Mastering Differentiation for CCEA A-Level Maths | CCEA A-Level 数学：微分考点精讲

📚 Mastering Differentiation for CCEA A-Level Maths | CCEA A-Level 数学：微分考点精讲

Differentiation is one of the cornerstones of A-Level Mathematics, forming the basis for understanding rates of change, curve sketching, and optimisation. In the CCEA specification, a solid grasp of differentiation techniques and their applications is essential for success in both the AS and A2 units. This guide breaks down every key topic, from first principles to parametric equations, with clear explanations and examples that mirror the style of exam questions.

微分是 A-Level 数学的核心板块之一，是理解变化率、曲线描绘和优化问题的基础。在 CCEA 的考纲中，扎实掌握各种微分技巧及其应用是在 AS 和 A2 单元中取得高分的关键。本文逐点拆解从第一性原理到参数方程等所有重要考点，辅以贴近真题风格的清晰说明与示例。

1. Power Rule and Basic Derivatives | 幂函数及基本导数公式

The power rule is the first and most fundamental differentiation rule you will encounter. If f(x) = xⁿ, then f'(x) = n xⁿ⁻¹. This rule works for any real exponent n, including negative and fractional powers, provided we rewrite the expression accordingly. You must also memorise the derivatives of common functions such as sin x, cos x, eˣ, and ln x.

幂函数法则是你会遇到的第一条、也是最基础的微分法则。若 f(x) = xⁿ，则 f'(x) = n xⁿ⁻¹。该法则对任何实数指数 n 都适用，包括负指数和分数指数，只要先把表达式改写为幂函数形式即可。你还必须熟记一些基本函数的导数，例如 sin x、cos x、eˣ 和 ln x。

d/dx (xⁿ) = n xⁿ⁻¹

d/dx (c) = 0

d/dx (x⁵) = 5x⁴
d/dx (1/x) = d/dx (x⁻¹) = -x⁻² = -1/x²
d/dx (√x) = d/dx (x½) = ½ x⁻½ = 1/(2√x)

d/dx (x⁵) = 5x⁴
d/dx (1/x) = d/dx (x⁻¹) = -x⁻² = -1/x²
d/dx (√x) = d/dx (x½) = ½ x⁻½ = 1/(2√x)

When differentiating a sum of terms, differentiate each term separately. Constant multiples remain unchanged: d/dx [k·f(x)] = k·f'(x).

在求多项之和的导数时，逐项求导即可。常数倍保持不变：d/dx [k·f(x)] = k·f'(x)。

2. The Chain Rule | 链式法则

The chain rule is used to differentiate composite functions, i.e. functions of the form f(g(x)). In its simplest form: dy/dx = (dy/du) × (du/dx). For example, to differentiate (3x² + 2)⁵, set u = 3x² + 2, then differentiate u⁵ with respect to u and multiply by du/dx = 6x. Mastery of the chain rule is critical because it underpins implicit differentiation and related rates.

链式法则用于求复合函数的导数，即形如 f(g(x)) 的函数。其最基本的形式为：dy/dx = (dy/du) × (du/dx)。例如，求 (3x² + 2)⁵ 的导数时，设 u = 3x² + 2，先对 u⁵ 求导（关于 u）再乘以 du/dx = 6x。掌握链式法则极为关键，因为它是隐函数微分和相关变化率问题的基础。

d/dx [f(g(x))] = f'(g(x)) · g'(x)

For practical differentiation, you can think ‘differentiate the outside function, keeping the inside unchanged, then multiply by the derivative of the inside’.

在实际求导时，可以这样想：“先对外层函数求导，保持内层不变，再乘以内层函数的导数”。

y = sin(2x) → dy/dx = cos(2x) × 2 = 2cos(2x)
y = ln(5x + 1) → dy/dx = 1/(5x+1) × 5 = 5/(5x+1)
y = e^(4x³) → dy/dx = e^(4x³) × 12x² = 12x² e^(4x³)

y = sin(2x) → dy/dx = cos(2x) × 2 = 2cos(2x)
y = ln(5x + 1) → dy/dx = 1/(5x+1) × 5 = 5/(5x+1)
y = e^(4x³) → dy/dx = e^(4x³) × 12x² = 12x² e^(4x³)

3. The Product Rule | 乘法法则

When you need to differentiate the product of two functions, use the product rule: d/dx (uv) = u dv/dx + v du/dx. In words, it is ‘first times derivative of second plus second times derivative of first’. You must be able to identify u and v clearly and always remember to express the final answer in a simplified form.

