📚 IGCSE Edexcel Mathematics: Differentiation – Key Concepts | IGCSE Edexcel 数学：微分 考点精讲
In IGCSE Edexcel Mathematics, differentiation is the process of finding the rate of change of a function. Mastering the basics of differentiation allows you to find gradients of curves, equations of tangents and normals, identify increasing and decreasing functions, and locate stationary points like maxima and minima. This article summarises all essential differentiation topics you need for the exam.
在 IGCSE Edexcel 数学中，微分是求函数变化率的过程。掌握微分基础，就能求解曲线的梯度、切线与法线方程、判断函数的增减性，并找出极大值、极小值等驻点。本文总结了考试所需的所有微分核心考点。
1. What is Differentiation? | 什么是微分？
he gradient of a curve at any point. The derivative of a function y = f(x) is written as f' ve, the gradient varies and differentiation provides a formula for this variable gradient.
微分给出曲线在任意点的梯度。函数 y = f(x) 的导数记作 f'(x) 或 dy/dx，表示 y 关于 x 的变化率。对于直线，梯度是常数；而对于曲线，梯度是变化的，微分给出了这一变化梯度的公式。
2. The Power Rule | 幂法则
emember that x¹ becomes x⁰ = 1, so the derivative of x is 1. Constants differentiate to 0.
对 x 的幂函数求导最重要的法则是：如果 y = xⁿ，那么 dy/dx = n·xⁿ⁻¹。乘以指数，指数减 1。例如 y = x⁵，dy/dx = 5x⁴。如果 y = 3x²，常数系数不变，dy/dx = 3 × 2x¹ = 6x。记住 x¹ 求导得 x⁰ = 1，所以 x 的导数是 1。常数项求导为 0。
If y = xⁿ, then dy/dx = n·xⁿ⁻¹
3. Differentiating Polynomials Term by Term | 多项式逐项求导
powers, the power rule still applies. For instance, y = 1/x = x⁻¹ → dy/dx = -1x⁻² = -1/x².
对多项式求导，只需逐项求导。例如 y = 4x³ - 2x² + 5x - 7，其导数为 dy/dx = 12x² - 4x + 5。注意常数项 -7 求导为 0。求导后应合并同类项。当幂为负数或分数时，幂法则同样适用。比如 y = 1/x = x⁻¹，求导得 dy/dx = -1x⁻² = -1/x²。
4. Finding the Gradient at a Specific Point | 求曲线上特定点的梯度
his tells you the slope of the tangent. If the gradient is positive, the curve is rising at that point; if negative, falling.
将点的 x 坐标代入导数即可求出该点的梯度。例如 y = x³ - 2x + 1，dy/dx = 3x² - 2。在 x = 2 处，梯度 m = 3(2)² - 2 = 10。这就给出了切线的斜率。梯度为正表示曲线在该点上升，为负表示下降。
5. Equations of Tangents and Normals | 切线与法线方程
. Tangent: y - 2 = 2(x - 1) → y = 2x. Normal gradient = -1/2, equation: y - 2 = -½(x - 1).
曲线在某点的切线斜率等于该点的导数值。使用点斜式 y - y₁ = m(x - x₁)。法线垂直于切线，其斜率 m(法) = -1/m(切)。例如曲线 y = x² + 1，在 x = 1 处，y = 2，切线斜率 m = 2。切线方程：y - 2 = 2(x - 1) → y = 2x。法线斜率 = -½，方程：y - 2 = -½(x - 1)。
6. Increasing and Decreasing Functions | 函数的递增与递减
A function f(x) is increasing where f'(x) > 0, and decreasing where f'(x) < 0. To find intervals, solve f'(x) = 0 for critical values, then test the sign of f'(x) in each interval. If f'(x) > 0, function is increasing; if f'(x) < 0, decreasing. For f(x) = 2x³ - 9x² + 12x, f' = 6(x - 1)(x - 2). Test intervals x < 1, 1 < x < 2, x > 2 to determine increase/decrease.
