A-Level数学：微分技巧完全指南 | A-Level Maths: Complete Guide to Differentiation Techniques

微分（Differentiation）是A-Level数学中最基础的微积分工具之一。掌握微分不仅对Pure Mathematics考试至关重要，在Mechanics中计算速度和加速度、在Statistics中处理优化问题也同样不可或缺。本文系统梳理A-Level阶段需要掌握的微分技巧和常见题型。

Differentiation is one of the most fundamental calculus tools in A-Level Mathematics. Mastering differentiation is not only essential for the Pure Mathematics exam, but also indispensable in Mechanics for calculating velocity and acceleration, and in Statistics for optimization problems. This article systematically covers the differentiation techniques and common question types you’ll encounter at A-Level.

一、基础微分规则 | Basic Differentiation Rules

1.1 幂函数法则 | The Power Rule

这是最基础的微分规则。对于函数 f(x) = x^n，其导数为 f'(x) = nx^(n-1)。这个规则适用于任何实数指数n。例如：f(x) = x^5 的导数是 f'(x) = 5x^4；f(x) = sqrt(x) = x^(1/2) 的导数是 f'(x) = (1/2)x^(-1/2) = 1/(2*sqrt(x))。需要注意常数的导数为零，因为 f(x) = c = c*x^0，所以 f'(x) = 0。

This is the most basic differentiation rule. For a function f(x) = x^n, its derivative is f'(x) = nx^(n-1). This rule applies to any real exponent n. For example: the derivative of f(x) = x^5 is f'(x) = 5x^4; the derivative of f(x) = sqrt(x) = x^(1/2) is f'(x) = (1/2)x^(-1/2) = 1/(2*sqrt(x)). Note that the derivative of a constant is zero.

1.2 线性组合法则 | Sum and Constant Multiple Rules

微分的两大基本性质：常数倍法则 d/dx [c*f(x)] = c*f'(x)；和差法则 d/dx [f(x) +/- g(x)] = f'(x) +/- g'(x)。这两个性质意味着我们可以逐项微分多项式。例如：d/dx (3x^4 – 2x^3 + 5x – 7) = 12x^3 – 6x^2 + 5。注意常数项-7的导数为0。

Two fundamental properties of differentiation: the constant multiple rule d/dx [c*f(x)] = c*f'(x); and the sum/difference rule d/dx [f(x) +/- g(x)] = f'(x) +/- g'(x). These properties mean we can differentiate polynomials term by term. For example: d/dx (3x^4 – 2x^3 + 5x – 7) = 12x^3 – 6x^2 + 5.

二、进阶微分法则 | Advanced Differentiation Rules

2.1 链式法则 | The Chain Rule

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)。这是A-Level考试中最高频的微分解法之一。关键技巧是先识别”外层函数”和”内层函数”。经典例题：求 y = (2x + 3)^5 的导数。设 u = 2x + 3（内层），则 y = u^5（外层）。dy/du = 5u^4，du/dx = 2。由链式法则得 dy/dx = 5u^4 * 2 = 10(2x + 3)^4。

The chain rule is used for differentiating composite functions: if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). This is one of the most frequently tested differentiation methods in A-Level exams. Classic example: find the derivative of y = (2x + 3)^5. Let u = 2x + 3 (inner), then y = u^5 (outer). dy/du = 5u^4, du/dx = 2. By the chain rule, dy/dx = 5u^4 * 2 = 10(2x + 3)^4.

更复杂的链式法则例题——三角函数复合：求 y = sin(3x^2 + 1) 的导数。设 u = 3x^2 + 1，则 y = sin(u)。dy/du = cos(u)，du/dx = 6x。所以 dy/dx = cos(3x^2 + 1) * 6x = 6x*cos(3x^2 + 1)。同理，对于 y = ln(cos x)，设 u = cos x，则 dy/dx = (1/u) * (-sin x) = -sin x / cos x = -tan x。

More complex chain rule example: find the derivative of y = sin(3x^2 + 1). Let u = 3x^2 + 1, then y = sin(u). dy/du = cos(u), du/dx = 6x. So dy/dx = cos(3x^2 + 1) * 6x = 6x*cos(3x^2 + 1). Similarly, for y = ln(cos x), let u = cos x, then dy/dx = (1/u) * (-sin x) = -sin x / cos x = -tan x.

