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

📚 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 ‘(x) is often written as dy/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

Consider two points on the curve y = f(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

Subtracting xⁿ and dividing by h leaves n xⁿ⁻¹ plus terms that 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

This is the power rule. In an exam, you may be asked 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

If f(x) = ax + b, then 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| is continuous at x = 0 but 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

One common mistake is substituting h = 0 too early, before simplifying 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

In Edexcel A-Level questions, a first-principles derivative question usually asks for a specific 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。

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