GCSE OCR Maths: Differentiation – Key Points | GCSE OCR 数学：微分考点精讲

📚 GCSE OCR Maths: Differentiation – Key Points | GCSE OCR 数学：微分考点精讲

Differentiation is a powerful tool that allows us to understand how functions change. For GCSE OCR Maths, mastering the basics of differentiation will help you tackle gradient of curves, equations of tangents, and turning points with confidence. This guide covers all the key concepts and exam techniques you need.

微分是一个强大的工具，能帮助我们理解函数如何变化。在 GCSE OCR 数学中，掌握微分的基础知识将助你自信地处理曲线的梯度、切线方程和极值点。本指南涵盖了你所需的所有关键概念和考试技巧。

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 gradient of the tangent to the curve at any point. In simple terms, it tells you the rate of change of y with respect to x.

微分是求一个函数导数的过程。导数通常写作 dy/dx 或 f ‘(x)，表示曲线在任意一点切线的梯度。简而言之，它告诉你 y 相对于 x 的变化率。

For a straight line, the gradient is constant. For a curve, the gradient changes at every point, and differentiation provides a rule to find it instantly.

对于直线，梯度是恒定的。对于曲线，每一点的梯度都不同，而微分提供了一种规则来立即求出它。

2. The Gradient of a Curve | 曲线的梯度

Consider the curve y = x². At x = 1, the curve is not very steep; at x = 3, it rises more sharply. The derivative dy/dx = 2x matches this: when x = 1, gradient = 2; when x = 3, gradient = 6. So differentiation gives us a formula for the gradient function.

考虑曲线 y = x²。在 x = 1 处，曲线不算陡峭；在 x = 3 处，上升得更急。导数 dy/dx = 2x 符合这一点：当 x = 1 时，梯度为 2；当 x = 3 时，梯度为 6。可见，微分给出了梯度函数的公式。

Always remember: the derivative is the gradient function. You can substitute any x-value to find the steepness at that exact point.

务必记住：导数就是梯度函数。代入任意 x 值，你就能求出该点的陡峭程度。

3. Differentiating from First Principles (Brief) | 从第一性原理微分（简述）

Although exams rarely ask for full first-principles derivation, understanding the concept helps. The derivative is defined as the limit as h → 0 of [f(x+h) – f(x)] / h. For f(x) = x², this simplifies to 2x. This limit formalises the idea of measuring the slope of a tiny secant line that becomes a tangent.

虽然考试很少要求完整的第一性原理推导，但理解概念是有帮助的。导数定义为当 h → 0 时 [f(x+h) – f(x)] / h 的极限。对于 f(x) = x²，这会简化为 2x。这个极限将测量微小割线斜率直至变成切线的想法形式化了。

For GCSE, you only need to apply the rules that result from this definition. But knowing where the rules come from can prevent silly errors.

在 GCSE 阶段，你只需应用从这个定义得出的规则。但知道规则从何而来能防止低级错误。

4. The Power Rule (xⁿ) | 幂法则 (xⁿ)

The most important differentiation rule is the power rule: if y = xⁿ, then dy/dx = n xⁿ⁻¹. Multiply by the power, then subtract 1 from the power.

最重要的微分法则是幂法则：若 y = xⁿ，则 dy/dx = n xⁿ⁻¹。乘以指数，再将指数减 1。

Example: y = x⁵ → dy/dx = 5x⁴.
示例： y = x⁵ → dy/dx = 5x⁴.
Example: y = x → dy/dx = 1x⁰ = 1.
示例： y = x → dy/dx = 1x⁰ = 1.
Example: y = √x = x½ → dy/dx = ½ x⁻½ = 1/(2√x).
示例： y = √x = x½ → dy/dx = ½ x⁻½ = 1/(2√x).

Practice this rule with positive, negative, and fractional powers. It works for any real n.

用正指数、负指数和分数指数来练习这个规则。它对任何实数 n 都适用。

5. Sum/Difference and Constant Multiple Rules | 和/差与常数倍法则

When a function has several terms, differentiate each term separately. The derivative of a sum is the sum of the derivatives. Also, constants multiplying a term can be brought outside: d/dx [k· f(x)] = k· f ‘(x).

当一个函数有多项时，对每一项分别求导。和的导数等于导数的和。另外，乘以项的常数可以提到外面：d/dx [k· f(x)] = k· f ‘(x)。

For example, if y = 4x³ – 2x² + 7x – 9, then dy/dx = 12x² – 4x + 7. The constant term –9 differentiates to 0.

