📚 Differentiation Techniques and Applications | 微分技巧与应用
Differentiation is one of the central pillars of Edexcel A-Level Mathematics. It measures how a function changes at any instant and underpins curve sketching, optimisation, kinematics and mathematical modelling. This revision guide walks through the key rules, the standard derivatives you must know by heart, and the applications that examiners test most often.
微分是爱德思 A-Level 数学的核心支柱之一。它用于度量函数在任意瞬间的变化情况，是曲线作图、优化问题、运动学与数学建模的基础。本复习指南将系统梳理关键求导法则、必须熟记的标准导数公式，以及考试中最常出现的应用题型。
1. The Gradient of a Curve and First Principles | 曲线的斜率与第一原理
The derivative of a function f at a point x is defined as the limit of the average gradient between two nearby points as the distance between them tends to zero. Geometrically this is the gradient of the tangent to the curve at that point, and physically it is the instantaneous rate of change.
函数 f 在点 x 处的导数，定义为两点间平均斜率在两点距离趋于零时的极限。从几何上看，它是曲线在该点处切线的斜率；从物理上看，它是瞬时变化率。
f'(x) = lim(h→0) [f(x + h) − f(x)] / h
h², subtract x², divide by h to obtain 2x + h, and then let h → 0 to reach the result 2x.
在爱德思考试中，可能会要求你对简单多项式（如 f(x) = x²）完成这一极限过程。先展开 (x + h)² = x² + 2xh + h²，减去 x²，再除以 h 得到 2x + h，最后令 h → 0，即得结果 2x。
This formal definition underpins every differentiation rule. Even when exam questions focus on applying rules, understanding first principles helps you justify the methods, handle unfamiliar functions, and answer proof-style questions where a rule must be established from scratch.
这一形式定义是所有求导法则的基础。即便考试题目侧重于应用法则，理解第一原理也有助于你说明方法的合理性、处理陌生函数，并完成从零推导某条法则的证明题。
2. Basic Differentiation Rules: Power, Constant Multiple and Sum | 基本求导法则：幂函数、常数倍与加法法则
fractional powers and the special case n = 0, where the derivative of a constant is zero.
对任意实数常量 n，幂函数求导法则为：xⁿ 的导数是 n xⁿ⁻¹。这一法则涵盖了正幂、负幂、分数幂，以及 n = 0 的特殊情形，即常数的导数为零。
If y = xⁿ, then dy/dx = n xⁿ⁻¹
The constant multiple rule and the sum rule allow us to differentiate linear combinations term by term, which is why polynomial differentiation is so quick once the power rule is secure.
常数倍法则与加法法则允许我们逐项求导线性组合，这也是掌握幂函数法则后多项式求导如此迅速的原因。
s because the gradient of a horizontal line is zero, and each term is differentiated independently of the others.
例如，若 y = 5x³ − 2x + 7，则 dy/dx = 15x² − 2。常数项 7 消失，因为水平直线的斜率为零；每一项都独立于其他项求导。
- Constant multiple rule: d/dx [k f(x)] = k f'(x) — 常数倍法则：常数可以提到导数符号外面。
- Sum and difference rule: d/dx [f(x) ± g(x)] = f'(x) ± g'(x) — 加减法则：逐项求导后相加减。
- Constant rule: d/dx (c) = 0 — 常数法则：常数的导数为零。
3. Rewriting Expressions Before Differentiating | 先化简再求导
Many exam expressions cannot be differentiated in the form given. Rewriting roots and fractions as powers of x is usually the fastest route, and examiners frequently award a method mark for this step alone.
许多考试表达式无法直接按原形式求导。把根式和分式改写为 x 的幂，通常是最快的途径，而且考官常常仅对这一步骤就给出方法分。
Useful conversions include √x = x1/2, 1/x = x⁻¹, 1/x² = x⁻², and 1/√x = x−1/2. Once in index form, the power rule applies directly and no quotient rule is needed.
常用的转换包括 √x = x1/2、1/x = x⁻¹、1/x² = x⁻²、1/√x = x−1/2。一旦化为指数形式，幂函数法则即可直接使用，无需商法则。
For instance, y = (2x³ − 5x) / √x can be rewritten as y = 2x5/2 − 5x1/2. Differentiating term by term gives dy/dx = 5x3/2 − (5/2)x−1/2, which may then be rewritten back into surd form if the question demands it.
