Differentiation Techniques for Edexcel A-Level Maths | Edexcel A-Level 数学微分技巧

📚 Differentiation Techniques for Edexcel A-Level Maths | Edexcel A-Level 数学微分技巧

This revision guide covers the main differentiation techniques required for Edexcel A-Level Mathematics, including differentiation from first principles, standard derivatives, the product, quotient and chain rules, implicit and parametric differentiation, second derivatives, stationary points and connected rates of change. Each section pairs the key theory with worked examples and exam tips so that you can move from understanding to exam-ready fluency.

本复习指南涵盖 Edexcel A-Level 数学要求的主要微分技巧，包括第一性原理微分、标准导数、乘法法则、除法法则、链式法则、隐函数微分、参数方程微分、二阶导数、驻点以及相关变化率。每一节都把核心理论与例题、考试提示配对呈现，帮助你从"理解"过渡到"考场熟练"。

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1. Differentiation from First Principles | 第一性原理微分

t. Formally it is defined as a limit: as the change in x, written h, tends to zero, the gradient of the chord joining x and x + h tends to the gradient of the tangent.

函数 f(x) 在某一点的导数，衡量曲线 y = f(x) 在该点处切线的斜率。其正式定义是一个极限：当 x 的改变量 h 趋于零时，连接 x 与 x + h 两点的割线斜率趋于切线的斜率。

f′(x) = lim_h→0 ( f(x + h) − f(x) ) / h

tracting f(x) and dividing by h gives 2x + h, and as h → 0 the limit is 2x. So f′(x) = 2x.

例如，若 f(x) = x²，则 f(x + h) = (x + h)² = x² + 2xh + h²。减去 f(x) 再除以 h 得 2x + h，令 h → 0，极限为 2x，所以 f′(x) = 2x。

/ (x(x + h)), dividing by h gives −1 / (x(x + h)), and letting h → 0 gives f′(x) = −1/x².

第二个典型例子是 f(x) = 1/x。写出 f(x + h) − f(x) = 1/(x + h) − 1/x = (x − (x + h)) / (x(x + h)) = −h / (x(x + h))，除以 h 得 −1 / (x(x + h))，令 h → 0 得 f′(x) = −1/x²。

Exam tip: in Edexcel papers you may be asked to prove the derivative of x², a simple polynomial or sin x from first principles. Write every step, keep the algebraic fraction clear, and always include the limit statement before you conclude.

考试提示：在 Edexcel 试卷中，你可能被要求从第一性原理证明 x²、简单多项式或 sin x 的导数。请写清每一步，保持代数分式清晰，并在得出结论前务必写出极限语句。

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2. Standard Derivatives and Basic Rules | 标准导数与基本法则

Knowing the standard derivatives by heart saves time and reduces errors. Two structural rules make differentiation of combinations straightforward: the sum rule, where each term is differentiated separately, and the constant multiple rule, where a constant factor can be taken outside the derivative.

熟记标准导数可以节省时间并减少错误。两条结构性法则让组合函数的求导变得简单：加法法则，即逐项求导；常数倍法则，即常数因子可以提到导数符号外面。

) = f′(x) ± g′(x), and d/dx (k f(x)) = k f′(x). Also, the derivative of any constant is 0.

用符号表示：d/dx (f(x) ± g(x)) = f′(x) ± g′(x)，并且 d/dx (k f(x)) = k f′(x)。此外，任何常数的导数都是 0。

f(x)	f′(x)
xⁿ	nxⁿ⁻¹
constant k	0
eˣ	eˣ
aˣ	aˣ ln a
ln x	1/x
sin x	cos x
cos x	−sin x
tan x	sec² x
sec x	sec x tan x
cosec x	−cosec x cot x
cot x	−cosec² x

Worked example: differentiate y = 4x³ − 2 sin x + eˣ. Differentiating term by term gives dy/dx = 12x² − 2 cos x + eˣ ers, including fractional and negative indices, follow the same power rule: for y = 5x⁻² + √x = 5x⁻² + x^1/2, dy/dx = −10x⁻³ + ½x^−1/2.

