📚 Differentiation from First Principles: Example 3.3.1 | 从第一性原理求导：例 3.3.1
irst principles - to find the derivative of f(x) = x² − 4x + 1 at the point x = 2. This topic tests algebraic accuracy as much as the concept of a limit, and examiners expect clear method working from first principles.
例题 3.3.1 是一道标准的 AQA 纯数学题。它要求你使用正式的极限定义求 f(x) = x² − 4x + 1 在 x = 2 处的导数。这个主题既考查极限概念，也考查代数运算的准确性，考官希望看到清晰的第一性原理解题步骤。
First-principles questions appear in every specification - AQA, Edexcel, OCR and WJEC all examine them - and they cannot be answered by quoting a formula. This guide works through the full solution line by line, then explains the mark scheme, the most common algebraic errors, and the variations of the same question you are likely to meet.
第一性原理的题目出现在所有考试局的大纲中——AQA、Edexcel、OCR 和 WJEC 都会考查——而且它们不能通过直接引用公式来作答。本文逐行讲解完整解法，然后说明评分方案、最常见的代数错误，以及你可能遇到的同类变式题。
1. What the Question Asks | 题目要求
The command 'using first principles' means you must begin with the gradient of a chord and take a limit. You should not quote the power rule until the question allows you to check your answer.
"使用第一性原理"意味着你必须从弦的斜率出发并取极限。在题目允许你检查答案之前，不应该直接引用幂法则。
The given function is f(x) = x² − 4x + 1, and the target is f '(2). You need to find the derived function f '(x) first, and then substitute x = 2.
给定的函数是 f(x) = x² − 4x + 1，目标是求 f '(2)。你需要先求出导函数 f '(x)，再代入 x = 2。
The marking structure rewards three separate skills, so it pays to keep them physically separate on the page:
评分结构奖励三项彼此独立的能力，所以在卷面上把它们分开书写是值得的：
- Write the limit definition clearly.
- Substitute f(x + h) and f(x) without sign errors.
- Simplify, cancel h, and evaluate the limit.
- Substitute the given value of x only after the limit has been taken.
- 清楚写出极限定义；
- 正确代入 f(x + h) 和 f(x)，避免符号错误；
- 化简、约去 h 并求出极限；
- 只有在取完极限之后，才代入给定的 x 值。
Notice that the question never asks for the equation of the tangent or the coordinates of the point. It asks for a single number, f '(2), but every mark on the way to that number comes from the method.
注意，题目并没有要求切线方程或该点坐标。它只要求一个数值 f '(2)，但通向这个数值的每一分都来自方法步骤。
2. Start with the Limit Definition | 从极限定义开始
For a function f, the derivative at a general point x is defined by:
对于函数 f，导数在一般点 x 处的定义为：
f '(x) = lim (h → 0) [ f(x + h) − f(x) ] / h
This quotient is the slope of the chord from x to x + h. The limit converts that chord slope into the tangent slope.
这个商是从 x 到 x + h 的弦的斜率。取极限后，弦的斜率就转化为切线的斜率。
In AQA mark schemes, quoting this definition correctly in the first line is often worth one method mark. It also makes the rest of your solution self-explanatory to the examiner, who can see exactly which structure you are following.
在 AQA 评分方案中，第一行正确写出这个定义通常可以获得一个方法分。它还能让考官清楚看到你遵循的是哪一套结构，使后续解答一目了然。
The definition contains three moving parts: the increment h added to x, the difference f(x + h) − f(x) in the numerator, and the division by h. Every line of the solution is simply a careful handling of those three parts.
这个定义包含三个组成部分：加在 x 上的增量 h、分子中的差 f(x + h) − f(x)，以及除以 h。解题的每一行，都只是对这三部分的小心处理。
3. Why You Cannot Simply Put h = 0 | 为什么不能直接令 h = 0
o the quotient. Instead, we simplify the numerator until the h in the denominator cancels.
因为除以零没有定义，所以不能直接把 h = 0 代入这个商。相反，我们要先化简分子，直到分母中的 h 被约去。
The phrase 'as h tends to zero' is not the same as 'h equals zero'. The limit describes the behaviour of the chord gradient as h becomes arbitrarily small, without ever actually reaching zero.
"当 h 趋于零"与"h 等于零"并不相同。极限描述的是当 h 任意小时弦斜率的趋势，而 h 从未真正取到零。
ingle most common conceptual error in first-principles questions. A candidate who writes f ' = 0 / 0 has abandoned the method, and normally loses every mark after the definition mark.