当你需要对两个函数的乘积求导时，使用乘法法则：d/dx (uv) = u dv/dx + v du/dx。用口语记忆就是 “第一乘以第二的导数加上第二乘以第一的导数”。你必须能够清晰地识别 u 和 v，并记住最后要将答案化为最简形式。

d/dx (u(x) · v(x)) = u'(x)v(x) + u(x)v'(x)

Example: differentiate y = x² sin x.

示例：求 y = x² sin x 的导数。

Let u = x² and v = sin x. Then u’ = 2x, v’ = cos x.
dy/dx = (2x)(sin x) + (x²)(cos x) = 2x sin x + x² cos x.

设 u = x²，v = sin x。则 u’ = 2x，v’ = cos x。
dy/dx = (2x)(sin x) + (x²)(cos x) = 2x sin x + x² cos x。

For products of three functions, extend the idea: d/dx (uvw) = u’vw + uv’w + uvw’.

对于三个函数的乘积，可将法则推广：d/dx (uvw) = u’vw + uv’w + uvw’。

4. The Quotient Rule | 除法法则

The quotient rule handles fractions where both numerator and denominator are functions of x: d/dx (u/v) = (v du/dx − u dv/dx) / v². The order of terms is crucial: ‘bottom times derivative of top minus top times derivative of bottom, all over bottom squared’. This rule is particularly tested with rational functions and combinations of polynomials with trig or exponential functions.

除法法则处理分子和分母都是 x 的函数的分式：d/dx (u/v) = (v du/dx − u dv/dx) / v²。各项的顺序至关重要：“分母乘以分子的导数减去分子乘以分母的导数，整体再除以分母的平方”。这个法则在有理函数以及多项式与三角或指数函数相结合的表达式中经常考查。

d/dx [u(x)/v(x)] = [u'(x)v(x) − u(x)v'(x)] / [v(x)]²

Example: differentiate y = (x² + 1)/cos x.

示例：求 y = (x² + 1)/cos x 的导数。

u = x² + 1, v = cos x; u’ = 2x, v’ = −sin x.
dy/dx = [(2x)(cos x) − (x²+1)(−sin x)] / cos²x = (2x cos x + (x²+1) sin x) / cos²x.

u = x² + 1，v = cos x；u’ = 2x，v’ = −sin x。
dy/dx = [(2x)(cos x) − (x²+1)(−sin x)] / cos²x = (2x cos x + (x²+1) sin x) / cos²x。

5. Differentiating Exponential and Logarithmic Functions | 指数与对数函数的微分

Exponential and logarithmic functions have unique derivatives that you must know by heart. The derivative of eˣ is simply eˣ. As soon as the exponent becomes a function of x, the chain rule applies: d/dx (e^(g(x))) = g'(x) e^(g(x)). For natural logarithms, d/dx (ln x) = 1/x, and the chain rule gives d/dx (ln(g(x))) = g'(x)/g(x).

指数函数和对数函数拥有独特的导数公式，必须牢记。eˣ 的导数就是 eˣ 自身。一旦指数变成了 x 的函数，就需要用到链式法则：d/dx (e^(g(x))) = g'(x) e^(g(x))。对于自然对数，d/dx (ln x) = 1/x，而链式法则给出 d/dx (ln(g(x))) = g'(x)/g(x)。

d/dx (eˣ) = eˣ

d/dx (ln x) = 1/x, x > 0

Function	Derivative
eˣ	eˣ
e^(ax+b)	a e^(ax+b)
ln x	1/x
ln(ax+b)	a/(ax+b)

CCEA often asks you to differentiate expressions like xeˣ or ln(x)/x, requiring a combination of product/quotient rules with these functions. Always simplify answers by factorising where possible.

