📚 pdfjoiner_(4)-075 | A-Level Edexcel数学微分专项复习
Welcome to this comprehensive revision guide on differentiation for the Edexcel A Level Mathematics specification. This article consolidates key concepts, rules, and applications of differentiation, essential for success in the pure mathematics components. Whether you are preparing for the AS or A2 examinations, mastering these techniques is vital for tackling gradient-based problems, optimisation, and curve sketching.
欢迎阅读本针对Edexcel A Level数学考试大纲的微分综合复习指南。本文整合了微分的核心概念、法则与应用，这些知识对于纯数部分取得优异成绩至关重要。无论您正在准备AS还是A2考试，掌握这些技巧对于解决梯度问题、最优化及曲线草图绘制都十分重要。
1. Introduction to Differentiation | 微分简介
that measures how a function changes as its input changes. Geometrically, the derivative f' x. It is sometimes denoted as dy/dx, and the process of finding the derivative is called differentiating with respect to x. Understanding the rate of change is crucial for physics, engineering, and economics applications.
微分是微积分中的基本工具，用于衡量函数随自变量变化而变化的快慢。从几何角度看，导数 f'(x) 表示曲线 y = f(x) 在任意点 x 处切线的斜率，有时记作 dy/dx，求导过程称为对 x 求导。理解变化率对于物理、工程和经济学应用至关重要。
2. Differentiation from First Principles | 第一原理求导
The derivative can be defined using limits. The formula from first principles is:
f'(x) = limₕ→₀ (f(x+h) – f(x)) / h
For example, to differentiate f(x) = x² from first principles:
f'(x) = limₕ→₀ ((x+h)² – x²) / h = limₕ→₀ (2xh + h²) / h = limₕ→₀ (2x + h) = 2x
This technique, while not always used in routine differentiation, underpins the definition of the derivative and is examinable in Edexcel A Level Maths.
导数可以用极限来定义。第一原理的公式为：
f'(x) = limₕ→₀ (f(x+h) – f(x)) / h
例如，用第一原理对 f(x) = x² 求导：
f'(x) = limₕ→₀ ((x+h)² – x²) / h = limₕ→₀ (2xh + h²) / h = limₕ→₀ (2x + h) = 2x
虽然在常规求导中不常用，但这一方法为导数的定义奠定了理论基础，也是Edexcel A Level数学的考试内容。
3. Basic Derivative Rules | 基本求导公式
Once the limit definition is understood, we can build a set of standard derivatives that make differentiation much faster. The table below summarises the essential rules for common functions:
一旦理解了极限定义，就可以建立一套标准求导公式，大大加快求导速度。下表总结了常见函数的基本求导规则：
| f(x) | f'(x) |
|---|---|
| c (constant 常数) | 0 |
| xⁿ | n xⁿ⁻¹ |
| eˣ | eˣ |
| ln x | 1/x |
| sin x | cos x |
| cos x | −sin x |
| tan x | sec² x |
can be combined with constant multiples and sums: if y = a u(x) + b v(x), then dy/dx = a u'(x) + b v' ys remember to rewrite roots and reciprocals using exponents before differentiating, e.g., √x = x½ and 1/x² = x⁻².
这些规则可以与常数倍及和差法则结合使用：若 y = a u(x) + b v(x)，则 dy/dx = a u'(x) + b v'(x)。求导前切记将根式和倒数形式改写为指数形式，例如 √x = x½，1/x² = x⁻²。
4. Chain Rule | 链式法则
composite functions where one function is inside another. If y = f(u) and u = g(x), then:
dy/dx = dy/du × du/dx
u = 5u⁴, du/dx = 6x, so dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴. Practice identifying the 'inner' and 'outer' functions to apply the chain rule correctly in Edexcel exams.
链式法则用于求复合函数的导数，即一个函数包含在另一个函数内部。若 y = f(u) 且 u = g(x)，则：
dy/dx = dy/du × du/dx
例如，对 y = (3x² + 1)⁵ 求导，令 u = 3x² + 1，则 y = u⁵。dy/du = 5u⁴，du/dx = 6x，因此 dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴。在Edexcel考试中，需练习识别“内层”和“外层”函数以正确运用链式法则。
5. Product Rule | 乘积法则
When differentiating the product of two functions, y = u(x) v(x), we use the product rule:
dy/dx = u (dv/dx) + v (du/dx)
Alternatively written as y' = u v' + v u'. For instance, if y = x² sin x, let u = x², v = sin x. Then u' = 2x, v' uct rule is essential and often appears alongside the chain rule in more complex problems.
当求两个函数乘积 y = u(x) v(x) 的导数时，我们使用乘积法则：
dy/dx = u (dv/dx) + v (du/dx)
也可写为 y' = u v' + v u'。例如，若 y = x² sin x，令 u = x², v = sin x，则 u' = 2x, v' = cos x，因此 dy/dx = x² cos x + sin x × 2x = x² cos x + 2x sin x。乘积法则至关重要，并常常与链式法则结合出现在复杂问题中。
6. Quotient Rule | 商法则
tion given as a quotient y = u(x) / v(x), the derivative is found using the quotient rule:
dy/dx = (v du/dx – u dv/dx) / v²
example: y = tan x = sin x / cos x. Let u = sin x, v = cos x, then du/dx = cos x, dv/dx = –sin x. Substituting: dy/dx = (cos x × cos x – sin x × (– n derivative. Remember to square the denominator and keep the subtraction in the numerator's order.
