📚 Differentiation from First Principles: Example 7.2.1 | 从第一原理求导：例题 7.2.1
entiate f(x)=x² from first principles and explain every line of the limit algebra clearly.
在这道 AQA A-level 数学例题中，我们将从第一原理出发对 f(x)=x² 求导，并逐步清晰地解释极限运算的每一步。
1. The Example Question | 例题
The question is: Show from first principles that if f(x)=x², then f'(x)=2x.
题目是：从第一原理证明，若 f(x)=x²，则 f'(x)=2x。
2. What First Principles Means | 什么是第一原理
In AQA pure mathematics, differentiation from first principles uses the limit of the gradient of a chord as the two points get infinitely close. It is the formal definition of the derivative, so you cannot just apply the power rule in the working.
在 AQA 纯数学中，从第一原理求导使用弦的斜率在两个点无限接近时的极限。它是导数的正式定义，因此不能在解题过程中直接套用幂法则。
3. The Definition to Use | 使用的定义
Write the definition with f(x+h) and f(x). The letter h represents a small change in x.
写出包含 f(x+h) 和 f(x) 的定义。字母 h 表示 x 的微小变化。
f'(x) = lim (h → 0) [f(x + h) − f(x)] / h
4. Substituting into the Formula | 代入公式
Because f(x)=x², we replace f(x+h) with (x+h)² and f(x) with x².
因为 f(x)=x²，所以将 f(x+h) 替换为 (x+h)²，将 f(x) 替换为 x²。
f'(x) = lim (h → 0) [(x + h)² − x²] / h
5. Expanding and Simplifying | 展开与化简
Expand the bracket, then cancel the x² terms. The numerator simplifies to 2xh + h².
展开括号，然后消去 x² 项。分子化简为 2xh + h²。
(x + h)² − x² = x² + 2xh + h² − x² = 2xh + h²
6. Dividing by h | 除以 h
Factorise the numerator and divide by h. Since h is not exactly zero during the limiting process, this division is valid.
将分子因式分解并除以 h。由于在极限过程中 h 不真正为零，因此这个除法是有效的。
[f(x + h) − f(x)] / h = (2xh + h²) / h = 2x + h
7. Letting h Tend to 0 | 令 h 趋近于 0
As h approaches 0, the term 2x is unchanged and the term h vanishes, leaving 2x.
当 h 趋近于 0 时，2x 项不变，而 h 项消失，结果为 2x。
f'(x) = lim (h → 0) (2x + h) = 2x
8. Final Answer and Notation | 最终答案与记号
Therefore f' ion includes dy/dx=2x if y=x². For example, when x=3, the gradient of the curve y=x² is 6.
因此 f'(x)=2x。若 y=x²，等价记号为 dy/dx=2x。例如，当 x=3 时，曲线 y=x² 的斜率为 6。
9. Interpreting the Gradient Function | 理解导函数
The result f' le number: it is a gradient function. It gives the gradient of y=x² at any chosen x-value.
结果 f'(x)=2x 不是单一数值，而是一个导函数。它给出 y=x² 在任意选定 x 值处的斜率。
At x = 0, the gradient is 2(0) = 0, so the curve has a stationary point at the origin.
当 x = 0 时，斜率为 2(0) = 0，因此曲线在原点有驻点。
At x = 1, the gradient is 2, and at x = 3, the gradient is 6.
当 x = 1 时，斜率为 2；当 x = 3 时，斜率为 6。
10. Common Mistakes in AQA Exams | AQA 考试常见错误
Examiners often see the following errors in first-principles questions. Avoid them carefully.
阅卷官在导函数第一原理题中常看到以下错误。请仔细避免。
Forgetting to write the limit sign at every step before the final line.
在最终一行之前的每一步都忘记写极限符号。
Treating h as exactly zero too early, before cancelling.
在约分之前就把 h 当作真正的零。
Expanding (x+h)² incorrectly, for example writing x²+h² without the 2xh term.
展开 (x+h)² 时出错，例如漏掉 2xh 项而写成 x²+h²。
Using the power rule as your main working when the question explicitly says 'from first principles'.
当题目明确要求“从第一原理”时，却用幂法则作为主要解题过程。
11. Checking with the Power Rule | 用幂法则检验
Although the working must use the limit method, you can check your answer by differentiating x² using the standard rule.
尽管解题必须使用极限方法，你仍然可以用标准法则对 x² 求导来检查答案。
If y = xⁿ, then dy/dx = nxⁿ⁻¹. For n = 2: dy/dx = 2x¹ = 2x.
12. Exam Technique and Practice | 考试技巧与自主练习
In AQA mark schemes, the complete limiting argument must be shown. A direct answer with no method earns no method marks.
在 AQA 评分标准中，必须展示完整的极限论证。没有方法步骤的直接答案无法获得方法分。
Try the same method to prove that if f(x)=x³, then f'(x)=3x². Start with f(x+h)=(x+h)³.
请用同样的方法证明：若 f(x)=x³，则 f'(x)=3x²。从 f(x+h)=(x+h)³ 开始。
f(x) = x³ ⇒ f(x + h) = (x + h)³ = x³ + 3x²h + 3xh² + h³
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