CCEA 经常会要求对 xeˣ 或 ln(x)/x 这样的表达式求导，这就需要将乘法/除法法则与这些函数结合运用。尽量通过提取公因式来化简答案。

6. Differentiating Trigonometric Functions | 三角函数的微分

The derivatives of the six basic trigonometric functions, especially sin x, cos x and tan x, are essential. In CCEA exams, questions usually involve compound angles, so you will constantly use the chain rule: d/dx (sin(kx + b)) = k cos(kx + b). Also be comfortable with differentiating sec x = 1/cos x, cosec x = 1/sin x and cot x using the quotient rule.

六个基本三角函数的导数，尤其是 sin x、cos x 和 tan x，都是必备知识。在 CCEA 的试题中，通常会涉及复合角，因此你会频繁使用链式法则：d/dx (sin(kx + b)) = k cos(kx + b)。同时，要能熟练地运用除法法则对 sec x = 1/cos x、cosec x = 1/sin x 以及 cot x 进行求导。

d/dx (sin x) = cos x

d/dx (cos x) = −sin x

d/dx (tan x) = sec² x

Key extensions:
d/dx (sec x) = sec x tan x
d/dx (cosec x) = −cosec x cot x
d/dx (cot x) = −cosec² x

重要延伸：
d/dx (sec x) = sec x tan x
d/dx (cosec x) = −cosec x cot x
d/dx (cot x) = −cosec² x

Remember that CCEA often embeds these inside more complex functions, e.g. y = cos(3x²) requires chain rule twice or a double chain rule approach.

请记住，CCEA 经常将这些函数嵌入更复杂的表达式中，例如 y = cos(3x²) 就需要两次链式法则，或者采用双重链式法则处理。

7. Implicit Differentiation | 隐函数微分

When a relationship between x and y is given implicitly rather than as y = f(x), use implicit differentiation. The key idea is to differentiate every term with respect to x, treating y as a differentiable function of x. Each time you differentiate a term in y, multiply by dy/dx due to the chain rule. This technique is frequently tested and leads to expressions for dy/dx in terms of both x and y.

当 x 与 y 之间的关系是以隐函数形式给出、而非 y = f(x) 的显式形式时，使用隐函数微分法。核心思想是：逐项对 x 求导，将 y 视为 x 的可微函数。每次对含 y 的项求导时，都要根据链式法则乘以 dy/dx。这个技巧经常考查，并且会导出含有 x 和 y 的 dy/dx 表达式。

d/dx (yⁿ) = n yⁿ⁻¹ · dy/dx

Example: Find dy/dx if x³ + y³ = 6xy.

示例：若 x³ + y³ = 6xy，求 dy/dx。

Differentiating: 3x² + 3y² dy/dx = 6y + 6x dy/dx. Collect dy/dx terms: 3y² dy/dx − 6x dy/dx = 6y − 3x². Factor: dy/dx (3y² − 6x) = 6y − 3x². Hence dy/dx = (6y − 3x²)/(3y² − 6x) = (2y − x²)/(y² − 2x).

两边对 x 求导：3x² + 3y² dy/dx = 6y + 6x dy/dx。整理含 dy/dx 的项：3y² dy/dx − 6x dy/dx = 6y − 3x²。提取公因式：dy/dx (3y² − 6x) = 6y − 3x²。因此，dy/dx = (6y − 3x²)/(3y² − 6x) = (2y − x²)/(y² − 2x)。

Be methodical: after differentiating, isolate dy/dx terms on one side, factor, and divide.

要有条不紊：求导后，将含 dy/dx 的项移到一边，提取公因式，再相除。

8. Parametric Differentiation | 参数方程微分

Parametric equations express x and y in terms of a third variable, usually t. The gradient dy/dx is found by dy/dx = (dy/dt) ÷ (dx/dt). For second derivatives, use d²y/dx² = d/dx (dy/dx) = [d/dt (dy/dx)] ÷ (dx/dt). This topic often appears alongside concavity or finding the equation of a tangent at a specific parameter value.