2.2 乘积法则 | The Product Rule

当函数是两个因式的乘积时使用。若 y = u(x) * v(x)，则 dy/dx = u’v + uv’。记忆口诀：”第一项的导数乘第二项，加上第一项乘第二项的导数”。例题：求 y = x^2*e^x 的导数。u = x^2，u’ = 2x；v = e^x，v’ = e^x。dy/dx = 2x*e^x + x^2*e^x = e^x(2x + x^2) = x*e^x(x + 2)。

Used when a function is the product of two factors. If y = u(x) * v(x), then dy/dx = u’v + uv’. Example: find the derivative of y = x^2*e^x. u = x^2, u’ = 2x; v = e^x, v’ = e^x. dy/dx = 2x*e^x + x^2*e^x = e^x(2x + x^2) = x*e^x(x + 2).

乘积法则在含有对数函数和三角函数的题目中尤其常见。例如 y = x*ln x：u = x，u’ = 1；v = ln x，v’ = 1/x。dy/dx = 1*ln x + x*(1/x) = ln x + 1。再如 y = e^x*sin x：y’ = e^x*sin x + e^x*cos x = e^x(sin x + cos x)。

The product rule is especially common in questions involving logarithmic and trigonometric functions. For example, y = x*ln x: u = x, u’ = 1; v = ln x, v’ = 1/x. dy/dx = 1*ln x + x*(1/x) = ln x + 1. Another example, y = e^x*sin x: y’ = e^x*sin x + e^x*cos x = e^x(sin x + cos x).

2.3 商法则 | The Quotient Rule

处理分式函数的微分：若 y = u(x) / v(x)，则 dy/dx = (u’v – uv’) / v^2。注意分子的顺序：是 u’v – uv’，不是反过来！例题：y = (x^2 + 1) / (x – 2)。u = x^2 + 1，u’ = 2x；v = x – 2，v’ = 1。dy/dx = [2x(x – 2) – (x^2 + 1)*1] / (x – 2)^2 = [2x^2 – 4x – x^2 – 1] / (x – 2)^2 = (x^2 – 4x – 1) / (x – 2)^2。

For differentiating rational functions: if y = u(x) / v(x), then dy/dx = (u’v – uv’) / v^2. Be careful with the numerator order: it’s u’v – uv’, not the other way around! Example: y = (x^2 + 1) / (x – 2). u = x^2 + 1, u’ = 2x; v = x – 2, v’ = 1. dy/dx = [2x(x – 2) – (x^2 + 1)*1] / (x – 2)^2 = (x^2 – 4x – 1) / (x – 2)^2.

常见错误：很多同学会把分子的顺序写成 uv’ – u’v，导致整个结果符号错误。另一个常见错误是忘记分母平方。考试中商法则常与三角恒等式结合——例如 y = sin x / cos x = tan x，用商法则求导可得 y’ = (cos x*cos x – sin x*(-sin x)) / cos^2 x = (cos^2 x + sin^2 x) / cos^2 x = 1/cos^2 x = sec^2 x，与 tan x 的标准导数一致。

Common mistake: many students write the numerator as uv’ – u’v, flipping the sign. Another common error is forgetting to square the denominator. The quotient rule frequently combines with trig identities — e.g., y = sin x / cos x = tan x. Using the quotient rule gives y’ = sec^2 x, matching the standard derivative of tan x.

三、隐函数微分 | Implicit Differentiation

当y不能明确表达为x的函数时（例如圆的方程 x^2 + y^2 = r^2 或包含x和y混合项的方程），就需要使用隐函数微分。基本方法：对等式两边同时对x求导，遇到y的项时使用链式法则（d/dx [y^n] = n*y^(n-1)*dy/dx），最后解出 dy/dx。

When y cannot be expressed explicitly as a function of x (e.g., the equation of a circle x^2 + y^2 = r^2), we need implicit differentiation. Basic method: differentiate both sides with respect to x, apply the chain rule whenever differentiating terms involving y (d/dx [y^n] = n*y^(n-1)*dy/dx), and finally solve for dy/dx.

经典例题：求曲线 x^2 + y^2 = 25 在点 (3, 4) 处的切线斜率。对两边求导：2x + 2y*dy/dx = 0, so dy/dx = -x/y。代入点 (3, 4) 得斜率 = -3/4。切线方程：y – 4 = (-3/4)(x – 3), so 3x + 4y = 25。

Classic example: Find the slope of the tangent to x^2 + y^2 = 25 at (3, 4). Differentiate: 2x + 2y*dy/dx = 0, so dy/dx = -x/y. At (3, 4): slope = -3/4. Tangent: y – 4 = (-3/4)(x – 3), so 3x + 4y = 25.

