的且容易求得。对于曲线，陡峭程度随点变化——微分会给出一个新的函数，称为导数，它可以告诉我们任意给定 x 坐标处的斜率。
If we write the equation of a curve as y = f(x), the derivative is written as f'(x) or dy/dx. It describes the instantaneous rate of change of y with respect to x.
如果我们把曲线的方程写作 y = f(x)，导数记作 f'(x) 或 dy/dx，它描述了 y 对 x 的瞬时变化率。
2. The Power Rule for Differentiation | 幂函数求导法则
The most essential tool in differentiation is the power rule. For any term of the form xⁿ, where n is a constant, the derivative is n xⁿ⁻¹. This rule applies to positive and negative powers, fractions, and roots once they are written in index form.
微分中最基本的工具是幂函数法则。对于任何形式为 xⁿ 的项（n 是常数），导数为 n xⁿ⁻¹。该法则适用于正指数、负指数、分数和根式，只要它们先写成指数形式。
If y = xⁿ, then dy/dx = n xⁿ⁻¹
如果 y = xⁿ，那么 dy/dx = n xⁿ⁻¹
For example: y = x⁵ gives dy/dx = 5x⁴. y = x¹ (just x) gives 1. A constant term such as y = 7 differentiates to 0, since the graph of a constant is a horizontal line with zero gradient.
例如：y = x⁵ 得到 dy/dx = 5x⁴；y = x¹（就是 x）得到 1；常数项例如 y = 7 求导为 0，因为常数的图像是一条斜率为零的水平线。
3. Differentiating Polynomials & the Sum Rule | 多项式微分与加减法则
Most functions you will meet are polynomials made up of several terms. You can differentiate term by term – the constant multiple rule says you can multiply by the constant, and the sum rule says you can differentiate each term separately and add the results.
你会遇到的大多数函数都是由多个项组成的多项式。你可以逐项求导——常数倍法则允许你乘以常数，加减法则意味着你可以分别对每一项求导然后把结果相加。
Example: y = 3x⁴ − 5x³ + 2x − 9. Differentiating term by term gives dy/dx = 12x³ − 15x² + 2. Note that the constant −9 vanishes.
例子：y = 3x⁴ − 5x³ + 2x − 9。逐项求导得到 dy/dx = 12x³ − 15x² + 2。注意常数 −9 消失了。
This works equally well when some powers are negative or fractional. For instance, y = 2/x² = 2x⁻² differentiates to dy/dx = −4x⁻³, which can be rewritten as −4/x³.
当某些幂是负数或分数时同样适用。例如 y = 2/x² = 2x⁻² 求导得到 dy/dx = −4x⁻³，也可以写为 −4/x³。
4. Finding the Gradient at a Specific Point | 求特定点的斜率
Once you have the derivative, you can find the gradient of the curve at any point by substituting the x‑coordinate into the derivative function. This is often a simple two‑step process: differentiate, then substitute.
一旦你得到了导数，就可以通过将 x 坐标代入导函数求出曲线上任意一点的斜率。这通常是一个简单的两步过程：先求导，再代入。
Example: For y = x³ − 4x + 1, find the gradient at x = 2. First, dy/dx = 3x² − 4. Then substitute x = 2: gradient = 3(2)² − 4 = 3×4 − 4 = 8. So the curve has a steepness of 8 at that point.
例子：对于 y = x³ − 4x + 1，求 x = 2 处的斜率。首先，dy/dx = 3x² − 4。然后代入 x = 2：斜率 = 3(2)² − 4 = 3×4 − 4 = 8。所以该点处曲线的陡峭程度为 8。
Always present the gradient as a number or an algebraic expression, and make sure you have differentiated correctly before substituting.
始终以数字或代数表达式的形式给出斜率，并确保代入前求导正确。
5. Equations of Tangents to a Curve | 曲线切线方程
A tangent is a straight line that touches a curve at exactly one point and has the same gradient as the curve at that point. To find its equation, you need the point (x₁, y₁) and the gradient m (found by differentiation).
切线是一条恰好与曲线交于一点并且在该点与曲线斜率相同的直线。要求切线方程，你需要已知点 (x₁, y₁) 和斜率 m（通过微分求得）。
Use the point‑gradient form: y − y₁ = m(x − x₁). Example: Curve y = x² + 1 at x = 3. First, find y when x = 3: y = 9 + 1 = 10, so the point is (3, 10). Then dy/dx = 2x, so at x = 3, m = 6. The tangent equation is y − 10 = 6(x − 3), which simplifies to y = 6x − 8.
