📚 Mastering Differentiation for WJEC A-Level Mathematics | WJEC A-Level 数学微分考点精讲
Differentiation is the cornerstone of calculus, allowing us to measure instantaneous rates of change, analyse motion, and solve optimisation problems. For WJEC A-Level Mathematics, a deep understanding of differentiation techniques – from first principles to parametric and implicit methods – is essential. This guide covers every key concept, providing clear explanations, examples, and exam-focused insights to help you master the topic with confidence.
微分是微积分的基石，使我们能够衡量瞬时变化率、分析运动并解决优化问题。对 WJEC A-Level 数学而言，深入理解微分技巧——从第一原理到参数方程与隐函数求导——至关重要。本指南涵盖每一个关键概念，提供清晰的解释、例题和考试重点，帮助你自信掌握这一主题。
1. The Definition of the Derivative and First Principles | 导数的定义与第一原理
The derivative of a function f(x) at a point x is defined as the limit of the difference quotient: f'(x) = lim_{h→0} [f(x+h) – f(x)] / h. This limit, if it exists, gives the gradient of the tangent to the curve at that point. In WJEC exams, you may be asked to differentiate simple functions like x² or 1/x directly from this definition.
函数 f(x) 在点 x 处的导数定义为差商的极限：f'(x) = lim_{h→0} [f(x+h) – f(x)] / h。如果该极限存在，则给出曲线在该点切线的斜率。在 WJEC 考试中，可能要求你直接从这一定义出发对 x² 或 1/x 等简单函数求导。
To apply first principles for f(x) = x², expand f(x+h) = (x+h)² = x² + 2xh + h², then subtract f(x), divide by h, and let h → 0. This yields f'(x) = 2x. Similarly, for f(x) = √x, algebraic manipulation with rationalisation is required. Mastering this technique deepens your understanding of what differentiation truly represents.
对于 f(x) = x² 应用第一原理，先展开 f(x+h) = (x+h)² = x² + 2xh + h²，然后减去 f(x)，除以 h，令 h → 0，得到 f'(x) = 2x。类似地，对 f(x) = √x 求导时需要进行有理化操作。掌握这一技巧能加深你对微分真正含义的理解。
2. Basic Differentiation Rules | 基本求导法则
The power rule states that if f(x) = xⁿ, then f'(x) = n xⁿ⁻¹, where n is any real number. This rule, combined with the sum, difference, and constant multiple rules, allows rapid differentiation of polynomials. For WJEC, you must be able to differentiate expressions like 4x⁵ – 3x² + 2x – 7 instantly.
幂函数法则指出，若 f(x) = xⁿ，则 f'(x) = n xⁿ⁻¹，其中 n 是任意实数。该法则与和差法则、常数倍法则结合使用，可快速对多项式求导。在 WJEC 考试中，你必须能够迅速求出诸如 4x⁵ – 3x² + 2x – 7 这类表达式的导数。
The constant rule (d/dx[c] = 0), sum rule (d/dx[u ± v] = u’ ± v’), and constant multiple rule (d/dx[c u] = c u’) form the backbone of differentiation. Recognising when to rewrite a term, for example writing 1/x² as x⁻², is crucial for applying the power rule correctly.
常数法则 (d/dx[c] = 0)、和差法则 (d/dx[u ± v] = u’ ± v’) 以及常数倍法则 (d/dx[c u] = c u’) 构成了微分的基础。正确识别何时需将项改写——例如将 1/x² 写作 x⁻²——对正确应用幂法则至关重要。
3. The Chain Rule | 链式法则
The chain rule is used to differentiate composite functions, i.e., functions of the form y = f(g(x)). It states that dy/dx = (dy/du) × (du/dx), where u = g(x). This is one of the most heavily tested skills in WJEC, appearing in everything from trigonometric differentiation to rates of change.
链式法则用于对复合函数求导，即形如 y = f(g(x)) 的函数。该法则指出 dy/dx = (dy/du) × (du/dx)，其中 u = g(x)。这是 WJEC 中考查最频繁的技能之一，从三角函数求导到相关变化率问题都会出现。
For example, to differentiate y = (2x³ – 5)⁴, let u = 2x³ – 5, so y = u⁴. Then dy/du = 4u³ and du/dx = 6x², giving dy/dx = 4(2x³ – 5)³ × 6x² = 24x²(2x³ – 5)³. Always remember to multiply by the derivative of the inner function.
