📚 Example 5.6.1: Differentiation from First Principles | 例题 5.6.1：从第一性原理求导
first principles, using the formal limit definition of the derivative. It is a core AQA A-Level Pure Maths skill that is often tested in AS and A-Level exams.
本例题展示如何利用导数的正式极限定义，从第一性原理求函数 f(x) = x² 的导数。这是 AQA A-Level 纯数学的核心技能，经常在 AS 和 A-Level 考试中出现。
Understanding first principles is more than a routine calculation: it proves that you know what a derivative actually measures. The AQA mark scheme rewards clear algebraic steps and correct limit notation.
理解第一性原理不仅仅是一种常规计算：它证明你真正理解导数所衡量的含义。AQA 评分方案奖励清晰的代数步骤和正确的极限符号。
1. What This Example Demonstrates | 本例题演示内容
The question asks you to find f'(x) for f(x) = x² using the definition f'(x) = limₕ→₀ (f(x+h) − f(x))/h, not by applying the power rule.
题目要求你使用定义 f'(x) = limₕ→₀ (f(x+h) − f(x))/h 来求 f(x) = x² 的导数 f'(x)，而不是直接套用幂法则。
This example tests three key skills: correct substitution into the difference quotient, algebraic expansion of brackets, and evaluation of a limit as h approaches 0.
本例题考查三项关键技能：正确代入差商、括号的代数展开，以及计算 h 趋向于 0 时的极限。
On AQA papers, first-principles questions often carry 4 to 6 marks. You must show each step to gain full credit.
在 AQA 试卷中，第一性原理题通常占 4 到 6 分。你必须展示每一步才能获得满分。
2. The Formal Definition of the Derivative | 导数的正式定义
The derivative of a function f at a point x is defined as the limit of the average rate of change over an interval of width h, as h tends to zero.
函数 f 在点 x 处的导数定义为区间宽度为 h 上的平均变化率当 h 趋于零时的极限。
f'(x) = limₕ→₀ (f(x+h) − f(x))/h
In this formula, f(x+h) is the value of the function at a point slightly to the right of x, and f(x) is the value at the starting point. The denominator h is the horizontal step.
在这个公式中，f(x+h) 是函数在 x 右侧稍远处点的值，f(x) 是起点处的函数值。分母 h 是水平步长。
The expression (f(x+h) − f(x))/h is called the difference quotient. Geometrically, it represents the gradient of a chord between two points on the curve.
表达式 (f(x+h) − f(x))/h 称为差商。从几何上看，它表示曲线上两点之间弦的斜率。
As h becomes smaller, the chord gets closer to the tangent line at x, and the difference quotient approaches the gradient of that tangent.
当 h 越来越小，弦越来越接近 x 处的切线，差商就趋近于该切线的斜率。
3. Step 1: Substitute into the Difference Quotient | 步骤1：代入差商
nction f(x) = x². Write the expression for f(x+h) by replacing every x in f(x) with (x+h).
从给定函数 f(x) = x² 开始。将 f(x) 中的每个 x 都替换为 (x+h)，写出 f(x+h) 的表达式。
f(x+h) = (x+h)²
Then form the difference quotient exactly as the definition states.
然后严格按照定义写出差商。
f'(x) = limₕ→₀ ((x+h)² − x²)/h
Do not skip this substitution step. AQA examiners look for explicit evidence that you have replaced f(x+h) correctly, even before simplifying.
不要跳过这一步代入。AQA 考官会寻找你正确替换 f(x+h) 的明确证据，即使还没有化简。
4. Step 2: Expand (x+h)² | 步骤2：展开 (x+h)²
bracket using the binomial square identity (a+b)² = a² + 2ab + b², where a = x and b = h.
使用完全平方公式 (a+b)² = a² + 2ab + b² 展开括号，其中 a = x，b = h。
(x+h)² = x² + 2xh + h²
Substitute this expansion into the numerator of the difference quotient.
将这个展开式代入差商的分子中。
f'(x) = limₕ→₀ (x² + 2xh + h² − x²)/h
ential because it removes the singularity that would make the fraction undefined at h = 0.
x² 项和 −x² 项立即抵消。这个抵消是必要的，因为它消除了使分式在 h = 0 处无意义的奇点。
5. Step 3: Simplify the Numerator | 步骤3：化简分子
After cancelling x², the numerator becomes 2xh + h². Notice that both remaining terms contain a factor of h.
抵消 x² 之后，分子变为 2xh + h²。注意剩下的两项都含有因子 h。
f'(x) = limₕ→₀ (2xh + h²)/h
step is important because you cannot simply set h = 0 while h is still in the denominator.
从分子中提出因子 h。这一步很重要，因为当 h 仍然在分母中时，你不能直接将 h 设为零。
2xh + h² = h(2x + h)
The difference quotient can now be written in a form that allows cancellation.
差商现在可以写成允许约分的形式。
6. Step 4: Divide by h | 步骤4：除以 h
Divide the numerator h(2x + h) by the denominator h. Since h is not zero during the limiting process, the cancellation is valid.
