Differentiation in A-Level Edexcel Pure Mathematics | A-Level Edexcel 纯数学中的微分

📚 Differentiation in A-Level Edexcel Pure Mathematics | A-Level Edexcel 纯数学中的微分

Differentiation is one of the most fundamental topics in the Edexcel A-Level Pure Mathematics specification. It gives us a precise way to describe how a function changes at any instant, and it underpins applications such as tangents, normals, stationary points, optimisation and modelling. This article revises the key definitions, rules and problem types you need for the exam.

微分是 Edexcel A-Level 纯数学考试中最基础的主题之一。它让我们能够精确地描述函数在任意瞬间的变化情况，并且是切线、法线、驻点、优化和建模等应用的基础。本文复习考试所需的关键定义、法则和常见题型。

1. What Differentiation Measures | 导数衡量什么

Differentiation measures the instantaneous rate of change of a function. If y = f(x), then the derivative f'(x) gives the gradient of the tangent to the curve y = f(x) at any point where the function is differentiable. It is defined as a limit of average rates of change over smaller and smaller intervals.

微分衡量的是函数的瞬时变化率。如果 y = f(x)，那么导数 f'(x) 给出曲线 y = f(x) 在可导点处切线的斜率。它被定义为越来越小区间上平均变化率的极限。

f'(x) = lim (h → 0) [f(x + h) − f(x)] / h

The notation dy/dx is used when y is given as an explicit function of x. The operator d/dx means ‘differentiate with respect to x’. For example, dy/dx = 2x means that at any point x, the gradient of the tangent is 2x.

当 y 表示为 x 的显函数时，使用记号 dy/dx。算子 d/dx 表示“对 x 求导”。例如，dy/dx = 2x 表示在任意点 x 处，切线的斜率为 2x。

2. Differentiation from First Principles | 从第一性原理求导

The derivative from first principles uses the limit definition. For f(x) = x², we compute f(x + h) = (x + h)² = x² + 2xh + h². The difference quotient is [f(x + h) − f(x)] / h = [(x² + 2xh + h²) − x²] / h = 2x + h. Taking the limit as h → 0 gives 2x.

从第一性原理求导使用极限定义。对于 f(x) = x²，计算 f(x + h) = (x + h)² = x² + 2xh + h²。差商为 [f(x + h) − f(x)] / h = [(x² + 2xh + h²) − x²] / h = 2x + h。令 h → 0，得到 2x。

d/dx (x²) = 2x

Edexcel exam papers often include a first-principles proof for powers of x, such as x² or x³. You must show the expansion, simplify the quotient, and state the limit clearly. Avoid jumping straight to the answer without justification.

Edexcel 试卷经常要求对 x 的幂，如 x² 或 x³，进行第一性原理证明。你必须展示展开、化简差商，并清楚写出极限。不要不写过程直接跳到答案。

3. Standard Derivatives and the Power Rule | 标准导数与幂法则

The power rule is the most common differentiation tool. For any real power n, the derivative of xⁿ is n xⁿ⁻¹. This works for positive powers, negative powers and fractional powers.

幂法则是最常用的求导工具。对于任意实数指数 n，xⁿ 的导数是 n xⁿ⁻¹。该法则适用于正指数、负指数和分数指数。

d/dx (xⁿ) = n xⁿ⁻¹

x⁵ → 5x⁴
1/x = x⁻¹ → −x⁻²
√x = x¹/² → (1/2)x⁻¹/²

You must also memorise the standard derivatives of trigonometric, exponential and logarithmic functions. These appear constantly in chain, product and quotient rule problems.

你还必须牢记三角函数、指数函数和对数函数的标准导数。它们在链式法则、乘积法则和商法则问题中经常出现。

d/dx (sin x) = cos x, d/dx (cos x) = −sin x, d/dx (eˣ) = eˣ, d/dx (ln x) = 1/x

4. Linear Combinations and Higher Derivatives | 线性组合与高阶导数

Differentiation is linear, so constants multiply and sums split. To differentiate y = 3x⁴ − 5x² + 2x − 7, differentiate each term separately: the constant −7 differentiates to 0, 2x gives 2, −5x² gives −10x, and 3x⁴ gives 12x³.

