Differentiation from First Principles: AS Level AQA Pure Maths Guide — 从第一原理出发的微分：AQA AS 数学纯数指南

一、什么是微分？曲线的斜率 | What Is Differentiation? The Gradient of a Curve

在 AS 数学的纯数部分（AQA 单元一）中，微分（differentiation）是微积分的第一块基石。它的核心任务是回答一个非常直观的问题：一条曲线在某一个点上的”陡峭程度”到底是多少？对于一条直线，斜率是固定的，你可以任取两点用”纵坐标变化量除以横坐标变化量”来计算。但对于一条曲线，比如 y = x²，斜率在每一点都不同：在 x = 0 处曲线是水平的，而在 x 越大的地方曲线越陡。微分提供了一套系统的方法，让我们能够精确地算出曲线在任意一点的斜率。

In the pure mathematics section of AS Maths (AQA Unit 1), differentiation is the first cornerstone of calculus. Its central task is to answer a very intuitive question: how steep is a curve at a particular point? For a straight line the gradient is fixed, and you can pick any two points and divide the vertical change by the horizontal change. But for a curve such as y = x², the gradient is different at every point: at x = 0 the curve is flat, and the further x gets from zero, the steeper the curve becomes. Differentiation gives us a systematic method to work out the gradient of a curve at any point exactly.

我们把这一”曲线上某一点的斜率”称为导数（derivative），记作 dy/dx 或 f'(x)。这个符号本身就在提醒我们它的含义：dy 与 dx 分别表示 y 与 x 的微小变化量，dy/dx 就是这两个微小变化量之比，也就是”瞬时变化率”。理解这个概念，比死记求导公式重要得多，因为后续的切线、驻点和最优化问题全部建立在这个基础之上。

We call this “gradient at a point on the curve” the derivative, written as dy/dx or f'(x). The notation itself hints at its meaning: dy and dx stand for tiny changes in y and x respectively, and dy/dx is the ratio of those two tiny changes, that is, the instantaneous rate of change. Understanding this idea matters far more than memorising differentiation formulas, because everything that follows, tangents, stationary points and optimisation problems, is built on this foundation.

二、从第一原理出发的微分：割线斜率的极限 | Differentiation from First Principles: The Limit of a Secant Line

求导最严格的起点，是”第一原理”（first principles），也就是极限定义。设想曲线上有两点：点 A(x, f(x)) 和点 B(x+h, f(x+h))，其中 h 是一个很小的增量。连接 A 与 B 的直线叫割线（chord），它的斜率是 [f(x+h) – f(x)] / h。当我们让 h 越来越小、越来越接近 0 时，点 B 会沿着曲线向点 A 无限靠近，这条割线也越来越接近曲线在点 A 处的切线（tangent）。因此，切线的斜率就是当 h 趋于 0 时割线斜率的极限。

The most rigorous starting point for differentiation is first principles, the limit definition. Imagine two points on a curve: point A(x, f(x)) and point B(x+h, f(x+h)), where h is a small increment. The straight line joining A and B is called a chord, and its gradient is [f(x+h) – f(x)] / h. As we let h get smaller and smaller, approaching 0, point B slides along the curve infinitely close to point A, and the chord approaches the tangent to the curve at A. The gradient of the tangent is therefore the limit of the chord’s gradient as h tends to 0.

用数学符号写出来就是：f'(x) = lim (h → 0) [f(x+h) – f(x)] / h。AQA 的 AS 试卷经常要求考生用这个定义对简单函数（如 y = x² 或 y = x³）完成一整套第一原理求导，因此你必须熟练展开 (x+h)² 或 (x+h)³，消去可以约分的 h，再让 h 趋近于 0 得到结果。这个过程的每一步都要写清楚，考试会按步骤给分。

Written in symbols this is: f'(x) = lim (h → 0) [f(x+h) – f(x)] / h. AQA AS papers frequently ask candidates to work through a full first-principles differentiation of a simple function such as y = x² or y = x³, so you must be comfortable expanding (x+h)² or (x+h)³, cancelling the common factor of h, and then letting h approach 0 to reach the answer. Write every step clearly, because marks are awarded step by step in the exam.