使用点斜式：y − y₁ = m(x − x₁)。例子：曲线 y = x² + 1 在 x = 3 处。首先，当 x = 3 时 y = 9 + 1 = 10，所以点是 (3, 10)。然后 dy/dx = 2x，所以在 x = 3 处 m = 6。切线方程为 y − 10 = 6(x − 3)，化简得 y = 6x − 8。
The tangent is a common exam question – always check that your gradient is obtained accurately and that you have used the correct coordinates.
切线是常见的考试题目——务必确保准确求得斜率，并且使用了正确的坐标。
6. Equations of Normals | 法线方程
The normal to a curve at a point is the line perpendicular to the tangent at that same point. If the tangent gradient is m (and m ≠ 0), the normal gradient is −1/m. When m = 0, the normal is vertical with undefined gradient.
曲线上一点的法线是指在同一点上垂直于切线的直线。如果切线斜率为 m（且 m ≠ 0），法线斜率为 −1/m。当 m = 0 时，法线为竖直线，斜率无定义。
Example: Using the previous curve y = x² + 1 at x = 3, the tangent gradient was 6, so the normal gradient is −1/6. The point is still (3, 10). The normal equation is y − 10 = −1/6 (x − 3), which can be written as 6y − 60 = −x + 3, or x + 6y = 63.
例子：沿用之前曲线 y = x² + 1 在 x = 3 处，切线斜率为 6，因此法线斜率为 −1/6。点仍然是 (3, 10)。法线方程为 y − 10 = −1/6 (x − 3)，可写成 6y − 60 = −x + 3，即 x + 6y = 63。
Watch out for whole‑number normal equations: multiplying through to avoid fractions earns method marks and a neater final answer.
注意整理法线方程：通过去分母避免分数，可以获得步骤分并使最终答案更整洁。
7. The Second Derivative | 二阶导数
Differentiating a function once gives the first derivative, dy/dx. If you differentiate dy/dx again, you obtain the second derivative, written as d²y/dx² or f”(x). It measures the rate at which the gradient itself is changing – in other words, it tells you how the slope is curving.
对函数求一次导得到一阶导数 dy/dx。如果你再对 dy/dx 求导，就得到二阶导数，记作 d²y/dx² 或 f”(x)。它衡量的是斜率本身的变化率——换言之，它告诉你斜率的弯曲情况。
Example: y = 3x⁴ − 2x² + 5. First derivative: dy/dx = 12x³ − 4x. Then second derivative: d²y/dx² = 36x² − 4. The second derivative is essential for classifying the nature of stationary points.
例子：y = 3x⁴ − 2x² + 5。一阶导数：dy/dx = 12x³ − 4x。二阶导数：d²y/dx² = 36x² − 4。二阶导数对于判断驻点的性质至关重要。
8. Stationary Points and Their Nature | 驻点及其性质
A stationary point occurs where the curve’s gradient is zero, i.e. dy/dx = 0. These points are where the graph has a flat tangent, and they can be a local maximum, a local minimum, or a point of inflection.
驻点出现在曲线斜率为零的位置，即 dy/dx = 0。这些点处图形有一条水平切线，它们可能是局部极大值点、局部极小值点或拐点。
To find stationary points, solve dy/dx = 0 for x, then find the corresponding y‑values. To determine the nature, use the second derivative test: substitute the x‑value into d²y/dx². If d²y/dx² < 0, it is a local maximum; if d²y/dx² > 0, a local minimum; if d²y/dx² = 0, the test is inconclusive and you should check the sign of dy/dx on either side.
要求驻点，解 dy/dx = 0 得到 x，再求出相应的 y 值。要判断性质，使用二阶导数检验法：将 x 值代入 d²y/dx²。如果 d²y/dx² < 0，为局部极大值；如果 d²y/dx² > 0，为局部极小值；如果 d²y/dx² = 0，则无法确定，应检查 dy/dx 在两侧的正负。
Example: For y = x³ − 3x, dy/dx = 3x² − 3. Setting this to 0 gives x = ±1. At x = 1, y = −2; at x = −1, y = 2. The second derivative is d²y/dx² = 6x. At x = 1, d²y/dx² = 6 > 0 → minimum. At x = −1, d²y/dx² = −6 < 0 → maximum. You can then sketch the curve marking these features.
