📚 Edexcel A-Level Mathematics: Core Differentiation Techniques | Edexcel A-Level 数学核心微分技巧
This revision guide covers the essential differentiation methods required for the Edexcel A-Level Mathematics specification. From first principles to implicit and parametric differentiation, each topic is explained with step-by-step reasoning. Whether you are preparing for Pure Mathematics Paper 1 or tackling applied problems involving rates of change, mastery of these techniques is crucial for achieving a top grade.
本复习指南涵盖 Edexcel A-Level 数学大纲所要求的全部核心微分方法。从第一原理到隐函数微分和参数微分，每个主题都配有逐步推理。无论你是在准备纯数试卷 1，还是解决涉及变化率的应用题，掌握这些技巧对取得高分至关重要。
1. The Gradient Function from First Principles | 从第一原理推导梯度函数
ifferentiation begins with the concept of a limit. Given a function f(x), its derivative f'(x) is defined as the limit of the difference quotient as h → 0: f'(x) = lim(h→0) [f(x+h) – = x² + 2xh + h² leads to [2xh + h²]/h = 2x + h, and as h approaches zero, the limit is 2x.
微分始于极限的概念。给定函数 f(x)，其导数 f'(x) 定义为当 h → 0 时差商的极限：f'(x) = lim(h→0) [f(x+h) – f(x)] / h。该公式捕捉了函数的瞬时变化率。对于 f(x) = x²，展开 (x+h)² = x² + 2xh + h² 得到 [2xh + h²]/h = 2x + h，当 h 趋于零时，极限为 2x。
+ 3xh² + h³ yields a difference quotient of 3x² + 3xh + h², whose limit is 3x². This foundational understanding helps you appreciate why the power rule works and prepares you for more rigorous analysis.
你需要利用这个极限定义证明简单函数（如 x² 和 x³）的导数。对于 f(x) = x³，展开式 (x+h)³ = x³ + 3x²h + 3xh² + h³ 给出差商 3x² + 3xh + h²，其极限为 3x²。这种基础理解有助于你领会幂法则的缘由，并为更严格的分析做好准备。
f'(x) = lim(h→0) [f(x+h) – f(x)] / h
2. Power Rule and Basic Derivatives | 幂法则与基本导数
ny real constant n, including negative and fractional powers. Thus d/dx (x⁵) = 5x⁴, d/dx (√x) = d/dx (x½) = ½ x⁻½ = 1/(2√x), and d/dx (1/x²) = d/dx (x⁻²) = –2x⁻³ = –x³. You must be comfortable rewriting expressions using index laws before differentiating.
最常用的工具是幂法则：若 y = xⁿ，则 dy/dx = nxⁿ⁻¹。这适用于任何实数常数 n，包括负指数和分数指数。因此 d/dx (x⁵) = 5x⁴，d/dx (√x) = d/dx (x½) = ½ x⁻½ = 1/(2√x)，d/dx (1/x²) = d/dx (x⁻²) = –2x⁻³ = –2/x³。在做微分之前，你必须善于利用指数律改写表达式。
the constant multiple rule and sum rule, you can differentiate any polynomial. For y = 4x⁶ – 3x² + 5x – 7, the derivative is dy/dx = 24x⁵ – also memorise the derivatives of standard functions: d/dx (sin x) = cos x, d/dx (cos x) = –sin x, d/dx (tan x) = sec² x, d/dx (eˣ) = eˣ, d/dx (ln x) = 1/x for x > 0.
结合常数倍法则与和法则，你就可以对任意多项式求导。对于 y = 4x⁶ – 3x² + 5x – 7，导数为 dy/dx = 24x⁵ – 6x + 5。常数的导数为零。你还应熟记标准函数的导数：d/dx (sin x) = cos x，d/dx (cos x) = –sin x，d/dx (tan x) = sec² x，d/dx (eˣ) = eˣ，d/dx (ln x) = 1/x（x > 0）。
d/dx (xⁿ) = nxⁿ⁻¹
3. The Chain Rule | 链式法则
(u) and u = g(x), then dy/dx = (dy/du) × (du/dx). In function notation, d/dx [f(g(x))] = f'(g(x)) g' = u⁴, dy/du = 4u³, du/dx = 6x. Multiplying gives dy/dx = 4(3x² + 5)³ × 6x = 24x(3x² + 5)³.
