📚 Example 3.4.1: Differentiation from First Principles for a Quadratic Function | 二次函数的第一性原理求导
quadratic function f(x) = x² from first principles. The question asks for the derivative f' ires careful use of limits together with accurate algebraic expansion. This article walks through the full solution step by step, explains the reasoning behind every line, and shows how to present the work so that it earns full marks.
在这个 AQA A-Level 数学例题中，我们要从第一性原理求二次函数 f(x) = x² 的导数。题目要求求出导数 f'(x)，以及曲线在 x = 3 处的梯度。该方法在 AQA 纯数学考纲中有明确考查要求，需要仔细使用极限和代数展开。本文会逐步完整地讲解解答过程，解释每一步背后的理由，并说明怎样书写才能拿到满分。
1. Understanding the First-Principles Definition | 理解第一性原理定义
The derivative of a function f(x) is defined as the limit of the average rate of change as the interval h approaches 0. This definition is called differentiation from first principles, or differentiation from the definition, and it is the foundation from which every shortcut rule is later built.
函数 f(x) 的导数定义为：当区间 h 趋近于 0 时，平均变化率的极限。这个定义被称为第一性原理求导（或从定义出发求导），它是后面所有快捷求导法则的基础。
f'(x) = limₕ→₀ [ f(x + h) − f(x) ] / h
ole fraction as h becomes arbitrarily small, not to substitute h = 0 immediately. You must not treat 'lim' as a plug-in step until after the fraction has been simplified.
这里 f(x + h) 表示函数在 x + h 处的值，h 是 x 的一个小增量。记号 limₕ→₀ 是要你考察当 h 任意小时整个分式的变化趋势，而不是立刻把 h = 0 代入。在化简分式之前，绝不可以把 'lim' 当成简单的代入步骤。
The whole method follows a fixed three-stage routine, and every first-principles question in the specification is marked against these same stages.
整套方法遵循固定的三步流程，考纲中所有第一性原理的题目都按同样的步骤给分。
- Stage 1: find f(x + h) by replacing every x in the formula with (x + h).
第一步：把公式中每一个 x 都替换成 (x + h)，求出 f(x + h)。 - Stage 2: form the difference quotient [f(x + h) − f(x)] / h and simplify it until the h in the denominator cancels.
第二步：写出差商 [f(x + h) − f(x)] / h，并一直化简到分母中的 h 被约掉。 - 0 in the simplified expression and write the answer with the limit notation still visible.
第三步：在化简后的表达式中令 h → 0，并保留极限记号写出答案。
2. The Worked Example: Setting Up the Limit | 例题：建立极限表达式
ce brackets around the whole expression (x + h) before doing anything else, because the entire binomial must be squared.
对于 f(x) = x²，将 x 替换为 (x + h)，得到 f(x + h) = (x + h)²。最安全的做法是先把 (x + h) 整个加上括号再处理，因为平方作用在整个二项式上。
Substitute f(x + h) and f(x) into the first-principles formula to obtain the difference quotient for this particular function:
把 f(x + h) 和 f(x) 代入第一性原理公式，得到本题的差商：
f'(x) = limₕ→₀ [ (x + h)² − x² ] / h
(x) = x² has simply been written back in place, and the numerator is the difference between the new value and the old value of the function.
此时还没有做任何化简，这是正常的。评分者希望先看到正确的代入，再进行代数运算。注意 f(x) = x² 只是被原样写回，分子是函数新值与旧值之差。
A common presentation error is to write 'f'(x) = (x + h)² − x²' ut the denominator there is no average rate of change, and the whole definition collapses.
常见的书写错误是写成 'f'(x) = (x + h)² − x²' 而忘记除以 h。没有分母就没有平均变化率，整个定义就不成立了。
3. Expanding (x + h)² and Forming the Difference Quotient | 展开 (x + h)² 并写出差商
Expand the binomial carefully. Squaring (x + h) means multiplying it by itself, and the result contains three terms, not two:
仔细展开这个二项式。对 (x + h) 平方就是把它与自身相乘，结果包含三项而不是两项：
(x + h)² = x² + 2xh + h²
Students who forget the middle term 2xh lose the whole question, because the derivative of x² comes entirely from that term. Write the expansion on its own line before substituting, then replace (x + h)² inside the numerator:
忘记中间项 2xh 的学生会丢掉整道题的分数，因为 x² 的导数完全来自这一项。先把展开式单独写一行，再把它代入分子：
[ (x² + 2xh + h²) − x² ] / h
Now simplify the numerator by collecting like terms. The x² and the −x² cancel exactly, because subtraction is applied to the whole expression in the brackets. What remains is:
现在合并同类项来化简分子。x² 与 −x² 恰好抵消，因为减号作用于括号内的整个表达式。剩下的部分为：
( 2xh + h² ) / h
e. The expression is still undefined when h = 0, and the limit has not yet been evaluated.