进阶例题：已知 e^(xy) + x^2*y = 5，求 dy/dx。对x求导：e^(xy)*(y + x*dy/dx) + 2x*y + x^2*dy/dx = 0。整理得 dy/dx = -(y*e^(xy) + 2xy) / (x*e^(xy) + x^2)。

Advanced example: Given e^(xy) + x^2*y = 5, find dy/dx. Differentiate: e^(xy)*(y + x*dy/dx) + 2x*y + x^2*dy/dx = 0. Collect terms: dy/dx = -(y*e^(xy) + 2xy) / (x*e^(xy) + x^2).

四、标准函数的导数公式 | Derivatives of Standard Functions

A-Level考试中必须熟记以下标准导数公式：

d/dx (e^x) = e^x
d/dx (a^x) = a^x*ln a
d/dx (ln x) = 1/x
d/dx (sin x) = cos x
d/dx (cos x) = -sin x
d/dx (tan x) = sec^2 x
d/dx (cot x) = -csc^2 x
d/dx (sec x) = sec x*tan x
d/dx (csc x) = -csc x*cot x
d/dx (arcsin x) = 1/sqrt(1 – x^2)
d/dx (arccos x) = -1/sqrt(1 – x^2)
d/dx (arctan x) = 1/(1 + x^2)

重点提醒：cos x 的导数是 -sin x，负号千万不要漏掉！另外，反三角函数的导数在Edexcel和CIE考纲中都会出现，务必记住 arcsin、arccos、arctan 的三个公式。

Key reminder: The derivative of cos x is -sin x — don’t forget the negative sign! Additionally, derivatives of inverse trigonometric functions appear in both Edexcel and CIE syllabi.

五、指数与对数微分 | Exponential and Logarithmic Differentiation

这部分内容在A-Level Pure Mathematics 3 (P3) 中占据重要地位。核心结论：e^x 是唯一导函数等于自身的函数，即 d/dx(e^x) = e^x。当指数部分是x的函数时，使用链式法则：d/dx(e^(f(x))) = e^(f(x))*f'(x)。例如 d/dx(e^(3x^2)) = e^(3x^2)*6x。

This section occupies an important position in A-Level Pure Mathematics 3 (P3). The core conclusion: e^x is the only function whose derivative equals itself. When the exponent is a function of x, use the chain rule: d/dx(e^(f(x))) = e^(f(x))*f'(x). Example: d/dx(e^(3x^2)) = e^(3x^2)*6x.

对数微分技巧：面对形如 y = (f(x))^(g(x)) 的复杂指数函数，可以先取自然对数再求导。例题：求 y = x^x 的导数。两边取ln：ln y = x*ln x。对x求导：(1/y)*dy/dx = ln x + 1。所以 dy/dx = y*(ln x + 1) = x^x*(ln x + 1)。

Logarithmic differentiation technique: for functions of the form y = (f(x))^(g(x)), take ln first. Example: find dy/dx for y = x^x. ln y = x*ln x. Differentiate: (1/y)*dy/dx = ln x + 1. So dy/dx = x^x*(ln x + 1).

六、二阶导数与驻点分类 | Second Derivatives and Stationary Point Classification

一阶导数 f'(x) 表示函数的梯度（变化率），令 f'(x) = 0 可求得驻点（stationary points）。二阶导数 f”(x) 则用于判断驻点的性质：f”(x) > 0 时为局部极小值点（local minimum）；f”(x) < 0 时为局部极大值点（local maximum）；f''(x) = 0 时需进一步判断（可能是拐点inflection point）。

The first derivative f'(x) represents the gradient; setting f'(x) = 0 finds stationary points. The second derivative f”(x) classifies them: f”(x) > 0 indicates a local minimum; f”(x) < 0 indicates a local maximum; f''(x) = 0 requires further investigation (possible inflection point).

完整例题：求函数 f(x) = x^3 – 3x^2 – 9x + 5 的驻点并分类。求一阶导数：f'(x) = 3x^2 – 6x – 9 = 3(x – 3)(x + 1)。令 f'(x) = 0, x = -1 或 x = 3。二阶导数：f”(x) = 6x – 6 = 6(x – 1)。在 x = -1：f”(-1) = -12 < 0，局部极大值，f(-1) = 10。在 x = 3：f''(3) = 12 > 0，局部极小值，f(3) = -22。

Complete example: Find and classify stationary points of f(x) = x^3 – 3x^2 – 9x + 5. f'(x) = 3x^2 – 6x – 9 = 3(x – 3)(x + 1) = 0, x = -1 or x = 3. f”(x) = 6x – 6. At x = -1: f”(-1) = -12 < 0, local max, f(-1) = 10. At x = 3: f''(3) = 12 > 0, local min, f(3) = -22.