例子：对于 y = x³ − 3x，dy/dx = 3x² − 3。令其为零得 x = ±1。在 x = 1 处，y = −2；在 x = −1 处，y = 2。二阶导数为 d²y/dx² = 6x。在 x = 1 处，d²y/dx² = 6 > 0 → 极小值。在 x = −1 处，d²y/dx² = −6 < 0 → 极大值。然后你可以根据这些特征绘制草图。
9. Maximum and Minimum Problems | 极大值与极小值应用题
Differentiation is often used to solve optimisation problems, such as finding the maximum area of a shape or the minimum surface area of a container given a constraint. You will need to express the quantity to be maximised or minimised in terms of one variable, then differentiate and set the derivative to zero.
微分常常用于解决最优化问题，例如在给定约束下求形状的最大面积或容器的最小表面积。你需要将待最大化或最小化的量表示为一个变量的函数，然后求导并令导数为零。
Example: A rectangular field borders a river on one side and has 200 m of fencing for the other three sides. If the two perpendicular sides are of length x, the side parallel to the river is 200 − 2x. The area is A = x(200 − 2x) = 200x − 2x². Differentiating: dA/dx = 200 − 4x. Set to zero: x = 50. The second derivative d²A/dx² = −4 < 0, so it is a maximum. Dimensions: 50 m by 100 m, maximum area 5000 m².
例子：一块矩形场地一边靠河，另外三边用 200 米围栏。设两条垂直边长为 x，则平行河岸的边长为 200 − 2x。面积 A = x(200 − 2x) = 200x − 2x²。求导：dA/dx = 200 − 4x。令导数为零得 x = 50。二阶导数 d²A/dx² = −4 < 0，因此为极大值。尺寸为 50 m 乘以 100 m，最大面积 5000 m²。
Always state the practical meaning of the answer and check that it makes sense within the constraints (e.g. x must be positive and less than 100).
务必说明答案的实际意义，并检查其是否符合约束条件（例如 x 必须为正且小于 100）。
10. Application to Kinematics: Velocity and Acceleration | 运动学应用：速度与加速度
In kinematics, if displacement s is given as a function of time t, the velocity v is the first derivative: v = ds/dt. Acceleration a is the derivative of velocity, or the second derivative of displacement: a = dv/dt = d²s/dt². This links differentiation directly to motion.
在运动学中，如果位移 s 是关于时间 t 的函数，那么速度 v 就是一阶导数：v = ds/dt。加速度 a 是速度的导数，或位移的二阶导数：a = dv/dt = d²s/dt²。这就把微分和运动直接联系了起来。
Example: s = t³ − 6t² + 9t (in metres). Velocity v = ds/dt = 3t² − 12t + 9. Acceleration a = d²s/dt² = 6t − 12. The particle is at rest when v = 0, giving 3(t² − 4t + 3) = 0 → t = 1 or t = 3. At those times you can find the displacement and acceleration to describe the motion fully.
例子：s = t³ − 6t² + 9t（单位米）。速度 v = ds/dt = 3t² − 12t + 9。加速度 a = d²s/dt² = 6t − 12。当 v = 0 时物体静止，解 3(t² − 4t + 3) = 0 得 t = 1 或 t = 3。在这些时刻你可以求出位移和加速度以完整描述运动。
Be ready to interpret the physical meaning: a negative acceleration means deceleration if velocity is positive, but check the sign carefully.
准备好解释物理意义：如果速度为正面加速度为负意味着减速，但务必仔细检查正负号。
11. Common Mistakes & Exam Tips | 常见错误与备考建议
Achieving full marks in differentiation questions relies on avoiding frequent pitfalls. Common errors include forgetting to multiply by the original power when applying the power rule, mishandling constant terms, substituting incorrectly into derivative expressions, and mixing up tangent and normal gradients.
想在微分题中获得满分需要避开常见陷阱。常见错误包括：应用幂函数法则时忘记乘以原来的幂；错误处理常数项；代入导数表达式时出错；混淆切线和法线的斜率等。
Always write down your derivative clearly before substituting values. Check that your final answer is in the requested form (e.g. simplified fraction, equation in ax + by = c). For stationary points, show both the x‑coordinate solutions and the full coordinates, and clearly state the nature using a second derivative test or a gradient sign table.
在代入数值之前务必先清楚写出导数。检查最终答案是否符合题目要求的形式（如化简分数、方程为 ax + by = c）。对于驻点，既要给出 x 坐标解也要给出完整坐标，并用二阶导数检验或斜率符号表清楚说明性质。
Practise a wide variety of past paper questions, especially those combining differentiation with geometry or physics. Time yourself to improve both speed and accuracy. With systematic revision and careful working, differentiation will become one of your most reliable topics.
练习各式各样的历年真题，尤其是那些将微分与几何或物理结合的题目。给自己计时以提高速度和准确性。通过系统复习和细致运算，微分将成为你最得心应手的主题之一。
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