例如，若 y = 4x³ – 2x² + 7x – 9，则 dy/dx = 12x² – 4x + 7。常数项 –9 的导数为 0。

Always write the derivative in its simplest form, combining like terms where possible.

始终将导数化为最简形式，尽可能合并同类项。

6. Finding the Derivative at a Point | 求一点处的导数

To find the gradient of a curve at a specific x-value, differentiate the function, then substitute the x-coordinate. This gives the numerical value of the gradient at that point.

要找出曲线上特定 x 值处的梯度，先对函数求导，再代入 x 坐标。这样就能得到该点梯度的数值。

For y = 2x³ – 5x + 1, we have dy/dx = 6x² – 5. At x = 2, the gradient is 6(2)² – 5 = 24 – 5 = 19.

对于 y = 2x³ – 5x + 1，我们有 dy/dx = 6x² – 5。在 x = 2 处，梯度为 6(2)² – 5 = 24 – 5 = 19。

This is a common exam requirement, sometimes followed by finding the equation of the tangent or normal.

这是常见的考试要求，有时后面还会要求求切线或法线方程。

7. Equation of a Tangent | 切线方程

Once you know the gradient at a point (x₁, y₁), the tangent line equation is given by y – y₁ = m (x – x₁), where m = dy/dx evaluated at that point. Make sure you have the y-coordinate from the original function.

一旦知道点 (x₁, y₁) 处的梯度，切线方程就可以用 y – y₁ = m (x – x₁) 给出，其中 m 是该点的 dy/dx 值。确保你从原函数求出了 y 坐标。

Example: For y = x² + 3x at x = 1, y₁ = 1² + 3(1) = 4. dy/dx = 2x + 3, so m = 2(1) + 3 = 5. Tangent: y – 4 = 5(x – 1) → y = 5x – 1.

例如：对 y = x² + 3x 在 x = 1 处，y₁ = 1² + 3(1) = 4。dy/dx = 2x + 3，故 m = 2(1) + 3 = 5。切线方程：y – 4 = 5(x – 1) → y = 5x – 1。

If asked for the normal, use the negative reciprocal of m as the gradient for the perpendicular line.

如果要求法线方程，用 m 的负倒数作为垂直线的梯度。

8. Stationary Points | 驻点

Stationary points occur where the derivative equals zero: dy/dx = 0. At these points, the tangent is horizontal, and the function may have a maximum, minimum, or point of inflection.

驻点出现在导数等于零的地方：dy/dx = 0。在这些点处，切线是水平的，函数可能存在最大值、最小值或拐点。

To find stationary points, solve the equation f ‘(x) = 0. Then substitute those x-values back into the original function to find the corresponding y-values.

要找驻点，解方程 f ‘(x) = 0。然后将这些 x 值代回原函数，求出相应的 y 值。

For y = x³ – 3x, dy/dx = 3x² – 3 = 3(x² – 1) = 0 ⇒ x = ±1. Points: (1, –2) and (–1, 2).

对于 y = x³ – 3x，dy/dx = 3x² – 3 = 3(x² – 1) = 0 ⇒ x = ±1。驻点为 (1, –2) 和 (–1, 2)。

9. Determining Nature of Stationary Points | 判断驻点的性质

To classify a stationary point as a maximum or minimum, you can use the second derivative or examine the gradient either side of the point. Both methods are accepted in OCR exams.

要判定驻点是极大值还是极小值，你可以使用二阶导数，或者检查该点两侧的梯度。OCR 考试中两种方法都接受。

Method 1 (gradient either side): Pick x-values just left and right of the stationary point. Evaluate dy/dx. If gradient changes from positive to negative, it’s a local maximum; from negative to positive, a local minimum.
方法 1（两侧梯度）： 选取驻点稍左和稍右的 x 值，计算 dy/dx。如果梯度由正变负，则为局部最大值；由负变正，则为局部最小值。
Method 2 (second derivative): Find d²y/dx². If at the stationary point d²y/dx² < 0, it's a maximum; if > 0, it’s a minimum. If d²y/dx² = 0, the test is inconclusive (use Method 1).
方法 2（二阶导数）： 求 d²y/dx²。如果在驻点处 d²y/dx² < 0，则为最大值；若 > 0，则为最小值。若 d²y/dx² = 0，则此检验不确定（用方法 1）。

Always clearly state your reasoning and conclusion.