此阶段要把分母上的 h 保留着。当 h = 0 时该表达式仍然没有意义，极限还没有被求出。
4. Simplifying the Quotient and Cancelling h | 化简分式并约去 h
Both terms in the numerator contain a factor of h, so factorise the numerator as h(2x + h). The quotient then becomes:
分子中的两项都含有因子 h，因此把分子因式分解为 h(2x + h)。于是分式变为：
h(2x + h) / h = 2x + h
This cancellation is the crucial algebraic step of the whole method, and it is only legal because h is not equal to zero. The limit process allows h to be as small as we like, but never exactly zero, so dividing numerator and denominator by h is valid.
这一步约分是整套方法中最关键的代数步骤，它之所以合法，是因为 h 不等于零。极限过程允许 h 任意小，但永远不等于零，所以分子分母同时除以 h 是合法的。
ch row is the gradient of a chord joining the point (3, 9) to a nearby point on the curve.
差商的值依赖于 h。下表以 x = 3 为例说明，每一行都是连接点 (3, 9) 与曲线上邻近点的一条弦的梯度。
| h | 2x + h with x = 3 / 当 x = 3 时的值 | Meaning / 含义 |
| 1 | 7 | Chord gradient over an interval of 1 / 区间长度为 1 的弦梯度 |
| 0.1 | 6.1 | Closer to the tangent / 更接近切线 |
| 0.01 | 6.01 | Very close to the tangent / 非常接近切线 |
| 0.001 | 6.001 | Approaching the limit 6 / 趋于极限值 6 |
The table makes the idea of a limit concrete: the chord gradients converge towards a single number, and that number is the gradient of the tangent.
这张表让极限的概念变得具体：弦的梯度收敛到一个唯一的数，这个数就是切线的梯度。
5. Taking the Limit as h → 0 | 取 h → 0 的极限
The simplified difference quotient is 2x + h. As h approaches 0, the term h disappears and the expression approaches 2x. Therefore the derivative function is:
化简后的差商是 2x + h。当 h 趋近于 0 时，h 项消失，表达式趋近于 2x。因此导函数为：
f'(x) = limₕ→₀ (2x + h) = 2x
inal line, even though the substitution is now trivial. AQA mark schemes award a specific mark for correct limit notation and for clearly stating the limiting process, so omitting it can cost marks even when the answer 2x is correct.
最后一行要写出极限记号 limₕ→₀，即使此时代入已经显而易见。AQA 评分标准对正确的极限记号和明确说明极限过程单独给分，所以即使答案 2x 正确，漏写也可能失分。
The result f' verts any x-coordinate on the curve y = x² into the gradient of the tangent at that point.
结果 f'(x) = 2x 是一个梯度函数。它相当于一台机器，把曲线 y = x² 上任意一点的横坐标转换成该点处切线的梯度。
6. Finding the Gradient at x = 3 | 求 x = 3 处的梯度
e at x = 3. Substitute x = 3 into the derivative function, not into the original function:
题目第二部分要求曲线在 x = 3 处的梯度。应把 x = 3 代入导函数，而不是原函数：
f'(3) = 2(3) = 6
nd from the original function: when x = 3, y = 3² = 9, so the point of tangency is (3, 9).
所以 y = x² 在 x = 3 处切线的梯度恰好是 6。要记住曲线上的点由原函数给出：当 x = 3 时，y = 3² = 9，因此切点是 (3, 9)。
hrough the point (3, 9), the equation of the tangent line follows from y − y₁ = m(x − x₁):
已知梯度为 6 且过点 (3, 9)，由 y − y₁ = m(x − x₁) 可得切线方程：
y − 9 = 6(x − 3), so y = 6x − 9
You can verify this from the chord calculation in Section 4: the secant from (3, 9) to (3.01, 9.0601) has gradient 6.01, which is very close to 6, and the secant gradients decrease towards 6 as h shrinks.
你可以用第 4 节的弦计算来验证：从 (3, 9) 到 (3.01, 9.0601) 的割线梯度为 6.01，非常接近 6，而且随着 h 变小，割线梯度不断向 6 靠拢。
7. Checking the Answer with the Power Rule | 用幂法则检验
The power rule states that if f(x) = xⁿ then f'(x) = n·xⁿ⁻¹. Applying it to f(x) = x² gives f' -principles result. Checking your work this way is quick and catches most algebraic slips.