对于商形式的函数 y = u(x) / v(x)，求导需使用商法则：
dy/dx = (v du/dx – u dv/dx) / v²
一个经典示例：y = tan x = sin x / cos x。令 u = sin x, v = cos x，则 du/dx = cos x, dv/dx = –sin x。代入得：dy/dx = (cos x × cos x – sin x × (–sin x)) / cos² x = (cos² x + sin² x) / cos² x = 1 / cos² x = sec² x，验证了已知导数。务必注意分母平方以及分子中相减的先后顺序。
7. Implicit Differentiation | 隐函数求导
on of x. Every time we differentiate a term involving y, we multiply by dy/dx using the chain rule. For instance, d/dx (y²) = 2y dy/dx.
当 y 并未显式表达为 x 的函数时（例如 x² + y² = 25），我们对方程每一项关于 x 求导，将 y 视为 x 的隐函数。每次对包含 y 的项求导时，需依据链式法则乘以 dy/dx。例如，d/dx (y²) = 2y dy/dx。
Differentiating x² + y² = 25 gives 2x + 2y (dy/dx) = 0, so dy/dx = – annot be easily expressed as y = f(x), and it appears regularly in Edexcel A Level papers.
对 x² + y² = 25 求导可得 2x + 2y (dy/dx) = 0，因此 dy/dx = –x/y。隐函数求导对于难以写成 y = f(x) 形式的曲线尤为有用，是Edexcel A Level考试中的常见内容。
8. Parametric Differentiation | 参数方程求导
the derivative dy/dx is obtained by differentiating both with respect to the parameter t:
dy/dx = (dy/dt) / (dx/dt)
For example, a curve is given by x = t² + 1, y = t³ – 3t. Then dx/dt = 2t, dy/dt = 3t² – 3. Hence dy/dx = (3t² – chnique allows us to find tangents, normals, and stationary points in parametric contexts.
当曲线由参数方程 x = f(t), y = g(t) 定义时，导数的求法是分别对参数 t 求导，然后相除：
dy/dx = (dy/dt) / (dx/dt)
例如，一条曲线的参数方程为 x = t² + 1, y = t³ – 3t，则 dx/dt = 2t, dy/dt = 3t² – 3，因此 dy/dx = (3t² – 3) / 2t。该方法使我们能够求解参数曲线的切线、法线和驻点。
9. Second Derivatives & Concavity | 二阶导数与凹凸性
The second derivative, denoted d²y/dx² or f''(x), is the derivative of the first derivative. It measures the rate of change of the gradient and helps determine the concavity of a function. If f''(x) > 0, the graph curves upwards (convex), and if f''(x) < 0, it curves downwards (concave).
二阶导数记作 d²y/dx² 或 f''(x)，是一阶导数的导数。它衡量斜率的变化率，有助于判断函数的凹凸性。若 f''(x) > 0，图像向上弯曲（凸函数）；若 f''(x) < 0，则向下弯曲（凹函数）。
tionally, points of inflection occur where the concavity changes, often found by setting f'' dexcel, second derivatives are also used to classify stationary points (maxima or minima) via the second derivative test.
此外，拐点出现在凹凸性变化处，通常令 f''(x) = 0 并检查符号变化。在Edexcel考试中，二阶导数也用于通过二阶导数判定法对驻点进行分类（最大值或最小值）。
10. Applications: Tangents, Normals & Stationary Points | 应用：切线、法线与驻点
point. The tangent at (x₁, y₁) has gradient m = dy/dx at that point, and its equation is y – y₁ = m(x – x₁). The normal is perpendicular, so its gradient is –1/m.
微分的一大应用是求曲线在某点的切线或法线方程。切线在 (x₁, y₁) 处的斜率 m 等于 dy/dx 在该点的值，其方程为 y – y₁ = m(x – x₁)。法线与切线垂直，因此其斜率为 –1/m。
ints of inflection. The nature can be determined by checking the sign of dy/dx either side of the point or by evaluating the second derivative f''(x) if it is non-zero.
驻点出现在 dy/dx = 0 处。它们可以是极大值、极小值或拐点。其性质可通过检查 dy/dx 在驻点两侧的符号或通过计算非零二阶导数 f''(x) 来判定。
11. Modelling with Differentiation | 微分建模
Differentiation is widely used in real-world modelling problems, especially optimisation (maximising area, minimising surface area, etc.) and rates of change involving connected variables. For example, given a formula for volume V and a rate of change dV/dt, you might use the chain rule to find dr/dt, where r is radius.
微分广泛用于现实建模问题，特别是最优化（最大化面积、最小化表面积等）以及涉及相关变量的变化率。例如，已知体积 V 的公式和变化率 dV/dt，可利用链式法则求 dr/dt，其中 r 为半径。
In Edexcel A Level, these questions often combine multiple differentiation techniques. It is essential to identify the variables, express the quantity to be optimised in terms of one variable, differentiate, and then check second derivatives or values at boundaries to confirm maxima or minima.
在Edexcel A Level考试中，此类题目常结合多种微分技巧。关键步骤为：确定变量，将待优化量用一个变量表示，求导，并通过二阶导数或端值检验确认是否为最大值或最小值。
12. Summary of Key Differentiation Concepts | 微分核心概念总结
Mastering differentiation for Edexcel A Level Mathematics requires fluency in basic rules, chain, product, quotient, implicit, and parametric methods. Always remember to simplify algebraic expressions before differentiating, and keep an eye out for hidden composite functions. Practise past-paper questions to build speed and accuracy.
掌握Edexcel A Level数学的微分部分，需要熟练运用基本求导公式、链式、乘积、商、隐函数及参数方程求导法。务必在求导前化简代数表达式，并留意隐藏的复合函数。通过历年真题练习来提高解题速度和准确性。
Regularly revisiting these concepts will not only boost your confidence but also strengthen your overall calculus skills for further study in mathematics, physics, and engineering.
定期复习这些概念不仅能增强自信，还能巩固微积分整体技能，为未来数学、物理和工程领域的学习打下扎实基础。
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