参数方程用第三个变量（通常是 t）分别表示 x 和 y。梯度 dy/dx 由 dy/dx = (dy/dt) ÷ (dx/dt) 求得。对于二阶导数，使用 d²y/dx² = d/dx (dy/dx) = [d/dt (dy/dx)] ÷ (dx/dt)。这一知识点经常与函数凹凸性或求给定参数值处的切线方程一同出现。

dy/dx = (dy/dt) / (dx/dt)

Example: x = t² + 2t, y = t³ − 3t. Then dx/dt = 2t + 2, dy/dt = 3t² − 3.
dy/dx = (3t² − 3)/(2t + 2) = 3(t² − 1)/[2(t + 1)] = 3(t − 1)/2 (for t ≠ −1).

示例：x = t² + 2t，y = t³ − 3t。那么 dx/dt = 2t + 2，dy/dt = 3t² − 3。
dy/dx = (3t² − 3)/(2t + 2) = 3(t² − 1)/[2(t + 1)] = 3(t − 1)/2（当 t ≠ −1 时）。

To find d²y/dx², first write dy/dx as a function of t, differentiate it with respect to t, and then divide by dx/dt. Always simplify before substituting numerical values.

要求 d²y/dx² 时，先将 dy/dx 写成 t 的函数，对其关于 t 求导，然后再除以 dx/dt。务必先化简再代入具体数值。

9. Second Derivatives and Their Applications | 二阶导数及其应用

The second derivative, f”(x) or d²y/dx², provides information about the curvature (concavity) of a function. It is used to classify stationary points: if f”(a) > 0, the point is a local minimum; if f”(a) < 0, it is a local maximum. If f''(a) = 0, the test is inconclusive and you should use the first derivative test or consider a point of inflection.

二阶导数，即 f”(x) 或 d²y/dx²，提供了有关函数曲率（凹凸性）的信息。它可用于判定驻点的类型：若 f”(a) > 0，该点是局部极小点；若 f”(a) < 0，则为局部极大点。若 f''(a) = 0，该判别法失效，此时应使用一阶导数判别法，或考虑拐点的可能性。

Local minimum: f'(c) = 0 and f”(c) > 0

Local maximum: f'(c) = 0 and f”(c) < 0

Additionally, the sign change of f”(x) indicates points of inflection, where the concavity changes. In parametric contexts, finding d²y/dx² accurately is a common source of errors, so double-check your algebra.

此外，f”(x) 符号的改变标志着拐点——即凹凸性发生变化的点。在参数方程的背景下，准确求得 d²y/dx² 是最容易出错的地方，因此务必仔细检查代数运算。

10. Stationary Points and Points of Inflection | 驻点与拐点

A stationary point occurs where f'(x) = 0. To determine whether it is a maximum, minimum, or point of inflection, either examine the sign change of f'(x) (first derivative test) or use the second derivative f”(x). A point of inflection is where the concavity changes — mathematically, f”(x) changes sign. Note that f”(x) = 0 does not guarantee a point of inflection; you must verify the sign change.

驻点出现在 f'(x) = 0 处。要判断它是极大点、极小点还是拐点，可以通过观察 f'(x) 的符号变化（一阶导数判别法），或者使用二阶导数 f”(x) 来协助。拐点是指函数凹凸性发生变化的位置——从数学上看，即 f”(x) 改变符号。注意，f”(x) = 0 并不保证一定是拐点，你必须验证符号是否真的发生改变。

Example: f(x) = x⁴ − 2x³. f'(x) = 4x³ − 6x² = 2x²(2x − 3). Stationary points at x = 0 and x = 1.5. f”(x) = 12x² − 12x. At x = 1.5, f”(1.5) = 9 > 0 → minimum. At x = 0, f”(0) = 0. Check sign of f”(x) around 0: it is negative on both sides → not a point of inflection; it is a horizontal point of inflection only if f'(x) sign changes, but here f'(x) does not change sign, so it’s a point of inflection with stationary property? Actually f'(x) around 0 is positive both sides (since 2x² ≥ 0), so neither maximum nor minimum, it is a stationary point of inflection because f”(x) changes sign? Let’s test: f”(x) = 12x(x − 1). For x < 0, f''(x) positive; for 0 < x < 1, negative; so f''(x) changes sign at x=0 → point of inflection. Since f'(0)=0, it's a stationary point of inflection.