例如，对 y = (2x³ – 5)⁴ 求导，令 u = 2x³ – 5，则 y = u⁴。于是 dy/du = 4u³， du/dx = 6x²，因此 dy/dx = 4(2x³ – 5)³ × 6x² = 24x²(2x³ – 5)³。务必记住乘以内层函数的导数。
4. Product and Quotient Rules | 乘积法则与商法则
The product rule is used when differentiating the product of two functions: d/dx (u v) = u (dv/dx) + v (du/dx). The quotient rule handles division: d/dx (u/v) = (v du/dx – u dv/dx) / v². WJEC examiners expect you to apply these correctly, often in combination with the chain rule.
乘积法则用于对两个函数的乘积求导：d/dx (u v) = u (dv/dx) + v (du/dx)。商法则处理除法：d/dx (u/v) = (v du/dx – u dv/dx) / v²。WJEC 考官期望你能正确应用这些法则，并常常与链式法则组合使用。
For instance, to differentiate y = x² eˣ, let u = x², v = eˣ. Then u’ = 2x, v’ = eˣ, giving dy/dx = x² eˣ + 2x eˣ = x eˣ (x + 2). For y = (sin x) / x, using the quotient rule with u = sin x, v = x, we obtain (x cos x – sin x) / x².
例如，对 y = x² eˣ 求导，令 u = x², v = eˣ，则 u’ = 2x, v’ = eˣ，于是 dy/dx = x² eˣ + 2x eˣ = x eˣ (x + 2)。对于 y = (sin x) / x，应用商法则，u = sin x, v = x，得到 (x cos x – sin x) / x²。
5. Derivatives of Exponential and Logarithmic Functions | 指数函数与对数函数的导数
The derivative of the natural exponential function is remarkably straightforward: d/dx (eˣ) = eˣ. For a general base a, d/dx (aˣ) = aˣ ln a. The natural logarithm gives d/dx (ln x) = 1/x, and for logarithms with other bases, the change-of-base formula is used: d/dx (logₐ x) = 1/(x ln a).
自然指数函数的导数极其简洁：d/dx (eˣ) = eˣ。对于一般底数 a，d/dx (aˣ) = aˣ ln a。自然对数的导数为 d/dx (ln x) = 1/x，而对其他底数的对数，则使用换底公式：d/dx (logₐ x) = 1/(x ln a)。
When the exponent or argument is a function of x, the chain rule must be applied. For example, d/dx (e^(2x+1)) = e^(2x+1) × 2 = 2 e^(2x+1); and d/dx (ln(3x²)) = (1/(3x²)) × 6x = 2/x. These combinations appear frequently in WJEC problems involving growth, decay, and integration by reverse differentiation.
当指数或真数是 x 的函数时，必须应用链式法则。例如，d/dx (e^(2x+1)) = e^(2x+1) × 2 = 2 e^(2x+1)；而 d/dx (ln(3x²)) = (1/(3x²)) × 6x = 2/x。这些组合在涉及增长、衰减及通过逆向求导进行积分的问题中频繁出现。
6. Derivatives of Trigonometric Functions | 三角函数的导数
WJEC candidates must memorise the standard derivatives: d/dx (sin x) = cos x, d/dx (cos x) = –sin x, and d/dx (tan x) = sec² x. The reciprocal functions sec, cosec, and cot are also examined; their derivatives are d/dx (sec x) = sec x tan x, d/dx (cosec x) = –cosec x cot x, and d/dx (cot x) = –cosec² x.
WJEC 考生必须熟记标准三角函数的导数：d/dx (sin x) = cos x，d/dx (cos x) = –sin x，d/dx (tan x) = sec² x。倒数函数 sec、cosec 和 cot 也在考试范围内，其导数分别为 d/dx (sec x) = sec x tan x，d/dx (cosec x) = –cosec x cot x，d/dx (cot x) = –cosec² x。
When the argument is a linear function, such as sin(ax+b), the chain rule gives a cos(ax+b). For more complex compositions, like cos² 3x, rewrite as (cos 3x)² and differentiate carefully: d/dx cos² 3x = 2 cos 3x × (–sin 3x) × 3 = –6 cos 3x sin 3x = –3 sin 6x (using double-angle identity).
