📚 A-Level Mathematics: Basic Differentiation | A-Level数学：导数的基本求法
In A-Level Mathematics, differentiation is one of the most fundamental tools in calculus. It measures the instantaneous rate at which a function changes, and forms the foundation for solving optimisation problems, motion analysis, and curve sketching. This article covers the core rules for differentiating basic functions, including the definition from first principles, polynomial rules, product and quotient rules, and the chain rule.
在A-Level数学中，微分（求导）是微积分中最基础的工具之一。它衡量函数在某一点上的瞬时变化率，是解决优化问题、运动分析以及曲线作图的基础。本文将系统地介绍导数的基本求法，包括一阶原理定义、多项式函数求导、乘积法则、商法则以及链式法则。
1. The Definition of the Derivative | 导数的定义
The derivative of a function f(x) at a point x is defined as the limit of the average rate of change over an interval as that interval shrinks to zero. If the function is continuous, the limit of the difference quotient gives the exact slope of the tangent line at that point.
函数 f(x) 在某一点 x 的导数定义为：当区间长度趋近于零时，平均变化率的极限。如果函数连续，那么差分商的极限就是该点处切线的精确斜率。
f′(x) = limₕ→₀ [f(x + h) − f(x)] / h
This definition is sometimes called “differentiation from first principles.” It is often tested in A-Level papers as a short derivation for simple polynomials such as f(x) = x².
这一定义也被称为“一阶原理求导”。在A-Level考试中，经常要求用这种方法对简单多项式（例如 f(x) = x²）进行推导。
For f(x) = x², using first principles:
对于 f(x) = x²，用一阶原理求导：
f′(x) = limₕ→₀ [(x + h)² − x²] / h = limₕ→₀ [2xh + h²] / h = limₕ→₀ (2x + h) = 2x
2. The Power Rule | 幂函数求导法则
The power rule is the first and most frequently used differentiation rule. For any real constant n, the derivative of xⁿ is n·xⁿ⁻¹. This rule works for positive integers, negative integers, and fractional powers, provided the resulting expression is defined.
幂函数法则是求导中最常用、最基本的法则。对于任意实常数 n，函数 xⁿ 的导数为 n·xⁿ⁻¹。该法则适用于正整数、负整数以及分数次幂，前提是所得表达式有意义。
d/dx (xⁿ) = n·xⁿ⁻¹
Common examples include:
常见例子包括：
- d/dx (x³) = 3x²
- d/dx (x⁵) = 5x⁴
- d/dx (x¹ᐟ²) = ½ x⁻¹ᐟ²
- d/dx (x⁻²) = −2x⁻³
In A-Level exams, candidates are expected to rewrite roots and reciprocals as powers before differentiating. For example, √x must be written as x¹ᐟ² before applying the power rule.
在A-Level考试中，考生需要先将根式与倒数形式改写为幂的形式，再运用幂函数法则。例如，√x 需写作 x¹ᐟ²，然后再求导。
3. Constant Multiple Rule and Sum Rule | 常数倍法则与和差法则
The constant multiple rule states that the derivative of a constant times a function is the constant times the derivative of the function. The sum rule states that the derivative of a sum of functions is the sum of their individual derivatives. Together, they allow us to differentiate any polynomial by handling each term separately.
常数倍法则指出：常数乘以函数的导数等于该常数乘以函数的导数。和差法则指出：若干个函数之和的导数等于各函数导数之和。这两条法则结合使用，即可对任意多项式逐项求导。
d/dx [c·f(x)] = c·f′(x)
d/dx [f(x) ± g(x)] = f′(x) ± g′(x)
For instance, to differentiate f(x) = 4x³ − 2x + 7:
例如，对 f(x) = 4x³ − 2x + 7 求导：
f′(x) = 12x² − 2
Notice that the constant term 7 disappears, because its derivative is zero. This is a frequent area where students lose marks: forgetting that constants differentiate to zero.
注意常数项 7 的导数为零，因此求导后消失。这是学生常丢分之处：忘记常数项求导后为 0。
4. Differentiating Exponential Functions | 指数函数的求导
The derivative of the natural exponential function eˣ is particularly elegant: the function is its own derivative. For a more general exponential base a, the derivative is aˣ·ln(a). For composite expressions such as eᵏˣ, the chain rule gives k·eᵏˣ.
自然指数函数 eˣ 的导数非常简洁：其导数等于其自身。对于一般底数 a 的指数函数，导数为 aˣ·ln(a)。对于复合表达式如 eᵏˣ，利用链式法则可得 k·eᵏˣ。
d/dx (eˣ) = eˣ
d/dx (aˣ) = aˣ·ln(a)
d/dx (eᵏˣ) = k·eᵏˣ
For example, the derivative of e³ˣ is 3e³ˣ, and the derivative of 2ˣ is 2ˣ·ln2.
