📚 Mastering Differentiation from First Principles | 从第一原理掌握微分
Differentiation from first principles is a foundational skill in Edexcel A-Level Pure Mathematics. It tests your understanding of limits, algebraic simplification, and the formal definition of the derivative. In this article, we break down the method step by step, showing how to derive the gradient function from the limit of a chord’s slope, and we link this to standard differentiation rules used in later topics.
从第一原理求导是 Edexcel A-Level 纯数学的基础技能。它考察你对极限、代数化简以及导数正式定义的理解。本文将逐步拆解该方法，展示如何从弦的斜率极限推导出梯度函数，并将其与后续专题中使用的标准求导法则联系起来。
1. The Formal Definition of the Derivative | 导数的正式定义
At A-Level, the derivative f'(x) is defined as the limit of the difference quotient as h approaches zero.
在 A-Level 中，导数 f'(x) 被定义为当 h 趋近于零时差商的极限。
f'(x) = limₕ→₀ [f(x+h) − f(x)] / h
Here, f(x+h) − f(x) is the change in y, and h is the change in x. This ratio represents the gradient of the chord joining two close points on the curve.
这里 f(x+h) − f(x) 是 y 的变化量，h 是 x 的变化量。该比值表示曲线上两个邻近点之间弦的梯度。
2. The Geometric Idea Behind the Limit | 极限背后的几何思想
Imagine a secant line through the points (x, f(x)) and (x+h, f(x+h)). As h becomes smaller, this secant gets closer to the tangent at x, and its slope approaches the instantaneous rate of change.
想象一条经过点 (x, f(x)) 和 (x+h, f(x+h)) 的割线。当 h 变小时，这条割线越来越接近 x 处的切线，其斜率也趋近于瞬时变化率。
This is why the derivative is called the gradient of the tangent. It is not an approximation but the limit of the secant slopes.
这就是导数被称为切线梯度的原因。它不是近似值，而是割线斜率的极限。
3. Setting Up the First Principles Calculation | 建立第一原理计算
To differentiate a function y = f(x) from first principles, you must first write f(x+h), then form f(x+h) − f(x), and finally divide by h before taking the limit.
要从第一原理求函数 y = f(x) 的导数，你必须先写出 f(x+h)，然后构造 f(x+h) − f(x)，最后除以 h 再取极限。
It is essential to expand and simplify fully before substituting h = 0, because direct substitution usually gives the indeterminate form 0/0.
在代入 h = 0 之前必须完全展开并化简，因为直接代入通常会得到 0/0 的不定式。
4. Worked Example: f(x) = x² | 例题：f(x) = x²
Start with f(x+h) = (x+h)² = x² + 2xh + h². Subtract f(x) and divide by h.
先写出 f(x+h) = (x+h)² = x² + 2xh + h²。减去 f(x) 并除以 h。
[f(x+h) − f(x)] / h = (x² + 2xh + h² − x²) / h = (2xh + h²) / h = 2x + h
Now take the limit as h approaches zero: f'(x) = limₕ→₀ (2x + h) = 2x.
现在令 h 趋近于零取极限：f'(x) = limₕ→₀ (2x + h) = 2x。
This confirms the familiar rule that the derivative of x² is 2x.
这证实了熟悉的法则：x² 的导数是 2x。
5. Worked Example: f(x) = x³ | 例题：f(x) = x³
For f(x) = x³, expand f(x+h) = (x+h)³ = x³ + 3x²h + 3xh² + h³.
对于 f(x) = x³，展开 f(x+h) = (x+h)³ = x³ + 3x²h + 3xh² + h³。
[f(x+h) − f(x)] / h = (x³ + 3x²h + 3xh² + h³ − x³) / h = 3x² + 3xh + h²
Taking the limit h → 0 gives f'(x) = 3x², since the terms containing h vanish.
令 h → 0 取极限得到 f'(x) = 3x²，因为含有 h 的项趋于零。
6. Extending to the General Power Rule | 推广到一般幂法则
For f(x) = xⁿ, expanding (x+h)ⁿ by the binomial theorem gives xⁿ + n xⁿ⁻¹h + terms involving h² or higher.
对于 f(x) = xⁿ，用二项式定理展开 (x+h)ⁿ 得到 xⁿ + n xⁿ⁻¹h + 含有 h² 或更高次的项。
[f(x+h) − f(x)] / h = n xⁿ⁻¹ + h × (terms in h)
As h → 0, the extra terms vanish, leaving f'(x) = n xⁿ⁻¹. This is the standard power rule for positive integer powers.