当自变量为线性函数时，如 sin(ax+b)，链式法则给出 a cos(ax+b)。对于更复杂的复合函数，如 cos² 3x，先改写为 (cos 3x)² 再仔细求导：d/dx cos² 3x = 2 cos 3x × (–sin 3x) × 3 = –6 cos 3x sin 3x = –3 sin 6x（使用倍角公式）。
7. Implicit Differentiation | 隐函数求导
When an equation defines y implicitly as a function of x, we differentiate both sides with respect to x, treating y as a function of x and applying the chain rule to terms involving y. Each time we differentiate a y-term, we multiply by dy/dx. WJEC often tests this with circles, ellipses, and mixed polynomial-trigonometric relations.
当一个方程将 y 隐含地定义为 x 的函数时，我们将方程两边同时对 x 求导，把 y 视为 x 的函数，并对含 y 的项应用链式法则。每次对含 y 的项求导时，都要乘以 dy/dx。WJEC 常以圆、椭圆以及多项式-三角混合关系来考查这一知识点。
For example, for the circle x² + y² = 25, differentiating gives 2x + 2y (dy/dx) = 0, hence dy/dx = –x/y. For x³ + y³ = 6xy, we obtain 3x² + 3y² dy/dx = 6(y + x dy/dx), which requires algebraic rearrangement to isolate dy/dx. Implicit differentiation is indispensable for finding tangents to curves not given in explicit form.
例如，对于圆 x² + y² = 25，求导得 2x + 2y (dy/dx) = 0，故 dy/dx = –x/y。对于 x³ + y³ = 6xy，求导得 3x² + 3y² dy/dx = 6(y + x dy/dx)，需要代数整理以解出 dy/dx。隐函数求导对于寻找非显式形式曲线的切线是不可或缺的。
8. Parametric Differentiation | 参数方程求导
When x and y are given as functions of a parameter t, the derivative dy/dx is found by dividing dy/dt by dx/dt: dy/dx = (dy/dt) / (dx/dt). This is a key concept in WJEC Mechanics and Pure Mathematics for analysing curves and motion along a path.
当 x 和 y 均以参数 t 的函数形式给出时，导数 dy/dx 可通过 dy/dt 除以 dx/dt 求得：dy/dx = (dy/dt) / (dx/dt)。这是 WJEC 力学和纯数学中分析曲线与沿路径运动的关键概念。
If x = t² + 1 and y = t³ – 3t, then dx/dt = 2t and dy/dt = 3t² – 3, so dy/dx = (3t² – 3) / (2t). At t = 2, the gradient is (3×4 – 3)/(4) = 9/4. The second derivative d²y/dx² requires differentiating dy/dx with respect to t and then dividing by dx/dt: d²y/dx² = d(dy/dx)/dt ÷ dx/dt.
若 x = t² + 1, y = t³ – 3t，则 dx/dt = 2t，dy/dt = 3t² – 3，于是 dy/dx = (3t² – 3) / (2t)。在 t = 2 处，斜率为 (3×4 – 3)/4 = 9/4。二阶导数 d²y/dx² 需要对 dy/dx 关于 t 求导，再除以 dx/dt：d²y/dx² = d(dy/dx)/dt ÷ dx/dt。
9. Second Derivatives and Higher Order | 二阶导数与高阶导数
The second derivative, denoted f”(x) or d²y/dx², measures the rate of change of the gradient and reveals concavity. In WJEC, second derivatives are used to classify stationary points (minima, maxima, or points of inflection) and to verify the nature of turning points in optimisation.
二阶导数，记作 f”(x) 或 d²y/dx²，用于衡量斜率的变化率并揭示凹凸性。在 WJEC 考试中，二阶导数用于对驻点进行分类（极小值点、极大值点或拐点），并验证优化问题中极值点的性质。
For f(x) = 2x⁴ – 8x³ + 5, f'(x) = 8x³ – 24x² and f”(x) = 24x² – 48x. Evaluating f”(x) at stationary points helps determine their nature: if f”(a) > 0, the point is a local minimum; if f”(a) < 0, a local maximum; if f''(a) = 0, further investigation is required (e.g., using the first derivative test).
