📚 Edexcel A-Level Pure Maths: Differentiation – Rules, Stationary Points and Tangents | 爱德思 A-Level 纯数学：微分——法则、驻点与切线
Differentiation is one of the most heavily examined topics in Edexcel A-Level Mathematics. It underpins questions on gradients, tangents, normals, stationary points, optimisation and curve sketching, and it reappears inside later topics such as sequences, trigonometry and mechanics. This revision article covers first principles, the standard derivatives and rules, higher derivatives, stationary points and the equations of tangents and normals. Work through every example and pay close attention to the Edexcel command words 'show that', 'find' and 'justify'.
微分是爱德思 A-Level 数学中考查频率最高的主题之一。它是梯度、切线、法线、驻点、优化和曲线作图等问题的基础，并且会在数列、三角函数和力学等后续主题中反复出现。本复习文章涵盖第一原理、标准导数与求导法则、高阶导数、驻点以及切线和法线方程。请务必完成每一道例题，并注意 Edexcel 题干中 'show that'、'find' 和 'justify' 等指令词。
1. First Principles | 第一原理
Differentiation from first principles uses the limit of the gradient of a chord as the interval tends to zero. For a function f(x), the derivative f'(x) is defined by the limit below. The gradient of the chord joining (x, f(x)) and (x + h, f(x + h)) is the difference quotient, and the derivative is what that quotient approaches as h shrinks to zero.
第一原理求导使用弦的斜率在区间趋近于零时的极限。对于函数 f(x)，导数 f'(x) 由下面的极限定义。连接 (x, f(x)) 与 (x + h, f(x + h)) 两点的弦的斜率就是差商，而导数就是当 h 趋于零时该差商所趋近的值。
f'(x) = limₕ→₀ [ f(x + h) − f(x) ] / h
ets, simplifying and cancelling h before finally taking the limit. Generic phrases such as 'as h → 0, h → 0' are not enough; the algebra must be visible on the page.
Edexcel 题目常要求你用该极限证明 x² 或 x³ 的导数。你必须展示所有步骤，包括展开括号、化简、约去 h，最后再取极限。只写 'as h → 0' 这类笼统语句是不够的，代数过程必须清楚写在卷面上。
n below. Notice that h is only set to zero at the very end, after it has cancelled as a common factor.
对于 f(x) = x²，完整过程如下。注意 h 只有在最后被约去之后，才令其等于零。
f' = limₕ→₀ [ (x + h)² − x² ] / h = limₕ→₀ [ x² + 2xh + h² − x² ] / h = limₕ→₀ (2x + h) = 2x
a clearly stated final conclusion. A similar argument with (x + h)³ gives d/dx (x³) = 3x².
由此得到标准结果 d/dx (x²) = 2x。得分点在于正确展开、正确化简以及明确写出最终结论。对 (x + h)³ 做类似处理，可得到 d/dx (x³) = 3x²。
2. Standard Derivatives and Rules | 标准导数与求导法则
You must know the standard results for power functions, trigonometric functions, exponential functions and logarithms from memory, because the formula booklet does not supply them. Linear combinations can always be differentiated term by term, and a constant multiplier can be pulled outside the derivative operator.
你必须熟记幂函数、三角函数、指数函数和对数函数的标准导数结果，因为公式册不会提供这些内容。线性组合始终可以逐项求导，常数因子也可以提到求导符号外面。
d/dx [ a f(x) + b g(x) ] = a f'(x) + b g'(x)
The table below summarises the results most frequently tested in Edexcel Pure Mathematics papers. Learn them in both directions, because rearrangement questions ask you to recognise, for example, that a term of the form 5/x² should be rewritten as 5x⁻² before it can be differentiated.
下表汇总了爱德思纯数学试卷中最常考的结果。要能正反两向使用，因为有些题目要求你先识别并改写，例如把 5/x² 改写成 5x⁻² 之后才能求导。
| Function f(x) | Derivative f'(x) |
|---|---|
| k (constant) | 0 |
| xⁿ | n xⁿ⁻¹ |
| eˣ | eˣ |
| eᵏˣ | k eᵏˣ |
| 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 |
A quick worked example: differentiate y = 4x³ − 7/x + 3√x 4x³ − 7x⁻¹ + 3x^½, then differentiate term by term to obtain dy/dx = 12x² + 7x⁻² + (3/2)x−½.