例题：求 y = 4x³ − 2 sin x + eˣ 的导数。逐项求导得 dy/dx = 12x² − 2 cos x + eˣ。正指数、负指数以及分数指数都遵循同一幂法则：对 y = 5x⁻² + √x = 5x⁻² + x^1/2，有 dy/dx = −10x⁻³ + ½x^−1/2。

Exam tip: rewrite roots and reciprocals as powers before differentiating. A term such as 3/x³ should be written as 3x⁻³ so that the power rule applies directly, and final answers are usually expected in simplified form with positive indices.

考试提示：求导前先把根式和倒数改写为幂的形式。像 3/x³ 这样的项应先写成 3x⁻³，以便直接使用幂法则；最终答案通常要求化简，并尽量使用正指数。

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3. The Product Rule | 乘法法则

re both functions of x. The product rule states that the derivative is u′v + uv′: differentiate the first factor, multiply by the second, then add the first factor multiplied by the derivative of the second.

当一个函数是两个关于 x 的函数之积时，设 y = uv，其中 u 与 v 都是 x 的函数。乘法法则指出，导数为 u′v + uv′：先对第一个因子求导并乘以第二个因子，再加上第一个因子乘以第二个因子的导数。

(uv)′ = u′v + uv′

Example 1: y = x² eˣ. Take u = x², so u′ = 2x, and v = eˣ, so v′ = eˣ. Then dy/dx = 2x eˣ + x² eˣ = x eˣ (x + 2).

例 1：y = x² eˣ。取 u = x²，则 u′ = 2x；取 v = eˣ，则 v′ = eˣ。于是 dy/dx = 2x eˣ + x² eˣ = x eˣ (x + 2)。

= ln x with v′ = 1/x. So dy/dx = 3x² ln x + x³ × (1/x) = 3x² ln x + x² = x² (3 ln x + 1).

例 2：y = x³ ln x。此时 u = x³，u′ = 3x²；v = ln x，v′ = 1/x。所以 dy/dx = 3x² ln x + x³ × (1/x) = 3x² ln x + x² = x² (3 ln x + 1)。

Exam tip: the most common errors are forgetting the second term and mixing up which derivative belongs to which factor. Write u, u′, v and v′ out explicitly before substituting into the formula, and factorise your final answer whenever the question asks for it.

考试提示：最常见的错误是漏掉第二项，以及把两个因子的导数对应错。请先明确写出 u、u′、v、v′，再代入公式；若题目要求，最后一定要因式分解。

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4. The Quotient Rule | 除法法则

ro, the derivative is (u′v − uv′) / v². Note the order in the numerator: it is the derivative of the top times the bottom, minus the top times the derivative of the bottom.

对于商 y = u/v（v 不为零），其导数为 (u′v − uv′) / v²。注意分子的顺序：上面的导数乘以下面，减去上面乘以下面的导数。

dy/dx = ( u′v − uv′ ) / v²

(3x + 1)(2x) = 3x² + 6 − 6x² − 2x = −3x² − 2x + 6, so dy/dx = (−3x² − 2x + 6) / (x² + 2)².

例 1：y = (3x + 1) / (x² + 2)。取 u = 3x + 1，u′ = 3；v = x² + 2，v′ = 2x。分子为 3(x² + 2) − (3x + 1)(2x) = 3x² + 6 − 6x² − 2x = −3x² − 2x + 6，所以 dy/dx = (−3x² − 2x + 6) / (x² + 2)²。

= (1 − ln x) / x². This result is used in many optimisation and curve-sketching questions.