例如对 f(x) = x²，f(x+h) – f(x) = (x+h)² – x² = 2xh + h²，除以 h 得 2x + h，令 h 趋于 0 即得 f'(x) = 2x。这个结果就是下面要讲的”幂法则”的一个特例。

For example, with f(x) = x², we have f(x+h) – f(x) = (x+h)² – x² = 2xh + h², dividing by h gives 2x + h, and letting h tend to 0 yields f'(x) = 2x. This result is a special case of the power rule discussed next.

三、幂法则：快速求导 x^n | The Power Rule: Differentiating x^n Quickly

虽然第一原理是根基，但在实际解题中我们不会每次都用极限定义。最常用、也最需要烂熟于心的规则是幂法则（power rule）：若 y = x^n，则 dy/dx = n·x^(n-1)。也就是说，”把指数搬到前面当系数，指数本身减一”。例如 y = x³ 的导数是 3x²，y = x⁵ 的导数是 5x⁴。

Although first principles is the foundation, in practice we do not use the limit definition every time. The most common rule, and the one you must know by heart, is the power rule: if y = x^n, then dy/dx = n·x^(n-1). In words, bring the power down to the front as a coefficient, and reduce the power itself by one. For example, the derivative of y = x³ is 3x², and the derivative of y = x⁵ is 5x⁴.

幂法则同样适用于负指数和分数指数，只要先把根式和分式写成 x 的幂的形式。例如 1/x = x^(-1)，其导数为 -x^(-2)，即 -1/x²；而 √x = x^(1/2)，其导数为 (1/2)·x^(-1/2)，即 1/(2√x)。掌握这种”先改写为幂形式再求导”的技巧，是处理分数和根式函数的关键。

The power rule works equally well for negative and fractional powers, provided you first rewrite roots and reciprocals as powers of x. For instance 1/x = x^(-1) has derivative -x^(-2), that is -1/x², while √x = x^(1/2) has derivative (1/2)·x^(-1/2), that is 1/(2√x). Mastering this “rewrite as a power first, then differentiate” technique is the key to handling fractional and radical functions.

下面这张表总结了几个最常考的幂法则例子，建议你背下来，做到一看到原式就能立刻写出导数。

The table below summarises a few of the most frequently tested power-rule examples. Learn them so well that you can write the derivative instantly on sight.

原函数 Function y	导数 Derivative dy/dx
x²	2x
x³	3x²
x (即 x¹)	1
1/x (即 x⁻¹)	-1/x²
√x (即 x^(1/2))	1/(2√x)

四、加法法则、常数倍法则与常数法则 | The Sum, Constant-Multiple and Constant Rules

实际题目里的函数几乎从来不是单个 x^n，而是若干项的和或差，例如 y = 4x³ – 3x² + 2x – 5。处理这类函数需要三条配套法则。第一是加法法则（sum rule）：和的导数等于导数的和，也就是可以”逐项求导”。第二是常数倍法则（constant-multiple rule）：常数可以提到求导符号外面，例如 d(4x³)/dx = 4 · d(x³)/dx = 12x²。第三是常数法则（constant rule）：任何常数的导数都是 0，因为没有 x 变化的项，斜率恒为零。

Functions in real exam questions are almost never a single x^n, but a sum or difference of several terms, for example y = 4x³ – 3x² + 2x – 5. Three companion rules handle such functions. The first is the sum rule: the derivative of a sum is the sum of the derivatives, so you can differentiate term by term. The second is the constant-multiple rule: a constant can be moved outside the differentiation, for example d(4x³)/dx = 4 · d(x³)/dx = 12x². The third is the constant rule: the derivative of any constant is 0, because a term with no x in it has zero gradient everywhere.

把这些规则合起来，对 y = 4x³ – 3x² + 2x – 5 逐项求导，就得到 dy/dx = 12x² – 6x + 2。注意常数项 -5 求导后直接消失。考试中一个最常见的失分点，就是忘了对常数项求导（或者把它当成 1 而不是 0），务必小心。

Combining these rules, differentiating y = 4x³ – 3x² + 2x – 5 term by term gives dy/dx = 12x² – 6x + 2. Notice the constant term -5 vanishes on differentiation. One of the most common places to lose marks in the exam is forgetting to differentiate the constant term, or treating its derivative as 1 instead of 0, so be careful.

