📚 Differentiation: Key Concepts for Edexcel A-Level Maths | 微分：Edexcel A-Level 数学考点精讲
Differentiation is a cornerstone of A-Level mathematics, providing tools to analyse rates of change, gradients of curves, and the behaviour of functions. In the Edexcel specification, you will be expected to differentiate a wide range of functions, apply rules such as the chain, product and quotient rules, and use differentiation to solve problems involving tangents, normals, stationary points and optimisation. This article covers the essential knowledge and techniques you need to master differentiation, with clear explanations and examples.
微分是A-Level数学的基石，它为我们提供了分析变化率、曲线斜率以及函数行为的工具。在Edexcel考试大纲中，你需要熟练对各种函数求导，应用链式法则、乘积法则和商法则，并利用微分解决切线、法线、驻点和优化等问题。本文涵盖了掌握微分所需的核心知识与技巧，并配有清晰的解释和例题。
1. First Principles and Basic Derivatives | 导数定义与基本求导公式
The derivative of a function f(x) can be defined from first principles as f'(x) = limh→0 [f(x+h) – f(x)] / h. While you are not expected to use this limit definition in most exam questions, understanding it underpins the concept of instantaneous rate of change. For power functions, the rule is straightforward: if y = xⁿ, then dy/dx = n xⁿ⁻¹. This works for any real constant n, including negative and fractional powers.
函数 f(x) 的导数可以通过第一原理定义：f'(x) = limh→0 [f(x+h) – f(x)] / h。虽然考试中不常直接考极限定义，但理解它有助于把握瞬时变化率的概念。对于幂函数，规则很简单：若 y = xⁿ，则 dy/dx = n xⁿ⁻¹。这适用于任何实数 n，包括负指数和分数指数。
You must also know that the derivative of a constant is zero, and the derivative of k·f(x) is k·f'(x). Sums and differences differentiate term by term: d/dx [u ± v] = du/dx ± dv/dx.
还需记住常数的导数为零，k·f(x) 的导数为 k·f'(x)。和与差的求导是逐项进行的：d/dx [u ± v] = du/dx ± dv/dx。
Example: If y = 5x⁴ – 3x² + 2x – 7, then dy/dx = 20x³ – 6x + 2.
例：若 y = 5x⁴ – 3x² + 2x – 7，则 dy/dx = 20x³ – 6x + 2。
2. The Chain Rule | 链式法则
The chain rule is used to differentiate composite functions: if y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). In practice, you differentiate the outer function, leaving the inner function unchanged, and then multiply by the derivative of the inner function.
链式法则用于求复合函数的导数：若 y = f(g(x))，则 dy/dx = f'(g(x)) · g'(x)。实际操作时，先对外层函数求导并保持内层函数不变，再乘以内层函数的导数。
For example, y = (3x² + 5)⁴. Let u = 3x² + 5, then y = u⁴. So dy/du = 4u³, du/dx = 6x. Hence dy/dx = 4(3x² + 5)³ · 6x = 24x(3x² + 5)³. Edexcel often expects you to leave the answer in factorised form.
例如，y = (3x² + 5)⁴。令 u = 3x² + 5，则 y = u⁴，dy/du = 4u³，du/dx = 6x，因此 dy/dx = 4(3x² + 5)³ · 6x = 24x(3x² + 5)³。Edexcel 通常希望答案保留因式分解形式。
Special cases include y = eᵏˣ gives dy/dx = k eᵏˣ, and y = ln(f(x)) gives dy/dx = f'(x)/f(x). For trigonometric functions like y = sin(kx), dy/dx = k cos(kx).
常见特例有：y = eᵏˣ 的导数为 k eᵏˣ；y = ln(f(x)) 的导数为 f'(x)/f(x)；三角函数如 y = sin(kx) 的导数为 k cos(kx)。
3. The Product Rule | 乘积法则
When two functions u(x) and v(x) are multiplied, the derivative is given by d/dx (uv) = u dv/dx + v du/dx. The formula is easy to apply, but careful algebraic simplification often follows.
两个函数 u(x) 与 v(x) 相乘时，导数为 d/dx (uv) = u dv/dx + v du/dx。公式应用简单，但之后往往需要进行细致的代数化简。
Example: y = x² e³ˣ. Let u = x² (du/dx = 2x) and v = e³ˣ (dv/dx = 3e³ˣ). Then dy/dx = x²·3e³ˣ + e³ˣ·2x = e³ˣ x (3x + 2). Factorising like this helps in finding stationary points.
