📚 pdfjoiner_(4)-094 | A-Level Edexcel 数学微分精讲：从基础到应用
Welcome to this focused revision article on differentiation, a cornerstone of A-Level Edexcel Mathematics. Covering everything from the power rule to parametric second derivatives and real‑world optimisation, the techniques explained here will help you tackle both Pure and Applied questions with confidence. Each section pairs concise English explanations with equivalent Chinese paragraphs, ensuring you master concepts and terminology in both languages.
欢迎阅读这篇针对A-Level Edexcel数学微分专题的精讲复习文章。微分是Edexcel数学的核心之一，从最基本的幂法则到参数方程二阶导数以及实际优化问题，本文都将逐一剖析。每个知识点均采用英文说明与中文解释一一对应的方式，帮助你在掌握解题方法的同时熟悉双语表达。
1. The Power Rule and Basic Differentiation | 幂法则与基本求导
orms the bedrock of polynomial differentiation and extends naturally to negative and fractional powers.
对于形如 y = xⁿ 的函数，其中 n 为实常数，导数公式为 dy/dx = n xⁿ⁻¹。这一法则是多项式求导的基石，并可自然推广到负指数和分数指数。
hen dy/dx = 12x³ − 10x + 2. Constants vanish because the derivative of a constant is zero.
对多项式逐项求导即可。例如 y = 3x⁴ − 5x² + 2x − 7，则 dy/dx = 12x³ − 10x + 2。常数项导数为零，因此直接消失。
Remember to rewrite roots as fractional powers before differentiating: √x becomes x½, and 1/x² becomes x⁻². This makes it straightforward to apply the power rule every time.
求导前务必将根式改写为分数指数形式：√x 写作 x½，1/x² 写作 x⁻²。这样就能始终套用幂法则，避免出错。
2. The Chain Rule | 链式法则
e composite functions, i.e. functions of the form y = f(g(x)). The derivative is dy/dx = f'(g(x)) × g' leave the inner function unchanged, then multiply by the derivative of the inner function.
链式法则用于求复合函数的导数，即形如 y = f(g(x)) 的函数。其公式为 dy/dx = f'(g(x)) × g'(x)。通俗地说：先对外层函数求导并保留内层函数，再乘以内层函数的导数。
2x³ + 5. Then y = u⁴, dy/du = 4u³, du/dx = 6x², so dy/dx = 4(2x³+5)³ × 6x² = 24x²(2x³+5)³.
例如 y = (2x³ + 5)⁴，令 u = 2x³ + 5，则 y = u⁴，dy/du = 4u³，du/dx = 6x²，因此 dy/dx = 4(2x³+5)³ × 6x² = 24x²(2x³+5)³。
Chain rule is also vital when differentiating eᵏˣ, ln(kx), sin(kx) and similar functions. For y = e³ˣ, dy/dx = 3e³ˣ; for y = ln(5x), dy/dx = 1/x.
链式法则在求 eᵏˣ、ln(kx)、sin(kx) 等函数的导数时同样至关重要。如 y = e³ˣ，则 dy/dx = 3e³ˣ；对 y = ln(5x)，得 dy/dx = 1/x。
3. The Product and Quotient Rules | 积法则与商法则
lied, use the product rule: if y = uv, then dy/dx = u dv/dx + v du/dx. A handy mnemonic is "first times derivative of second plus second times derivative of first".
两函数相乘时，使用积法则：若 y = uv，则 dy/dx = u dv/dx + v du/dx。记忆口诀：“前导后不导加上后导前不导”。
n x gives u = x², v = sin x, du/dx = 2x, dv/dx = cos x. Hence dy/dx = x² cos x + 2x sin x.
例如 y = x² sin x，令 u = x²，v = sin x，则 du/dx = 2x，dv/dx = cos x，故 dy/dx = x² cos x + 2x sin x。
functions: y = u/v leads to dy/dx = (v du/dx − u dv/dx) / v². A common way to remember is "low d high minus high d low, over low squared".
两函数相除时使用商法则：y = u/v 的导数为 dy/dx = (v du/dx − u dv/dx) / v²。记忆口诀：“下导上不导减上导下不导，除以下平方”。
For y = eˣ / x, set u = eˣ, v = x. Then dy/dx = (x·eˣ − eˣ·1) / x² = eˣ(x − 1)/x².
