📚 Mastering Differentiation for Edexcel A-Level Pure Mathematics | 精通爱德思 A-Level 纯数学：微分
Differentiation is one of the most important pillars of Edexcel A-Level Pure Mathematics. It appears in Paper 1 and Paper 2, and it is essential for curve sketching, optimisation, kinematics and connected rates of change. This article provides a complete revision guide covering the key rules, standard results and common applications you need to master for the exam.
微分是爱德思 A-Level 纯数学最重要的支柱之一。它出现在试卷一和试卷二中，并且对于曲线草图、最优化、运动学以及相关变化率都至关重要。本文提供一份完整的复习指南，涵盖考试中必须掌握的关键法则、标准结论和常见应用。
1. The Derivative as a Limit | 作为极限的导数
The derivative of a function f(x) at a point x = a is defined as the limit of the average rate of change as the interval tends to zero. This idea is written formally as f'(a) = lim(h → 0) [f(a + h) – f(a)] / h.
函数 f(x) 在点 x = a 处的导数定义为当区间趋近于零时平均变化率的极限。这一概念形式化地写作 f'(a) = lim(h → 0) [f(a + h) – f(a)] / h。
In practice, you do not need to evaluate this limit every time, because the standard differentiation rules give the same result much faster. However, Edexcel may ask you to use the limit definition for a simple polynomial or to interpret the derivative as the gradient of the tangent.
实际上，你不需要每次都计算这个极限，因为标准微分法则可以更快地给出同样结果。不过，爱德思考试可能会要求你对简单多项式使用极限定义，或者将导数解释为切线的斜率。
dy/dx = lim(h → 0) [f(x + h) – f(x)] / h
2. Basic Differentiation Rules | 基本微分法则
The power rule is the foundation of polynomial differentiation. If f(x) = xⁿ, where n is any real number, then the derivative is f'(x) = n xⁿ⁻¹. This rule also works for negative and fractional powers, so expressions such as 1/x² and √x can be differentiated directly after rewriting them in index form.
幂函数法则是多项式微分的基础。如果 f(x) = xⁿ，其中 n 是任意实数，那么导数就是 f'(x) = n xⁿ⁻¹。这一法则同样适用于负指数和分数指数，因此像 1/x² 和 √x 这样的表达式在改写为指数形式后可以直接求导。
The derivative of a constant is zero, and differentiation is linear. This means d/dx [a f(x) + b g(x)] = a f'(x) + b g'(x), where a and b are constants.
常数的导数为零，并且微分是线性的。这意味着 d/dx [a f(x) + b g(x)] = a f'(x) + b g'(x)，其中 a 和 b 为常数。
- f(x) = x⁵ → f'(x) = 5x⁴
- f(x) = 3x² – 4x + 7 → f'(x) = 6x – 4
- f(x) = 4/x³ = 4x⁻³ → f'(x) = -12x⁻⁴
3. Chain Rule | 链式法则
The chain rule is used to differentiate composite functions, where one function is applied inside another. If y = f(u) and u = g(x), then dy/dx = dy/du × du/dx. This is one of the most frequently tested techniques in Edexcel A-Level Mathematics.
链式法则用于对复合函数求导，也就是一个函数嵌套在另一个函数内部的情况。如果 y = f(u) 且 u = g(x)，那么 dy/dx = dy/du × du/dx。这是爱德思 A-Level 数学中最常考查的技巧之一。
A common form is d/dx [f(g(x))] = f'(g(x)) g'(x). For example, to differentiate y = (3x² + 5)⁴, set u = 3x² + 5, so y = u⁴. Then dy/du = 4u³ and du/dx = 6x, giving dy/dx = 24x(3x² + 5)³.
