📚 Differentiation Techniques for Edexcel A Level Pure Mathematics | 爱德思A Level纯数学：微分技巧
Differentiation is one of the most important topics in Edexcel A Level Pure Mathematics. It allows you to find the gradient of a curve at any point and to solve real-world problems involving rates of change, optimisation and curve sketching. This article covers the key techniques you need to master for Paper 1 and Paper 2.
微分是爱德思A Level纯数学中最重要的主题之一。它使你能够求出曲线上任意一点的斜率，并解决涉及变化率、优化和曲线绘制的实际问题。本文涵盖你在Paper 1和Paper 2中需要掌握的关键技巧。
You should be confident with algebraic manipulation before studying differentiation, because simplifying an expression before differentiating often makes the process much easier. The methods below are examined regularly, so practise them with past-paper questions.
在学习微分之前，你应当熟练掌握代数变形，因为在求导之前先化简表达式通常会使过程简单得多。以下方法经常在考试中出现，因此请结合历年真题加以练习。
1. Differentiation from First Principles | 从第一性原理求导
The derivative of a function f(x) is defined as the limit of the average rate of change as the interval h tends to zero:
函数 f(x) 的导数定义为当区间 h 趋近于零时平均变化率的极限：
f'(x) = lim(h → 0) [f(x + h) − f(x)] / h
This limit gives the gradient of the tangent to the curve y = f(x) at the point where the x-coordinate is x. Edexcel often asks you to prove the derivative of a simple function, such as x² or x³, from first principles.
这个极限给出了曲线 y = f(x) 在横坐标为 x 的点处切线的斜率。爱德思考试经常要求你从第一性原理证明简单函数（例如 x² 或 x³）的导数。
For example, for f(x) = x², expand f(x + h) = x² + 2xh + h², subtract f(x), divide by h and let h tend to zero. The result is f'(x) = 2x.
例如，对于 f(x) = x²，展开 f(x + h) = x² + 2xh + h²，减去 f(x)，除以 h，再令 h 趋于零。结果是 f'(x) = 2x。
2. Standard Derivatives and the Power Rule | 标准导数与幂法则
The power rule is the most frequently used differentiation rule. For any real constant n, the derivative of xⁿ is n xⁿ⁻¹:
幂法则是最常用的微分法则。对于任意实常数 n，xⁿ 的导数为 n xⁿ⁻¹：
d/dx (xⁿ) = n xⁿ⁻¹
This rule works for positive integers, negative powers and fractional powers. For example, d/dx (x⁻²) = −2x⁻³ and d/dx (√x) = d/dx (x¹ᐟ²) = ½ x⁻¹ᐟ².
该法则适用于正整数幂、负整数幂和分数幂。例如，d/dx (x⁻²) = −2x⁻³，并且 d/dx (√x) = d/dx (x¹ᐟ²) = ½ x⁻¹ᐟ²。
You should memorise the following standard results, as they appear constantly in Edexcel papers:
你应当熟记以下标准结果，因为它们在爱德思试卷中经常出现：
- d/dx (xⁿ) = n xⁿ⁻¹ | d/dx (xⁿ) = n xⁿ⁻¹
- d/dx (eˣ) = eˣ | d/dx (eˣ) = eˣ
- d/dx (ln x) = 1/x | d/dx (ln x) = 1/x
- d/dx (sin x) = cos x | d/dx (sin x) = cos x
- d/dx (cos x) = −sin x | d/dx (cos x) = −sin x
- d/dx (tan x) = sec² x | d/dx (tan x) = sec² x
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:
链式法则用于求复合函数的导数，其中一个函数嵌套在另一个函数内。如果 y = f(u) 且 u = g(x)，那么：
dy/dx = dy/du × du/dx
A common exam example is differentiating y = (3x² + 5)⁴. Let u = 3x² + 5, so y = u⁴. Then dy/du = 4u³ and du/dx = 6x, giving dy/dx = 4(3x² + 5)³ × 6x = 24x(3x² + 5)³.
常见的考试示例是对 y = (3x² + 5)⁴ 进行微分。令 u = 3x² + 5，则 y = u⁴。于是 dy/du = 4u³ 且 du/dx = 6x，因此 dy/dx = 4(3x² + 5)³ × 6x = 24x(3x² + 5)³。
You can also apply the chain rule repeatedly for nested functions such as y = sin(ln x) or y = e^(2x³). Identify the outer function, differentiate it, then multiply by the derivative of the inner function.
你还可以对 y = sin(ln x) 或 y = e^(2x³) 这类嵌套函数反复应用链式法则。先识别外层函数，对其求导，再乘以内层函数的导数。
4. Product Rule | 乘积法则
When differentiating the product of two functions u(x) and v(x), use the product rule:
当对两个函数 u(x) 和 v(x) 的乘积进行微分时，使用乘积法则：
d/dx (uv) = u’v + uv’
Here u’ and v’ mean du/dx and dv/dx. For example, if y = x² sin x, let u = x² and v = sin x. Then u’ = 2x and v’ = cos x, so dy/dx = 2x sin x + x² cos x.