一个快速例题：对 y = 4x³ − 7/x + 3√x 求导。先改写为 y = 4x³ − 7x⁻¹ + 3x^½，再逐项求导得到 dy/dx = 12x² + 7x⁻² + (3/2)x−½。
3. The Chain Rule | 链式法则
z notation the rule is compact and easy to remember because the du terms appear to cancel.
当一个函数由内层函数和外层函数复合而成时，就要使用链式法则，例如 y = (3x + 1)⁵ 或 y = sin(2x²)。在莱布尼茨记号下，该法则形式紧凑、便于记忆，因为 du 看起来可以约掉。
dy/dx = dy/du × du/dx
ion is linear, the chain rule simply multiplies by the coefficient of x, which allows a quick check of your answer.
对于 y = (3x + 1)⁵，令 u = 3x + 1，则 y = u⁵。于是 dy/du = 5u⁴，du/dx = 3，所以 dy/dx = 15(3x + 1)⁴。当内层函数是一次函数时，链式法则只是乘以 x 的系数，这可以用来快速检验答案。
- y = e5x ⟹ dy/dx = 5e5x
- y = ln(4x + 3) ⟹ dy/dx = 4 / (4x + 3)
- y = cos(x³) ⟹ dy/dx = −3x² sin(x³)
- y = √x² + 9 ⟹ dy/dx = x / √x² + 9
In the last example, writing √x² + 9 as (x² + 9)^½ and applying the chain rule gives (1/2)(x² + 9)−½ × 2x, which simplifies to x / √x² + 9. Always simplify your final answer, since Edexcel mark schemes often require a particular unsimplified-to-simplified transition.
在最后一个例子中，把 √x² + 9 写成 (x² + 9)^½，再应用链式法则得到 (1/2)(x² + 9)−½ × 2x，化简后即 x / √x² + 9。一定要化简最终答案，因为爱德思评分标准常常要求从中间形式过渡到化简形式。
4. Product and Quotient Rules | 乘积法则与商法则
When two functions are multiplied or divided, you cannot simply differentiate each part and combine. The product rule and the quotient rule give the correct results, and you should be able to state both from memory.
当两个函数相乘或相除时，不能分别求导后再简单组合。乘积法则和商法则给出正确结果，你应当能够凭记忆写出这两个公式。
(uv)' = u'v + uv' (u/v)' = (u'v − uv') / v²
For y = x² eˣ, take u = x² and v = eˣ, so u' = 2x and v' = eˣ. Then dy/dx = 2x eˣ + x² eˣ = x eˣ it makes the signs of the derivative obvious when you later search for stationary points.
对于 y = x² eˣ，取 u = x²、v = eˣ，则 u' = 2x、v' = eˣ。于是 dy/dx = 2x eˣ + x² eˣ = x eˣ (x + 2)。提取公因式是好习惯，因为当你之后寻找驻点时，它能让你一眼看出导数的符号。
= (x cos x − sin x) / x². A very common error is to forget the minus sign, or to write uv' − u'v instead of u'v − uv'. Double-check the order every time.
对于 y = sin x / x，令商法则中 u = sin x、v = x，得到 dy/dx = (x cos x − sin x) / x²。极常见的错误是漏掉负号，或把分子写成 uv' − u'v 而不是 u'v − uv'。每次都要检查顺序。
- Differentiate ln x first: y = x² ln x ⟹ dy/dx = 2x ln x + x.
- Quotient with a constant numerator: y = 3 / (2x + 1) ⟹ dy/dx = −6 / (2x + 1)².
- Product of three factors: split into two products and apply the rule twice.
If a quotient has a constant numerator, the chain rule is often faster than the quotient rule, as the second bullet illustrates. Choosing the most efficient method saves time in a two-hour paper that contains around ten pure questions.
如果商的分子是常数，用链式法则往往比商法则更快，如上面第二条所示。选择最高效的方法能节省时间，因为一份两小时的试卷大约包含十道纯数学题。
5. Second and Higher Derivatives | 二阶与高阶导数
The second derivative is found by differentiating the first derivative, and it is written d²y/dx² or f''(x). It measures the rate of change of the gradient, which is exactly what you need in order to classify stationary points and to discuss concavity.
二阶导数通过对一阶导数再求导得到，写作 d²y/dx² 或 f''(x)。它衡量梯度的变化率，这正是判定驻点和讨论凹凸性所需要的信息。
y = x⁴ − 3x² ⟹ dy/dx = 4x³ − 6x ⟹ d²y/dx² = 12x² − 6
Interpretation matters as much as calculation. Where d²y/dx² > 0 the curve is convex and the gradient is increasing; where d²y/dx² < 0 the curve is concave and the gradient is decreasing. A point where the second derivative changes sign is a point of inflection, which may or may not be a stationary point.
