📚 Edexcel A-Level Maths: Differentiation - First Principles, Rules and Applications | 爱德思 A-Level 数学：微分——从第一原理到应用
Differentiation is one of the most heavily examined topics in Edexcel A-Level Mathematics, appearing in Pure Paper 1 and often linked with coordinate geometry, trigonometry, exponentials, logarithms and real-world modelling. This article gathers the key definitions, standard rules, applications and common errors so that you can approach the topic systematically.
微分是爱德思 A-Level 数学中考查频率最高的主题之一，出现在纯数 Paper 1 中，并经常与坐标几何、三角、指数、对数以及实际建模结合。本文整理关键定义、标准法则、应用和常见错误，帮助你系统掌握这一主题。
1. First Principles and the Derivative | 第一原理与导数
nition for simple functions such as f(x) = x² or f(x) = x³ before applying standard rules.
函数 f 在某一点 x 的导数度量曲线 y = f(x) 在该点切线的斜率。第一原理定义由差商的极限给出，你需要能对简单函数使用这一定义，例如 f(x) = x² 或 f(x) = x³，再过渡到标准法则。
f'(x) = lim (h→0) [f(x+h) − f(x)] / h
to 0, the derivative is 2x. This confirms that the gradient function of y = x² is linear.
对于 f(x) = x²，差商变为 [(x+h)² − x²] / h = 2x + h，因此当 h 趋于 0 时，导数为 2x。这验证了 y = x² 的斜率函数是线性的。
2. Standard Derivatives and the Power Rule | 基本导数与幂法则
The power rule states that the derivative of xⁿ is n xⁿ⁻¹. You must be able to handle negative and fractional powers, rewriting expressions such as 1/x² as x⁻² and √x as x^½ before differentiating.
幂法则指出 xⁿ 的导数是 n xⁿ⁻¹。你必须能够处理负指数和分数指数，在求导前将表达式改写，例如将 1/x² 写成 x⁻²，将 √x 写成 x^½。
d/dx (xⁿ) = n xⁿ⁻¹
| Function | Derivative |
| xⁿ | n xⁿ⁻¹ |
| eˣ | eˣ |
| ln x | 1/x |
| sin x | cos x |
| cos x | −sin x |
| tan x | sec²x |
Many exam mistakes come from forgetting to rewrite roots or reciprocals first. For example, differentiating 3/√x gives −3/(2x3/2), not 3/(2√x), so practise converting to index form carefully.
许多考试错误源于忘记先改写根式或倒数。例如，对 3/√x 求导得到 −3/(2x3/2)，而不是 3/(2√x)，因此需要仔细练习转换为指数形式。
3. Chain Rule | 链式法则
the derivative is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.
当一个函数嵌套在另一个函数中时使用链式法则，例如 y = (2x+3)⁵ 或 y = sin(3x)。它表示导数等于外层函数在内层函数处的导数乘以内层函数的导数。
dy/dx = dy/du × du/dx
giving dy/dx = 10(2x+3)⁴. The method is essential for any composite function, including ekx and ln(kx).
对于 y = (2x+3)⁵，设 u = 2x+3，则 y = u⁵。于是 dy/du = 5u⁴，du/dx = 2，得到 dy/dx = 10(2x+3)⁴。该方法适用于任何复合函数，包括 ekx 和 ln(kx)。
4. Product Rule | 乘积法则
ule is used when two functions of x are multiplied together, such as y = x² sin x or y = eˣ ormula keeps one function unchanged while differentiating the other, then swaps the roles.
当两个关于 x 的函数相乘时使用乘积法则，例如 y = x² sin x 或 y = eˣ ln x。该公式先保持一个函数不变而对另一个求导，然后交换角色。
d/dx (uv) = u'v + uv'
For y = x² sin x, let u = x² and v = sin x. Then u' = 2x and v' u should also be ready to factorise the result when a question asks for a simplified form.
对于 y = x² sin x，设 u = x²，v = sin x。则 u' = 2x，v' = cos x，所以 dy/dx = 2x sin x + x² cos x。你还需要在题目要求化简时对结果进行因式分解。
5. Quotient Rule | 商法则
of x is divided by another, such as y = x / (x² + 1). The numerator of the derivative is u'v − uv' and the denominator is v², so the order and subtraction sign are very important.
当一个关于 x 的函数除以另一个函数时使用商法则，例如 y = x / (x² + 1)。导数的分子是 u'v − uv'，分母是 v²，因此顺序和减号非常重要。
d/dx (u/v) = (u'v − uv') / v²
For y = x / (x² + 1), let u = x and v = x² + 1. Then u' = 1 and v' + 1) − x(2x)] / (x² + 1)² = (1 − x²) / (x² + 1)². Many candidates lose marks by writing uv' − u'v instead of u'v − uv'.
对于 y = x / (x² + 1)，设 u = x，v = x² + 1。则 u' = 1，v' = 2x，所以 dy/dx = [(x² + 1) − x(2x)] / (x² + 1)² = (1 − x²) / (x² + 1)²。许多考生因为写成 uv' − u'v 而不是 u'v − uv' 而失分。
6. Second Derivatives and Concavity | 二阶导数与凹凸性
The second derivative is obtained by differentiating the first derivative. It is written f''(x) or d²y/dx² and it tells you how the gradient itself is changing. If f''(x) > 0, the curve is convex or concave upward; if f''(x) < 0, the curve is concave or concave downward.