微分是线性的，因此常数倍可以保留，和可以拆分。要求 y = 3x⁴ − 5x² + 2x − 7 的导数，逐项求导：常数 −7 求导为 0，2x 求导为 2，−5x² 求导为 −10x，3x⁴ 求导为 12x³。

dy/dx = 12x³ − 10x + 2

The second derivative is obtained by differentiating again. It is written d²y/dx² and is read as ‘d two y by dx squared’. For the example above, d²y/dx² = 36x² − 10.

二阶导数由再次求导得到。它写作 d²y/dx²，读作“d two y by dx squared”。对于上面的例子，d²y/dx² = 36x² − 10。

5. Chain Rule | 链式法则

The chain rule is used for composite functions: if y = f(u) and u = g(x), then dy/dx = dy/du × du/dx. It is often described as ‘differentiate the outer function and multiply by the derivative of the inner function’.

链式法则用于复合函数：如果 y = f(u) 且 u = g(x)，那么 dy/dx = dy/du × du/dx。它通常被描述为“对外层函数求导，再乘以内层函数的导数”。

dy/dx = dy/du × du/dx

Example: y = (3x² + 1)⁵. Let u = 3x² + 1, so y = u⁵. Then dy/du = 5u⁴ and du/dx = 6x. Multiplying gives dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴.

例子：y = (3x² + 1)⁵。令 u = 3x² + 1，则 y = u⁵。于是 dy/du = 5u⁴，du/dx = 6x。相乘得到 dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴。

You can also apply the chain rule directly to standard forms. For example, d/dx [sin(kx)] = k cos(kx), d/dx [eᵏˣ] = keᵏˣ, and d/dx [ln(g(x))] = g'(x)/g(x).

你还可以将链式法则直接应用于标准形式。例如，d/dx [sin(kx)] = k cos(kx)，d/dx [eᵏˣ] = keᵏˣ，以及 d/dx [ln(g(x))] = g'(x)/g(x)。

6. Product Rule | 乘积法则

If y = uv, where u and v are both functions of x, the product rule states that dy/dx = u’v + uv’. This gives the derivative of a product without expanding it first.

如果 y = uv，其中 u 和 v 都是 x 的函数，乘积法则指出 dy/dx = u’v + uv’。这样无需先展开就可以求出乘积的导数。

d/dx (uv) = u’v + uv’

Example: y = x² sin x. Let u = x² and v = sin x, so u’ = 2x and v’ = cos x. Applying the product rule gives dy/dx = 2x sin x + x² cos x.

例子：y = x² sin x。令 u = x²，v = sin x，则 u’ = 2x，v’ = cos x。应用乘积法则得到 dy/dx = 2x sin x + x² cos x。

In exam questions, be careful to identify u and v clearly before differentiating. Leaving answers in factorised or expanded form is usually acceptable, but factorised forms are often easier to use in later parts of a question.

在考试题中，求导前要清楚地确定 u 和 v。答案写成因式分解形式或展开形式通常都可以，但因式分解形式在后续小题中往往更容易使用。

7. Quotient Rule | 商法则

If y = u/v, where u and v are functions of x, the quotient rule states that dy/dx = (u’v − uv’) / v². The order matters: numerator is ‘bottom times derivative of top minus top times derivative of bottom’.

如果 y = u/v，其中 u 和 v 是 x 的函数，商法则指出 dy/dx = (u’v − uv’) / v²。顺序很重要：分子是“下面乘上面求导减去上面乘下面求导”。

d/dx (u/v) = (u’v − uv’) / v²

Example: y = (x² + 1)/(x − 2). Let u = x² + 1 and v = x − 2, so u’ = 2x and v’ = 1. Then dy/dx = [(2x)(x − 2) − (x² + 1)(1)] / (x − 2)² = (x² − 4x − 1)/(x − 2)².

例子：y = (x² + 1)/(x − 2)。令 u = x² + 1，v = x − 2，则 u’ = 2x，v’ = 1。于是 dy/dx = [(2x)(x − 2) − (x² + 1)(1)] / (x − 2)² = (x² − 4x − 1)/(x − 2)²。

Common errors include forgetting to square the denominator, or writing uv’ − u’v in the numerator instead of u’v − uv’. Always double-check the sign before simplifying.

常见错误包括忘记给分母平方，或者在分子中写成 uv’ − u’v 而不是 u’v − uv’。化简前一定要检查符号。

8. Stationary Points and Classification | 驻点及分类

A stationary point occurs where the first derivative is zero: dy/dx = 0. To find stationary points, differentiate the function, set the derivative equal to zero, solve for x, and substitute each x into the original function to find the corresponding y-coordinate.

