📚 IGCSE OCR Mathematics: Differentiation – Key Points Review | IGCSE OCR 数学：微分 考点精讲
Differentiation is a fundamental calculus tool that measures how a function changes at any given point. In IGCSE OCR Higher Tier, you will learn to find the derivative, or gradient function, and apply it to sketch curves, solve motion problems, and locate maximum or minimum values. This article walks you through every critical concept with clear examples and bilingual explanations.
微分是微积分的基础工具，用来衡量函数在任意一点的变化率。在 IGCSE OCR 高等卷中，你将学习如何求导数（梯度函数），并将其应用于曲线草图绘制、运动问题求解以及寻找极大极小值。本文通过清晰实例和中英双语讲解，带你逐一掌握所有核心考点。
1. What is Differentiation? | 什么是微分？
antaneous rate of change of a function. If y = f(x), its derivative, written as dy/dx or f'(x), tells us the gradient of the curve at any point x.
微分是求函数瞬时变化率的过程。若 y = f(x)，其导数记为 dy/dx 或 f'(x)，表示曲线上任意一点的梯度。
The derivative is derived from the limit of the gradient of a chord as the two points get infinitely close. In practice, for IGCSE, you will use simple rules rather than first principles.
导数的概念源于弦的梯度极限，即当两点无限接近时的斜率。在 IGCSE 考试中，你只需要用简单的求导法则，而不需通过第一原理推导。
The derivative dy/dx can be interpreted as ‘the rate at which y changes with respect to x’. This idea is essential in later applications like kinematics.
导数 dy/dx 可解释为“y 随 x 变化的速率”。这一概念在后面的运动学等应用中至关重要。
2. The Power Rule | 幂函数求导法则
For any term y = xⁿ, the derivative is given by:
对于任意项 y = xⁿ，其导数为：
dy/dx = nxn−1
This is the most important rule in IGCSE differentiation. Multiply by the power, then subtract one from the power.
这是 IGCSE 微分中最重要的法则：乘以指数，再将指数减 1。
Example: y = x⁵ → dy/dx = 5x⁴. For y = x¹⁰⁰ → dy/dx = 100x⁹⁹.
例如：y = x⁵ → dy/dx = 5x⁴；y = x¹⁰⁰ → dy/dx = 100x⁹⁹。
If the power is a fraction or negative, the rule still holds, provided you rewrite the expression in the form xⁿ first.
若指数是分数或负数，该法则同样适用，但需要先将表达式改写为 xⁿ 的形式。
d/dx (√x) = d/dx (x½) = ½ x−½ = 1/(2√x)
d/dx (1/x) = d/dx (x−1) = −1 x−2 = −1/x²
3. Differentiating Polynomials | 多项式求导
A polynomial is a sum of terms. You can differentiate term by term. Any constant coefficient stays in place.
多项式是若干项的和，可以逐项求导，常数系数保持不变。
If y = axⁿ, then dy/dx = a × nxn−1. The sum rule: if y = u + v, then dy/dx = du/dx + dv/dx.
若 y = axⁿ，则 dy/dx = a × nxⁿ⁻¹。和的法则：若 y = u + v，则 dy/dx = du/dx + dv/dx。
Example: y = 3x⁴ − 5x² + 2x − 7. Differentiate each part:
例如：y = 3x⁴ − 5x² + 2x − 7，逐项求导得：
dy/dx = 12x³ − 10x + 2
Note that the derivative of a constant (e.g., −7) is 0. Always remember to include the ‘+ c’ when integrating, but not when differentiating.
注意常数的导数为 0（如 −7）。积分时才需要加常数 c，求导时不必。
When terms involve negative or fractional powers, first convert to the form kxⁿ, then apply the power rule.
当项含有负指数或分数指数时，先将其转换为 kxⁿ 的形式，再使用幂法则。
Example: y = 4/x² + 3√x = 4x−2 + 3x½ → dy/dx = −8x−3 + (3/2)x−½.
例如：y = 4/x² + 3√x = 4x⁻² + 3x½，得 dy/dx = −8x⁻³ + (3/2)x⁻½。
4. Finding the Gradient at a Point | 求某点的导数（梯度）
Once you have the derivative dy/dx, you can find the exact gradient at a specific x-coordinate by substituting that x-value into the derivative.
一旦求出导数 dy/dx，只需将特定的 x 值代入导数，即可得到该点的确切梯度。
xample: if y = 2x³ − x² + 5, then dy/dx = 6x² − 2x. At x = 1, gradient = 6(1)² − 2(1) = 4.
例如：y = 2x³ − x² + 5，dy/dx = 6x² − 2x，在 x = 1 处梯度为 4。
This gradient is the slope of the tangent line to the curve at that point. A positive gradient means the function is increasing; negative means decreasing.
此梯度正是该点切线的斜率。梯度为正表示函数递增，梯度为负表示函数递减。
This step is frequently tested before finding the equation of a tangent or normal.
这一步常作为求切线或法线方程的前置步骤来考查。
5. Equations of Tangents and Normals | 切线和法线方程
A tangent touches a curve at a point and has the same gradient as the curve there. A normal is perpendicular to the tangent.