五、二阶导数：斜率的变化率 | The Second Derivative: The Rate of Change of the Gradient

把导数 dy/dx 再求一次导，就得到二阶导数（second derivative），记作 d²y/dx² 或 f”(x)。如果说一阶导数告诉我们曲线是”向上走还是向下走”以及”有多陡”，那么二阶导数告诉我们的是”斜率本身在如何变化”，也就是曲线的弯曲方向。当 d²y/dx² > 0 时，曲线向上弯（下凹朝上，凹口向上），斜率在增大；当 d²y/dx² < 0 时，曲线向下弯（凹口向下），斜率在减小。

Differentiating dy/dx once more gives the second derivative, written as d²y/dx² or f”(x). If the first derivative tells us whether the curve is going up or down and how steeply, the second derivative tells us how the gradient itself is changing, that is, which way the curve is bending. When d²y/dx² > 0 the curve bends upward and the gradient is increasing; when d²y/dx² < 0 the curve bends downward and the gradient is decreasing.

例如 y = x³，一阶导数为 3x²，二阶导数为 6x。当 x > 0 时 6x > 0，曲线向上弯；当 x < 0 时 6x < 0，曲线向下弯。二阶导数在下一节的驻点分类中扮演关键角色，它提供了一种快速判断"极值点到底是极大还是极小"的方法。

For example, with y = x³, the first derivative is 3x² and the second derivative is 6x. When x > 0 we have 6x > 0 and the curve bends upward; when x < 0 we have 6x < 0 and the curve bends downward. The second derivative plays a key role in classifying stationary points in the next section, giving a fast way to decide whether a turning point is a maximum or a minimum.

六、切线与法线：求直线方程 | Tangents and Normals: Finding the Equation of a Line

导数最直接的应用之一，就是求曲线在某一点处的切线（tangent）与法线（normal）方程。切线是与曲线在该点相切、斜率等于导数 f'(a) 的直线；法线是过同一点且与切线垂直的直线，因此法线的斜率是切线斜率的负倒数，即 -1/f'(a)。解题的标准步骤是：先求出该点处的 y 坐标，再求导得到斜率，最后代入直线的点斜式方程 y – y₁ = m(x – x₁)。

One of the most direct applications of the derivative is finding the equation of the tangent and the normal to a curve at a given point. The tangent is the straight line touching the curve at that point, with gradient equal to the derivative f'(a); the normal passes through the same point but is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent’s gradient, that is -1/f'(a). The standard solving procedure is: first find the y-coordinate at the point, then differentiate to get the gradient, and finally substitute into the point-slope form y – y₁ = m(x – x₁).

举一个完整的例子。求曲线 y = x² + 3 在 x = 2 处的切线与法线方程。先算 y 坐标：y = 2² + 3 = 7，切点为 (2, 7)。再求导：dy/dx = 2x，在 x = 2 处斜率为 4。切线方程是 y – 7 = 4(x – 2)，化简得 y = 4x – 1。法线斜率为 -1/4，方程为 y – 7 = -(1/4)(x – 2)，化简得 y = -(1/4)x + 15/2。像这样把每一步都写全，是拿满过程分的关键。

Here is a complete worked example. Find the equations of the tangent and normal to y = x² + 3 at x = 2. First the y-coordinate: y = 2² + 3 = 7, so the point is (2, 7). Next differentiate: dy/dx = 2x, giving gradient 4 at x = 2. The tangent is y – 7 = 4(x – 2), which simplifies to y = 4x – 1. The normal has gradient -1/4, so its equation is y – 7 = -(1/4)(x – 2), simplifying to y = -(1/4)x + 15/2. Writing out every step like this is the key to scoring full method marks.