当 h → 0 时，额外项消失，剩下 f'(x) = n xⁿ⁻¹。这正是正整数指数时的标准幂法则。
7. Exam Pitfalls: Algebra and Limit Notation | 考试陷阱：代数与极限记号
A common mistake is to write the limit only at the final line but cancel h incorrectly earlier. Always show the full difference quotient and the explicit limₕ→₀ step.
一个常见错误是只在最后一步写极限，但在前面错误地约去 h。务必展示完整的差商以及明确的 limₕ→₀ 步骤。
Another pitfall is forgetting to expand f(x+h) correctly, especially with negative signs or brackets. For example, f(x+h) for f(x) = x² − 3x is not x² + h² − 3x + h; it must be (x+h)² − 3(x+h).
另一个陷阱是忘记正确展开 f(x+h)，特别是在负号或括号情况下。例如，f(x) = x² − 3x 的 f(x+h) 不是 x² + h² − 3x + h；必须是 (x+h)² − 3(x+h)。
8. First Principles with f(x) = 1/x | 用第一原理求 f(x) = 1/x 的导数
With algebraic fractions, the key is to combine the numerator f(x+h) − f(x) into a single fraction before dividing by h.
对于代数分式，关键是在除以 h 之前将分子 f(x+h) − f(x) 合并为一个分式。
f(x+h) = 1/(x+h), so [f(x+h) − f(x)] / h = [1/(x+h) − 1/x] / h = [ (x − (x+h)) / x(x+h) ] / h = −1 / [x(x+h)]
Taking h → 0 gives f'(x) = −1/x². This matches the power rule x⁻¹ → −x⁻².
令 h → 0 得到 f'(x) = −1/x²。这与幂法则 x⁻¹ → −x⁻² 一致。
9. First Principles with f(x) = √x | 用第一原理求 f(x) = √x 的导数
For square roots, rationalising the numerator is the standard trick. Use (√(x+h) − √x)(√(x+h) + √x) = h to simplify the difference quotient.
对于平方根函数，有理化分子是标准技巧。利用 (√(x+h) − √x)(√(x+h) + √x) = h 化简差商。
[√(x+h) − √x] / h = [ (√(x+h) − √x)(√(x+h) + √x) ] / [h(√(x+h) + √x)] = 1 / (√(x+h) + √x)
As h → 0, √(x+h) → √x, so f'(x) = 1/(2√x), valid for x > 0.
当 h → 0 时，√(x+h) → √x，因此 f'(x) = 1/(2√x)，在 x > 0 时成立。
10. Linking First Principles to Standard Differentiation Rules | 将第一原理与标准求导法则联系
First principles is not just an isolated exam topic; it justifies the derivative of any function before shortcuts are used. Once you have proved the power rule, linearity and the product rule follow in later topics.
第一原理不仅是一个孤立的考试专题；它证明了任何函数在使用简便方法之前的导数来源。一旦证明了幂法则，后续专题中的线性性和乘积法则便可相应推出。
Edexcel examiners often ask candidates to differentiate a function from first principles in the Pure 1 or Pure 2 paper, usually for a quadratic, reciprocal, or square root function.
Edexcel 考官经常在 Pure 1 或 Pure 2 试卷中要求考生从第一原理求函数导数，通常是二次函数、倒数函数或平方根函数。
11. Mark Scheme Must-Haves | 评分方案必备要点
- Write f(x+h) explicitly and expand correctly. / 明确写出 f(x+h) 并正确展开。
- Show the difference quotient [f(x+h) − f(x)] / h. / 展示差商 [f(x+h) − f(x)] / h。
- Simplify until you can cancel h safely. / 化简到能够安全约去 h。
- Write limₕ→₀ before substituting h = 0. / 在代入 h = 0 前写出 limₕ→₀。
- State the final derivative with correct notation, such as f'(x) or dy/dx. / 用正确记号写出最终导数，如 f'(x) 或 dy/dx。
12. Exam-Style Practice Question | 考试风格练习题
Question: Differentiate f(x) = x² + 5x from first principles.
题目：用第一原理求 f(x) = x² + 5x 的导数。
Solution outline: f(x+h) = (x+h)² + 5(x+h) = x² + 2xh + h² + 5x + 5h. Then f(x+h) − f(x) = 2xh + h² + 5h. Divide by h to get 2x + h + 5. Let h → 0 to obtain f'(x) = 2x + 5.
解题要点：f(x+h) = (x+h)² + 5(x+h) = x² + 2xh + h² + 5x + 5h。然后 f(x+h) − f(x) = 2xh + h² + 5h。除以 h 得到 2x + h + 5。令 h → 0 得到 f'(x) = 2x + 5。
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