二阶导数通过对一阶导数再次求导得到，写作 f''(x) 或 d²y/dx²，它告诉你斜率本身如何变化。如果 f''(x) > 0，曲线是下凸或向上凹；如果 f''(x) < 0，曲线是上凸或向下凹。
d²y/dx² = d/dx (dy/dx)
e is 3x² − 6x and the second derivative is 6x − 6. Points of inflection occur where the concavity changes, and the second derivative can be used to classify stationary points.
对于 y = x³ − 3x² + 2，一阶导数为 3x² − 6x，二阶导数为 6x − 6。拐点出现在凹凸性改变的位置，二阶导数可用于对驻点进行分类。
7. Tangents and Normals | 切线与法线
the straight-line formula y − y₁ = m(x − x₁). The normal is perpendicular to the tangent.
要找到某点切线的方程，先求导得到斜率函数，代入 x 坐标得到斜率 m，然后使用直线公式 y − y₁ = m(x − x₁)。法线垂直于切线。
mnormal = −1 / m_tangent
ngent is y − 9 = 6(x − 3), and the normal has gradient −1/6, giving y − 9 = −(1/6)(x − 3).
对于 y = x² 在 x = 3 处，导数为 2x，所以在点 (3, 9) 处 m = 6。切线为 y − 9 = 6(x − 3)，法线斜率为 −1/6，因此法线为 y − 9 = −(1/6)(x − 3)。
8. Stationary Points and Classification | 驻点及其分类
Stationary points occur where the first derivative is zero, so you should solve f' l maxima, local minima or points of inflection using either the second derivative test or a sign change table.
驻点出现在一阶导数为零的位置，因此你需要解方程 f'(x) = 0。驻点可以通过二阶导数检验或符号变化表分类为局部极大值、局部极小值或拐点。
f'(x) = 0 for a stationary point
- Maximum: f'(x) changes from positive to negative, or f''(x) < 0.
- Minimum: f'(x) changes from negative to positive, or f''(x) > 0.
- Point of inflection: f'(x) does not change sign, or f''(x) = 0 with further investigation.
For y = x³ − 3x, f'(x) = 3x² − 3 = 0 gives x = ±1. At x = −1, f''(x) = 6x = −6, so it is a maximum; at x = 1, f'' 6, so it is a minimum. Always quote the y-coordinates when asked for the full coordinates.
对于 y = x³ − 3x，f'(x) = 3x² − 3 = 0 得到 x = ±1。在 x = −1 处，f''(x) = 6x = −6，所以是极大值；在 x = 1 处，f''(x) = 6，所以是极小值。当题目要求完整坐标时，始终要写出 y 坐标。
9. Optimisation Problems | 优化问题
Optimisation questions ask you to maximise or minimise a real quantity such as area, volume, cost or distance. You should express the quantity in terms of one variable using the given constraints, then differentiate and set the derivative equal to zero.
优化问题要求你最大化或最小化某个实际量，例如面积、体积、成本或距离。你应该利用给定约束将该量表示为一个变量的函数，然后求导并令导数等于零。
dQ/dx = 0, then justify maximum or minimum
x = 10 gives the maximum area of 100 m², and the second derivative −2 confirms a maximum.
例如，一个矩形花园的周长为 40 m。如果一边长为 x，面积为 A = x(20 − x) = 20x − x²。求导得 dA/dx = 20 − 2x，因此 x = 10 时最大面积为 100 m²，二阶导数 −2 确认这是最大值。
10. Rates of Change and Kinematics | 变化率与运动学
ing at a known rate, the rate of change of area is found by writing dA/dt = dA/dr × dr/dt.
相关变化率问题使用链式法则将一个变化率与另一个变化率联系起来。如果圆的半径以已知速率增大，则面积变化率通过 dA/dt = dA/dr × dr/dt 求得。
dA/dt = dA/dr × dr/dt
In kinematics, if displacement s is a function of time t, the velocity v is ds/dt and the acceleration a is dv/dt or d²s/dt². You should be able to move between displacement, velocity and acceleration using differentiation and integration.
在运动学中，如果位移 s 是时间 t 的函数，则速度 v 为 ds/dt，加速度 a 为 dv/dt 或 d²s/dt²。你应该能够利用微分和积分在位移、速度和加速度之间转换。
11. Exam-Style Tips and Common Errors | 考试技巧与常见错误
In Edexcel exams, differentiation questions often combine several techniques in one expression. Before differentiating, simplify the expression, rewrite roots and reciprocals as powers, and decide which rule or combination of rules is required.
在爱德思考试中，微分题常常在一个表达式中结合多种技巧。求导前先化简表达式，将根式和倒数改写为幂，并判断需要使用哪种法则或哪几种法则的组合。
- Use the product rule only for products, not for terms added together.
- Use the quotient rule for divisions, or rewrite 1/v as v⁻¹ if the chain rule is easier.
- Remember the chain rule when differentiating ekx, sin(kx), cos(kx) and ln(kx).
- Always check whether a question asks for the derivative, the gradient, the equation of a tangent, or the coordinates of a stationary point.
- Justify whether a stationary point is a maximum or minimum using the second derivative or a sign test.
- Write final answers in exact form unless a decimal approximation is requested.
例如，对于 h(x) = (x² + 1)⁴ ln x，你需要先用乘积法则，同时在求导 (x² + 1)⁴ 时使用链式法则。这类混合题是 A-Level 考试中的典型高分题型。
oduct rule and also the chain rule when differentiating (x² + 1)⁴. This type of mixed question is typical of high-mark A-Level items.
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