例如，e³ˣ 的导数为 3e³ˣ；2ˣ 的导数为 2ˣ·ln2。
5. Differentiating Logarithmic Functions | 对数函数的求导
The derivative of the natural logarithm function ln(x) is 1/x. This result follows directly from the fact that y = ln(x) is the inverse of x = eʸ. For a logarithmic function with base a, the derivative is 1/(x·ln(a)).
自然对数函数 ln(x) 的导数为 1/x。这一结果可以直接由 y = ln(x) 与 x = eʸ 互为反函数而得出。对于以 a 为底的对数函数，其导数为 1/(x·ln(a))。
d/dx [ln(x)] = 1/x
d/dx [logₐ(x)] = 1/(x·ln(a))
When differentiating a composite logarithm, such as ln(2x + 1), the chain rule must be applied: the derivative is 2/(2x + 1). A common error is forgetting the inner derivative, which is exactly the factor generated by the chain rule.
当对复合对数函数求导时，例如 ln(2x + 1)，必须运用链式法则：其导数为 2/(2x + 1)。常见错误是忘记内层函数导数，而链式法则恰恰要求乘上这一因子。
6. Differentiating Trigonometric Functions | 三角函数的求导
For the six basic trigonometric functions, the derivatives are standard results that students must memorise. The most important are those of sin(x) and cos(x), measured in radians. The derivatives of tan(x), cot(x), sec(x), and csc(x) can be derived from the first two using quotient rules and identities.
对于六个基本三角函数，其导数属于必须记忆的标准结论。其中最重要的是 sin(x) 和 cos(x) 的导数，且角度以弧度为单位。tan(x)、cot(x)、sec(x) 和 csc(x) 的导数均可由前两者借助商法则和恒等式推导得出。
- d/dx [sin(x)] = cos(x)
- d/dx [cos(x)] = −sin(x)
- d/dx [tan(x)] = sec²(x)
- d/dx [cot(x)] = −csc²(x)
- d/dx [sec(x)] = sec(x)·tan(x)
- d/dx [csc(x)] = −csc(x)·cot(x)
All of these formulas assume that the angle is measured in radians. When using degrees, additional constant factors appear, so A-Level exams always operate in radians unless otherwise stated.
所有这些公式均假设角度以弧度为单位。若使用角度制，则会出现额外的常数因子。因此，A-Level考试中除特别说明外一律使用弧度制。
7. The Product Rule | 乘积求导法则
When a function is the product of two differentiable functions, the derivative is not simply the product of their individual derivatives. Instead, the product rule must be used. If y = u·v, where u and v are functions of x, then:
当一个函数是两个可导函数的乘积时，其导数并不是两个函数导数的简单乘积，而必须使用乘积法则。若 y = u·v，其中 u 和 v 均为 x 的函数，则：
dy/dx = u·(dv/dx) + v·(du/dx)
Consider y = x²·eˣ. Let u = x² and v = eˣ, then du/dx = 2x and dv/dx = eˣ. Therefore:
以 y = x²·eˣ 为例。令 u = x²，v = eˣ，则 du/dx = 2x，dv/dx = eˣ。由此可得：
dy/dx = x²·eˣ + 2x·eˣ = eˣ(x² + 2x)
A helpful mnemonic is “first times derivative of second plus second times derivative of first.” The order of the two terms does not matter algebraically, but consistency reduces careless mistakes.
记忆口诀可以是“前导后不导加后导前不导”。两项的先后顺序在代数上并无区别，但保持固定顺序有助于减少粗心错误。
8. The Quotient Rule | 商法则
When a function is written as the ratio of two differentiable functions, the quotient rule provides the correct derivative. If y = u/v, where both u and v are functions of x and v ≠ 0, then:
当一个函数表示为两个可导函数之比时，商法则给出了正确的导数结果。若 y = u/v，其中 u 和 v 均为 x 的函数且 v ≠ 0，则：
dy/dx = [v·(du/dx) − u·(dv/dx)] / v²
For example, take y = x / (x² + 1). Let u = x and v = x² + 1, then du/dx = 1 and dv/dx = 2x. Applying the quotient rule:
例如，求 y = x / (x² + 1) 的导数。令 u = x，v = x² + 1，则 du/dx = 1，dv/dx = 2x。代入商法则：
dy/dx = [(x² + 1)·1 − x·2x] / (x² + 1)² = (1 − x²) / (x² + 1)²
Students often confuse the order of the subtraction in the numerator. Remember that the numerator is “v·u′ − u·v′”; swapping the order changes the sign of the entire expression.
