📚 Differentiation from First Principles | 从第一性原理求导
Differentiation is one of the central ideas in A-Level Mathematics, and Edexcel exam papers often test whether you can derive a derivative directly from its limit definition. This article explains the method from first principles, also called differentiation using the limit of a chord gradient, with fully worked steps and exam advice.
微分是 A-Level 数学的核心内容之一，Edexcel 考试经常考查你是否能从极限定义直接推导导数。本文讲解从第一性原理求导的方法，也叫弦斜率极限法，并配有完整步骤和考试建议。
1. What Is Differentiation? | 什么是微分？
Differentiation gives the instantaneous rate of change of a function, or the gradient of the tangent to a curve at a point. In pure mathematics, the derivative f '(x) is defined as a limit of average rates of change over smaller and smaller intervals.
微分给出函数的瞬时变化率，也就是曲线在某一点处切线的斜率。在纯数学中，导数 f '(x) 被定义为平均变化率在越来越小区间上的极限。
For a curve y = f(x), the derivative f ' dx. Understanding the limit behind this notation is essential for A-Level proof questions.
对于曲线 y = f(x)，导数 f '(x) 通常写作 dy/dx。理解这个记号背后的极限，对 A-Level 证明题非常重要。
2. The Gradient of a Chord | 弦的斜率
x): one at x and one at x + h. The gradient of the straight line joining these points is called the chord gradient:
考虑曲线 y = f(x) 上的两个点：一个在 x，另一个在 x + h。连接这两点的直线斜率称为弦的斜率：
m₍chord₎ = (f(x+h) − f(x)) / h
This fraction is the average rate of change of f over the interval from x to x + h. The chord is not yet the tangent, but as h gets closer to 0, the chord approaches the tangent.
这个分式是 f 在 x 到 x + h 区间上的平均变化率。弦还不是切线，但随着 h 越来越接近 0，弦会逐渐接近切线。
3. The Limit Definition | 极限定义
The derivative is obtained by taking the limit of the chord gradient as h tends to 0:
导数通过令 h 趋近于 0 时取弦斜率的极限得到：
f '(x) = lim (h → 0) [(f(x+h) − f(x)) / h]
This definition is exactly what Edexcel calls differentiation from first principles. You must write 'lim' at every line until the value of h is actually substituted as 0.
这个定义正是 Edexcel 所说的从第一性原理求导。你必须每一行都写上“lim”，直到最后一步才真正将 h 代入为 0。
4. Computing f '(x) for f(x) = x² | 对 f(x) = x² 求导
Let f(x) = x². Then f(x+h) = (x+h)² = x² + 2xh + h². The difference is:
设 f(x) = x²。则 f(x+h) = (x+h)² = x² + 2xh + h²。差为：
f(x+h) − f(x) = 2xh + h²
Divide by h, assuming h ≠ 0:
除以 h，假设 h ≠ 0：
(f(x+h) − f(x)) / h = (2xh + h²) / h = 2x + h
Now take the limit as h → 0:
现在令 h → 0 取极限：
f '(x) = lim (h → 0) (2x + h) = 2x
This proves that the derivative of x² is 2x. The same process can be applied to any polynomial term.
这就证明了 x² 的导数是 2x。同样的过程可以应用于任何多项式项。
5. General Power Rule from First Principles | 从第一性原理推导幂函数法则
For f(x) = xⁿ, the binomial expansion gives:
对于 f(x) = xⁿ，二项式展开给出：
(x+h)ⁿ = xⁿ + n xⁿ⁻¹ h + ⁿC₂ xⁿ⁻² h² + ... + hⁿ
still contain h. After taking the limit as h → 0, every remaining term with h disappears:
减去 xⁿ 并除以 h 后，剩下 n xⁿ⁻¹ 以及仍含有 h 的项。令 h → 0 取极限后，所有含 h 的项都消失：
d/dx (xⁿ) = n xⁿ⁻¹
ked to prove it for a specific value such as n = 3 or n = 4, rather than the general case.
这就是幂函数法则。在考试中，你可能会被要求对特定值如 n = 3 或 n = 4 进行证明，而不是一般情形。
6. Constant and Linear Functions | 常数函数与一次函数
If f(x) = c, where c is a constant, then f(x+h) = c, so f(x+h) − f(x) = 0. Therefore:
如果 f(x) = c，其中 c 是常数，那么 f(x+h) = c，因此 f(x+h) − f(x) = 0。所以：
d/dx (c) = 0
f(x+h) = a(x+h) + b = ax + ah + b. The difference is ah, and dividing by h gives a. Thus:
如果 f(x) = ax + b，那么 f(x+h) = a(x+h) + b = ax + ah + b。差为 ah，除以 h 得到 a。因此：
d/dx (ax + b) = a
These simple cases are useful checks when applying the first-principles method to longer expressions.
这些简单情形在将第一性原理方法应用于较长的表达式时，是有用的检验。
7. Differentiability and Continuity | 可微性与连续性
A function is differentiable at a point if the two-sided limit in the derivative definition exists. This requires the left-hand and right-hand chord limits to give the same value.