七、微分在Mechanics中的应用 | Applications in Mechanics

在A-Level Mechanics中，微分用于描述运动学关系：位移s(t)的一阶导数是速度v(t) = ds/dt；速度的一阶导数是加速度a(t) = dv/dt = d^2s/dt^2。例如，s(t) = 2t^3 – 15t^2 + 24t + 8，则 v(t) = 6t^2 – 30t + 24，a(t) = 12t – 30。当 v(t) = 0 时物体静止，解 6t^2 – 30t + 24 = 0 得 t = 1 或 t = 4。

In A-Level Mechanics, differentiation describes kinematics: v(t) = ds/dt, a(t) = dv/dt = d^2s/dt^2. Example: if s(t) = 2t^3 – 15t^2 + 24t + 8, then v(t) = 6t^2 – 30t + 24, a(t) = 12t – 30. When v(t) = 0 the particle is at rest: t = 1 or t = 4.

八、考试技巧与常见陷阱 | Exam Tips and Common Pitfalls

1. 符号错误：三角函数导数中的负号最容易出错。cos x -> -sin x, cot x -> -csc^2 x, csc x -> -csc x*cot x, arccos x -> -1/sqrt(1 – x^2)。建议考前默写一遍。

1. Sign errors: Negative signs in trig derivatives are the most common mistakes. cos x -> -sin x, cot x -> -csc^2 x, csc x -> -csc x*cot x, arccos x -> -1/sqrt(1 – x^2). Rewrite them before the exam.

2. 链式法则遗漏：例如，d/dx(e^(2x)) 正确结果应该是 2e^(2x)，不是 e^(2x)。d/dx(sin(5x)) = 5cos(5x)，不是 cos(5x)。

2. Forgetting the chain rule: d/dx(e^(2x)) = 2e^(2x), not e^(2x). d/dx(sin(5x)) = 5cos(5x), not cos(5x).

3. 乘积/商法则顺序混淆：乘积法则中 u’v + uv’ 的顺序可以互换，但商法则中 (u’v – uv’)/v^2 绝对不能颠倒。

3. Confusing product/quotient rule order: In the quotient rule, (u’v – uv’)/v^2 must never be reversed — otherwise the entire question scores zero.

4. 隐函数微分中忘记 dy/dx：每次对含有y的项求导时，必须乘以 dy/dx。例如 d/dx(y^3) = 3y^2*dy/dx，不能写成 3y^2。

4. Forgetting dy/dx in implicit differentiation: Every time you differentiate a term containing y, multiply by dy/dx. d/dx(y^3) = 3y^2*dy/dx, not just 3y^2.

九、练习题目 | Practice Questions

试着用本文介绍的方法完成以下题目：

求 y = (3x^2 – 2x + 1)^4 的导数。
求 y = e^x*cos x 的导数。
求 y = ln(x^2 + 1) / x 的导数。
已知 x^2*y + x*y^2 = 6，求 dy/dx。
求 f(x) = 2x^3 – 9x^2 + 12x – 4 的所有驻点并分类。
质点位移 s(t) = t^3 – 6t^2 + 9t，求：a) 速度表达式 b) 静止时刻 c) 加速度为0的时刻。

Try completing the following questions:

Find the derivative of y = (3x^2 – 2x + 1)^4.
Find the derivative of y = e^x*cos x.
Find the derivative of y = ln(x^2 + 1) / x.
Given x^2*y + x*y^2 = 6, find dy/dx.
Find and classify all stationary points of f(x) = 2x^3 – 9x^2 + 12x – 4.
A particle’s displacement is s(t) = t^3 – 6t^2 + 9t. Find: a) velocity expression b) when at rest c) when acceleration is zero.

总结 | Summary

微分是A-Level数学的核心工具。从基础的幂函数法则到隐函数微分，再到Mechanics中的实际应用，每一步都需要扎实的理解和大量的练习。记住：链式法则是最常用的技巧，乘积法则和商法则是处理复杂函数组合的利器，隐函数微分则打开了一扇处理非线性关系的大门。考试中保持冷静，按步骤书写，尤其是商法则中保持 (u’v – uv’) 的严格顺序。祝你考试顺利！

Differentiation is a core tool in A-Level Mathematics. From the basic power rule to implicit differentiation, and to practical applications in Mechanics, every step requires solid understanding and ample practice. Remember: the chain rule is the most frequently used technique, the product and quotient rules are powerful tools for handling complex function combinations, and implicit differentiation opens the door to dealing with nonlinear relationships. Stay calm during the exam, work through each step methodically. Good luck with your exams!

A-Level数学：微分技巧完全指南 | A-Level Maths: Complete Guide to Differentiation Techniques