七、驻点：极大值、极小值与拐点 | Stationary Points: Maxima, Minima and Points of Inflection

当 dy/dx = 0 时，曲线在这一点的切线是水平的，这样的点称为驻点（stationary point）。驻点分为三类：极大值点（maximum，曲线先升后降）、极小值点（minimum，曲线先降后升）和拐点（point of inflection，曲线在此处改变弯曲方向，但并不改变升降方向）。求驻点的第一步永远是令 dy/dx = 0 并解出 x 的值。

When dy/dx = 0 the tangent at that point is horizontal, and such a point is called a stationary point. There are three kinds: a maximum (where the curve rises then falls), a minimum (where the curve falls then rises), and a point of inflection (where the curve changes its direction of bending without changing whether it is rising or falling). The first step in finding stationary points is always to set dy/dx = 0 and solve for x.

求出驻点的 x 坐标后，必须判断它的类型。最常用的方法是”二阶导数判别法”（second derivative test）：把该 x 代入 d²y/dx²，若结果为正则是极小值点，若为负则是极大值点，若恰好为零则此法失效，需要改看一阶导数在驻点两侧的符号。另一个更稳妥的方法是”梯度符号法”：分别在驻点左边和右边各取一个 x 值代入 dy/dx，观察斜率从正变负（极大）还是从负变正（极小）。

Once you have the x-coordinate of a stationary point, you must classify its type. The most common method is the second derivative test: substitute that x into d²y/dx². A positive result means a minimum, a negative result means a maximum, and if the result is exactly zero the test fails and you must look at the sign of the first derivative on either side. A more robust alternative is the gradient-sign method: pick an x value just left and just right of the stationary point, substitute into dy/dx, and observe whether the gradient changes from positive to negative (a maximum) or from negative to positive (a minimum).

完整例子：求 y = x³ – 3x 的驻点并分类。dy/dx = 3x² – 3 = 3(x² – 1)，令其为零得 x = 1 或 x = -1。二阶导数 d²y/dx² = 6x。在 x = 1 处为 6 > 0，是极小值点，y = 1 – 3 = -2，即 (1, -2)；在 x = -1 处为 -6 < 0，是极大值点，y = -1 + 3 = 2，即 (-1, 2)。这类题目几乎每份 AS 试卷都会出现，务必熟练。

Full worked example: find and classify the stationary points of y = x³ – 3x. We have dy/dx = 3x² – 3 = 3(x² – 1), and setting this to zero gives x = 1 or x = -1. The second derivative is d²y/dx² = 6x. At x = 1 it equals 6 > 0, a minimum point with y = 1 – 3 = -2, giving (1, -2); at x = -1 it equals -6 < 0, a maximum point with y = -1 + 3 = 2, giving (-1, 2). This type of question appears in almost every AS paper, so make sure you are fluent.

八、递增函数与递减函数 | Increasing and Decreasing Functions

导数还能告诉我们一个函数在哪个区间递增、在哪个区间递减。规则很简单：当 dy/dx > 0 时函数递增（increasing），当 dy/dx < 0 时函数递减（decreasing）。这其实只是把"斜率为正则上升、斜率为负则下降"的直观图像翻译成了代数语言。要找出递增或递减区间，先令 dy/dx = 0 求出临界点，再用这些点把 x 轴分成若干区间，并在每个区间里任取一个测试值判断导数的正负。

The derivative also tells us on which intervals a function is increasing or decreasing. The rule is simple: when dy/dx > 0 the function is increasing, and when dy/dx < 0 the function is decreasing. This is just the intuitive picture "positive gradient means rising, negative gradient means falling" translated into algebra. To find the intervals, first set dy/dx = 0 to locate the critical points, then use them to split the x-axis into intervals, and test the sign of the derivative with a sample value in each interval.

继续用 y = x³ – 3x 的例子。dy/dx = 3(x² – 1)，临界点为 x = -1 和 x = 1。在区间 x < -1 取 x = -2，得 dy/dx = 9 > 0，递增；在 -1 < x < 1 取 x = 0，得 dy/dx = -3 < 0，递减；在 x > 1 取 x = 2，得 dy/dx = 9 > 0，递增。可以看到，这个结论与上一节”(-1, 2) 是极大、 (1, -2) 是极小”完全一致。

Continuing with y = x³ – 3x, we have dy/dx = 3(x² – 1) with critical points x = -1 and x = 1. On x < -1 take x = -2, giving dy/dx = 9 > 0, so it is increasing; on -1 < x < 1 take x = 0, giving dy/dx = -3 < 0, so it is decreasing; on x > 1 take x = 2, giving dy/dx = 9 > 0, so it is increasing again. Notice this agrees exactly with the previous section, where (-1, 2) was a maximum and (1, -2) was a minimum.