切线在一点处接触曲线，且在该点的斜率与曲线相同。法线垂直于切线。
Step 1: Differentiate to get dy/dx, then plug in the x-coordinate to find the gradient of the tangent, mtangent.
第一步：求导得到 dy/dx，代入 x 坐标求切线斜率 m切线。
Step 2: The gradient of the normal is mnormal = −1 / mtangent, provided mtangent ≠ 0.
第二步：法线斜率 m法线 = −1 / m切线（若 m切线 ≠ 0）。
Step 3: Use the point-slope form y − y₁ = m(x − x₁) with the given coordinates.
第三步：利用点斜式 y − y₁ = m(x − x₁) 代入已知坐标。
Example: curve y = x² at (3, 9). dy/dx = 2x, so mT = 6. Tangent: y − 9 = 6(x − 3). Normal: mN = −1/6, so y − 9 = (−1/6)(x − 3).
例如：曲线 y = x² 在 (3,9)，dy/dx = 2x，切线斜率 6，切线方程 y − 9 = 6(x − 3)；法线斜率 −1/6，方程 y − 9 = (−1/6)(x − 3)。
Always simplify the equation into the form y = mx + c or ax + by + c = 0 as requested.
始终按要求将方程化简为 y = mx + c 或 ax + by + c = 0 形式。
6. Second Derivative | 二阶导数
The second derivative, written as d²y/dx² or f''(x), is found by differentiating the first derivative. It measures the rate of change of the gradient.
二阶导数记为 d²y/dx² 或 f''(x)，是再对一阶导数求导的结果，衡量梯度的变化率。
If f''(x) > 0 at a point, the gradient is increasing and the curve is convex (∪). If f''(x) < 0, the gradient is decreasing and the curve is concave (∩).
若某点 f''(x) > 0，则梯度在增大，曲线呈凹向上 (∪)；若 f''(x) < 0，则梯度在减小，曲线呈凸向下 (∩)。
− 6x² + 9x. First derivative: dy/dx = 3x² − 12x + 9. Second derivative: d²y/dx² = 6x − 12.
例如：y = x³ − 6x² + 9x，一阶导数：3x² − 12x + 9；二阶导数：6x − 12。
The second derivative is crucial for determining the nature of stationary points and for understanding concavity in kinematics (acceleration).
二阶导数对于判断驻点性质以及理解运动学中的凹凸性（加速度）至关重要。
7. Kinematics: Displacement, Velocity, Acceleration | 运动学：位移、速度、加速度
In a straight line motion, if displacement s (in metres) is given as a function of time t (in seconds), then:
在直线运动中，若位移 s（米）表示为时间 t（秒）的函数，则有：
Velocity: v = ds/dt, Acceleration: a = dv/dt = d²s/dt²
Differentiating displacement gives velocity; differentiating velocity gives acceleration. This is a direct application of the power rule.
对位移求导得速度，对速度求导得加速度。这是幂法则的直接应用。
Example: s = t³ − 4t² + 7t. Then v = 3t² − 8t + 7, and a = 6t − 8.
例如：s = t³ − 4t² + 7t，则 v = 3t² − 8t + 7，a = 6t − 8。
ed to find when a particle is at rest (v = 0), or to compute the speed at a specific time.
考题可能会让你求质点何时静止 (v = 0)，或在某时刻的速度大小。
Also, note that speed is the magnitude of velocity and is always non‑negative.
同时注意速率是速度的大小，总为非负数。
8. Stationary Points and Turning Points | 驻点与拐点
i.e., dy/dx = 0. These points can be local maxima, local minima, or points of inflection.
梯度为零的点称为驻点，即 dy/dx = 0。这些点可能是局部极大值、局部极小值或拐点（驻点处的平稳拐点）。
/dx = 0 for x, then substitute back into the original equation to find the corresponding y‑coordinate.
求驻点时，解方程 dy/dx = 0 得 x 值，再代回原方程求对应 y 坐标。
x² − 3x + 2 = 0 → (x−1)(x−2)=0, so x = 1 or 2. The stationary points are (1,5) and (2,4).
例如：y = 2x³ − 9x² + 12x，dy/dx = 6x² − 18x + 12 = 0，化简得 x²−3x+2=0，得 x=1 或 2。驻点为 (1,5) 和 (2,4)。
Not every stationary point is a turning point; it might be a horiontal inflection. We must test its nature.
并非所有驻点都是拐点（极值点），它也可能是水平拐点，需进一步判定性质。
9. Determining Nature of Stationary Points | 判定驻点性质
Two common methods: (1) using the second derivative, and (2) checking the sign of dy/dx on either side of the point.
两种常用方法：（1）利用二阶导数检验；（2）检查该点左右 dy/dx 的符号变化。
Method 1 – Second derivative test: Substitute the x-value into d²y/dx². If result > 0, the point is a local minimum (∪). If result < 0, it is a local maximum (∩). If d²y/dx² = 0, the test is inconclusive; use method 2.