此处 u’ 和 v’ 表示 du/dx 和 dv/dx。例如，如果 y = x² sin x，令 u = x²，v = sin x。则 u’ = 2x，v’ = cos x，所以 dy/dx = 2x sin x + x² cos x。
It is often useful to keep the original factors in the final answer rather than expanding everything, because the factorised form is usually easier to set equal to zero when finding stationary points.
在最终答案中保留原始因式通常比全部展开更有用，因为在求驻点时，因式分解形式通常更容易令其等于零。
5. Quotient Rule | 商法则
For a function written as y = u/v, where u and v are functions of x, the quotient rule states:
对于写作 y = u/v 的函数，其中 u 和 v 都是 x 的函数，商法则指出：
d/dx (u/v) = (u’v − uv’) / v²
For example, differentiate y = x / (x² + 1). Let u = x and v = x² + 1. Then u’ = 1 and v’ = 2x, so dy/dx = [(1)(x² + 1) − (x)(2x)] / (x² + 1)² = (1 − x²) / (x² + 1)².
例如，对 y = x / (x² + 1) 进行微分。令 u = x，v = x² + 1。则 u’ = 1，v’ = 2x，所以 dy/dx = [(1)(x² + 1) − (x)(2x)] / (x² + 1)² = (1 − x²) / (x² + 1)²。
Remember the order: u’v minus uv’, all over v². A common error is to reverse the subtraction, so check your working carefully.
记住顺序：u’v 减去 uv’，再除以 v²。常见的错误是减法顺序颠倒，因此请仔细检查运算过程。
6. Differentiating Trigonometric, Exponential and Logarithmic Functions | 三角、指数与对数函数的微分
Edexcel requires you to know the derivatives of the standard trigonometric, exponential and logarithmic functions. The key results are:
爱德思要求你掌握标准三角函数、指数函数和对数函数的导数。关键结果如下：
d/dx (sin x) = cos x, d/dx (cos x) = −sin x, d/dx (tan x) = sec² x
d/dx (eˣ) = eˣ, d/dx (ln x) = 1/x, d/dx (aˣ) = aˣ ln a
These can be combined with the chain rule. For example, d/dx (sin 3x) = 3 cos 3x and d/dx (e^(2x)) = 2 e^(2x).
这些公式可以与链式法则结合使用。例如，d/dx (sin 3x) = 3 cos 3x，并且 d/dx (e^(2x)) = 2 e^(2x)。
For y = ln(5x), apply the chain rule or use the logarithm law ln(5x) = ln 5 + ln x. Since ln 5 is constant, dy/dx = 1/x. Both approaches give the same result.
对于 y = ln(5x)，可应用链式法则，或使用对数法则 ln(5x) = ln 5 + ln x。由于 ln 5 是常数，dy/dx = 1/x。两种方法结果相同。
7. Implicit Differentiation | 隐函数微分
When a curve is defined by an equation involving x and y but not explicitly as y = f(x), you can use implicit differentiation. Differentiate both sides with respect to x, treating y as a function of x and using the chain rule for any y terms.
当曲线由包含 x 和 y 的方程定义，而不是明确写成 y = f(x) 时，可以使用隐函数微分。对等式两边关于 x 求导，将 y 视为 x 的函数，并对含 y 的项使用链式法则。
For example, the circle x² + y² = 25 gives:
例如，圆 x² + y² = 25 给出：
2x + 2y dy/dx = 0, so dy/dx = −x/y
This technique is essential for finding tangents and normals to curves that cannot easily be rearranged, such as x³ + y³ = 6xy.
对于难以重新整理的曲线（例如 x³ + y³ = 6xy），该技巧对于求切线和法线至关重要。
8. Parametric Differentiation | 参数方程微分
Some curves are described using a parameter t, with x = f(t) and y = g(t). To find dy/dx, differentiate each function with respect to t and divide:
有些曲线使用参数 t 描述，即 x = f(t) 和 y = g(t)。要求 dy/dx，先对每个函数关于 t 求导，然后相除：
dy/dx = (dy/dt) ÷ (dx/dt)
For example, if x = t² and y = t³, then dx/dt = 2t and dy/dt = 3t². Hence dy/dx = 3t² / 2t = 3t / 2, provided t ≠ 0.
例如，如果 x = t² 且 y = t³，则 dx/dt = 2t，dy/dt = 3t²。因此 dy/dx = 3t² / 2t = 3t / 2，前提是 t ≠ 0。
You can also find the second derivative d²y/dx² using d/dx (dy/dx) = d/dt (dy/dx) ÷ dx/dt. This is often tested in Edexcel parametric questions.