例题：y = x² e³ˣ。设 u = x² (du/dx = 2x)，v = e³ˣ (dv/dx = 3e³ˣ)，则 dy/dx = x²·3e³ˣ + e³ˣ·2x = e³ˣ x (3x + 2)。这样因式分解有助于后续寻找驻点。
You can also use the product rule for three functions by grouping: d/dx (uvw) = u’vw + uv’w + uvw’. However, it rarely appears in Edexcel exams; better to multiply two at a time.
对于三个函数的乘积，可采用分组方式：d/dx (uvw) = u’vw + uv’w + uvw’。不过这在Edexcel考试中很少见，建议两两相乘处理。
4. The Quotient Rule | 商法则
For a quotient y = u/v, where u and v are functions of x, the derivative is dy/dx = (v du/dx – u dv/dx) / v². Remember the numerator is “v du minus u dv”, and the denominator is always v². It is useful when the denominator is a simple function, but sometimes rewriting as a product with v⁻¹ and using product rule plus chain rule is an alternative.
对于商 y = u/v，其中 u 和 v 均为 x 的函数，导数为 dy/dx = (v du/dx – u dv/dx) / v²。牢记分子是“v du 减 u dv”，分母永远是 v²。当分母是简单函数时，该法则很实用；但有时也可将原式改写为 u·v⁻¹ 的乘积形式，结合乘积法则与链式法则求解。
Example: y = (2x + 1) / (x² – 3). Let u = 2x+1, du/dx=2; v = x²-3, dv/dx=2x. Then dy/dx = [(x²-3)·2 – (2x+1)·2x] / (x²-3)² = [2x² – 6 – 4x² – 2x] / (x²-3)² = [-2x² – 2x – 6] / (x²-3)². Simplify if possible.
例：y = (2x + 1) / (x² – 3)。令 u = 2x+1, du/dx=2; v = x²-3, dv/dx=2x。则 dy/dx = [(x²-3)·2 – (2x+1)·2x] / (x²-3)² = [2x² – 6 – 4x² – 2x] / (x²-3)² = [-2x² – 2x – 6] / (x²-3)²。如能化简则继续化简。
5. Differentiating Exponentials and Logarithms | 指数函数与对数函数的微分
The derivative of eˣ is eˣ itself. More generally, d/dx (eᶠ⁽ˣ⁾) = f'(x) eᶠ⁽ˣ⁾. This follows from the chain rule. For natural logarithms, d/dx (ln x) = 1/x, and d/dx (ln f(x)) = f'(x)/f(x), provided f(x) > 0.
eˣ 的导数就是其自身。更一般地，d/dx (eᶠ⁽ˣ⁾) = f'(x) eᶠ⁽ˣ⁾，这由链式法则推出。对于自然对数，d/dx (ln x) = 1/x，而 d/dx (ln f(x)) = f'(x)/f(x)，前提是 f(x) > 0。
You may also need to differentiate aˣ. Because aˣ = e^(x ln a), the derivative is aˣ ln a. This can appear in modelling contexts.
可能还需对 aˣ 求导。因为 aˣ = e^(x ln a)，所以导数为 aˣ ln a。这类问题可能出现在建模情境中。
Example: y = ln(5x² + 3). Then dy/dx = (10x) / (5x² + 3). No need to simplify further unless a specific form is requested.
例：y = ln(5x² + 3)，则 dy/dx = (10x) / (5x² + 3)。除非题目要求特定形式，一般无需进一步化简。
6. Trigonometric Differentiation | 三角函数的微分
Edexcel requires you to know the derivatives of sin x, cos x and tan x, and to apply them with the chain rule. The basic results are:
Edexcel要求掌握 sin x、cos x 和 tan x 的导数，并能结合链式法则使用。基本公式如下：
- d/dx (sin x) = cos x
- d/dx (cos x) = -sin x
- d/dx (tan x) = sec² x
For sin(kx), cos(kx) and tan(kx), multiply the derivative of the inner function (k) accordingly. Also, sec x, cosec x and cot x can be differentiated: d/dx (sec x) = sec x tan x, d/dx (cosec x) = -cosec x cot x, d/dx (cot x) = -cosec² x. These are less common but still examinable.
对于 sin(kx)、cos(kx) 和 tan(kx)，需乘上内层函数的导数 k。此外，sec x、cosec x 和 cot x 的导数也需要了解：d/dx (sec x) = sec x tan x，d/dx (cosec x) = -cosec x cot x，d/dx (cot x) = -cosec² x。虽不频繁出现，但仍属考点。
Example: y = 3 sin(2x) – 4 cos(5x). Then dy/dx = 3·2 cos(2x) – 4·(-5 sin(5x)) = 6 cos(2x) + 20 sin(5x).