许多 A-Level 函数都是简单函数的复合。链式法则指出，若 y = f(u) 且 u = g(x)，则 dy/dx = (dy/du) × (du/dx)。用函数记号表示为 d/dx [f(g(x))] = f'(g(x)) g'(x)。对于 y = (3x² + 5)⁴，令 u = 3x² + 5，于是 y = u⁴，dy/du = 4u³，du/dx = 6x。相乘得 dy/dx = 4(3x² + 5)³ × 6x = 24x(3x² + 5)³。
rivative of the inner function. For y = sin(2x + 1), the outer is sin, so derivative is cos(2x + 1) × 2 = 2 cos(2x + 1). For y = e5x³, derivative is e5x³ 15x². Make sure you can combine the chain rule with products and quotients when necessary.
经过练习，你将能快速应用链式法则：对外层函数求导，保持内层函数不变，再乘以内层函数的导数。对于 y = sin(2x + 1)，外层为正弦，导数为 cos(2x + 1) × 2 = 2 cos(2x + 1)。对于 y = e5x³，导数为 e5x³ × 15x²。确保你在必要时能将链式法则与乘积、商法则结合使用。
dy/dx = (dy/du) × (du/dx)
4. Product and Quotient Rules | 积法则与商法则
Then du/dx = 2x, dv/dx = cos x. Hence dy/dx = x² cos x + 2x sin x. Always present the answer in its simplest factored form where possible, e.g., x(x cos x + 2 sin x).
当函数是两个式子的乘积时，使用积法则：若 y = u(x)v(x)，则 dy/dx = u dv/dx + v du/dx。对于 y = x² sin x，令 u = x²，v = sin x。则 du/dx = 2x，dv/dx = cos x。因此 dy/dx = x² cos x + 2x sin x。尽量将答案写成最简因式分解形式，例如 x(x cos x + 2 sin x)。
The quotient rule handles divisions: if y = u(x) / v(x), then dy/dx = (v du/dx – u dv/dx) / v². A memorable mnemonic is "low d-high minus high d-low over low-squared" u = x³, v = x² + 1. Then du/dx = 3x², dv/dx = 2x. Substituting gives dy/dx = [(x²+1)(3x²) – (x³)(2x)] / (x²+1)² = [3x⁴ + 3x² – 2x⁴] / (x²+1)² = (x⁴ + 3x²) / (x²+1)².
商法则用于除法：若 y = u(x) / v(x)，则 dy/dx = (v du/dx – u dv/dx) / v²。一句好记的口诀是「低乘高导减高乘低导，除以低平方」。对于 y = (x³) / (x² + 1)，令 u = x³，v = x² + 1。则 du/dx = 3x²，dv/dx = 2x。代入得 dy/dx = [(x²+1)(3x²) – (x³)(2x)] / (x²+1)² = [3x⁴ + 3x² – 2x⁴] / (x²+1)² = (x⁴ + 3x²) / (x²+1)²。
Product Rule: d/dx (uv) = u dv/dx + v du/dx
Quotient Rule: d/dx (u/v) = (v du/dx – u dv/dx) / v²
5. Implicit Differentiation | 隐函数微分
tiate both sides term by term with respect to x, treating y as a function of x and using the chain rule for any y-terms: d/dx (yⁿ) = n yⁿ⁻¹ (dy/dx). For the circle equation x² + y² = 25, differentiating gives 2x + 2y (dy/dx) = 0, so dy/dx = –x/y.
有些方程不易写成 y = f(x) 的形式。隐函数微分法让你可以直接从关于 x 和 y 的方程中求出 dy/dx。对每一项关于 x 逐项求导，将 y 视为 x 的函数，并对含 y 的项使用链式法则：d/dx (yⁿ) = n yⁿ⁻¹ (dy/dx)。对于圆的方程 x² + y² = 25，求导得 2x + 2y (dy/dx) = 0，因此 dy/dx = –x/y。
elds (1·y + x·dy/dx) + cos y (dy/dx) = 0. Collect dy/dx terms: x (dy/dx) + cos y (dy/dx) = –y, so dy/dx = – he second derivative or to locate stationary points on curves that are defined implicitly.
