📚 Example 7.2.2: Differentiation from First Principles | 例 7.2.2：从第一性原理求导
erentiate f(x)=x³ from first principles, using algebraic expansion and a limiting process.
在 AQA A-Level 数学中，利用第一性原理求导是一项基本技能，它将割线斜率的极限与导数联系起来。本例展示如何通过代数展开和取极限过程，从第一性原理求 f(x)=x³ 的导数。
1. What Differentiation from First Principles Means | 什么是第一性原理求导
Differentiation from first principles means using the formal limit definition of the derivative rather than applying shortcut rules.
从第一性原理求导意味着使用导数的正式极限定义，而不是直接套用求导公式。
The derivative measures the instantaneous rate of change of f(x) at a point, found by making a small increment h tend to 0.
导数测量函数 f(x) 在某一点的瞬时变化率，通过让微小增量 h 趋于 0 来求得。
2. The Limit Definition of the Derivative | 导数的极限定义
For a function f(x), the derivative f '(x) is defined by the following limit, provided it exists:
对于函数 f(x)，导数 f '(x) 由以下极限定义（若该极限存在）：
f '(x) = lim (h→0) [ (f(x+h) − f(x)) / h ]
This expression is called the difference quotient. It represents the gradient of the chord from x to x+h.
这个表达式称为差商。它表示从 x 到 x+h 的割线斜率。
When h becomes arbitrarily small, the chord gradient approaches the tangent gradient at x.
当 h 变得任意小时，割线斜率趋向于 x 点处的切线斜率。
3. Setting Up the Difference Quotient for f(x)=x³ | 对 f(x)=x³ 构造差商
For the function f(x)=x³, we first calculate f(x+h):
对于函数 f(x)=x³，我们先计算 f(x+h)：
f(x+h) = (x+h)³
The difference quotient becomes:
差商变为：
[ (x+h)³ − x³ ] / h
Our task is to expand the numerator, simplify, and then let h tend to 0.
我们的任务是展开分子、化简，然后令 h 趋于 0。
4. Expanding (x+h)³ Using the Binomial Expansion | 用二项式展开法展开 (x+h)³
Use the binomial expansion or repeated distribution to expand (x+h)³:
使用二项式展开或逐项分配来展开 (x+h)³：
(x+h)³ = x³ + 3x²h + 3xh² + h³
You should recognise the coefficients 1, 3, 3, 1 from Pascal's triangle.
你应能看出系数 1、3、3、1 来自帕斯卡三角。
| Term 1 | x³ |
| Term 2 | 3x²h |
| Term 3 | 3xh² |
| Term 4 | h³ |
Make sure every term is written clearly before subtracting x³.
在减去 x³ 之前，请确保每一项都写清楚。
5. Simplifying the Numerator | 化简分子
Subtract f(x)=x³ from the expanded form:
从展开式中减去 f(x)=x³：
(x+h)³ − x³ = x³ + 3x²h + 3xh² + h³ − x³
The x³ and −x³ cancel, leaving:
x³ 与 −x³ 相互抵消，剩下：
3x²h + 3xh² + h³
This is the key simplification that allows division by h without a zero denominator.
这是关键的化简步骤，它使得除以 h 时不会出现零分母。
6. Dividing by h | 除以 h
Place the simplified numerator back over h:
将化简后的分子放回 h 上方：
[3x²h + 3xh² + h³] / h
Because h is not zero while we are still before the limit, we divide each term by h:
由于在取极限之前 h 不为零，我们可以将每一项除以 h：
(3x²h/h) + (3xh²/h) + (h³/h) = 3x² + 3xh + h²
The expression 3x² + 3xh + h² is now ready for the limiting process.
表达式 3x² + 3xh + h² 现在已准备好进行极限运算。
7. Taking the Limit as h Tends to 0 | 令 h 趋于 0 取极限
Apply the limit to each term:
对每一项取极限：
f
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