对于 y = eˣ / x，取 u = eˣ，v = x，则 dy/dx = (x·eˣ − eˣ·1) / x² = eˣ(x − 1)/x²。
4. Differentiating Exponential and Logarithmic Functions | 指数与对数函数求导
The natural exponential function eˣ is unique because it is its own derivative: d/dx eˣ = eˣ. More generally, d/dx eᵏˣ = k eᵏˣ, which follows directly from the chain rule.
自然指数函数 eˣ 的独特之处在于其导数等于自身：d/dx eˣ = eˣ。更一般地，d/dx eᵏˣ = k eᵏˣ，这可由链式法则直接得到。
l logarithm, d/dx ln x = 1/x for x > 0. When the argument is a function, d/dx ln(f(x)) = f'(x)/f(x). This is a highly tested result in Edexcel exams.
对自然对数而言，d/dx ln x = 1/x（x > 0）。当自变量为函数时，d/dx ln(f(x)) = f'(x)/f(x)。这是Edexcel考试中高频出现的结论。
If y = aˣ, where a > 0, we rewrite using e: aˣ = eˣ ln a, so dy/dx = aˣ ln a. Similarly, for y = logₐ x, the derivative is 1/(x ln a).
若 y = aˣ（a > 0），可借助 e 改写：aˣ = eˣ ln a，因此 dy/dx = aˣ ln a。类似地，y = logₐ x 的导数为 1/(x ln a)。
with the chain rule allows differentiation of complicated products and powers, e.g. y = xˣ. Take ln both sides: ln y = x ln x, then differentiate implicitly.
将对数求导法与链式法则结合，可以处理复杂乘积和幂指函数，例如 y = xˣ。两边取对数：ln y = x ln x，再隐函数求导即可。
5. Differentiating Trigonometric Functions | 三角函数的微分
sin x = cos x, d/dx cos x = −sin x, and d/dx tan x = sec² x. These assume x is in radians.
三角函数的基本导数公式必须熟记：d/dx sin x = cos x，d/dx cos x = −sin x，d/dx tan x = sec² x。以上公式均以弧度制为前提。
x, and d/dx cot x = −cosec² x. These can be derived using the quotient rule or chain rule.
倒数三角函数的导数：d/dx sec x = sec x tan x，d/dx cosec x = −cosec x cot x，d/dx cot x = −cosec² x。这些公式可利用商法则或链式法则推导得出。
in(ax). Typical Edexcel questions combine these with exponential or logarithmic functions.
当角度为 x 的倍数时，结合链式法则：d/dx sin(ax) = a cos(ax)，d/dx cos(ax) = −a sin(ax)。Edexcel 试题常将此类导数与指数或对数函数组合考查。
6. Implicit Differentiation | 隐函数微分
ith respect to x, treating y as a function of x and applying the chain rule: d/dx f(y) = f'(y) dy/dx.
当 y 并未显式表达为 x 的函数时，需要使用隐函数微分。对每一项关于 x 求导，将 y 视为 x 的函数并运用链式法则：d/dx f(y) = f'(y) dy/dx。
0, so dy/dx = −x/y. This expression can then be used to find gradients at specific points.
对于圆 x² + y² = 25，两边求导得 2x + 2y dy/dx = 0，故 dy/dx = −x/y。利用此式便可求出曲线上任意指定点的切线斜率。
8, differentiating yields 3x² + 6xy + 3x² dy/dx + 3y² dy/dx = 0, then collect dy/dx terms.
更复杂的隐函数问题会涉及 x²y 等乘积项或 sin(y) 等函数。例如 x³ + 3x²y + y³ = 8 求导得 3x² + 6xy + 3x² dy/dx + 3y² dy/dx = 0，然后合并 dy/dx 项。
Edexcel often asks for the equation of a tangent or normal to an implicitly defined curve, so you must be comfortable finding dy/dx and evaluating it at a given point.
Edexcel 常要求写出隐函数曲线在某点的切线或法线方程，因此务必能熟练求出 dy/dx 并代入点的坐标计算斜率。
7. Parametric Differentiation | 参数方程求导
dt) / (dx/dt), provided dx/dt ≠ 0. This converts the rate of change with respect to t into the rate with respect to x.