例 2：y = ln x / x。此时 u = ln x，u′ = 1/x；v = x，v′ = 1，所以 dy/dx = ((1/x)(x) − (ln x)(1)) / x² = (1 − ln x) / x²。这一结果常用于最优化与曲线描画类问题。

ith v = x⁻⁴. The quotient rule is still acceptable, but it costs more time and creates more chances to slip.

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考试提示：如果分母是单项式，例如 x⁴，通常把分式拆开用幂法则更快，或使用乘法法则取 v = x⁻⁴。使用除法法则也可以，但更耗时，也更容易出错。

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5. The Chain Rule | 链式法则

The chain rule is used for composite functions, where one function is inside another. If y is a function of u and u is a function of x, then the rate of change of y with respect to x is the product of the two linked rates.

链式法则用于复合函数，即一个函数嵌套在另一个函数内部。若 y 是 u 的函数，而 u 是 x 的函数，则 y 关于 x 的变化率等于两个相关联的变化率之积。

dy/dx = (dy/du) × (du/dx)

y = u⁷, dy/du = 7u⁶ = 7(2x + 5)⁶ and du/dx = 2. Hence dy/dx = 14(2x + 5)⁶. In the quicker "outside-inside" form: differentiate the bracket, then multiply by the derivative of what is inside.

例 1：y = (2x + 5)⁷。设 u = 2x + 5，则 y = u⁷，dy/du = 7u⁶ = 7(2x + 5)⁶，du/dx = 2。因此 dy/dx = 14(2x + 5)⁶。也可以使用更快的"外—内"形式：先对外层求导，再乘以括号内部函数的导数。

ine and the inner function is 3x², so dy/dx = cos(3x²) × 6x = 6x cos(3x²). Likewise, y = e^{x² + 1} gives dy/dx = 2x e^{x² + 1}.

例 2：y = sin(3x²)。外层函数是正弦，内层函数是 3x²，所以 dy/dx = cos(3x²) × 6x = 6x cos(3x²)。同样地，y = e^{x² + 1} 给出 dy/dx = 2x e^{x² + 1}。

, then multiply by the derivative of cos 2x, which is −2 sin 2x, giving dy/dx = −2 tan 2x.

链式法则可以嵌套多次。对于 y = ln(cos 2x)，由外向内计算：外层导数为 1/cos 2x，再乘以 cos 2x 的导数 −2 sin 2x，得到 dy/dx = −2 tan 2x。

Exam tip: most chain rule marks in Edexcel papers are lost by missing the inner derivative. Before simplifying, ask yourself: "What is inside, and what is its derivative?" Writing u and du/dx in the margin often prevents this error.

考试提示：Edexcel 试卷中链式法则的失分，大多是因为漏掉内层导数。化简之前先问自己："里面是什么？它的导数是什么？"在草稿上写出 u 与 du/dx，往往就能避免这一错误。

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6. Exponential and Logarithmic Functions | 指数函数与对数函数

The exponential function eˣ the most important function in calculus. Combining this with the chain rule gives d/dx (e^kx) = k e^kx, and d/dx (e^(f(x))) = f′(x) e^(f(x)).

指数函数 eˣ 的导数等于它本身，这使它成为微积分中最重要的函数。结合链式法则可得 d/dx (e^kx) = k e^kx，以及 d/dx (e^(f(x))) = f′(x) e^(f(x))。

For other bases, d/dx (aˣ) = aˣ ln a. For example, d/dx (2ˣ) = 2ˣ ln 2, and d/dx (5^3x) = 3 × 5^3x ln 5.

对于其他底数，d/dx (aˣ) = aˣ ln a。例如 d/dx (2ˣ) = 2ˣ ln 2，而 d/dx (5^3x) = 3 × 5^3x ln 5。

+ 1) gives dy/dx = 10x / (5x² + 1), and y = ln(sin x) gives dy/dx = cos x / sin x = cot x.