一种常见形式是 d/dx [f(g(x))] = f'(g(x)) g'(x)。例如，要对 y = (3x² + 5)⁴ 求导，令 u = 3x² + 5，则 y = u⁴。于是 dy/du = 4u³，du/dx = 6x，得到 dy/dx = 24x(3x² + 5)³。
dy/dx = (dy/du) × (du/dx)
4. Product and Quotient Rules | 乘法法则与除法法则
When a function is the product of two simpler functions, use the product rule: d/dx (u v) = u dv/dx + v du/dx. For y = x² sin x, let u = x² and v = sin x, so dy/dx = x² cos x + 2x sin x.
当一个函数是两个较简单函数的乘积时，使用乘法法则：d/dx (u v) = u dv/dx + v du/dx。对于 y = x² sin x，令 u = x²，v = sin x，则 dy/dx = x² cos x + 2x sin x。
For division, the quotient rule states d/dx (u/v) = [v du/dx – u dv/dx] / v². For y = x / (x + 1), this gives dy/dx = [(x + 1)(1) – x(1)] / (x + 1)² = 1 / (x + 1)².
对于除法，商法则为 d/dx (u/v) = [v du/dx – u dv/dx] / v²。对于 y = x / (x + 1)，得到 dy/dx = [(x + 1)(1) – x(1)] / (x + 1)² = 1 / (x + 1)²。
It is important to choose u and v carefully. The quotient rule should not be used if the denominator is a constant, because simple linearity is faster.
仔细选择 u 和 v 非常重要。如果分母是常数，则不应使用商法则，因为直接使用线性性质会更快捷。
5. Differentiating Standard Functions | 常见函数求导
You must memorise the derivatives of standard functions. These results are assumed in Edexcel exams and are often combined with the chain rule, product rule or quotient rule.
你必须熟记常见函数的导数。这些结论在爱德思考试中是默认已知的，并且经常与链式法则、乘法法则或商法则结合使用。
| f(x) | f'(x) |
| eˣ | eˣ |
| ln x | 1/x |
| sin x | cos x |
| cos x | -sin x |
| tan x | sec² x |
For exponential functions with other bases, d/dx (aˣ) = aˣ ln a. For trigonometric functions, angles must be in radians unless the question states otherwise.
对于其他底数的指数函数，d/dx (aˣ) = aˣ ln a。对于三角函数，除非题目另有说明，角度必须以弧度为单位。
6. Implicit Differentiation | 隐函数求导
Implicit differentiation is used when the relationship between x and y is given in a form that is not easily solved for y. Differentiate every term with respect to x, and multiply by dy/dx every time you differentiate a function of y.
当 x 与 y 之间的关系不容易写成 y = f(x) 的形式时，使用隐函数求导。对每一项关于 x 求导，并且每次对 y 的函数求导时都要乘以 dy/dx。
For example, if x² + y² = 25, differentiating gives 2x + 2y dy/dx = 0, so dy/dx = -x/y. This method is essential for finding tangents and normals to circles and other implicit curves.
例如，若 x² + y² = 25，求导得到 2x + 2y dy/dx = 0，因此 dy/dx = -x/y。这种方法对于求圆和其他隐式曲线的切线与法线至关重要。
d/dx (y²) = 2y dy/dx
7. Parametric Differentiation | 参数方程求导
When x and y are both given as functions of a parameter t, the derivative dy/dx is found by dividing dy/dt by dx/dt, provided dx/dt is not zero.
当 x 和 y 都作为参数 t 的函数给出时，导数 dy/dx 等于 dy/dt 除以 dx/dt，前提是 dx/dt 不为零。
If x = t² + 1 and y = 2t³ – t, then dx/dt = 2t and dy/dt = 6t² – 1. Therefore dy/dx = (6t² – 1) / (2t). This technique is often used for finding stationary points on parametric curves.
如果 x = t² + 1，y = 2t³ – t，则 dx/dt = 2t，dy/dt = 6t² – 1。因此 dy/dx = (6t² – 1) / (2t)。这种技巧常用于求参数曲线上的驻点。
The second derivative d²y/dx² is obtained from d/dt (dy/dx) divided by dx/dt. Do not simply divide d²y/dt² by d²x/dt².