九、应用：用微分解决最优化问题 | Applications: Solving Optimisation Problems

微分在 AS 考试中最有”应用题味”的部分，是最优化（optimisation）问题。典型题目会给出一个由某种材料或约束条件限定的几何量，比如一个固定周长的矩形，要求你找出使面积最大的边长。解题流程是：先用题目条件把目标量表示成单一变量的函数，再求导令其为零找出驻点，最后用二阶导数或梯度符号确认那是最大值（或最小值），并回到题意写出最终答案和单位。

The most “word-problem” style part of differentiation in the AS exam is optimisation. A typical question gives a geometric quantity constrained by some material or condition, such as a rectangle of fixed perimeter, and asks you to find the dimensions that maximise the area. The workflow is: express the target quantity as a function of a single variable using the given conditions, differentiate and set the derivative to zero to find the stationary point, confirm with the second derivative or gradient sign that it is a maximum (or minimum), then return to the context to write the final answer with its units.

完整例子：用 40 米长的篱笆围一个矩形，求面积最大时长与宽各是多少。设长为 x，则宽为 20 – x（因为周长 2x + 2w = 40）。面积 A = x(20 – x) = 20x – x²。求导得 dA/dx = 20 – 2x，令其为零得 x = 10。二阶导数 d²A/dx² = -2 < 0，所以这是极大值。于是长 10 米、宽 10 米（即正方形）时面积最大，为 100 平方米。这类题目的关键是从题意中正确写出约束条件，很多失分都发生在这个建模第一步。

Full worked example: a fence of 40 metres encloses a rectangle; find the length and width that maximise the area. Let the length be x, so the width is 20 – x (since the perimeter 2x + 2w = 40). The area is A = x(20 – x) = 20x – x². Differentiating gives dA/dx = 20 – 2x, and setting this to zero gives x = 10. The second derivative is d²A/dx² = -2 < 0, so this is a maximum. Therefore length 10 m and width 10 m (a square) give the maximum area of 100 m². The key to these problems is writing the constraint correctly from the wording; most marks are lost at this modelling first step.

十、常见错误与易错点 | Common Mistakes and Pitfalls

微分这一章虽然规则不多，但考试中的失分点却非常集中。第一个高频错误是把常数项的导数当成 1，例如把 y = x² + 5 的导数写成 2x + 1，而正确结果是 2x，因为常数 5 的导数是 0。第二个错误是在求法线时忘了取负倒数，只把切线斜率抄过去，导致整道题从中间就错了。第三个错误是在第一原理求导时忘记让 h 趋于 0，或者约分 h 时出错。

Although this chapter has few rules, the places where marks are lost in the exam are highly concentrated. The first common mistake is treating the derivative of a constant as 1, for example writing the derivative of y = x² + 5 as 2x + 1, when the correct answer is 2x because the derivative of the constant 5 is 0. The second mistake is forgetting to take the negative reciprocal when finding a normal, simply copying the tangent gradient and going wrong halfway through the question. The third is forgetting to let h tend to 0 in a first-principles differentiation, or making an error when cancelling the factor h.

还有两个更隐蔽的陷阱。一是在把分数和根式改写为幂形式时把符号搞反，例如误以为 1/x² = x² 而不是 x^(-2)。二是在对驻点分类时只依赖二阶导数，当二阶导数恰好等于零时不知道改用梯度符号法，结果白白丢掉分类这一步的分。把这些易错点单独列出来对照复习，比盲目刷题更高效。

There are two subtler traps as well. One is getting the sign wrong when rewriting fractions and roots as powers, for example thinking 1/x² = x² instead of x^(-2). The other is relying only on the second derivative when classifying stationary points and not knowing to fall back on the gradient-sign method when the second derivative is exactly zero, thereby throwing away the marks for the classification step. Listing these pitfalls separately and reviewing against them is far more efficient than grinding through practice questions blindly.