驻点发生在一阶导数为零处：dy/dx = 0。要求驻点，先对函数求导，令导数等于零，解出 x，再将每个 x 代入原函数求出对应的 y 坐标。

To classify a stationary point, find the second derivative d²y/dx². If d²y/dx² > 0, the point is a local minimum. If d²y/dx² < 0, the point is a local maximum. If d²y/dx² = 0, the test is inconclusive and you should use a first derivative sign test.

要判断驻点类型，求出二阶导数 d²y/dx²。如果 d²y/dx² > 0，该点是局部极小值。如果 d²y/dx² < 0，该点是局部极大值。如果 d²y/dx² = 0，该判别法无效，应使用一阶导数符号检验。

d²y/dx² > 0 ⇒ local minimum, d²y/dx² < 0 ⇒ local maximum

Example: y = x³ − 3x + 1. Differentiating gives dy/dx = 3x² − 3 = 3(x − 1)(x + 1), so stationary points occur at x = 1 and x = −1. The second derivative is d²y/dx² = 6x. At x = 1, d²y/dx² = 6 > 0, so (1, −1) is a minimum. At x = −1, d²y/dx² = −6 < 0, so (−1, 3) is a maximum.

例子：y = x³ − 3x + 1。求导得 dy/dx = 3x² − 3 = 3(x − 1)(x + 1)，因此驻点发生在 x = 1 和 x = −1。二阶导数为 d²y/dx² = 6x。在 x = 1 处，d²y/dx² = 6 > 0，所以 (1, −1) 是极小值。在 x = −1 处，d²y/dx² = −6 < 0，所以 (−1, 3) 是极大值。

9. Tangents and Normals | 切线与法线

A tangent to a curve at a point has the same gradient as the curve at that point. Once you know the gradient m and a point (x₁, y₁), the equation of the tangent is y − y₁ = m(x − x₁).

曲线在某点的切线具有与该点曲线相同的斜率。一旦知道斜率 m 和点 (x₁, y₁)，切线方程就是 y − y₁ = m(x − x₁)。

The normal is perpendicular to the tangent. If the tangent gradient is m, the normal gradient is −1/m. The normal equation is y − y₁ = (−1/m)(x − x₁).

法线垂直于切线。如果切线斜率为 m，法线斜率为 −1/m。法线方程为 y − y₁ = (−1/m)(x − x₁)。

Tangent: y − y₁ = m(x − x₁), Normal: y − y₁ = (−1/m)(x − x₁)

Example: Find the tangent to y = x² at x = 3. The derivative y’ = 2x gives m = 6. The point is (3, 9). The tangent is y − 9 = 6(x − 3), which simplifies to y = 6x − 9.

例子：求 y = x² 在 x = 3 处的切线。导数 y’ = 2x 给出 m = 6。点为 (3, 9)。切线为 y − 9 = 6(x − 3)，化简为 y = 6x − 9。

10. Second Derivative and Applications | 二阶导数与应用

The second derivative measures the rate of change of the first derivative, which tells us about the concavity of a function. If d²y/dx² > 0 on an interval, the graph is convex or ‘smiling’. If d²y/dx² < 0, the graph is concave or 'frowning'.

二阶导数衡量一阶导数的变化率，它告诉我们函数的凹凸性。如果在某区间上 d²y/dx² > 0，图像是凸的，像“微笑”的形状。如果 d²y/dx² < 0，图像是凹的，像“皱眉”的形状。

Differentiation is essential in optimisation problems. A typical Edexcel modelling question gives a real-world quantity such as surface area, volume, cost or distance. You must express the quantity as a function of one variable, differentiate, set the derivative to zero, confirm that the stationary point gives a maximum or minimum, and then answer the question in context.

微分在优化问题中至关重要。典型的 Edexcel 建模题会给出表面积、体积、成本或距离等实际量。你必须将该量表示为单一变量的函数，求导，令导数等于零，确认驻点是最大值或最小值，然后结合题目情境作答。

Always check the domain in modelling questions. For example, lengths and volumes must be positive, and endpoints may give extreme values even when the derivative is not zero inside the interval.

在建模题中一定要检查定义域。例如，长度和体积必须为正，而且即使区间内部导数不为零，端点也可能给出极值。

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Differentiation in A-Level Edexcel Pure Mathematics | A-Level Edexcel 纯数学中的微分