如果 f'(x) > 0，函数在该区间递增；如果 f'(x) < 0，则递减。为求区间，先解 f'(x) = 0 得临界值，然后在每个区间测试 f'(x) 的符号。例如 f(x) = 2x³ - 9x² + 12x，f'(x) = 6x² - 18x + 12 = 6(x - 1)(x - 2)。检验区间 x < 1, 1 < x < 2, x > 2 内 f'(x) 的正负，即可判断增减性。
7. Stationary Points and their Nature | 驻点及其性质
Stationary points occur where f' ion. To determine the nature, use either the first derivative test (check sign change of f'(x) around the point) or the second derivative test. If f''(x) > 0, it's a minimum; if f''(x) < 0, it's a maximum; if f''(x) = 0, use the first derivative test or further inspection. For y = x³, f'(0)=0, f'' oint is a point of inflection because gradient does not change sign (both sides positive).
驻点发生在 f'(x) = 0 处，可以是局部极大值、极小值或拐点。判断性质可用一阶导数检测（检视 f'(x) 在驻点两侧的符号变化），或二阶导数检测。若 f''(x) > 0，为极小值；若 f''(x) < 0，为极大值；若 f''(x) = 0，需用一阶导数检测或其他方法。例如 y = x³，f'(0)=0，f''(0)=0，该点是拐点，因为两侧梯度均为正（符号未变）。
8. The Second Derivative | 二阶导数
The second derivative, written as f'' The second derivative also helps classify stationary points and determine concavity. If f''(x) > 0, the curve is concave up (∪); if f''(x) < 0, concave down (∩).
二阶导数，记作 f''(x) 或 d²y/dx²，是对一阶导数再求导。它衡量梯度的变化率。在运动学中，若位移 s 是时间 t 的函数，则速度 v = ds/dt，加速度 a = dv/dt = d²s/dt²。二阶导数也可用于判断驻点性质和曲线的凹凸性。f''(x) > 0 时曲线上凹（∪）；f''(x) < 0 时曲线下凹（∩）。
9. Applications: Kinematics | 运动学应用
often applied to motion along a straight line. Given displacement s(t), velocity v(t) = s'(t), acceleration a(t) = v'(t) = s'' article is at rest. These questions test your ability to interpret derivatives in context.
IGCSE 考题常将微分应用于直线运动。已知位移 s(t)，则速度 v(t) = s'(t)，加速度 a(t) = v'(t) = s''(t)。常需求质点何时静止（v = 0）、最大位移或者某一时刻的加速度。例如 s = t³ - 6t² + 9t，则 v = 3t² - 12t + 9，a = 6t - 12。令 v = 0 可求静止时刻。此类问题考验在具体情境中解释导数的能力。
10. Common Mistakes and Tips | 常见错误与备考技巧
multiply by the original power, incorrectly reducing powers (e.g., writing derivative of x³ as 3x² is correct but sometimes mistakes like 3x³ occur), misapplying the constant rule, and sign errors with negative powers. Also, remember to set f' original function. When finding the equation of a normal, do not forget to use the negative reciprocal. In kinematics, distinguish between speed (magnitude of velocity) and velocity. Always show clear working, and if a question asks for the maximum area or volume, link the derivative to the problem.
常见错误包括忘记乘以原有指数、降幂出错（如把 x³ 的导数误写成 3x³）、常数求导未归零，以及负指数时符号错误。还须记住，求驻点时应令 f'(x) = 0，而非原函数。求法线方程时，别忘记斜率取负倒数。在运动学中，要区分速率（速度的大小）与速度。务必展示清晰的步骤，若题目要求最大面积或体积，应将导数与实际情境相联系。
11. Revision and Exam Practice | 复习与考题练习
egative indices. Solve past paper questions under timed conditions. Typical exam question: " oordinates of the stationary points of y = 2x³ - 3x² - 36x + 5 and determine their nature." ive test) to classify. Try similar questions, and check your answers against mark schemes.
通过练习不同函数的求导来巩固，包括多项式、分数和负指数表达式。在计时条件下做历年真题。典型考题："求 y = 2x³ - 3x² - 36x + 5 的驻点坐标，并判断其性质。" 解答步骤：dy/dx = 6x² - 6x - 36 = 6(x - 3)(x + 2) = 0 → x = 3, x = -2。求出 y 坐标，然后用二阶导数（或一阶导数检测）确定性质。请尝试类似题目，并根据评分方案检查答案。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图