GCSE OCR Maths: Second Order Differential Equations Key Exam Guide | GCSE OCR 数学：二阶微分方程考点精讲

📚 GCSE OCR Maths: Second Order Differential Equations Key Exam Guide | GCSE OCR 数学：二阶微分方程考点精讲

Second-order differential equations are typically studied at A-level, but the OCR GCSE Maths Higher Tier introduces the second derivative as a powerful tool for analysing curves. Understanding how to differentiate twice and apply the second derivative test to classify stationary points is essential for top exam marks. This guide covers everything you need to know about second derivatives and their role in solving optimisation and curve-sketching problems on the GCSE OCR paper.

二阶微分方程通常出现在 A-level 课程中，但 OCR GCSE 数学（高阶段）引入了二阶导数作为分析曲线的重要工具。熟练掌握两次求导并运用二阶导数判别法对驻点进行分类，对于在考试中取得高分至关重要。本考点精讲涵盖二阶导数及其在 OCR GCSE 试卷中优化与曲线草图问题中的应用，助你全面备考。

1. What Is a Second Derivative? | 什么是二阶导数？

The second derivative is simply the derivative of the first derivative. If y = f(x), then dy/dx is the rate of change of y with respect to x. Taking the derivative again gives d²y/dx², which tells us how the gradient itself is changing. In notation, if f'(x) = dy/dx, then f”(x) = d²y/dx².

二阶导数就是一阶导数的导数。若 y = f(x)，则 dy/dx 表示 y 关于 x 的变化率。再次求导得到 d²y/dx²，它反映了斜率自身的变化情况。符号上，若 f'(x) = dy/dx，则 f”(x) = d²y/dx²。

For GCSE, you only need to compute the second derivative of polynomial functions and use it for stationary point classification. You will not be solving actual differential equations, but the concept is foundational.

对于 GCSE 来说，你只需计算多项式函数的二阶导数，并利用它对驻点进行分类。你不需要求解真正的微分方程，但这一概念是基础。

2. Differentiating Twice: Step-by-Step | 两次求导：逐步讲解

Given a polynomial y = axⁿ, the first derivative is dy/dx = naxⁿ⁻¹. To find the second derivative, simply differentiate dy/dx again using the same power rule. For example, if y = 3x⁴, then dy/dx = 12x³, and d²y/dx² = 36x².

给定多项式 y = axⁿ，一阶导数为 dy/dx = naxⁿ⁻¹。要求二阶导数，只需再次使用幂法则对 dy/dx 求导。例如，若 y = 3x⁴，则 dy/dx = 12x³，d²y/dx² = 36x²。

Every term follows the same pattern: multiply by the power and reduce the power by 1. Constants become zero on the first differentiation, so they vanish completely when differentiating twice.

每一项都遵循相同规律：乘以指数并将指数减 1。常数在一阶求导时变为 0，因此两次求导后完全消失。

Example: y = 2x⁵ − 3x³ + 4x² − 7

dy/dx = 10x⁴ − 9x² + 8x
d²y/dx² = 40x³ − 18x + 8

示例： y = 2x⁵ − 3x³ + 4x² − 7
dy/dx = 10x⁴ − 9x² + 8x
d²y/dx² = 40x³ − 18x + 8

3. Finding Stationary Points from the First Derivative | 利用一阶导数求驻点

A stationary point occurs where the gradient is zero, i.e., dy/dx = 0. Solve this equation for x to find the x-coordinates of all turning points or points of inflection. For a function y = x³ − 3x, we set dy/dx = 3x² − 3 = 0 → x² = 1 → x = 1 or x = −1.

当梯度为零，即 dy/dx = 0 时，会出现驻点。解此方程求出 x，即可得到所有转折点或拐点的横坐标。对于函数 y = x³ − 3x，令 dy/dx = 3x² − 3 = 0 → x² = 1 → x = 1 或 x = −1。

Substitute these x-values back into the original y = f(x) to obtain the corresponding y-coordinates: when x = 1, y = (1)³ − 3(1) = −2; when x = −1, y = (−1)³ − 3(−1) = 2. So the stationary points are (1, −2) and (−1, 2).

将这些 x 值代回原函数 y = f(x)，求得对应的 y 坐标：当 x = 1 时，y = (1)³ − 3(1) = −2；当 x = −

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