例如，y = (2x³ − 5x) / √x 可改写为 y = 2x5/2 − 5x1/2。逐项求导得 dy/dx = 5x3/2 − (5/2)x−1/2；若题目要求，还可以再改写回根式形式。
Always check whether a bracket can be expanded first. Expanding (x + 3)(x − 5) into x² − 2x − 15 before differentiating is far safer than attempting a product rule with avoidable algebra.
务必先检查括号能否展开。把 (x + 3)(x − 5) 先展开为 x² − 2x − 15 再求导，比硬套乘积法则安全得多，也能避免不必要的代数错误。
4. Tangents, Normals and the Gradient at a Point | 切线、法线与某点处的斜率
e x-coordinate to find the gradient m, then use the straight-line form y − y₁ = m(x − x₁).
要求曲线在某给定点处的切线，先求导得到 dy/dx，代入该点的横坐标求出斜率 m，再用直线方程 y − y₁ = m(x − x₁)。
is not zero. If the tangent is horizontal, the normal is vertical and has equation x = x₁.
法线与切线垂直，因此其斜率为负倒数 −1/m（要求 m 不为零）。若切线水平，则法线竖直，方程为 x = x₁。
s 8, the point is (2, 0), the tangent is y = 8x − 16, and the normal is y = −(1/8)(x − 2).
典型题目给出 y = x³ − 4x，要求在 x = 2 处的切线与法线。此时 dy/dx = 3x² − 4，斜率是 8，点为 (2, 0)，切线为 y = 8x − 16，法线为 y = −(1/8)(x − 2)。
Remember that the phrase "gradient of the curve" always means dy/dx at that point, and that setting dy/dx equal to a given value is the standard way to locate points with a specified gradient.
牢记"曲线的斜率"始终指该点处的 dy/dx；而令 dy/dx 等于某个给定值，是求指定斜率对应点的标准方法。
5. Stationary Points and the Second Derivative Test | 驻点与二阶导数判别法
nima, or points of inflection, and classifying them correctly is worth marks in almost every calculus question.
驻点出现在 dy/dx = 0 之处。它可能是极大值点、极小值点或拐点，正确判定其性质在几乎所有微积分题目中都占分。
The second derivative test uses d²y/dx², obtained by differentiating dy/dx again. If d²y/dx² > 0 at a stationary point, the curve is concave up and the point is a minimum. If d²y/dx² < 0, the curve is concave down and the point is a maximum.
二阶导数判别法使用 d²y/dx²，即对 dy/dx 再求导一次。若某驻点处 d²y/dx² > 0，曲线开口向上，该点为极小值点；若 d²y/dx² < 0，曲线开口向下，该点为极大值点。
ive, so you must examine the sign of dy/dx on either side of the point. This is the standard fallback and examiners accept a clearly presented sign table as full justification.
当 d²y/dx² = 0 时，判别法失效，必须检查该点两侧 dy/dx 的符号。这是标准备选方法，考官也认可清晰呈现的符号表作为完整论证。
nce d²y/dx² = 6x − 6, the point at x = 0 is a maximum and the point at x = 2 is a minimum.
对 y = x³ − 3x²，dy/dx = 3x² − 6x = 3x(x − 2)，故驻点在 x = 0 与 x = 2。因 d²y/dx² = 6x − 6，x = 0 处为极大值点，x = 2 处为极小值点。
6. Optimisation Problems and Modelling | 优化问题与建模
Optimisation questions ask for the greatest or least value of a quantity such as area, volume, cost or surface area. The method is always the same: define variables, form a single equation in one variable, differentiate, set the derivative to zero, and justify the nature of the result.
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。方法始终一致：定义变量，建立只含一个变量的方程，求导，令导数为零，并论证结果的极值性质。
, so h = 500/x². The surface area is A = x² + 4xh = x² + 2000/x, and dA/dx = 2x − 2000/x².