如果导数定义中的双侧极限存在，则函数在该点可微。这要求左侧和右侧的弦斜率极限给出相同的值。
For example, f(x) = |x| ut not differentiable there, because the left gradient is −1 and the right gradient is +1.
例如，f(x) = |x| 在 x = 0 处连续但不可微，因为左侧斜率为 −1，而右侧斜率为 +1。
Edexcel questions usually focus on functions that are differentiable, but understanding this distinction helps you avoid claiming a derivative exists when it does not.
Edexcel 的题目通常关注可微函数，但理解这一区别有助于避免在导数不存在时错误地认为它存在。
8. First Principles for f(x) = √x | 用第一性原理求 √x 的导数
For f(x) = √x, the difference quotient is:
对于 f(x) = √x，差商为：
(√x+h − √x) / h
Rationalise the numerator by multiplying top and bottom by √x+h + √x:
分子有理化，将分子和分母同时乘以 √x+h + √x：
( (x+h) − x ) / [ h(√x+h + √x) ] = h / [ h(√x+h + √x) ]
Cancel h and take the limit as h → 0:
约去 h，并令 h → 0 取极限：
f '(x) = 1 / (√x+0 + √x) = 1 / (2√x)
This is a standard result and shows that rationalising can be a key step when radicals are involved.
这是一个标准结果，也说明当涉及根式时，有理化可能是关键步骤。
9. Common Mistakes | 常见错误
ying the fraction. If you do this, you may obtain 0/0 and lose the structure of the limit.
一个常见错误是在化简分式之前过早代入 h = 0。这样做可能会得到 0/0，并丢失极限的结构。
- Forgetting to write the limit symbol after the first line.
- Incorrectly expanding (x+h)² or (x+h)³.
- Dividing by h without keeping the numerator fully expanded.
- Treating f(x+h) as f(x) + h, which is not generally true.
中文要点：
- 第一行之后忘记写极限符号。
- 错误展开 (x+h)² 或 (x+h)³。
- 除以 h 时分子没有完全展开。
- 把 f(x+h) 误当作 f(x) + h，这通常不成立。
Checking each line carefully is essential because examiners award method marks for correct expansion, division and limiting.
仔细检查每一步很重要，因为考官会根据正确的展开、除法以及取极限给方法分。
10. Exam Technique | 考试技巧
c function such as f(x) = x², f(x) = x³ or f(x) = 1/x. Follow a clear four-step structure:
在 Edexcel A-Level 试题中，第一性原理求导题通常会指定一个具体函数，如 f(x) = x²、f(x) = x³ 或 f(x) = 1/x。请遵循清晰的三步结构：
- Find f(x+h) and expand carefully.
- Form and simplify the difference quotient f(x+h) − f(x) over h.
- Take the limit as h → 0 and state f '(x).
中文要点：
- 求出 f(x+h) 并仔细展开。
- 构造并化简差商 [f(x+h) − f(x)] / h。
- 令 h → 0 取极限，并写出 f '(x)。
Always write the limit symbol until the final substitution, and state clearly that h ≠ 0 when cancelling h.
始终写上极限符号直到最后代入，并在约去 h 时明确说明 h ≠ 0。
11. Worked Example: f(x) = 3x² − 5x + 2 | 例题：f(x) = 3x² − 5x + 2
Find f '(x) from first principles for f(x) = 3x² − 5x + 2.
从第一性原理求 f(x) = 3x² − 5x + 2 的导数 f '(x)。
First, compute f(x+h):
首先计算 f(x+h)：
f(x+h) = 3(x+h)² − 5(x+h) + 2 = 3x² + 6xh + 3h² − 5x − 5h + 2
Subtract f(x):
减去 f(x)：
f(x+h) − f(x) = 6xh + 3h² − 5h
Divide by h:
除以 h：
(f(x+h) − f(x)) / h = 6x + 3h − 5
Take the limit as h → 0:
令 h → 0 取极限：
f '(x) = lim (h → 0) (6x + 3h − 5) = 6x − 5
This matches the result obtained using the power rule term by term.
这与逐项使用幂函数法则得到的结果一致。
12. Summary and Key Points | 总结与核心要点
The first-principles definition connects the derivative to the gradient of a tangent through a limit of chord slopes. The key formula is:
第一性原理定义通过弦斜率的极限将导数与切线斜率联系起来。核心公式为：
f '(x) = lim (h → 0) [(f(x+h) − f(x)) / h]
| Function f(x) | Derivative f '(x) |
|---|---|
| x² | 2x |
| xⁿ | n xⁿ⁻¹ |
| √x | 1 / (2√x) |
| ax + b | a |
Mastering this method gives you confidence with proof questions and strengthens your understanding of differentiation before moving on to rules such as the product rule and chain rule.
掌握这种方法能让你在证明题中更有信心，并在学习乘法法则和链式法则之前加深对微分的理解。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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