📚 Worked Example 4.3.5: Differentiation from First Principles | 例题 4.3.5：从第一性原理求导
principles. The method is fundamental for understanding what a derivative actually measures.
本文讲解 AQA A-Level 数学例题 4.3.5：从第一性原理求 f(x) = x² 的导数。这一方法是理解导数本质的基础。
1. What Is Differentiation from First Principles? | 什么是第一性原理求导
First principles differentiation uses the limit of a gradient between two nearby points on a curve as the distance between them shrinks to zero.
第一性原理求导利用曲线上两个邻近点之间割线斜率的极限，当两点距离趋近于零时得到切线斜率。
For AQA A-Level Maths, you must be able to apply this definition to simple functions such as x², x³ and 1/x without relying on the standard power rule.
在 AQA A-Level 数学中，你必须能够在不依赖标准幂法则的情况下，将这个定义应用于 x²、x³ 和 1/x 等简单函数。
f'(x) = limh→0 [f(x+h) − f(x)] / h
This expression is the derivative from first principles. It is often called the delta method because h is a small change in x.
这个表达式就是从第一性原理求导的公式。它常被称为 delta 方法，因为 h 表示 x 的一个微小变化。
2. Understanding the Limit Definition | 理解极限定义
The quotient [f(x+h) − f(x)] / h gives the gradient of the secant line joining the point (x, f(x)) to the point (x+h, f(x+h)).
商 [f(x+h) − f(x)] / h 给出连接点 (x, f(x)) 与点 (x+h, f(x+h)) 的割线斜率。
As h approaches 0, the second point moves closer to the first, and the secant line becomes the tangent line at x.
当 h 趋近于 0 时，第二个点越来越靠近第一个点，割线就变成 x 处的切线。
We use lim because we never actually set h equal to 0; division by zero is undefined. Instead, we analyse the limiting value.
我们使用 lim 是因为我们永远不会真正让 h 等于 0；除以零没有定义。相反，我们分析极限值。
3. Setting Up Example 4.3.5 | 建立例题 4.3.5
In Example 4.3.5, the function is f(x) = x². We need to find f'(x) directly from the limit definition.
在例题 4.3.5 中，函数为 f(x) = x²。我们需要直接从极限定义求出 f'(x)。
Start by writing f(x+h). Since f(x) = x², we replace every x with x+h.
首先写出 f(x+h)。因为 f(x) = x²，我们把每个 x 替换为 x+h。
f(x+h) = (x+h)²
The next step is to form the difference f(x+h) − f(x).
下一步是构造差 f(x+h) − f(x)。
f(x+h) − f(x) = (x+h)² − x²
4. Step 1: Expand the Brackets | 第 1 步：展开括号
Expand (x+h)² carefully. Many errors occur when students forget the middle term 2xh.
仔细展开 (x+h)²。如果忘记中间项 2xh，就会出错。
(x+h)² = x² + 2xh + h²
Then subtract x² from this expansion.
然后从这个展开式中减去 x²。
x² + 2xh + h² − x² = 2xh + h²
The x² terms cancel, which is essential for the limit to exist.
x² 项相互抵消，这对于极限存在至关重要。
5. Step 2: Form the Difference Quotient | 第 2 步：构造差商
Place the simplified difference over h to form the difference quotient.
将简化后的差除以 h，构造差商。
[f(x+h) − f(x)] / h = (2xh + h²) / h
Factorise h from the numerator. This is safe because h is not zero, only approaching zero.
从分子中提取公因式 h。这样做是可行的，因为 h 并不等于零，只是趋近于零。
(2xh + h²) / h = h(2x + h) / h
Cancelling h gives 2x + h.
约去 h 后得到 2x + h。
[f(x+h) − f(x)] / h = 2x + h
6. Step 3: Take the Limit as h → 0 | 第 3 步：取 h → 0 时的极限
Now apply the limit. As h tends to 0, the term h disappears, leaving only 2x.
现在应用极限。当 h 趋近于 0 时，h 项消失，只剩下 2x。
f'(x) = limh→0 (2x + h) = 2x
Therefore, the derivative of x² with respect to x is 2x.
因此，x² 对 x 的导数是 2x。
This matches the standard power rule: d/dx (x²) = 2x²⁻¹ = 2x.
这与标准幂法则一致：d/dx (x²) = 2x²⁻¹ = 2x。
7. Notation and Interpretation | 符号与解释
The derivative can be written in several equivalent forms: f'(x), dy/dx, or d/dx [f(x)].
导数可以写成几种等价形式：f'(x)、dy/dx 或 d/dx [f(x)]。
In Example 4.3.5, the result is f'(x) = 2x. This means the gradient of the curve y = x² at any point x is exactly 2x.