经典例子是底为正方形、体积为 500 cm³ 的无盖盒子。设底边长为 x、高为 h，则 500 = x²h，故 h = 500/x²。表面积为 A = x² + 4xh = x² + 2000/x，且 dA/dx = 2x − 2000/x²。
ce d²A/dx² = 2 + 4000/x³ > 0 for positive x, this stationary value is indeed a minimum.
令 dA/dx = 0 得 2x³ = 2000，故 x³ = 1000，x = 10 cm。代回得 h = 5 cm。因 x > 0 时 d²A/dx² = 2 + 4000/x³ > 0，该驻点确实为极小值。
Always state units in your final answer and always include a justification sentence. An unsupported maximum or minimum claim typically loses the final accuracy mark in Edexcel mark schemes.
最终答案务必写出单位，并始终附上一句极值性质的论证。在爱德思评分标准中，未经论证的最大值或最小值结论通常会失去最后一个准确度分。
7. The Chain Rule for Composite Functions | 复合函数的链式法则
The chain rule differentiates a function of a function. If y is a function of u and u is a function of x, then the derivative is the product of the two rates of change.
链式法则用于对"函数的函数"求导。若 y 是 u 的函数，而 u 是 x 的函数，则导数等于两个变化率的乘积。
dy/dx = (dy/du) × (du/dx)
e outer power gives 5(3x² + 1)⁴ and the inside derivative is 6x, so dy/dx = 30x(3x² + 1)⁴.
实际操作中：先对外层函数求导、内层保持不变，再乘以内层函数的导数。对 y = (3x² + 1)⁵，外层幂给出 5(3x² + 1)⁴，内层导数为 6x，故 dy/dx = 30x(3x² + 1)⁴。
ln versions of the chain rule appear constantly: d/dx sin(ax + b) = a cos(ax + b), d/dx eax + b = a eax + b, and d/dx ln(ax + b) = a/(ax + b).
链式法则与 sin、cos、e、ln 结合的形式极为常见：d/dx sin(ax + b) = a cos(ax + b)，d/dx eax + b = a eax + b，d/dx ln(ax + b) = a/(ax + b)。
Forgetting the inner derivative is the single most common chain rule error. If your answer to a composite function question looks "too simple", check whether a factor is missing.
漏乘内层导数是链式法则中最常见的错误。如果一道复合函数题的答案看起来"过于简单"，请检查是否漏了某个因式。
8. The Product Rule and the Quotient Rule | 乘积法则与商法则
The product rule handles two functions multiplied together, while the quotient rule handles one function divided by another. Both must be memorised accurately, including the sign in the quotient rule numerator.
乘积法则处理两个函数相乘，商法则处理一个函数除以另一个函数。两者都必须准确记忆，尤其是商法则分子中的符号。
(uv)' = u'v + uv' and (u/v)' = (u'v − uv') / v²
For y = x³ ln x, take u = x³ and v = ln x, so u' = 3x² and v' = 1/x. Then dy/dx = 3x² ln x + x³ × (1/x) = 3x² ln x + x².
对 y = x³ ln x，取 u = x³、v = ln x，则 u' = 3x²、v' = 1/x。于是 dy/dx = 3x² ln x + x³ × (1/x) = 3x² ln x + x²。
− 3)². Notice that simplifying the numerator immediately is part of the expected working.
对 y = (2x + 1)/(x − 3)，由商法则得 dy/dx = [2(x − 3) − (2x + 1)] / (x − 3)² = −7 / (x − 3)²。注意立即化简分子本身就是预期步骤的一部分。
If a quotient can be rewritten as a product using negative indices, the product rule is often quicker than the quotient rule and reduces the risk of sign slips in the numerator.
若商式可以用负指数改写为乘积形式，乘积法则往往比商法则更快，也能降低分子符号出错的风险。
9. Exponential and Logarithmic Derivatives | 指数函数与对数函数的导数
The exponential function ex is its own derivative, which is why it models growth and decay so naturally. The natural logarithm has derivative 1/x, valid for x > 0.