当出现 x 与 y 的乘积时，应用积法则。对于 xy + sin y = 3，求导得 (1·y + x·dy/dx) + cos y (dy/dx) = 0。合并 dy/dx 项：x (dy/dx) + cos y (dy/dx) = –y，所以 dy/dx = –y / (x + cos y)。隐函数微分还可用于求二阶导数，或定位隐式曲线上的驻点。
d/dx (yⁿ) = n yⁿ⁻¹ (dy/dx)
6. Parametric Differentiation | 参数微分
ined by dividing dy/dt by dx/dt: dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0. For x = t² – 1, y = 4t + 3, we compute dx/dt = 2t, dy/dt = 4, so dy/dx = 4 / (2t) = 2/t.
当 x 和 y 都以第三个变量 t 给出时，我们得到参数方程 x = f(t)，y = g(t)。导数 dy/dx 由 dy/dt 除以 dx/dt 求得：dy/dx = (dy/dt) / (dx/dt)，前提是 dx/dt ≠ 0。对于 x = t² – 1，y = 4t + 3，计算 dx/dt = 2t，dy/dt = 4，从而 dy/dx = 4 / (2t) = 2/t。
(dy/dx) = [d/dt (dy/dx)] / (dx/dt). For the same example, dy/dx = 2t⁻¹, so d/dt (dy/dx) = –2 t⁻². Then d²y/dx² = (–2/t²) / (2t) = – tion is common in mechanics problems where t represents time and x, y represent positions.
要通过参数求二阶导数 d²y/dx²，使用 d²y/dx² = d/dx (dy/dx) = [d/dt (dy/dx)] / (dx/dt)。对上例，dy/dx = 2t⁻¹，所以 d/dt (dy/dx) = –2 t⁻²。于是 d²y/dx² = (–2/t²) / (2t) = –1/t³。参数微分在力学问题中很常见，其中 t 表示时间，x、y 表示位置。
dy/dx = (dy/dt) ÷ (dx/dt)
d²y/dx² = [d/dt (dy/dx)] / (dx/dt)
7. Gradients, Tangents and Normals | 梯度、切线与法线
tangent to the curve at that point. For y = f(x), the tangent at (a, f(a)) has equation y – f(a) = f'(a)(x – a). The normal is perpendicular to the tangent, so its gradient is –1 / f'(a), provided f'(a) ≠ 0. For y = x² at x = 3, f'(3) = 6, so the tangent equation is y – 9 = 6(x – 3) and the normal equation is y – 9 = –1/6 (x – 3).
函数在某点的导数值给出了曲线在该点处切线的梯度。对于 y = f(x)，切点 (a, f(a)) 处的切线方程为 y – f(a) = f'(a)(x – a)。法线垂直于切线，因此其梯度为 –1 / f'(a)，前提是 f'(a) ≠ 0。对于 y = x² 在 x = 3 处，f'(3) = 6，故切线方程为 y – 9 = 6(x – 3)，法线方程为 y – 9 = –1/6 (x – 3)。
These concepts appear frequently in coordinate geometry questions combined with calculus. You may need to find where a tangent meets the axes, or determine the equation of a normal that passes through a given external point. Always start by finding the derivative, evaluate it at the required x-coordinate, and then construct the straight line equation using point-gradient form.
这些概念经常在结合微积分的坐标几何题中出现。你可能需要求切线与坐标轴的交点，或确定通过给定外点的法线方程。始终从求导开始，在所需 x 坐标处计算导数值，然后用点斜式建立直线方程。
Tangent: y – f(a) = f'(a)(x – a)
Normal: y – f(a) = (–1/f'(a))(x – a)
8. Stationary Points and Their Nature | 驻点及其性质
it is a local maximum; if d²y/dx² = 0, the test is inconclusive and you should examine the sign of the first derivative on either side. For y = x³ – 3x, dy/dx = 3x² – es x = ±1. The second derivative d²y/dx² = 6x. At x = 1, it is 6 (>0), so a minimum at (1, –2); at x = –1, it is –6 (<0), so a maximum at (–1, 2).