例：y = 3 sin(2x) – 4 cos(5x)，则 dy/dx = 3·2 cos(2x) – 4·(-5 sin(5x)) = 6 cos(2x) + 20 sin(5x)。
7. Implicit Differentiation | 隐函数微分
When a relation between x and y is given implicitly (e.g. x² + y² = 25), you can differentiate both sides with respect to x, treating y as a function of x. For each term involving y, use the chain rule: d/dx (yⁿ) = n yⁿ⁻¹ (dy/dx).
当 x 与 y 的关系以隐式给出时（例如 x² + y² = 25），可对方程两边关于 x 求导，并将 y 视为 x 的函数。凡含 y 的项都使用链式法则：d/dx (yⁿ) = n yⁿ⁻¹ (dy/dx)。
Example: x³ + 2xy + y² = 10. Differentiating: 3x² + 2y + 2x (dy/dx) + 2y (dy/dx) = 0. Collect dy/dx terms: (2x + 2y) dy/dx = -3x² – 2y, so dy/dx = -(3x² + 2y) / (2x + 2y).
例：x³ + 2xy + y² = 10。求导得：3x² + 2y + 2x (dy/dx) + 2y (dy/dx) = 0。合并含 dy/dx 的项：(2x + 2y) dy/dx = -3x² – 2y，因此 dy/dx = -(3x² + 2y) / (2x + 2y)。
Implicit differentiation is essential for finding tangents and normals to curves defined implicitly, and for related rates problems.
隐函数微分对于求解隐式曲线的切线与法线、以及相关变化率问题至关重要。
8. Parametric Differentiation | 参数方程微分
When x and y are both given as functions of a parameter t, the derivative dy/dx is found via dy/dx = (dy/dt) / (dx/dt). This provides the gradient of the curve at a specific point determined by the parameter value.
当 x 和 y 均为参数 t 的函数时，导数 dy/dx 通过 dy/dx = (dy/dt) / (dx/dt) 求得。这给出了曲线在参数值对应点的梯度。
Example: x = t² + 1, y = 2t³ – t. Then dx/dt = 2t, dy/dt = 6t² – 1. Hence dy/dx = (6t² – 1) / (2t). To find the equation of a tangent, substitute the appropriate t to get the gradient and coordinates.
例：x = t² + 1, y = 2t³ – t。则 dx/dt = 2t，dy/dt = 6t² – 1。于是 dy/dx = (6t² – 1) / (2t)。求切线方程时，代入对应 t 即可得到梯度和坐标。
You can also find the second derivative d²y/dx² parametrically by using d²y/dx² = d/dx (dy/dx) = [d/dt (dy/dx)] / (dx/dt). This is a common exam question.
还可通过参数式求二阶导数：d²y/dx² = d/dx (dy/dx) = [d/dt (dy/dx)] / (dx/dt)。这是常见考点。
9. Second Derivatives and Their Applications | 二阶导数及其应用
The second derivative, denoted d²y/dx² or f”(x), measures the rate of change of the gradient. It helps determine the nature of stationary points: if f”(x) > 0 at a stationary point, the point is a local minimum; if f”(x) < 0, it is a local maximum. If f''(x) = 0, the test is inconclusive and you should use the first derivative test.
二阶导数记作 d²y/dx² 或 f”(x)，衡量梯度的变化率。它有助于判断驻点的性质：若驻点处 f”(x) > 0，则该点为局部极小点；若 f”(x) < 0，则为局部极大点。如果 f''(x) = 0，则二阶导数检验失效，需使用一阶导数检验。
Second derivatives also indicate concavity and points of inflection. A point where the curvature changes from convex to concave (or vice versa) and f”(x) = 0 is a point of inflection.
二阶导数还可指示凹凸性与拐点。若曲线凹凸性发生变化且 f”(x) = 0，该点即为拐点。
Example: y = x³ – 6x² + 9x. First derivative: dy/dx = 3x² – 12x + 9. Stationary points at x = 1, 3. Second derivative: d²y/dx² = 6x – 12. At x=1, f”(1) = -6 (<0) max; at x=3, f''(3) = 6 (>0) min.
例：y = x³ – 6x² + 9x。一阶导数：dy/dx = 3x² – 12x + 9，驻点为 x = 1, 3。二阶导数：d²y/dx² = 6x – 12。当 x=1 时 f”(1) = -6 (<0) 极大值；x=3 时 f''(3) = 6 (>0) 极小值。
10. Equations of Tangents and Normals | 切线与法线方程
At a point (x₁, y₁) on a curve y = f(x), the gradient of the tangent is m = f'(x₁). The tangent equation is y – y₁ = m (x – x₁). The normal is perpendicular to the tangent, so its gradient is -1/m, provided m ≠ 0. The normal equation is y – y₁ = (-1/m) (x – x₁).
