Differentiation | 微分

📚 Differentiation | 微分

Welcome to our GCSE OCR Maths revision guide on differentiation. In higher-tier GCSE, you need to know how to find the gradient of a curve at a point, apply the power rule, and solve problems involving stationary points and kinematics. This guide walks you through every essential skill, from the meaning of the derivative to real-life applications.

欢迎来到 GCSE OCR 数学微分考点精讲。在高等层级中，你需要掌握如何求曲线上某点的梯度、运用幂法则，并解决涉及驻点和运动学的问题。本指南将带你逐一掌握从导数含义到实际应用的所有核心技能。

1. What Is Differentiation? | 什么是微分？

Differentiation is a branch of calculus that measures how a function changes as its input changes. The derivative f'(x) or dy/dx gives the gradient of the tangent to the curve y = f(x) at any point. In simpler terms, it tells you the instantaneous rate of change.

微分是微积分的一个分支，用来度量函数随自变量变化而变化的快慢。导数 f'(x) 或 dy/dx 表示曲线 y = f(x) 在任意一点处切线的斜率。简单说，它告诉你瞬时变化率。

2. The Power Rule | 幂法则

For a monomial function of the form y = xⁿ, the derivative is given by dy/dx = nxⁿ⁻¹. This rule works for any real constant n. You must multiply by the power and then reduce the power by one.

对于形如 y = xⁿ 的单项式函数，导数由 dy/dx = nxⁿ⁻¹ 给出。这条法则对任何实常数 n 都成立。你需要先乘以指数，再将指数减 1。

For example, if y = x⁵, then dy/dx = 5x⁴. If y = 3x², then dy/dx = 3 × 2x¹ = 6x.

例如，若 y = x⁵，则 dy/dx = 5x⁴。若 y = 3x²，则 dy/dx = 3 × 2x¹ = 6x。

3. Differentiating Constant Terms | 常数项的微分

The derivative of any constant is zero. This is because a constant function has zero gradient everywhere. When differentiating a sum of terms, apply the rule term by term.

任何常数的导数为零。这是因为常数函数处处梯度为零。当对多项式逐项微分时，可以逐项应用法则。

If y = 4x³ + 7, then dy/dx = 12x² + 0 = 12x².

若 y = 4x³ + 7，则 dy/dx = 12x² + 0 = 12x²。

4. Sum and Difference Rule | 和差法则

You can differentiate polynomials by splitting into individual terms. The derivative of a sum (or difference) is the sum (or difference) of the derivatives.

你可以将多项式拆分为单项式分别求导。和（或差）的导数等于导数的和（或差）。

For y = 2x⁵ − 3x³ + 4x − 9, we get dy/dx = 10x⁴ − 9x² + 4.

对于 y = 2x⁵ − 3x³ + 4x − 9，可得 dy/dx = 10x⁴ − 9x² + 4。

5. Gradient at a Specific Point | 特定点的梯度

To find the gradient of a curve at a given x-value, first differentiate to obtain dy/dx, then substitute the x-coordinate into the derivative expression.

要找出特定 x 值处曲线的梯度，先求导得到 dy/dx，然后将 x 坐标代入导数表达式。

Example: Find the gradient of y = x³ − 2x at x = 2.

例题：求 y = x³ − 2x 在 x = 2 处的梯度。

dy/dx = 3x² − 2; substituting x = 2 gives 3(4) − 2 = 10.

dy/dx = 3x² − 2；代入 x = 2 得 3(4) − 2 = 10。

6. Equation of a Tangent | 切线方程

Once you have the gradient m at a point (x₁, y₁), you can write the equation of the tangent using y − y₁ = m(x − x₁). Ensure you calculate y₁ from the original curve equation.

一旦你得到某点 (x₁, y₁) 处的梯度 m，就可以用 y − y₁ = m(x − x₁) 写出切线方程。记得用原曲线方程算出 y₁。

For the previous example at x = 2, y = 2³ − 4 = 4. So tangent is y − 4 = 10(x − 2) ⇒ y = 10x − 16.

