📚 Differentiation from First Principles Example 7.1.1 | 从第一性原理求导（例 7.1.1）
ferentiation from first principles. This method is fundamental for understanding what a derivative actually represents.
在本复习指南中，我们详细讲解 AQA A-Level 数学大纲中的例 7.1.1：利用从第一性原理求导的方法，求函数 f(x) = x² 在 x = 3 处的导数。这一方法是理解导数本质的基础。
1. What Is Differentiation from First Principles? | 什么是从第一性原理求导？
Differentiation from first principles, also called the definition of the derivative, uses the limit of a chord's gradient to find the gradient of a curve at a single point. For a function f(x), the derivative f'(x) is defined by the following limit:
从第一性原理求导，又称导数的定义，利用割线斜率的极限来求曲线在某一点处的切线斜率。对于函数 f(x)，其导数 f'(x) 由以下极限定义：
f'(x) = limh→0 [f(x + h) − f(x)] / h
Here, h represents a small change in x. As h tends to zero, the chord becomes a tangent, and its gradient becomes the derivative.
其中 h 表示 x 的一个微小变化量。当 h 趋向于 0 时，割线变成切线，其斜率即为导数。
2. Example 7.1.1: The Question | 例 7.1.1：题目
², use differentiation from first principles to find the gradient of the tangent at x = 3.
已知函数 f(x) = x²，请用从第一性原理求导的方法，求出在 x = 3 处切线的斜率。
We will solve this step by step, making every algebraic manipulation clear.
我们将一步一步地求解，确保每一步代数运算都清晰明了。
3. Step 1: Substitute into the Formula | 第一步：代入公式
Write the definition of the derivative at x = 3. Since we want f'(3), we replace x with 3 in the formula:
写出在 x = 3 处的导数定义。因为我们要求 f'(3)，所以将公式中的 x 替换为 3：
f'(3) = limh→0 [f(3 + h) − f(3)] / h
ow calculate f(3 + h) and f(3). Since f(x) = x², we have f(3) = 9 and f(3 + h) = (3 + h)².
现在计算 f(3 + h) 和 f(3)。因为 f(x) = x²，所以 f(3) = 9，f(3 + h) = (3 + h)²。
4. Step 2: Expand and Simplify the Numerator | 第二步：展开并化简分子
Expand (3 + h)² using the distributive property:
使用分配律展开 (3 + h)²：
(3 + h)² = 9 + 6h + h²
Therefore, the numerator becomes (9 + 6h + h²) − 9. The constant 9 cancels with −9, leaving 6h + h².
因此，分子变为 (9 + 6h + h²) − 9。常数项 9 与 −9 相消，剩下 6h + h²。
5. Step 3: Divide by h and Simplify | 第三步：除以 h 并化简
We now write the full quotient:
现在写出完整的差商：
(6h + h²) / h = 6 + h
This simplification is valid because h ≠ 0 while we are taking the limit. We are not setting h equal to zero, only letting it approach zero.
这个化简在 h ≠ 0 时成立，因为我们在求极限的过程中并不是令 h 等于 0，而是让 h 无限趋近于 0。
6. Step 4: Take the Limit as h → 0 | 第四步：取 h → 0 的极限
Finally, we take the limit of 6 + h as h tends to 0:
最后，我们取 6 + h 当 h 趋向于 0 时的极限：
f'(3) = limh→0 (6 + h) = 6
= 3 is 6. This means the tangent line rises 6 units for every 1 unit it runs to the right.
因此，曲线 y = x² 在 x = 3 处的切线斜率为 6。这意味着切线每向右平移 1 个单位，就上升 6 个单位。
7. Generalising: The Derivative of x² | 推广：x² 的导数
The same method works for any value of x. Use f(x) = x² in the general formula:
同样的方法适用于任意 x 值。在一般公式中代入 f(x) = x²：
f'(x) = limh→0 [(x + h)² − x²] / h
cting x², we have 2xh + h². Dividing by h gives 2x + h. Taking the limit as h → 0 yields f'(x) = 2x.
展开 (x + h)² 得到 x² + 2xh + h²。减去 x² 后得到 2xh + h²。除以 h 得 2x + h。取 h → 0 的极限，得到 f'(x) = 2x。
So for f(x) = x², f'(x) = 2x. At x = 3, f'(3) = 6, confirming our answer.