对于曲线 y = f(x) 上一点 (x₁, y₁)，切线的梯度为 m = f'(x₁)。切线方程为 y – y₁ = m (x – x₁)。法线与切线垂直，因此其梯度为 -1/m（假设 m ≠ 0）。法线方程为 y – y₁ = (-1/m) (x – x₁)。
With implicit or parametric curves, you find the gradient at the point and then proceed similarly. Always simplify your final equation into the form ax + by + c = 0 or y = mx + c as requested.
对于隐式或参数方程给出的曲线，先求出该点处的梯度，然后同样代入方程。最终答案应按题目要求化简为 ax + by + c = 0 或 y = mx + c 的形式。
Example: Find the tangent to y = x² + 3x at x=1. y(1)=4, dy/dx = 2x+3, at x=1 m=5. Tangent: y – 4 = 5(x – 1) → y = 5x – 1. Normal: y – 4 = -1/5 (x – 1).
例：求曲线 y = x² + 3x 在 x=1 处的切线。y(1)=4, dy/dx = 2x+3，当 x=1 时 m=5。切线：y – 4 = 5(x – 1) → y = 5x – 1。法线：y – 4 = -1/5 (x – 1)。
11. Stationary Points and Curve Sketching | 驻点与曲线草图
Stationary points occur where dy/dx = 0. They can be maxima, minima or points of inflection. To sketch a curve, you need to find intercepts, stationary points (and their nature), and the behaviour as x → ±∞ (asymptotes if any).
驻点出现在 dy/dx = 0 处，可能是极大值、极小值或拐点。绘制曲线草图时，需要求出截距、驻点及其性质，以及 x → ±∞ 时的变化趋势（如果有渐近线）。
Consider y = x³ – 3x. dy/dx = 3x² – 3 = 0 → x = ±1. At x=-1, d²y/dx² = 6(-1) = -6 (max); at x=1, d²y/dx² = 6 (min). The curve crosses the axes at (0,0), (√3,0), (-√3,0). Sketch shows a cubic shape.
考虑 y = x³ – 3x。dy/dx = 3x² – 3 = 0 → x = ±1。在 x=-1 处 d²y/dx² = 6(-1) = -6（极大值）；在 x=1 处 d²y/dx² = 6（极小值）。曲线与坐标轴交于 (0,0)、(√3,0)、(-√3,0)。草图为典型三次曲线形状。
For rational functions, use differentiation to locate turning points and combine with other features such as vertical asymptotes (where denominator zero) and horizontal/oblique asymptotes.
对于有理函数，利用微分确定极值点，并结合其他特征如垂直渐近线（分母为零处）和水平/斜渐近线。
12. Optimisation Problems | 最优化问题
Differentiation is a powerful tool for solving real-world optimisation problems: maximising area or volume, minimising cost or surface area, etc. The general approach is to express the quantity to be optimised as a function of one variable, differentiate, set the derivative to zero, and confirm the nature of the stationary point using the second derivative or sign change of the first derivative.
微分是解决实际最优化问题的有力工具：最大化面积或体积，最小化成本或表面积等。一般步骤为：将待优化量表示为单一变量的函数，求导，令导数为零，并通过二阶导数或一阶导数符号变化确认驻点性质。
Example: A container with an open top has a square base of side x cm and height h cm. Its volume is 256 cm³. Find the dimensions that minimise the surface area. Since V = x²h = 256, h = 256/x². Surface area A = x² + 4xh = x² + 1024/x. Differentiate: dA/dx = 2x – 1024/x² = 0 → x³ = 512 → x = 8 cm. Then h = 256/64 = 4 cm. Second derivative: d²A/dx² = 2 + 2048/x³ > 0 at x=8, confirming minimum.
例：一个无盖容器底面为边长 x cm 的正方形，高 h cm，体积为 256 cm³。求使表面积最小的尺寸。由 V = x²h = 256 得 h = 256/x²。表面积 A = x² + 4xh = x² + 1024/x。求导：dA/dx = 2x – 1024/x² = 0 → x³ = 512 → x = 8 cm。则 h = 256/64 = 4 cm。二阶导数 d²A/dx² = 2 + 2048/x³，在 x=8 时大于 0，确认为极小值。
Always check that your answer lies within the feasible domain, and explicitly interpret the result in the context of the problem, as Edexcel often awards marks for conclusion statements.
务必检查所得答案在合理定义域内，并明确在问题情境中解释结果，因为Edexcel常对结论性陈述给予分数。
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