在前例中 x = 2 时，y = 2³ − 4 = 4。因此切线为 y − 4 = 10(x − 2)，即 y = 10x − 16。

7. Stationary Points | 驻点

A stationary point occurs where dy/dx = 0. At this point the tangent is horizontal. Solve the equation dy/dx = 0 to find the x-coordinates, then substitute back to find y.

驻点出现在 dy/dx = 0 的位置。此时切线为水平线。解方程 dy/dx = 0 求得 x 坐标，再代回求出 y。

Example: y = x³ − 3x. dy/dx = 3x² − 3 = 0 ⇒ x² = 1 ⇒ x = 1 or x = −1. The stationary points are (1, −2) and (−1, 2).

例题：y = x³ − 3x。dy/dx = 3x² − 3 = 0 ⇒ x² = 1 ⇒ x = 1 或 x = −1。驻点为 (1, −2) 和 (−1, 2)。

8. Nature of Stationary Points | 驻点的类型

To determine whether a stationary point is a maximum, minimum or point of inflection, you can use the second derivative d²y/dx². If d²y/dx² > 0, it is a local minimum; if d²y/dx² < 0, it is a local maximum; if d²y/dx² = 0, further investigation is needed.

要判断驻点是极大值、极小值还是拐点，可以使用二阶导数 d²y/dx²。若 d²y/dx² > 0，则为局部极小值；若 d²y/dx² < 0，则为局部极大值；若 d²y/dx² = 0，则需要进一步分析。

Using the previous example: d²y/dx² = 6x. At x = 1, 6 > 0 → minimum; at x = −1, −6 < 0 → maximum.

用上例：d²y/dx² = 6x。在 x = 1 处，6 > 0 → 极小值；在 x = −1 处，−6 < 0 → 极大值。

9. Second Derivative | 二阶导数

The second derivative is obtained by differentiating dy/dx again. It measures the rate of change of the gradient and is used to check the nature of stationary points.

二阶导数由 dy/dx 再次求导得到。它衡量梯度的变化率，用于检验驻点类型。

If y = 2x⁴, dy/dx = 8x³, then d²y/dx² = 24x².

若 y = 2x⁴，dy/dx = 8x³，那么 d²y/dx² = 24x²。

10. Differentiation in Kinematics | 运动学中的微分

If displacement s is given as a function of time t, velocity v is the first derivative ds/dt, and acceleration a is the second derivative d²s/dt². You can differentiate polynomial expressions for motion directly.

如果位移 s 是时间 t 的函数，那么速度 v 是一阶导数 ds/dt，加速度 a 是二阶导数 d²s/dt²。你可以直接对运动方程的多项式进行微分。

Example: s = t³ − 3t² + 2t. Then v = 3t² − 6t + 2 and a = 6t − 6.

例题：s = t³ − 3t² + 2t。则 v = 3t² − 6t + 2，a = 6t − 6。

11. Key Exam Tips | 考试关键技巧

Always remember to write ‘dy/dx =’ before differentiating. When solving for stationary points, show clear steps and state the coordinates fully. Label max/min clearly. For tangents, double-check that the point used lies on the curve.

进行微分前务必先写“dy/dx =”。求驻点时，展示清晰步骤并完整给出坐标。清楚标注极大/极小值。求切线时，再次确认所用点确实在曲线上。

Common mistakes include forgetting to reduce the power by one, not treating constants correctly, and mixing up signs when substituting. Practice with past OCR questions to avoid these.

常见错误包括忘记将指数减 1、没有正确处理常数项，以及代入时符号混乱。多做 OCR 历年真题以避免这些失误。

12. Summary | 总结

Master the power rule, find gradients, tangents, and stationary points, and apply differentiation to kinematics. These skills form the foundation of calculus at GCSE Higher level and will prepare you well for A-level mathematics.

掌握幂法则，会求梯度、切线和驻点，并将微分应用于运动学。这些技能构成 GCSE 高等层级微积分的基础，也将为你学习 A-level 数学做好充分准备。

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Differentiation | 微分