所以对于 f(x) = x²，f'(x) = 2x。在 x = 3 处，f'(3) = 6，与之前的结果一致。
8. Visual Interpretation | 几何直观理解
− 9, which touches the curve exactly at (3, 9) and has the same instantaneous direction as the curve.
导数在某一点表示切线的斜率。在抛物线 y = x² 上 x = 3 处的点是 (3, 9)。切线方程为 y = 6x − 9，它恰好与曲线在点 (3, 9) 相切，并且具有与曲线相同的瞬时方向。
As h shrinks, the chord through (3, f(3)) and (3 + h, f(3 + h)) becomes the tangent. This limiting process is the heart of calculus.
当 h 缩小时，经过 (3, f(3)) 和 (3 + h, f(3 + h)) 两点的割线逐渐变成切线。这一极限过程正是微积分的核心。
9. Common Mistakes to Avoid | 常见错误
Students often make errors when using first principles. Here are the most frequent pitfalls:
学生在使用第一性原理时经常会犯错误。以下是最常见的几类问题：
- Setting h = 0 too early: You cannot divide by zero. Always simplify the quotient before taking the limit.
- 错误地过早令 h = 0： 不能除以零。在取极限之前一定要先化简差商。
- Incorrect expansion: (x + h)² is x² + 2xh + h², not x² + h² or (x + h)(x − h).
- 展开错误： (x + h)² 是 x² + 2xh + h²，而不是 x² + h² 或 (x + h)(x − h)。
- Dropping the limit notation: Keep limh→0 until you actually substitute h = 0.
- 忽略极限符号： 在真正代入 h = 0 之前，要一直保留 limh→0。
- Mixing up x and 3: For f'(3), use f(3 + h) and f(3), not f(x + h) and f(x) mixed.
- 混淆 x 和 3： 求 f'(3) 时，应使用 f(3 + h) 和 f(3)，不要将 x 和 3 混用。
10. Practice Questions | 练习题
Try these similar questions to build confidence:
尝试以下类似问题，以增强信心：
- Use first principles to find f'(x) for f(x) = x² + 1. 用第一性原理求 f(x) = x² + 1 的 f'(x)。
- Use first principles to find the gradient of y = x² at x = −2. 用第一性原理求 y = x² 在 x = −2 处的斜率。
- Given f(x) = x², find the equation of the tangent at x = 4. 已知 f(x) = x²，求 x = 4 处的切线方程。
- Use first principles to find f'(x) for f(x) = 2x². 用第一性原理求 f(x) = 2x² 的 f'(x)。
Solutions: (1) f'(x) = 2x. (2) −4. (3) y = 8x − 16. (4) f'(x) = 4x.
答案：(1) f'(x) = 2x。(2) −4。(3) y = 8x − 16。(4) f'(x) = 4x。
11. Why This Example Matters for AQA A-Level Maths | 为什么这个例子对 AQA A-Level 数学很重要
Example 7.1.1 is a classic assessment question. AQA examiners expect you to know the formal limit definition of a derivative and to be able to apply it to simple polynomials. This skill also appears in questions about tangents, normals and stationary points.
例 7.1.1 是一道典型的考试题。AQA 考官期望你掌握导数的形式极限定义，并能将其应用于简单的多项式函数。这一技能也会出现在有关切线、法线和驻点的问题中。
vative is nxⁿ⁻¹, but understanding the limiting process ensures you can derive it if asked.
掌握第一性原理能让你更深入地理解为什么幂函数法则成立。对于 f(x) = xⁿ，其导数为 nxⁿ⁻¹，但理解极限过程能确保你在被要求时能够推导出它。
12. Summary | 总结
We have used Example 7.1.1 to show the full process of differentiation from first principles:
我们通过例 7.1.1 演示了从第一性原理求导的完整过程：
| Step 步骤 | Expression 表达式 |
| Start 开始 | f'(3) = limh→0 [(3 + h)² − 9] / h |
| Expand 展开 | = limh→0 (9 + 6h + h² − 9) / h |
| Simplify 化简 | = limh→0 (6h + h²) / h = limh→0 (6 + h) |
| Limit 极限 | = 6 |
By practising this method, you will be well prepared for any first-principles derivative question on the AQA A-Level Mathematics paper.
通过练习这一方法，你将能够从容应对 AQA A-Level 数学试卷中任何关于第一性原理求导的问题。
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