对于对数函数，当 x > 0 时 d/dx (ln x) = 1/x；更一般地 d/dx (ln f(x)) = f′(x) / f(x)。所以 y = ln(5x² + 1) 给出 dy/dx = 10x / (5x² + 1)，y = ln(sin x) 给出 dy/dx = cos x / sin x = cot x。

hen dy/dx = 2x/(x² + 1) − 1/(x − 3). This is much faster than differentiating the quotient directly.

一个实用技巧是先利用对数运算法则再求导。要对 x > 3 时的 y = ln( (x² + 1) / (x − 3) ) 求导，先写成 y = ln(x² + 1) − ln(x − 3)，于是 dy/dx = 2x/(x² + 1) − 1/(x − 3)。这比直接对商求导快得多。

Exam tip: questions often ask for the equation of a tangent or normal to a curve involving eˣ or ln x, or for the coordinates of a stationary point. Always evaluate the derivative at the given point rather than at a general x, and keep exact values such as ln 2 rather than rounding unless told otherwise.

考试提示：涉及 eˣ 或 ln x 的题目常要求求切线或法线方程，或求驻点坐标。请务必在给定点处求导数值，而不是保留一般形式 x；除非另有说明，答案应保留 ln 2 这样的精确值，不要四舍五入。

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7. Trigonometric Differentiation | 三角函数的微分

nd d/dx (tan x) = sec² x. The negative sign for cosine is a frequent source of lost marks.

基本三角函数导数为 d/dx (sin x) = cos x，d/dx (cos x) = −sin x，d/dx (tan x) = sec² x。余弦求导时出现的负号是常见失分点。

ce, tan x = sin x / cos x, and applying the quotient rule gives (cos² x + sin² x) / cos² x = 1 / cos² x = sec² x.

倒三角函数的导数可由除法法则推出：d/dx (sec x) = sec x tan x，d/dx (cosec x) = −cosec x cot x，d/dx (cot x) = −cosec² x。例如 tan x = sin x / cos x，用除法法则得 (cos² x + sin² x) / cos² x = 1 / cos² x = sec² x。

os x)³, dy/dx = 3 cos² x × (−sin x) = −3 sin x cos² x. For y = tan(x²), dy/dx = 2x sec²(x²).

配合链式法则，复合三角函数变得直接。对于 y = sin(4x − 1)，dy/dx = 4 cos(4x − 1)；对于 y = cos³ x，即 (cos x)³，dy/dx = 3 cos² x × (−sin x) = −3 sin x cos² x；对于 y = tan(x²)，dy/dx = 2x sec²(x²)。

ians first, and remember the small-angle approximations sin x ≈ x and cos x ≈ 1 − x²/2 for small x in radians.

一个关键考试要点是：这些导数结果只在角度以弧度为单位时成立。若题目涉及角度制，请先转换为弧度；同时记住小角度近似 sin x ≈ x 与 cos x ≈ 1 − x²/2（x 为小弧度角）。

Exam tip: in more demanding questions you may need to differentiate something like sin x cos 2x, which requires the product rule applied to two trigonometric factors, or sin(2x)/x, which requires the quotient rule. Identify the structure first, then choose the rule.

考试提示：在较难的题目中，你可能需要求导如 sin x cos 2x 这样的函数，需要用乘法法则处理两个三角因子；或 sin(2x)/x，需要除法法则。先判断结构，再选择法则。

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8. Implicit Differentiation | 隐函数微分

nes y implicitly, and to differentiate it we differentiate both sides with respect to x, treating y as a function of x. Every time we differentiate a term in y, the chain rule contributes a factor of dy/dx.

并非所有曲线都能写成 y = f(x)。像 x² + y² = 25 这样的方程以隐式方式定义 y，对其求导时，我们对等式两边同时关于 x 求导，并把 y 视为 x 的函数。每当我们对含 y 的项求导时，链式法则都会带来一个因子 dy/dx。

ing gives 2x + 2y (dy/dx) = 0, so dy/dx = −x/y. At the point (3, 4), the gradient is −3/4.

例 1：x² + y² = 25

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