驻点出现在 dy/dx = 0 处。为分类它们，使用二阶导数判别法：若 d²y/dx² > 0，该点为局部极小点；若 d²y/dx² < 0，为局部极大点；若 d²y/dx² = 0，该判别法无效，应检查一阶导数在点两侧的符号。对于 y = x³ – 3x，dy/dx = 3x² – 3 = 0 给出 x = ±1。二阶导数 d²y/dx² = 6x。在 x = 1 处，其值为 6 (>0)，故点 (1, –2) 为极小点；在 x = –1 处，其值为 –6 (<0)，故点 (–1, 2) 为极大点。
In optimisation problems, you model a real-world quantity as a function of one variable, find its stationary points and justify the maximum or minimum using the second derivative or by considering endpoints. Common examples include maximising enclosed area given a fixed perimeter, or minimising surface area for a given volume. Clearly defining variables and stating the domain is essential for full marks.
在优化问题中，你将现实量建模为单一变量的函数，求其驻点，并利用二阶导数或考虑端点来证明极大或极小。常见例子包括在给定周长下使围成面积最大化，或在给定体积下使表面积最小化。清楚地定义变量并陈述定义域是获得满分的必要条件。
Stationary point: dy/dx = 0
Second derivative test: d²y/dx² > 0 ⇒ minimum; < 0 ⇒ maximum
9. Connected Rates of Change | 相关变化率
nd you know dV/dt, you can find dr/dt using dV/dt = (dV/dr)(dr/dt). For a sphere, V = (4/3)π r³, so dV/dr = 4π r². Thus dr/dt = (1/(4π r²)) dV/dt. Such problems require you to identify the relevant geometric formula and differentiate with respect to the linking variable.
当两个或多个量随时间变化时，它们的变化率通过链式法则关联。例如，一个球形气球正在充气，若已知 dV/dt，可利用 dV/dt = (dV/dr)(dr/dt) 求得 dr/dt。对于球体，V = (4/3)π r³，故 dV/dr = 4π r²。于是 dr/dt = (1/(4π r²)) dV/dt。这类问题要求你识别相关的几何公式，并对关联变量求导。
You may also encounter problems where one rate is given and you must find the rate of change of an angle or length at a specific instant. Set up an equation linking the quantities, differentiate with respect to time t, substitute the known values and solve for the unknown rate. Make sure to state the units (e.g., cm/s, m²/s) and note whether a quantity is increasing or decreasing.
你也可能遇到给出一个变化率，而需求出某一瞬间角度或长度的变化率的问题。建立联系各量的方程，关于时间 t 求导，代入已知值并解出未知变化率。务必注明单位（如 cm/s、m²/s），并注意该量是增大还是减小。
Chain rule for rates: dA/dt = (dA/dr) × (dr/dt)
10. Second Derivatives and Concavity | 二阶导数与凹凸性
wards), while a negative second derivative means it is concave (curving downwards). Points where the concavity changes are called points of inflection. At an inflection point, d²y/dx² = 0 (or is undefined), but the second derivative must change sign.
二阶导数提供有关图形弯曲情况的信息。正的二阶导数表明曲线是凸的（向上弯曲），负的二阶导数则表明曲线是凹的（向下弯曲）。凹凸性发生改变的点称为拐点。在拐点处，d²y/dx² = 0（或不存在），但二阶导数必须变号。
Consider y = x⁴ – 6x² + 8x. The second derivative is d²y/dx² = 12x² – ting it to zero gives x = ±1. Testing intervals reveals that concavity changes at both x = – ncavity helps with accurate curve sketching and complements the stationary point analysis.
考虑 y = x⁴ – 6x² + 8x。其二阶导数为 d²y/dx² = 12x² – 12。令其为零得到 x = ±1。检验区间可知凹凸性在 x = –1 和 x = 1 处均发生改变，故两者均为拐点。理解凹凸性有助于精确绘制函数图像，并与驻点分析相辅相成。
Convex: d²y/dx² > 0; Concave: d²y/dx² < 0
Point of inflection: d²y/dx² = 0 and changes sign
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