二阶导数 d²y/dx² 等于 d/dt (dy/dx) 除以 dx/dt。不要简单地将 d²y/dt² 除以 d²x/dt²。
8. Second Derivative and Concavity | 二阶导数与凹凸性
The second derivative f”(x) measures the rate of change of the gradient. If f”(x) > 0, the curve is concave up, and if f”(x) < 0, the curve is concave down. Points where f''(x) changes sign are points of inflection.
二阶导数 f”(x) 衡量斜率的变化率。如果 f”(x) > 0，曲线向上凹；如果 f”(x) < 0，曲线向下凹。f''(x) 改变符号的点是拐点。
To classify a stationary point using the second derivative, find f”(a) where f'(a) = 0. If f”(a) > 0, the point is a local minimum; if f”(a) < 0, it is a local maximum. If f''(a) = 0, the test is inconclusive and you should use the gradient-change test.
要使用二阶导数对驻点进行分类，在 f'(a) = 0 处求 f”(a)。如果 f”(a) > 0，该点是局部极小值；如果 f”(a) < 0，该点是局部极大值。如果 f''(a) = 0，该检验无法得出结论，应使用梯度变化检验。
9. Stationary Points and Curve Sketching | 驻点与曲线草图
Stationary points occur where the first derivative is zero, that is f'(x) = 0. These points can be maxima, minima or points of inflection. To sketch a curve, you should find intercepts, stationary points, asymptotes and the behaviour as x tends to positive or negative infinity.
驻点出现在一阶导数为零的位置，即 f'(x) = 0。这些点可以是极大值点、极小值点或拐点。要绘制曲线草图，应找出截距、驻点、渐近线以及 x 趋向正无穷或负无穷时的行为。
For example, f(x) = x³ – 3x has f'(x) = 3x² – 3 = 3(x – 1)(x + 1). The stationary points are at x = 1 and x = -1. Since f”(x) = 6x, x = -1 gives a local maximum and x = 1 gives a local minimum.
例如，f(x) = x³ – 3x 的导数为 f'(x) = 3x² – 3 = 3(x – 1)(x + 1)。驻点在 x = 1 和 x = -1。因为 f”(x) = 6x，所以 x = -1 处是局部极大值，x = 1 处是局部极小值。
When sketching, label all key features clearly. Edexcel mark schemes reward the correct shape, coordinates of stationary points and axis intercepts.
绘制草图时，要清楚标注所有关键特征。爱德思评分标准会奖励正确的形状、驻点坐标和坐标轴截距。
10. Connected Rates of Change and Optimisation | 相关变化率与最优化
Connected rates of change problems involve two or more variables that change with time. The chain rule links these rates: dV/dt = dV/dr × dr/dt, where V is volume and r is radius. You must identify the quantities, state the relation, and substitute the given numerical information.
相关变化率问题涉及两个或两个以上随时间变化的变量。链式法则将这些变化率联系起来：dV/dt = dV/dr × dr/dt，其中 V 是体积，r 是半径。你必须识别各个量，说明关系式，并代入已知数值。
Optimisation problems ask you to find the maximum or minimum value of a quantity such as area, volume, cost or distance. First express the quantity as a function of one variable, then differentiate, set the derivative equal to zero, and justify that the stationary point is a maximum or minimum.
最优化问题要求你求出面积、体积、成本或距离等量的最大值或最小值。首先将该量表示为单个变量的函数，然后求导，令导数等于零，并证明该驻点是极大值或极小值。
For example, a rectangular box with a square base of side x and height h has volume V = x²h. If the surface area is fixed, you can eliminate one variable and differentiate V with respect to x to find the dimensions that maximise the volume.
例如，一个底面为边长 x 的正方形、高为 h 的长方体，其体积为 V = x²h。如果表面积固定，你可以消去一个变量，并对 V 关于 x 求导，以求出使体积最大化的尺寸。
dV/dt = (dV/dx) × (dx/dt)
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