幂法则指出：若 f(x) = xⁿ，则 f'(x) = n·xⁿ⁻¹。把它用于 f(x) = x² 得到 f'(x) = 2x¹ = 2x，与第一性原理的结果一致。这样检验既快捷，又能发现大部分代数失误。
| Function / 函数 | Power rule result / 幂法则结果 | First-principles result / 第一性原理结果 |
| f(x) = x² | 2x | 2x |
| f(x) = x³ | 3x² | 3x² |
| f(x) = 5x² | 10x | 10x |
Even though the power rule is faster, examiners still set first-principles questions because they test understanding of the limit definition, algebraic manipulation and correct notation all at once. Never replace a 'from first principles' instruction with the power rule in your written solution, however tempting that may be.
尽管幂法则更快，考官仍然会出第一性原理的题，因为它同时考查对极限定义的理解、代数运算能力和正确书写记号。无论多么想偷懒，都不要在要求 'from first principles' 的题目中用幂法则替代过程。
8. Why Substituting h = 0 Too Early Fails | 为什么过早代入 h = 0 会失败
rence quotient [ (x + h)² − x² ] / h, the numerator becomes 0 and the denominator becomes 0, producing the meaningless expression 0/0. This is called an indeterminate form, and it carries no information about the gradient.
如果把 h = 0 直接代入差商 [ (x + h)² − x² ] / h，分子变成 0，分母也变成 0，得到没有意义的表达式 0/0。这被称为不定式，它不包含任何关于梯度的信息。
h, which is defined at h = 0 (and everywhere else nearby). Only after that cancellation is it safe to let h → 0.
第 3、4 节的代数运算解决了这个问题：约去 h 正是把不定式转变成性质良好的表达式 2x + h 的关键，而后者在 h = 0 处（以及其附近的任何地方）都有定义。只有在完成约分之后，才可以安全地令 h → 0。
Two related presentation traps are worth memorising. First, writing 'f'(x) = 2x + h' without taking the limit, because the derivative must not contain h. Second, writing 'h = 0' if it were an ordinary substitution, which loses the distinction between limit and value.
有两个相关的书写陷阱值得记住。第一，写成 'f'(x) = 2x + h' 而不取极限，因为导数中不能含有 h。第二，在极限那一行里把 'h = 0' 当成普通代入来写，这会丢失极限与函数值之间的区别。
9. Geometric Interpretation: Secants, Tangents and Chords | 几何解释：割线与切线
Geometrically, [f(x + h) − f(x)] / h is the gradient of the chord (the secant line) joining the points A(x, f(x)) and B(x + h, f(x + h)) on the curve. As h shrinks to zero, the point B slides along the curve towards A, and the secant rotates towards a limiting position.
从几何上看，[f(x + h) − f(x)] / h 是曲线上两点 A(x, f(x)) 与 B(x + h, f(x + h)) 之间弦（割线）的梯度。当 h 缩小到 0 时，点 B 沿曲线滑向 A，割线也旋转到一个极限位置。
That limiting position is the tangent at A, and its gradient is f'(x). Differentiation from first principles is therefore a precise description of what a tangent gradient means for a curve, rather than a separate algebraic trick.
这个极限位置就是 A 点处的切线，其梯度即 f'(x)。因此第一性原理求导是对曲线切线梯度含义的精确描述，而不仅仅是一个代数技巧。
- For f(x) = x², the chord gradient over [3, 3 + h] is 6 + h.
对 f(x) = x²，在区间 [3, 3 + h] 上的弦梯度为 6 + h。 - The chord gradient is always larger than 6 for positive h, because the curve bends upwards.
当 h 为正时，弦梯度总大于 6，因为曲线向上弯曲。 - The tangent gradient at (3, 9) is the limit 6, and the tangent lies below the curve everywhere except at the point of contact.
点 (3, 9) 处切线梯度为极限值 6，而切线除切点外处处位于曲线下方。
10. Common Exam Errors and How to Avoid Them | 常见考试错误及避免方法
Most lost marks in first-principles questions come from a small set of repeated mistakes. The table below lists them alongside the correction, and it doubles as a final checklist before you hand in your paper.
第一性原理题目中丢分的原因大多来自少数几个反复出现的错误。下表列出这些错误及其纠正方法，也可作为交卷前的最后检查清单。
| Error / 错误 | Correction / 纠正 |
| Expanding (x + h)² as x² + h² 把 (x + h)² 展开成 x² + h² |
Include the middle term: x² + 2xh + h² 必须包含中间项：x² + 2xh + h² |
| Forgetting to divide by h 忘记除以 h |
Write the full difference quotient first 先把完整差商写出来 |