始终清楚地说明你的推理和结论。

10. Second Derivative Test | 二阶导数检验

The second derivative is the derivative of the derivative. For y = 4x³ – 6x², dy/dx = 12x² – 12x, and d²y/dx² = 24x – 12. Evaluate at the stationary point. If the result is positive, the curve is concave up (∪), indicating a minimum; if negative, concave down (∩), indicating a maximum.

二阶导数是导数的导数。对 y = 4x³ – 6x²，dy/dx = 12x² – 12x，d²y/dx² = 24x – 12。在驻点处求值。若结果为正，曲线凹向上 (∪)，指示极小值；若为负，凹向下 (∩)，指示极大值。

Let’s test x = 0: dy/dx = 0? 12(0) – 12(0) = 0, so stationary point. d²y/dx² = 24(0) – 12 = –12 < 0, therefore maximum at (0,0). Test x = 1: d²y/dx² = 24(1)–12 = 12 > 0, minimum.

检验 x = 0：dy/dx = 0? 12(0) – 12(0) = 0，所以是驻点。d²y/dx² = 24(0) – 12 = –12 < 0，因此在 (0,0) 处是极大值。检验 x = 1：d²y/dx² = 24(1)–12 = 12 > 0，极小值。

11. Optimisation (Max/Min Problems) | 优化问题（最大/最小值）

Differentiation is often used to solve real-world problems: finding the maximum volume, minimum surface area, or greatest profit. Form an equation for the quantity you want to optimise in terms of one variable, then find its derivative and set it to zero.

微分常被用来解决现实问题：求最大体积、最小表面积或最大利润。用单一变量建立希望优化的量的方程，然后求导并令其为零。

Steps: 1) Write the quantity Q as a function of x, using given constraints. 2) Differentiate to get dQ/dx. 3) Solve dQ/dx = 0 to find critical values. 4) Use the second derivative test to confirm maximum or minimum. 5) Answer in context, with units.

步骤：1) 用给定的约束条件将量 Q 写成 x 的函数。2) 求导得到 dQ/dx。3) 解 dQ/dx = 0 求临界值。4) 用二阶导数检验确认是最大值还是最小值。5) 结合实际情境作答，并带上单位。

Example: A farmer wants a rectangular field of area 200 m². Find dimensions to minimise fencing. Let width = x, length = 200/x. Perimeter P = 2x + 400/x. dP/dx = 2 – 400/x² = 0 ⇒ x² = 200 ⇒ x ≈ 14.14 m. Second derivative positive, so minimum.

例如：一位农民想要一块面积为 200 m² 的矩形田地，求使围栏最短的尺寸。设宽度为 x，长度为 200/x。周长 P = 2x + 400/x。dP/dx = 2 – 400/x² = 0 ⇒ x² = 200 ⇒ x ≈ 14.14 m。二阶导数为正，故为最小值。

12. Common Mistakes & Exam Tips | 常见错误与考试技巧

Mistake 1: Forgetting to reduce the power by 1. If y = x⁴, dy/dx is 4x³ not 4x⁴. Double-check the exponent.

错误 1：忘记将指数减 1。若 y = x⁴，dy/dx 应为 4x³ 而非 4x⁴。请反复检查指数。

Mistake 2: Misapplying the constant rule – constants that are multiplied remain, while constants added become zero. y = 5x² gives dy/dx = 10x; the 5 stays. y = x² + 5 gives dy/dx = 2x; the 5 vanishes.

错误 2：错误运用常数规则——相乘的常数保留，相加的常数导数为零。y = 5x² 得 dy/dx = 10x，5 保留。y = x² + 5 得 dy/dx = 2x，5 消失。

Mistake 3: Not simplifying before differentiating. For y = (2x + 1)(x – 3), expand first: y = 2x² – 5x – 3, then differentiate.

错误 3：求导前未化简。若 y = (2x + 1)(x – 3)，先展开：y = 2x² – 5x – 3，再求导。

Mistake 4: Confusing stationary points with roots. dy/dx = 0 finds turning points, not where the curve crosses the x-axis.

错误 4：混淆驻点与根。dy/dx = 0 求的是极值点，而非曲线与 x 轴的交点。

Exam tip: Always show all steps, including the derivative formula and substitution. If the question asks for a tangent, give the final equation in its simplest form.

考试技巧：务必展示所有步骤，包括导数公式和代入过程。如果题目要求求切线，以最简形式给出最终方程。

Finally, check the domain – if the question asks for x > 0, don’t give a negative stationary point unless it’s valid.

最后，检查定义域——如果题目要求 x > 0，不要给出负的驻点，除非它有效。

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GCSE OCR Maths: Differentiation – Key Points | GCSE OCR 数学：微分 考点精讲