在例题 4.3.5 中，结果为 f'(x) = 2x。这意味着曲线 y = x² 在任意点 x 处的切线斜率恰好为 2x。
or example, at x = 3, the gradient is 2 × 3 = 6; at x = −1, the gradient is 2 × (−1) = −2.
例如，在 x = 3 处，斜率为 2 × 3 = 6；在 x = −1 处，斜率为 2 × (−1) = −2。
8. Graphical Meaning of the Result | 结果的图形意义
The derivative f'(x) = 2x tells us how steep the parabola y = x² is at any chosen value of x.
导数 f'(x) = 2x 告诉我们抛物线 y = x² 在任意给定 x 值处的陡峭程度。
When x is negative, the gradient is negative, so the curve is decreasing. When x is positive, the gradient is positive, so the curve is increasing.
当 x 为负时，斜率为负，因此曲线下降。当 x 为正时，斜率为正，因此曲线上升。
x = 0, the gradient is 0, which corresponds to the minimum point of y = x² at the origin.
在 x = 0 处，斜率为 0，这对应于 y = x² 在原点处的最小值点。
9. Generalising to f(x) = xⁿ | 推广到 f(x) = xⁿ
, but the algebra becomes longer. For f(x) = x³, expand (x+h)³ using the binomial theorem.
第一性原理可以推广到更高次幂，但代数运算会更长。对于 f(x) = x³，需要使用二项式定理展开 (x+h)³。
(x+h)³ = x³ + 3x²h + 3xh² + h³
btracting x³ and dividing by h gives 3x² + 3xh + h². Taking the limit as h → 0 yields 3x².
减去 x³ 并除以 h 得到 3x² + 3xh + h²。取 h → 0 时的极限得到 3x²。
| Function f(x) | Derivative f'(x) |
|---|---|
| x² | 2x |
| x³ | 3x² |
| xⁿ | nxⁿ⁻¹ |
However, AQA requires you to show the full first principles working for specific examples, not just quote the power rule.
但是 AQA 要求你展示具体例题的完整第一性原理过程，而不仅仅是套用幂法则。
10. Common Mistakes in First Principles | 第一性原理中的常见错误
mistake is writing f(x+h) = x² + h² immediately, missing the 2xh term from the expansion.
一个常见错误是直接写成 f(x+h) = x² + h²，遗漏了展开式中的 2xh 项。
Another mistake is forgetting to form the difference f(x+h) − f(x) before dividing by h.
另一个错误是在除以 h 之前忘记构造差 f(x+h) − f(x)。
ubstitute h = 0 before cancelling, because this creates 0/0 and is mathematically invalid.
切勿在约分之前代入 h = 0，因为这会产生 0/0，在数学上是无效的。
Finally, always use lim notation at the correct stage. Leaving it out may lose marks in an AQA exam.
最后，一定要在正确的步骤使用 lim 符号。在 AQA 考试中省略它可能会失分。
11. Exam-Style Practice Question | 考试型练习题
Use first principles to differentiate f(x) = x² + 3x. Show each step clearly.
使用第一性原理求 f(x) = x² + 3x 的导数。清晰地展示每一步。
First, write f(x+h) = (x+h)² + 3(x+h). Expand to get x² + 2xh + h² + 3x + 3h.
首先写出 f(x+h) = (x+h)² + 3(x+h)。展开得到 x² + 2xh + h² + 3x + 3h。
Subtract f(x) = x² + 3x to leave 2xh + h² + 3h. Factorise h to get h(2x + h + 3).
减去 f(x) = x² + 3x 后剩下 2xh + h² + 3h。提取公因式 h 得到 h(2x + h + 3)。
Divide by h and take the limit as h → 0 to obtain f'(x) = 2x + 3.
除以 h 并取 h → 0 时的极限，得到 f'(x) = 2x + 3。
12. Summary and Key Takeaways | 总结与关键要点
First principles differentiation is the foundation of calculus. The limit definition f'(x) = limh→0 [f(x+h) − f(x)] / h must be memorised.
第一性原理求导是微积分的基础。极限定义 f'(x) = limh→0 [f(x+h) − f(x)] / h 必须记住。
For f(x) = x², the derivative is f'(x) = 2x. Always expand brackets fully, cancel h correctly, and only then take the limit.
对于 f(x) = x²，导数为 f'(x) = 2x。务必完整展开括号、正确约去 h，然后才能取极限。
Practise first principles with x², x³, and linear combinations so you can answer any AQA exam question confidently.
练习使用第一性原理处理 x²、x³ 以及线性组合，这样你就能自信地回答任何 AQA 考试题目。
Published by TutorHao | Maths 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