学生在公式分子中两个乘积的先后顺序上容易出错。需要注意：分子是 v·u′ − u·v′。若颠倒顺序，则整个表达式符号会发生改变。
9. The Chain Rule | 链式法则
The chain rule handles composite functions, where one function is nested inside another. If y = f(g(x)), then the derivative with respect to x is the derivative of the outer function evaluated at g(x), multiplied by the derivative of the inner function. In Leibniz notation, if y = f(u) and u = g(x), then:
链式法则用于处理复合函数的情形，即一个函数嵌套在另一个函数内部。若 y = f(g(x))，则对 x 求导时，等于外层函数在 g(x) 处的导数乘以内层函数的导数。采用莱布尼茨记号，若 y = f(u) 且 u = g(x)，则：
dy/dx = dy/du × du/dx
For example, consider y = sin(3x²). Here, the outer function is sin(u), and the inner function is u = 3x². The derivative of sin(u) is cos(u), and the derivative of 3x² is 6x. Therefore:
例如，求 y = sin(3x²) 的导数。此处外层函数为 sin(u)，内层函数为 u = 3x²。sin(u) 的导数为 cos(u)，3x² 的导数为 6x。因此：
dy/dx = cos(3x²) × 6x = 6x·cos(3x²)
The chain rule is sometimes called the “function of a function” rule. It is widely examined in A-Level papers, especially combined with trigonometric, exponential, and logarithmic functions.
链式法则也常被称为“函数的函数”法则。它在A-Level考试中考查频率很高，尤其常与三角函数、指数函数和对数函数结合出题。
10. Higher-Order Derivatives | 高阶导数
After differentiating a function once, the result can itself be differentiated again. The second derivative, written f″(x) or d²y/dx², measures the rate of change of the gradient and is used to classify stationary points and analyse concavity.
对一个函数求一次导数后，所得结果还可以继续求导。二阶导数记作 f″(x) 或 d²y/dx²，它衡量斜率的变化率，用于判断驻点的性质以及研究曲线的凹凸性。
For example, if f(x) = 2x⁴ − 3x² + 5, then:
例如，若 f(x) = 2x⁴ − 3x² + 5，则：
f′(x) = 8x³ − 6x
f″(x) = 24x² − 6
Higher-order derivatives beyond the second are used less frequently in basic A-Level papers, but third and fourth derivatives may appear in questions on polynomial functions or motion in a straight line.
在A-Level基础考试中，高于二阶的导数较少直接用，但三次或四次导数有时会出现在多项式函数或直线运动的相关题目中。
11. Applications: Tangents and Normals | 应用：切线方程与法线方程
One of the most direct applications of differentiation is finding the equation of the tangent line to a curve at a given point. The derivative at that point gives the slope of the tangent. Once the slope m and the point (x₁, y₁) are known, the equation of the tangent is given by y − y₁ = m(x − x₁). The normal line is perpendicular to the tangent, so its slope is the negative reciprocal: −1/m.
微分最直接的应用之一就是求曲线在某一点处的切线方程。该点处的导数值即为切线的斜率。已知斜率 m 和点 (x₁, y₁) 后，切线方程为 y − y₁ = m(x − x₁)。法线垂直于切线，因此其斜率为切线斜率的负倒数：−1/m。
For instance, find the tangent and normal to f(x) = x² at x = 3. Since f(3) = 9 and f′(3) = 6, the tangent equation is y − 9 = 6(x − 3), or y = 6x − 9. The normal has slope −1/6, so y − 9 = (−1/6)(x − 3).
例如，求曲线 f(x) = x² 在 x = 3 处的切线和法线方程。由于 f(3) = 9，f′(3) = 6，切线方程为 y − 9 = 6(x − 3)，即 y = 6x − 9。法线斜率为 −1/6，故法线方程为 y − 9 = (−1/6)(x − 3)。
12. Common Mistakes and Revision Tips | 常见错误与复习建议
The following table summarises common errors in basic differentiation and how to avoid them.
下表总结了基本求导中的常见错误以及应对方法。
| Common Mistake | 常见错误 | Correction | 正确做法 |
| Differentiating a constant term and leaving it unchanged | 对常数项求导后保留该常数 | The derivative of any constant is 0 | 任何常数的导数均为 0 |
| Forgetting the inner derivative in composite functions | 忘记复合函数中的内层函数导数 | Always multiply by the derivative of the inner term | 始终乘以内层函数的导数 |
| Using the product rule instead of applying linearity | 该用和差法则时误用乘积法则 | Recognise sums of terms before choosing a rule | 先判断是否为若干项之和，再选择对应法则 |
| Writing the quotient rule with the wrong sign order | 商法则分子中符号顺序写错 | Memorise: v·u′ − u·v′ | 牢记公式：v·u′ − u·v′ |
| Confusing the derivative of sin(x) and cos(x) | 混淆 sin(x) 与 cos(x) 的导数 | sin → cos, cos → −sin | sin 的导数是 cos，cos 的导数是 −sin |
To build fluency, practise differentiating the same function using different methods and check results by expanding first when possible. For example, y = x²(x + 1) can be expanded into x³ + x² and differentiated using linearity, which is helpful for verifying the product rule.
为提升熟练度，建议用多种方法对同一函数求导，并在可能时先展开再求导以进行检验。例如，y = x²(x + 1) 可展开为 x³ + x²，再用和差法则求导，这样可以有效检验乘积法则的结果。
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