指数函数 ex 的导数就是它自身，这正是它能如此自然地描述增长与衰减的原因。自然对数的导数为 1/x，在 x > 0 时成立。
d/dx (eˣ) = eˣ d/dx (ln x) = 1/x d/dx (aˣ) = aˣ ln a
The result for ln x follows from first principles. If y = ln x then x = ey, so dx/dy = ey technique of differentiating an inverse relationship is a standard exam-style derivation.
ln x 的结论可由第一原理推出。若 y = ln x，则 x = ey，故 dx/dy = ey = x，从而 dy/dx = 1/x。这种对反函数关系求导的技巧，是标准的考试推导题型。
Combined with the chain rule, these results extend to d/dx e^(f(x)) = f'(x) e^(f(x)) and d/dx ln(f(x)) = f' especially useful when a function must first be logged to simplify a product or quotient.
结合链式法则，这些结论可推广为 d/dx e^(f(x)) = f'(x) e^(f(x)) 以及 d/dx ln(f(x)) = f'(x)/f(x)。后一种形式在需要先取对数以化简乘积或商时特别有用。
o gradients of exponential curves or to the stationary points of functions such as y = x e−x, which has a single maximum at x = 1.
相关题目常把这些导数与指数曲线的斜率，或与 y = x e−x 这类函数的驻点联系起来；后者在 x = 1 处取得唯一极大值。
10. Trigonometric Derivatives and Second Derivatives | 三角函数导数与二阶导数
The standard trigonometric derivatives must be quoted exactly. In calculus the angle must always be measured in radians, and using degrees is a recognised source of lost marks.
标准三角函数导数必须准确引用。在微积分中角度始终以弧度为单位，使用角度制是会失分的常见原因。
x) = cos x d/dx (cos x) = −sin x d/dx (tan x) = sec²x = 1/cos²x
Note the negative sign in the cosine derivative; omitting it produces a sign error that propagates through the rest of the solution. For tan x the result is often quoted as 1/cos²x, which is easier to apply in chain rule questions.
注意余弦导数中的负号；遗漏它会引发符号错误，并影响后续全部计算。tan x 的结论通常写作 1/cos²x，在链式法则题目中更便于使用。
. This cyclic behaviour makes it easy to identify maxima and minima of oscillating models.
二阶导数 d²y/dx² 就是 dy/dx 的导数。对三角函数而言它呈循环：sin x → cos x → −sin x → −cos x。这种循环特性让我们很容易判断振荡模型的极大值与极小值。
Higher derivatives are written d³y/dx³ and so on, and occasionally appear in questions about the nature of a stationary point where the second derivative vanishes.
更高阶导数写作 d³y/dx³ 等，偶尔出现在二阶导数为零、需要进一步判断驻点性质的题目中。
11. Kinematics and Connected Rates of Change | 运动学与相关变化率
locity into acceleration. If displacement is s(t), then v = ds/dt and a = dv/dt = d²s/dt².
在运动学中，微分把位移转化为速度，把速度转化为加速度。若位移为 s(t)，则 v = ds/dt，a = dv/dt = d²s/dt²。
rticle is instantaneously at rest at t = 1 and t = 3, and the acceleration is a = 6t − 12.
对 s = t³ − 6t² + 9t 求导，得 v = 3t² − 12t + 9 = 3(t − 1)(t − 3)。质点在 t = 1 与 t = 3 时瞬时静止，加速度为 a = 6t − 12。
Maximum or minimum velocity is found by setting the derivative of v to zero, that is by setting the acceleration to zero. This distinction between "maximum displacement", "maximum velocity" and "maximum acceleration" is a frequent source of confusion.
最大或最小速度通过令 v 的导数为零求得，也就是令加速度为零。区分"最大位移""最大速度"与"最大加速度"是常见的混淆点。
ange of radius. Substituting the geometry of the shape, such as V = (4/3)πr³ so that dV/dr = 4πr², completes the solution.
相关变化率把链式法则推广到实际建模：dV/dt = (dV/dr) × (dr/dt) 把体积变化率与半径变化率联系起来。代入几何关系，例如 V = (4/3)πr³、dV/dr = 4πr²，即可完成求解。
12. Exam Technique, Common Errors and Calculator Checks | 考试技巧、常见错误与计算器检查
Most marks in differentiation questions are lost through algebra and notation rather than through misunderstanding of the calculus itself. Writing each step clearly, and never skipping the substitution of a coordinate, protects your accuracy marks.
微分题失分大多源于代数与书写规范，而不是
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