示例：f(x) = x⁴ − 2x³。f'(x) = 4x³ − 6x² = 2x²(2x − 3)。驻点为 x = 0 和 x = 1.5。f”(x) = 12x² − 12x。在 x = 1.5 处，f”(1.5) = 9 > 0 → 极小点。在 x = 0 处，f”(0) = 0。检查 f”(x) 在 0 附近的符号：两侧由正变负，因此是拐点，又因为 f'(0)=0，所以是驻点拐点。这表明单靠二阶导数为零不足以得出结论，必须检查符号变化。

11. Tangents and Normals | 切线与法线

The derivative at a point gives the gradient of the tangent to the curve. Therefore, the equation of the tangent at (x₁, y₁) can be written as y − y₁ = m (x − x₁), where m = dy/dx at that point. The normal is perpendicular to the tangent, so its gradient is −1/m. CCEA exam questions frequently ask for both equations, often at points derived from stationary points or intersections with axes.

函数在某点的导数值等于曲线在该点切线的斜率。因此，点 (x₁, y₁) 处的切线方程可写作 y − y₁ = m (x − x₁)，其中 m 为该点的 dy/dx 值。法线与切线垂直，故其斜率为 −1/m。CCEA 考题经常要求同时求出切线和法线方程，这些点通常是从驻点或与坐标轴的交点推导出来的。

Tangent: y − y₁ = m (x − x₁)

Normal: y − y₁ = (−1/m) (x − x₁)

When tackling these problems, first find the point coordinates and the derivative. If the point is given directly, evaluate dy/dx; if only x is given, find y from the original equation. Be careful with trigonometric and implicit functions where the gradient expression may involve both x and y.

处理此类问题时，先求出点的坐标和导数值。如果直接给了点，就求 dy/dx；若只给出了 x 值，则通过原方程求出 y。对于含有三角或隐函数的情形要格外小心，因为梯度表达式可能同时含有 x 和 y。

12. Optimisation Problems | 优化问题

Optimisation is a practical application of calculus where you find maximum or minimum values of a quantity under given constraints. Typically, you express the quantity to be optimised as a function of one variable using a constraint equation, differentiate, find stationary points, and verify their nature with the second derivative test. Common contexts include maximising area, volume, or profit, and minimising surface area, distance, or cost.

优化问题是微积分的实际应用，要求在给定约束条件下求某个量的最大值或最小值。通常的做法是：利用约束方程把待优化的量表示为单一变量的函数，求导以找到驻点，再用二阶导数判别法验证其类型。常见情境包括求面积、体积或利润的最大化，以及求表面积、距离或成本的最小化。

Steps for optimisation:
1. Identify the quantity Q to be maximised/minimised.
2. Express Q in terms of one variable (say x) using given constraints.
3. Find dQ/dx and set it to zero to locate stationary points.
4. Use d²Q/dx² or a sign test to confirm the nature.
5. Answer the question with the correct value and units.

优化问题的步骤：
1. 确定要最大化/最小化的量 Q。
2. 借助已知约束条件，将 Q 表示为单个变量（比如 x）的函数。
3. 求出 dQ/dx 并令其为零，以找到驻点。
4. 使用 d²Q/dx² 或符号检验，确认极值类型。
5. 用正确的值和单位回答实际问题。

Always re-read the question to ensure you’ve addressed the physical context (e.g. rejecting negative lengths or checking endpoints for closed intervals).

务必重新审题，确保你考虑了实际情境（例如排除负的长度，或在闭区间上检查端点值）。

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Mastering Differentiation for CCEA A-Level Maths | CCEA A-Level 数学：微分考点精讲