方法一（二阶导数检验）：将 x 值代入二阶导数。若结果 > 0，则为极小值 (∪)；若 < 0，则为极大值 (∩)；若等于 0，则检验无效，改用方法二。
Method 2 – Gradient sign change: Pick x-values slightly smaller and larger than the stationary point, and evaluate dy/dx. If the gradient changes from positive to negative, it is a maximum. From negative to positive, it is a minimum. If the sign does not change, it is a point of inflection.
方法二（梯度符号变化）：选取略小于和略大于驻点 x 的值，计算 dy/dx。若梯度由正变负，则为极大值；由负变正，则为极小值；符号不变则为拐点。
Using the above example: at x = 1, y'' = 12x − 18 → 12(1)−18 = −6 < 0 → maximum; at x = 2, y'' = 12(2)−18 = 6 > 0 → minimum.
接上例：x = 1 处，y'' = 12x−18 = −6 < 0，极大值；x = 2 处，y'' = 6 > 0，极小值。
10. Increasing and Decreasing Functions | 递增与递减函数
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.
若在某区间内 f'(x) > 0，则函数在该区间递增；若 f'(x) < 0，则函数递减。
This helps to describe behaviour of a function without plotting the full graph. It is often combined with stationary point analysis.
这有助于在不绘制完整图像的情况下描述函数性态，常与驻点分析结合考查。
Example: f(x) = x³ − 3x. f'(x) = 3x² − 3. Solve f'(x) = 0 → x = ±1. Test intervals: x < −1, f'(x) > 0 (increasing); −1 < x < 1, f'(x) < 0 (decreasing); x > 1, f'(x) > 0 (increasing). So a maximum at x = −1, minimum at x = 1.
例如：f(x) = x³ − 3x，f'(x) = 3x²−3。解 f'(x)=0 得 x=±1。检验区间：x < −1 时 f' >0，递增；−1 < x < 1 时 f' <0，递减；x > 1 时 f' >0，递增。故 x=−1 处为极大值，x=1 处为极小值。
11. Real-World Applications (Optimisation) | 实际应用（优化问题）
Differentiation helps solve optimisation problems: finding the maximum area, minimum surface area for a given volume, or maximum profit, etc.
微分可帮助解决优化问题，如求最大面积、给定体积下的最小表面积、最大利润等。
The standard approach: (1) Express the quantity to be maximised/minimised as a function of one variable using given constraints. (2) Differentiate and set derivative to zero to find stationary points. (3) Use the second derivative or sign test to confirm a maximum or minimum. (4) Answer in context.
标准解法：（1）利用约束条件将要优化的量表示为单变量函数；（2）求导并令导数为零，找到驻点；（3）用二阶导数或符号检验确认极大/极小值；（4）在情境中作答。
Example: A rectangle has perimeter 20 m. What dimensions give maximum area? x². dA/dx = 10 − 2x = 0 → x = 5. So square 5 m × 5 m gives max area 25 m² (confirmed by A'' = −2 < 0).
例如：矩形周长为 20 m，求面积最大时的尺寸。设宽 = x，长 = 10−x。面积 A = 10x − x²，dA/dx = 10 − 2x = 0 → x = 5，即正方形 5m × 5m，最大面积 25 m²（由 A'' = −2 < 0 确认为极大）。
Always check that your solution makes sense in the context (e.g., positive dimensions).
始终检验答案在语境中的合理性（例如尺寸须为正值）。
12. Common Exam Mistakes and Tips | 常见考试错误与技巧
Forgetting to rewrite roots and fractions: Always convert √x to x½, 1/x² to x−2 before differentiating. Errors often occur when students try to apply the power rule directly to 1/x².
忘记先将根号和分式变形：求导前务必将 √x 写成 x½，将 1/x² 写成 x⁻²。许多错误源于直接对分式使用幂法则时未变形。
Missing negative signs: Pay extra care with negative powers. Differentiating x−3 gives −3x−4, not +3x−4.
丢失负号：处理负指数时特别留意。对 x⁻³ 求导得 −3x⁻⁴，而非 +3x⁻⁴。
Tangent vs normal confusion: Remember that normal gradient is the negative reciprocal of the tangent gradient. Many candidates use the same slope for the normal.
混淆切线与法线：法线斜率是切线斜率的负倒数，不少考生会用法线时仍沿用原斜率。
Incomplete stationary point coordinates: , always substitute back to find the y-value. The answer is a coordinate pair, not just x.
驻点坐标不完整：解 dy/dx = 0 后务必代回原式求 y 值。答案应是坐标对，而不仅是 x 值。
Second derivative test misuse: automatically a point of inflection; always check sign of first derivative on either side.
二阶导数检验误用：若 d²y/dx² = 0，不要直接判定为拐点，务必检查该点两侧一阶导数的符号。
Unit and variable awareness: atics, differentiate with respect to time t, not x. Clearly state v = ds/dt and a = dv/dt.
注意单位和变量：运动学中，必须对时间 t 求导，而非 x。清楚标明 v = ds/dt、a = dv/dt。
By reviewing these tips regularly, you can avoid the most frequent pitfalls and boost your exam performance.
常回顾这些要点，可避开最常见失分陷阱，有效提升考试表现。
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