当曲线用参数方程 x = f(t)、y = g(t) 表示时，导数 dy/dx 由 dy/dx = (dy/dt) / (dx/dt) 得出，前提是 dx/dt ≠ 0。这样便将关于 t 的变化率转化为关于 x 的变化率。
o dy/dx = (3t² − 3)/(2t). To find the gradient at a point, determine the value of t first.
例如 x = t² + 1，y = t³ − 3t，则 dx/dt = 2t，dy/dt = 3t² − 3，因此 dy/dx = (3t² − 3)/(2t)。求某点梯度时，须先确定对应的 t 值。
x) / (dx/dt). This often appears in questions testing concavity and the nature of stationary points on parametric curves.
关于 x 的二阶导数由 d²y/dx² = d/dt(dy/dx) / (dx/dt) 求得。这类计算常见于考察参数曲线凹凸性和驻点性质的问题中。
8. Second Derivatives and Concavity | 二阶导数与凹凸性
The second derivative, d²y/dx², measures the rate of change of the gradient. It tells us how the slope of a curve is changing and is central to classifying stationary points and determining concavity.
二阶导数 d²y/dx² 衡量的是梯度的变化率。它反映曲线斜率如何变化，是判定驻点性质和确定曲线凹凸性的核心工具。
If d²y/dx² > 0 over an interval, the curve is convex (concave up) and any stationary point there is a local minimum. If d²y/dx² < 0, the curve is concave (concave down) and a stationary point is a local maximum.
若某个区间上 d²y/dx² > 0，曲线为凸（开口向上），该区间内的驻点为局部极小值点；若 d²y/dx² < 0，曲线为凹（开口向下），驻点为局部极大值点。
ed are potential points of inflection, provided the concavity changes sign. However, you must check there is an actual sign change — sometimes d²y/dx² can be zero without an inflection.
当 d²y/dx² = 0 或无定义，且凹凸性在两侧改变时，该点可能为拐点。但必须验证符号确实发生改变，因为二阶导数为零并不一定对应拐点。
9. Stationary Points and Curve Sketching | 驻点与曲线草图
axima, local minima or points of inflection using either the second derivative test or the first derivative sign test.
驻点出现在 dy/dx = 0 处。可借助二阶导数检验法或一阶导数符号检验法，将驻点划分为局部极大值、局部极小值或拐点。
determine their nature. Also examine the behaviour as x → ±∞ and identify any asymptotes.
绘制曲线草图时，先求出与坐标轴的交点，再确定驻点位置及性质，同时考察 x → ±∞ 的极限行为，并找出渐近线。
For rational functions, consider the denominator to find vertical asymptotes. For oblique asymptotes, polynomial long division may be needed. Combine all this information to produce a labelled sketch.
对于有理函数，通过分母为零寻找铅直渐近线；斜渐近线可能需要进行多项式长除法。汇总所有信息后，即可画出标注清晰的函数草图。
Edexcel often embeds stationary point analysis in modelling questions, where you must interpret the practical meaning of a maximum or minimum within context.
Edexcel 常在建模题中融入驻点分析，要求你根据实际情境解释极大值或极小值的含义。
10. Optimisation Problems | 优化问题
sed as a function of one variable using the constraint, differentiate, set dy/dx = 0, and verify the nature of the stationary point.
优化问题要求在给定约束条件下求某个量的最大值或最小值。关键步骤为：利用约束条件将该量表示为单变量函数，求导并令 dy/dx = 0，最后验证驻点性质。
ferentiate, set dA/dx = 0, solve for x, and confirm a maximum using the second derivative.
经典的例子是固定周长下最大化矩形面积。设边长为 x 和 y，2x + 2y = P，面积 A = xy = x(P/2 − x)。求导，令 dA/dx = 0，解出 x，并用二阶导数确认极大值。
Three-dimensional problems, such as maximising the volume of a box cut from a sheet by removing squares from corners, require careful algebraic formulation. Always check that the solution lies within the feasible domain.
立体几何问题——例如从矩形板材四角裁去正方形以制作无盖盒并最大化体积——需要细致地建立代数模型。一定要检验解是否落在合理的取值范围内。
In Edexcel exams, you should explicitly justify that a stationary point is indeed a maximum (or minimum) using d²y/dx² or a sign change table. Providing a concise conclusion with units is essential for full marks.
在Edexcel考试中，必须明确使用二阶导数或符号变化表说明所得驻点确为极大（或极小）值。给出带有单位的简要结论是获得满分的必要条件。
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