对于 f(x) = 2x⁴ – 8x³ + 5，f'(x) = 8x³ – 24x²，f”(x) = 24x² – 48x。在驻点处计算 f”(x) 有助于确定其性质：若 f”(a) > 0，该点为局部极小值；若 f”(a) < 0，则为局部极大值；若 f''(a) = 0，则需要进一步检验（例如使用一阶导数判别法）。
10. Tangents and Normals | 切线与法线
The equation of a tangent to the curve y = f(x) at point (x₁, y₁) is y – y₁ = m (x – x₁), where m = f'(x₁). The normal is perpendicular to the tangent, so its gradient mₙ = –1/m, provided m ≠ 0. WJEC questions often ask for both equations and may involve parallel or perpendicular lines.
曲线 y = f(x) 在点 (x₁, y₁) 处的切线方程为 y – y₁ = m (x – x₁)，其中 m = f'(x₁)。法线与切线垂直，因此其斜率 mₙ = –1/m（若 m ≠ 0）。WJEC 的题目经常要求同时求出两者，并可能涉及平行或垂直的关系。
For example, find the tangent and normal to y = ln x at x = 1. Since y(1) = 0, f'(x) = 1/x, m = 1, so tangent: y = x – 1; normal gradient = –1, normal: y = –(x – 1). Understanding the geometric interpretation reinforces the connection between algebraic differentiation and graphical behaviour.
例如，求曲线 y = ln x 在 x = 1 处的切线与法线。由于 y(1) = 0，f'(x) = 1/x，m = 1，因此切线为 y = x – 1；法线斜率为 –1，法线为 y = –(x – 1)。理解几何解释能强化代数微分与图形特征之间的联系。
11. Stationary Points and Curve Sketching | 驻点与函数作图
Stationary points occur where f'(x) = 0. WJEC expects you to find these and determine their nature using the second derivative test or a sign-change table for f'(x). This information, combined with intercepts and asymptotic behaviour, allows a detailed sketch of the function.
驻点出现在 f'(x) = 0 的位置。WJEC 要求考生找出这些点，并通过二阶导数检验或 f'(x) 的符号变化表来确定其性质。结合截距和渐近行为等信息，便可绘制出函数的详细草图。
Consider f(x) = x³ – 3x + 2. f'(x) = 3x² – 3 = 0 gives x = ±1. f”(x) = 6x; at x = 1, f”(1) = 6 > 0 → local minimum; at x = –1, f”(–1) = –6 < 0 → local maximum. The y-intercept is 2, and by finding sign of f'(x) in intervals, you can confirm the shape. Curve sketching questions are common in WJEC Pure Mathematics papers.
考虑函数 f(x) = x³ – 3x + 2。f'(x) = 3x² – 3 = 0 得 x = ±1。f”(x) = 6x；在 x = 1 处，f”(1) = 6 > 0 → 局部极小值；在 x = –1 处，f”(–1) = –6 < 0 → 局部极大值。y 轴截距为 2，通过考察各区间的 f'(x) 符号，可确认曲线形状。在 WJEC 纯数试卷中，函数作图题十分常见。
12. Optimisation Problems | 优化问题
Differentiation is a powerful tool for solving real-world maximum and minimum problems. Typical WJEC problems involve minimising surface area for a given volume, maximising enclosed area, or finding optimal dimensions in geometric contexts. The general approach: express the quantity to be optimised as a function of one variable, find its derivative, set it to zero, and verify the nature of stationary points.
微分是解决实际最大值与最小值问题的有力工具。WJEC 的典型问题包括给定体积下最小化表面积、最大化围成面积，或求几何情境下的最优尺寸。一般方法是：用单一变量表示待优化的量，求其导数，令导数为零，并验证驻点的性质。
Example: A cylindrical can with volume 1000 cm³ is to be made from minimal sheet metal. Let radius = r, height = h. V = πr²h = 1000 ⇒ h = 1000/(πr²). Surface area A = 2πr² + 2πrh = 2πr² + 2000/r. Then dA/dr = 4πr – 2000/r² = 0 ⇒ r³ = 500/π ⇒ r = (500/π)^(1/3). Second derivative test confirms minimum. Such problems test your ability to model, differentiate, and conclude correctly.
例题：一个容积为 1000 cm³ 的圆柱形罐头要用最少的金属材料制成。设半径为 r，高为 h。V = πr²h = 1000 ⇒ h = 1000/(πr²)。表面积 A = 2πr² + 2πrh = 2πr² + 2000/r。求导得 dA/dr = 4πr – 2000/r² = 0 ⇒ r³ = 500/π ⇒ r = (500/π)^(1/3)。二阶导数检验确认该点为最小值。此类问题考查建模、求导和正确总结的能力。
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