这是第一性原理题目中最常见的概念性错误。写出 f '(x) = 0 / 0 的考生实际上已经放弃了该方法，通常在定义分之后就再也拿不到分。
he numerator, cancel that h with the denominator, and only then let h → 0 in what remains.
正确的步骤始终如一：展开、化简、从分子每一项中提出公因式 h、与分母中的 h 约去，最后才让剩下的部分取 h → 0。
4. Building f(x + h) Without Sign Errors | 正确写出 f(x + h)
Replace every x in the function with the bracket (x + h). The function contains both an x² term and a −4x term, so the substitution must be made in both places.
把函数中的每一个 x 都用括号 (x + h) 替换。函数中既有 x² 项又有 −4x 项，所以两处都必须代入。
f(x + h) = (x + h)² − 4(x + h) + 1
The bracket around (x + h) in the second term is essential. Without it, candidates frequently expand −4(x + h) as −4x + 4h instead of −4x − 4h.
第二项中 (x + h) 外面的括号至关重要。没有它，考生常常把 −4(x + h) 展开成 −4x + 4h，而不是 −4x − 4h。
Expanding gives the full expression:
展开后得到完整表达式：
f(x + h) = x² + 2xh + h² − 4x − 4h + 1
Check the middle term: (x + h)² expands to x² + 2xh + h², not x² + h². Missing the 2xh term is the classic error, and it destroys the whole solution.
检查中间项：(x + h)² 展开为 x² + 2xh + h²，而不是 x² + h²。漏掉 2xh 是典型的错误，它会毁掉整个解答。
5. Forming the Difference f(x + h) − f(x) | 构造差 f(x + h) − f(x)
Now subtract the original function. Write f(x) inside its own bracket so that every sign is handled correctly:
现在减去原函数。把 f(x) 单独放进括号中，这样每一个符号都能被正确处理：
f(x + h) − f(x) = (x² + 2xh + h² − 4x − 4h + 1) − (x² − 4x + 1)
Subtracting requires all three signs inside the second bracket to change: −x² + 4x − 1. This is where most marks are lost in the whole question.
相减时，第二个括号内的三项符号都要改变：−x² + 4x − 1。整道题中大部分失分都发生在这里。
After removing the bracket and collecting like terms:
去掉括号并合并同类项之后：
f(x + h) − f(x) = 2xh + h² − 4h
The x² terms cancel, the −4x and +4x terms cancel, and the +1 and −1 cancel. Every surviving term still contains at least one factor of h, which is exactly what the next step requires.
x² 项相消，−4x 与 +4x 相消，+1 与 −1 相消。剩下的每一项都仍然含有至少一个因子 h，这正是下一步所需要的。
Two useful sanity checks: the constant term must always vanish, and if no term contains h, you have made an algebra error somewhere.
两个有用的自检：常数项必须总是被消掉；如果没有任何一项含有 h，说明你在某处出现了代数错误。
6. Dividing by h and Cancelling | 除以 h 并约分
Divide the difference by h to form the chord gradient, then cancel:
用差值除以 h 得到弦的斜率，然后约分：
[ f(x + h) − f(x) ] / h = (2xh + h² − 4h) / h
h, the division can be performed term by term: 2xh ÷ h = 2x, h² ÷ h = h, and −4h ÷ h = −4.
因为分子中每一项都含有因子 h，所以可以逐项相除：2xh ÷ h = 2x，h² ÷ h = h，−4h ÷ h = −4。
[ f(x + h) − f(x) ] / h = 2x + h − 4 (h ≠ 0)
The cancellation is legitimate precisely because h is not zero during this stage. We are working with secants whose second point is distinct from the first.
这次约分之所以合法，正是因为在这一阶段 h 不等于零。我们处理的是第二个点与第一个点不重合的割线。
If a candidate writes the fraction with a common denominator first - for example when the function is 1/x - the same principle applies: simplify until the h factor disappears from the denominator.
如果考生先用通分处理分式——例如函数为 1/x 时——原理仍然相同：一直化简到 h 因子从分母中消失为止。
7. Taking the Limit as h → 0 | 取 h → 0 的极限
Now that h appears only in a single term and no longer in a denominator, the limit can be evaluated safely:
现在 h 只出现在单独一项中，且不再位于分母，就可以安全地求极限：
f '(x) = lim (h → 0) (2x + h − 4) = 2x − 4
As h becomes arbitrarily small, the term h tends to 0, while 2x and −4 are unaffected because they do not contain h. The limit is therefore 2x − 4.