理解与计算同样重要。当 d²y/dx² > 0 时曲线下凸、梯度递增；当 d²y/dx² < 0 时曲线上凸、梯度递减。二阶导数变号的点称为拐点，它可能是驻点，也可能不是。
- Third derivative notation: d³y/dx³, used in some extension questions about rates of change of acceleration.
- For a polynomial of degree n, the (n + 1)th derivative is zero.
- If f'' stationary point, the second derivative test fails and you must use a sign table instead.
at the second derivative test is only a test, not a definition. A stationary point where f'(x) = 0 and f'' ould be a minimum, a maximum or a point of inflection depending on the higher derivatives.
记住，二阶导数判别法只是一种判别方法，而不是定义。若某驻点满足 f'(x) = 0 且 f''(x) = 0，它可能是极小值点、极大值点或拐点，取决于更高阶的导数。
6. Stationary Points and Their Nature | 驻点及其性质
izontal. To locate stationary points, differentiate, set the derivative equal to zero, and solve for x. Then substitute back into the original equation of the curve, not into the derivative, to obtain the y-coordinates.
驻点出现在 dy/dx = 0 处，此时曲线切线水平。求驻点的步骤是：求导、令导数等于零、解出 x。然后把 x 代回曲线原方程（而不是导数）求出 y 坐标。
f'(x) = 0 at a stationary point
To classify the point, use the second derivative test. If f''(x) > 0 at the point, it is a local minimum; if f''(x) < 0, it is a local maximum; if f''(x) = 0, the test is inconclusive and you should examine the sign of f'(x) immediately to the left and right of the point.
判定驻点性质时使用二阶导数判别法。若该点处 f''(x) > 0，则为局部极小值；若 f''(x) < 0，则为局部极大值；若 f''(x) = 0，判别法失效，应检查该点左右两侧 f'(x) 的符号。
| Condition at the point | Nature |
|---|---|
| f'(x) = 0 and f''(x) > 0 | Local minimum |
| f'(x) = 0 and f''(x) < 0 | Local maximum |
| f'(x) = 0, sign of f'(x) unchanged | Stationary point of inflection |
| f''(x) = 0, sign changes | Non-stationary point of inflection |
12, which is −6 at x = 1 and +6 at x = 3, so (1, 4) is a maximum and (3, 0) is a minimum.
例题：对于 y = x³ − 6x² + 9x，dy/dx = 3x² − 12x + 9 = 3(x − 1)(x − 3)，所以 x = 1 与 x = 3。二阶导数为 6x − 12，在 x = 1 处为 −6，在 x = 3 处为 +6，因此 (1, 4) 是极大值点，(3, 0) 是极小值点。
7. Equations of Tangents and Normals | 切线与法线方程
The tangent to a curve at a point has the same gradient as the curve there, while the normal is perpendicular to the tangent at that point. If the gradient of the tangent is m, the gradient of the normal is −1/m, provided m is not zero.
曲线在某点的切线具有与该点处曲线相同的梯度；法线在该点与切线垂直。若切线梯度为 m，则法线梯度为 −1/m（m 不为零时）。
y − y₁ = m(x − x₁) y − y₁ = −(1/m)(x − x₁)
straight-line form and rearrange into the requested form, usually y = mx + c or ax + by + c = 0.
步骤：把 x₁ 代入原方程求出 y₁；求导；把 x₁ 代入导数得到 m；再代入直线方程形式并整理成题目要求的形式，通常为 y = mx + c 或 ax + by + c = 0。
s gradient −1/5, so y − 4 = −(1/5)(x − 1), that is y = −x/5 + 21/5, or x + 5y − 21 = 0 after clearing fractions.
例题：求 y = x² + 3x 在点 (1, 4) 处的切线与法线。因 dy/dx = 2x + 3，故 m = 5。切线为 y − 4 = 5(x − 1)，即 y = 5x − 1。法线梯度为 −1/5，故 y − 4 = −(1/5)(x − 1)，即 y = −x/5 + 21/5，去分母后为 x + 5y − 21 = 0。
- If m = 0, the tangent is horizontal and the normal is the vertical line x = x₁.
- Questions often add a twist: the tangent at a second point is parallel, so set the two gradients equal.
- A tangent passing through the origin means the tangent equation must be satisfied by (0, 0).