将分子 h(2x + h) 除以分母 h。由于在极限过程中 h 不为零，这种约分是有效的。
f'(x) = limₕ→₀ (2x + h)
The expression inside the limit is now a simple linear polynomial in h. There is no longer a denominator that vanishes as h approaches zero.
极限内的表达式现在是关于 h 的简单线性多项式。不再存在当 h 趋近于零时消失的分母。
ho forget to factor h often try to substitute h = 0 too early and get an invalid 0/0 form.
这是使极限易于计算的关键代数步骤。忘记提出 h 的学生往往过早代入 h = 0，得到无效的 0/0 形式。
7. Step 5: Take the Limit as h→0 | 步骤5：取 h→0 时的极限
Now evaluate the limit by letting h approach 0 in the expression 2x + h. The term 2x is independent of h, so it remains unchanged.
现在通过让 h 趋近于 0 来计算表达式 2x + h 的极限。项 2x 与 h 无关，因此保持不变。
f'(x) = 2x + 0 = 2x
Therefore, the derivative of f(x) = x² from first principles is f'(x) = 2x. This matches the result given by the power rule d/dx(xⁿ) = nxⁿ⁻¹.
因此，由第一性原理求得 f(x) = x² 的导数为 f'(x) = 2x。这与幂法则 d/dx(xⁿ) = nxⁿ⁻¹ 给出的结果一致。
The agreement between the two methods confirms that the first-principles calculation has been carried out correctly.
两种方法结果一致，说明第一性原理的计算过程是正确的。
8. Full Worked Solution Summary | 完整解题过程总结
The complete solution can be presented as a single chain of equalities, which is the style AQA examiners expect in a "show that" or "prove from first principles" question.
完整解答可以呈现为一个连续的等式链，这也是 AQA 考官在“证明”或“从第一性原理推导”类题目中所期望的书写方式。
| Step | Working |
| Definition | f'(x) = limₕ→₀ (f(x+h) − f(x))/h |
| Substitute | = limₕ→₀ ((x+h)² − x²)/h |
| Expand | = limₕ→₀ (x² + 2xh + h² − x²)/h |
| Simplify | = limₕ→₀ (2xh + h²)/h |
| Factor and cancel | = limₕ→₀ (2x + h) |
| Take limit | = 2x |
When writing your final answer, state the result clearly: "Hence, f'(x) = 2x." This ensures the examiner sees that you have completed the argument.
书写最终答案时，请清楚地写出：“因此，f'(x) = 2x。” 这样可以确保考官看到你已经完成了整个论证。
9. Common Mistakes and How to Avoid Them | 常见错误及避免方法
Many students lose marks on first-principles questions by making avoidable algebraic or notational errors. The table below lists the most common ones.
许多学生在第一性原理题中因可避免的代数或符号错误而失分。下表列出了最常见的错误。
| Common mistake | Why it is wrong | Correct approach |
| Setting h = 0 before cancelling | Gives 0/0, which is undefined | Factor h first, then cancel |
| Wrong expansion of (x+h)² | Missing the 2xh term | Use (a+b)² = a² + 2ab + b² |
| Not writing limₕ→₀ at each step | Inaccurate mathematical notation | Keep the limit symbol until the final evaluation |
| Confusing f(x+h) with f(x) + h | Incorrect substitution | Replace x with (x+h) everywhere |
The AQA mark scheme often awards separate marks for correct substitution, correct expansion, simplification, and the final limit. Showing all steps protects you from losing several marks for one small slip.
AQA 评分方案通常对正确代入、正确展开、化简和最终极限分别给分。展示所有步骤可以防止因一个小失误而丢失多分。
10. AQA Exam Tips and Practice | AQA 考试技巧与练习
To master this type of question, practise differentiating simple powers and polynomial terms from first principles, not just x².
要掌握这类题目，请练习从第一性原理求简单幂函数和多项式项的导数，而不仅仅是 x²。
ing variation: use the same limit definition to show that the derivative of f(x) = x³ is f'(x) = 3x². Start by expanding (x+h)³ = x³ + 3x²h + 3xh² + h³.
尝试以下变式：使用相同的极限定义证明 f(x) = x³ 的导数为 f'(x) = 3x²。从展开 (x+h)³ = x³ + 3x²h + 3xh² + h³ 开始。
exercise is to find the derivative of the constant function f(x) = 5. You should obtain f'(x) = 0, which reinforces the geometric idea that a horizontal line has zero gradient.
另一个有用的练习是求常数函数 f(x) = 5 的导数。你应该得到 f'(x) = 0，这强化了水平线斜率为零的几何概念。
In the exam, allocate one line per algebraic step. If you make an error, the examiner can still award method marks for the later steps that follow correctly from your working.
在考试中，每个代数步骤分配一行。如果你出现错误，考官仍然可以根据你后续正确的解题步骤给予方法分。
Finally, remember that first principles is not just a standalone topic: it underpins the proof of differentiation rules, and AQA may ask you to explain this connection in words.
最后，请记住第一性原理不仅仅是一个独立的主题：它是微分法则证明的基础，AQA 可能会要求你用文字解释这一联系。
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