你还可以使用 d/dx (dy/dx) = d/dt (dy/dx) ÷ dx/dt 来求二阶导数 d²y/dx²。这经常在爱德思参数方程题目中考查。
9. 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 you whether a curve is concave upwards or concave downwards at a point.
二阶导数 d²y/dx² 衡量斜率的变化率。它告诉你曲线在某点是向上凹还是向下凹。
If d²y/dx² > 0, the curve is convex (or concave upwards). If d²y/dx² < 0, the curve is concave (or concave downwards).
如果 d²y/dx² > 0，曲线是凸的（或向上凹）。如果 d²y/dx² < 0，曲线是凹的（或向下凹）。
For y = x³ − 3x² + 2, dy/dx = 3x² − 6x and d²y/dx² = 6x − 6. The second derivative changes sign at x = 1, which helps identify the nature of any stationary points.
对于 y = x³ − 3x² + 2，dy/dx = 3x² − 6x，d²y/dx² = 6x − 6。二阶导数在 x = 1 处变号，这有助于判断驻点的性质。
10. Stationary Points and Curve Sketching | 驻点与曲线草图
Stationary points occur where dy/dx = 0. There are three main types: local maximum, local minimum and point of inflection. To classify them, you can use the second derivative test or examine the sign of dy/dx on either side.
驻点出现在 dy/dx = 0 处。主要有三种类型：局部最大值、局部最小值和拐点。要判断其类型，可以使用二阶导数判别法，或检查两侧 dy/dx 的符号。
If d²y/dx² > 0 at a stationary point, the point is a local minimum. If d²y/dx² < 0, it is a local maximum. If d²y/dx² = 0, use the first derivative sign test.
如果在驻点处 d²y/dx² > 0，则该点是局部最小值。如果 d²y/dx² < 0，则该点是局部最大值。如果 d²y/dx² = 0，则使用一阶导数符号判别法。
For example, y = x² − 4x + 3 has dy/dx = 2x − 4, so the stationary point occurs at x = 2. Since d²y/dx² = 2 > 0, it is a minimum. The minimum value is y = 4 − 8 + 3 = −1.
例如，y = x² − 4x + 3 的 dy/dx = 2x − 4，所以驻点出现在 x = 2。由于 d²y/dx² = 2 > 0，因此它是极小值。极小值为 y = 4 − 8 + 3 = −1。
11. Optimisation Problems | 优化问题
Optimisation problems ask you to find the maximum or minimum value of a quantity such as area, volume or cost. The general method is:
优化问题要求你求出面积、体积或成本等量的最大值或最小值。一般方法如下：
- Express the quantity to be optimised as a function of one variable. | 将需要优化的量表示为单变量函数。
- Differentiate and set the derivative equal to zero. | 求导并令导数等于零。
- Solve to find stationary points. | 解方程求出驻点。
- Use the second derivative or sign test to confirm maximum or minimum. | 使用二阶导数或符号判别法确认最大值或最小值。
- Answer in the context of the problem, including units where appropriate. | 结合问题背景作答，包括适当的单位。
For example, a rectangle has length x and width y, with perimeter fixed at 20 m, so 2x + 2y = 20, giving y = 10 − x. The area A = xy = x(10 − x) = 10x − x². Then dA/dx = 10 − 2x, so A has a maximum at x = 5, giving A = 25 m².
例如，一个矩形的长为 x，宽为 y，周长固定为 20 m，因此 2x + 2y = 20，得到 y = 10 − x。面积 A = xy = x(10 − x) = 10x − x²。则 dA/dx = 10 − 2x，所以 A 在 x = 5 处取得最大值，A = 25 m²。
12. Rates of Change and Connected Rates | 变化率与相关变化率
Many Edexcel questions involve variables that change with time. If a quantity V depends on r, and r depends on t, then the chain rule links the rates:
许多爱德思题目涉及随时间变化的变量。如果量 V 依赖于 r，而 r 依赖于 t，则链式法则将变化率联系起来：
dV/dt = dV/dr × dr/dt
For example, if the volume of a sphere is V = 4/3 π r³ and the radius increases at a constant rate of 2 cm s⁻¹, then dV/dt = 4π r² × 2 = 8π r² cm³ s⁻¹.
例如，如果球的体积为 V = 4/3 π r³，且半径以 2 cm s⁻¹ 的恒定速率增加，则 dV/dt = 4π r² × 2 = 8π r² cm³ s⁻¹。
Always identify which rate is given and which rate is required before substituting values. Make sure you convert units correctly, especially when comparing cm³ and m³.
在代入数值之前，始终先确定已知的变化率和要求的变化率。确保单位换算正确，特别是在比较 cm³ 和 m³ 时。
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