Always check whether the question asks for the tangent, the normal, or both. Reading the command word carefully prevents the most avoidable lost marks in this topic.
务必看清题目要求的是切线、法线还是两者都要。仔细阅读指令词，可以避免本主题中最不该丢的分数。
8. Increasing and Decreasing Functions | 增函数与减函数
A function is increasing on an interval where f'(x) ≥ 0 and decreasing where f'(x) ≤ 0. Strictly increasing means f' gives a direct method for solving inequality questions about where a curve rises or falls.
函数在 f'(x) ≥ 0 的区间上递增，在 f'(x) ≤ 0 的区间上递减。严格递增指在该区间上始终有 f'(x) > 0。这为求解曲线上升或下降区间的不等式问题提供了直接方法。
Example: show that f(x) = x³ + 3x² + 6x is increasing for all real x. Here f' g the square is therefore a powerful technique for proving a quadratic is always positive.
例题：证明 f(x) = x³ + 3x² + 6x 对所有实数 x 递增。因 f'(x) = 3x² + 6x + 6 = 3[(x + 1)² + 1]，而 (x + 1)² ≥ 0，故导数至少为 3，对任意 x 均为正。由此可见，配方法是证明二次式恒正的有力技巧。
- Sketch a sign diagram of f'(x) to find the intervals of increase and decrease.
- Use open or closed intervals carefully: Edexcel accepts either, but be consistent within your answer.
- For trigonometric functions, restrict the solution to the given domain in radians or degrees.
If a curve is increasing, its tangent has positive gradient and it points uphill as x increases. Translating the algebra back into this visual picture is excellent preparation for curve-sketching questions.
若曲线递增，其切线梯度为正，随着 x 增大曲线向上延伸。把代数结论还原为这一图像，对作答曲线作图题极有帮助。
9. Rates of Change and Connected Rates | 变化率与相关变化率
If a quantity y depends on time, dy/dt is its rate of change with respect to time. When two quantities are linked by an equation, the chain rule connects their rates, which is exactly the idea behind connected rates of change questions.
若量 y 随时间变化，则 dy/dt 是它关于时间的变化率。当两个量由某个方程相联系时，链式法则把它们的变率联系起来，这正是相关变化率问题的核心思想。
dy/dt = dy/dx × dx/dt
V/dt = 4πr² × dr/dt = 4π(100)(0.5) = 200π cm³ s⁻¹, and you should state the units clearly.
典型题目：球形气球的体积 V = (4/3)πr³，半径以 0.5 cm s⁻¹ 增大。求 r = 10 cm 时的 dV/dt。因 dV/dr = 4πr²，故 dV/dt = 4πr² × dr/dt = 4π(100)(0.5) = 200π cm³ s⁻¹，并应清楚写出单位。
- Write down every quantity you know and the one you want before starting.
- Differentiate the given relation with respect to the linking variable, not directly with respect to time.
- Substitute numerical values only at the end, to reduce rounding errors.
l geometric relations such as this are frequently used, so know the area and volume formulae for circles, spheres, cylinders and cones.
对于边长为 x 的膨胀正方形，面积 A = x² 给出 dA/dt = 2x dx/dt。这类小几何关系经常被使用，因此请熟记圆、球、圆柱和圆锥的面积与体积公式。
10. Optimisation Problems | 最优化问题
Optimisation questions ask for the maximum or minimum value of a quantity such as area, volume, cost or surface area. The method is always the same: form an expression in one variable, differentiate, set the derivative to zero, solve, and justify that the answer is a maximum or a minimum.
最优化问题要求求出面积、体积、成本或表面积等量的最大值或最小值。方法始终相同：建立只含一个变量的表达式，求导，令导数等于零，求解，并论证所得结果是最大值还是最小值。
hen x³ = 1000, so x = 10 cm and A = 100 + 200 = 300 cm². The second derivative 2 + 4000/x³ is positive at x = 10, confirming a minimum.
例题：一个无盖方底盒的底边长为 x cm，体积为 500 cm³。表面积 A = x² + 2000/x。则 dA/dx = 2x − 2000/x²，当 x³ = 1000 时为零，故 x = 10 cm，A = 100 + 200 = 300 cm²。二阶导数 2 + 4000/x³ 在 x = 10 处为正，确认为最小值。
- Always state the domain: lengths must be positive, and some constraints impose an upper bound.
- Edexcel awards a mark for the justification step, so never omit it.
- Read whether the question says 'prove that' or 'show that' - both demand a full argument, not just a result.