十一、如何备考：练习建议与答题策略 | How to Revise: Practice Tips and Exam Strategy

要把微分这一章练到熟练，建议按”理解、熟练、应用”三个阶段推进。理解阶段，用第一原理亲手推导 x² 和 x³ 的导数各一遍，真正明白”割线趋近于切线”的含义；熟练阶段，反复练习幂法则与逐项求导，目标是看到 4x³ – 3x² + 2x – 5 这样多项式能在十秒内写出导数；应用阶段，集中攻克切线法线、驻点分类和最优化这三类大题，因为它们几乎覆盖了微分在 AS 试卷上的全部分值。

To master this chapter, it is best to progress through three stages: understanding, fluency and application. In the understanding stage, derive the derivatives of x² and x³ from first principles by hand once each, so you genuinely grasp what “the chord approaches the tangent” means. In the fluency stage, drill the power rule and term-by-term differentiation until you can write down the derivative of a polynomial such as 4x³ – 3x² + 2x – 5 within ten seconds. In the application stage, focus on the three long-question types, tangents and normals, stationary-point classification, and optimisation, since together they cover nearly all the marks differentiation attracts in an AS paper.

答题时有一个通用策略：永远先明确题目要的是什么量，再决定用哪条规则。看到”切线”就想到先求导得斜率；看到”驻点”就想到令 dy/dx = 0；看到”最大/最小”就想到最优化流程。每一步都写出算式，因为 AQA 的评分标准是”过程分 + 答案分”分开给的，即使最终答案算错，清晰的过程依然能保住大部分分数。练习时给自己计时，模拟真实的考试压力。

One universal strategy when answering: always be clear about what quantity the question is asking for, then decide which rule to use. Seeing “tangent” should trigger “differentiate to get the gradient”; seeing “stationary point” should trigger “set dy/dx = 0”; seeing “maximum/minimum” should trigger the optimisation workflow. Write out every step of working, because the AQA mark scheme awards method marks and answer marks separately, so even if the final answer is wrong, clear working still secures most of the marks. When practising, time yourself to simulate real exam pressure.

Summary | 总结

微分是 AS 数学纯数单元一的核心工具，它把”曲线在某一点的斜率”这个几何直觉转化成了精确的代数运算。本文从第一原理的极限定义出发，系统梳理了幂法则、加法与常数倍法则、二阶导数、切线与法线、驻点分类、函数的增减性，以及最优化应用这一整套技能链。考试中的高分，靠的不是孤立地背公式，而是能把这套流程连贯地运用：先改写函数，再求导，令导数为零找驻点，用二阶导数或梯度符号分类，最后回到题意作答。

Differentiation is the core tool of the AS Maths Pure Unit 1, turning the geometric intuition of “the gradient at a point on a curve” into precise algebraic computation. Starting from the limit definition of first principles, this article has walked through the full chain of skills: the power rule, the sum and constant-multiple rules, the second derivative, tangents and normals, classification of stationary points, increasing and decreasing functions, and optimisation applications. High marks in the exam come not from memorising formulas in isolation, but from applying this flow smoothly: rewrite the function, differentiate, set the derivative to zero to find stationary points, classify them with the second derivative or gradient sign, and finally answer in the context of the question.

建议的复习顺序是：先用第一原理亲手求一遍 x² 和 x³ 的导数以建立信心，再大量练习幂法则与逐项求导直到形成肌肉记忆，然后集中攻克切线法线和驻点分类这两类高频大题，最后用最优化问题检验自己”把文字建模成函数”的能力。每一步都写清楚过程，过程分在 AQA 的评分标准里占很大比重。

A suggested revision order: first differentiate x² and x³ by hand from first principles to build confidence, then drill the power rule and term-by-term differentiation until they become second nature, then focus on the two most common long questions, tangents and normals, and stationary-point classification, and finally test your ability to “model words as a function” with optimisation problems. Show every step clearly; method marks carry a large share of the AQA mark scheme.

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