当 h 任意小时，h 项趋于 0，而 2x 与 −4 不含 h，因此不受影响。所以极限为 2x − 4。
This result is the derived function (the gradient function), valid for all real values of x. It is a new function, not a number, and it is worth a separate accuracy mark.
这个结果就是导函数（斜率函数），对所有实数 x 都成立。它是一个新函数，而不是一个数，因此值得单独一个准确分。
= −4, and d/dx (1) = 0, giving 2x − 4 - exactly what the first-principles method produced.
幂法则可以快速验证答案：d/dx (x²) = 2x，d/dx (−4x) = −4，d/dx (1) = 0，得到 2x − 4——与第一性原理方法得到的完全一致。
8. Substituting x = 2 | 代入 x = 2
Only now, with the derived function established, do we substitute the given value:
只有在求出导函数之后，我们才代入题目给出的数值：
f '(2) = 2(2) − 4 = 4 − 4 = 0
The gradient of the curve y = x² − 4x + 1 at the point where x = 2 is exactly zero.
曲线 y = x² − 4x + 1 在 x = 2 处的斜率恰好为零。
riginal function before differentiating would give a meaningless number and would normally score zero for the final mark.
请特别注意运算顺序：先求导，再代入。如果在求导之前就把 x = 2 代入原函数，得到的将是一个毫无意义的数值，通常会在最后一步得零分。
A zero gradient is a perfectly acceptable answer. Candidates sometimes panic and assume they have made a mistake; in fact a zero answer here is a strong signal that the algebra has been done correctly.
斜率为零是完全可接受的答案。考生有时会慌张，以为自己算错了；实际上，这里出现零恰恰是代数运算正确的一个强烈信号。
9. Geometry Check: the Tangent at the Turning Point | 几何检验：驻点处的切线
Evaluate the function itself at x = 2 to locate the point of tangency:
在 x = 2 处求出函数值，以确定切点位置：
f(2) = 2² − 4(2) + 1 = 4 − 8 + 1 = −3
2, −3), and because the gradient there is zero, the tangent is the horizontal line y = −3.
切点为 (2, −3)，由于该处斜率为零，切线就是水平直线 y = −3。
We can verify the geometry by completing the square:
我们可以通过配方法验证几何图像：
x² − 4x + 1 = (x − 2)² − 3
This is the vertex form of an upward parabola with its turning point at (2, −3). The graph has a minimum there, so a horizontal tangent is exactly what we should expect - the first-principles answer is consistent with the shape of the curve.
这是开口向上的抛物线的顶点式，顶点为 (2, −3)。图像在该点取得最小值，因此出现水平切线完全符合预期——第一性原理得到的答案与曲线形状一致。
Such a check costs a few seconds and can rescue a whole question if an arithmetic slip has occurred earlier. Examiners do not require it, but strong candidates build it into their routine.
这样的检验只需几秒钟，却能在之前出现算术失误时挽救一整道题。考官并不要求这一步，但优秀考生会把它变成习惯。
10. Common Errors and the AQA Mark Scheme | 常见错误与 AQA 评分方案
The table below lists the errors that appear most often in candidate scripts, together with the mark consequence and the correction.
下表列出考生答卷中最常见的错误，以及相应的失分后果与改正方法。
| Error | Consequence | Fix |
| Writing (x + h)² as x² + h² | Loses all accuracy marks | Always expand as x² + 2xh + h² |
| Expanding −4(x + h) as −4x + 4h | Wrong numerator, no cancellation | Keep the bracket until you multiply through |
| Dropping the bracket when subtracting f(x) | Sign errors in every term | Change all three signs explicitly |
| Substituting h = 0 too early | 0 / 0, method abandoned | Cancel h first, take the limit second |
| Substituting x = 2 before differentiating | No final answer mark | Find f '(x) first, then substitute |
on and expansion, one accuracy mark for simplifying to 2x + h − 4, one accuracy mark for the derived function f '(x) = 2x − 4, and one mark for the final value f '(2) = 0.
这类题目的典型分值分配是：写出极限定义得 1 个方法分，正确代入与展开得 1 个方法分，化简到 2x + h − 4 得 1 个准确分，得到导函数 f '(
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图