📚 Worked Example 2.3.1: Finding the Derivative of f(x)=x² from First Principles | 例题 2.3.1：由第一性原理求 f(x)=x² 的导数
In this worked example, we use the limit definition of the derivative to find f ' A A-Level Pure Mathematics skill that tests your ability to manipulate algebraic expressions and handle limits correctly. The method shown here appears frequently in Paper 1 and forms the foundation for all differentiation techniques.
在这个例题中，我们使用导数的极限定义来求 f(x)=x² 的导数 f '(x)。这是 AQA A-Level 纯数学的一项标准技能，考查你处理代数式以及正确使用极限的能力。这里展示的方法经常出现在试卷 1 中，并且是所有求导技巧的基础。
1. Learning Objectives | 学习目标
By the end of this worked example, you should be able to state the first principles definition of the derivative, substitute a quadratic function into the difference quotient, simplify the algebra correctly, and evaluate the limit as h approaches zero.
学完这个例题后，你应当能够写出导数的第一性原理定义，将二次函数代入差商，正确化简代数式，并求 h 趋于零时的极限。
the same method extends to any polynomial function, making it a foundational skill for AQA Paper 1. Mastery of this example will also help you understand why the power rule works rather than simply memorising it.
本例以 f(x)=x² 为重点，但同样的方法可推广到任意多项式函数，因此它是 AQA 试卷 1 的基础技能。掌握这个例题还能帮助你理解幂函数法则为什么成立，而不只是机械记忆。
- Write the limit definition of the derivative accurately.
- Expand and simplify a quadratic difference quotient.
- Correctly cancel the common factor h and evaluate the limit.
- Avoid common algebraic and notation errors.
这些学习目标包括：准确写出导数的极限定义；展开并化简二次差商；正确约去公因子 h 并计算极限；避免常见的代数与符号错误。
2. The First Principles Definition | 第一性原理定义
The derivative of a function f at a point x is defined as the limit of the average rate of change over a small interval h, as h tends to zero. Geometrically, the difference quotient represents the gradient of a secant line through the points (x, f(x)) and (x+h, f(x+h)). As h approaches zero, this secant becomes the tangent line at x.
函数 f 在点 x 处的导数定义为当 h 趋于零时，小区间 h 上平均变化率的极限。从几何上看，差商表示经过点 (x, f(x)) 和 (x+h, f(x+h)) 的割线斜率。当 h 趋于零时，这条割线就变成点 x 处的切线。
f '(x) = limh→0 [ f(x+h) − f(x) ] / h
This definition is called differentiation from first principles. The expression inside the limit is the difference quotient, and h represents a small change in x. It is essential to keep the limit symbol in every line until the actual evaluation takes place.
这个定义称为从第一性原理求导。极限内的表达式就是差商，h 表示 x 的一个微小变化。在真正代入求值之前，每一行都必须保留极限符号。
In AQA exams, you may be asked to derive the derivative of simple functions such as x², x³, or 1/x using this definition, so it is essential to memorise it and write it correctly at the start of your answer.
在 AQA 考试中，你可能需要利用这个定义推导 x²、x³ 或 1/x 等简单函数的导数，因此必须牢记该定义并在解题一开始就正确写出。
3. Step 1: Write the Difference Quotient | 第1步：写出差商
iable x is treated as a constant when we vary h, so (x+h)² is an expression in both x and h.
从一般的第一性原理公式出发，写出给定函数的差商。这里 f(x)=x²，因此 f(x+h)=(x+h)²。当我们让 h 变化时，变量 x 被视作常数，因此 (x+h)² 是一个既含 x 又含 h 的表达式。
f '(x) = limh→0 [ (x+h)² − x² ] / h
Notice that we keep the limit symbol throughout the calculation. Losing the limit is one of the most common errors in AQA mark schemes. The expression inside the brackets is the difference quotient, and it is not yet simplified.
注意，在整个计算过程中我们始终保留极限符号。丢失极限符号是 AQA 评分标准中最常见的错误之一。括号内的表达式就是差商，目前尚未化简。
d become zero and the expression would be undefined. Instead, we need to manipulate the numerator algebraically to remove the h in the denominator.
这一步为代数运算搭建了结构。先不要把 h=0 代入，因为分母会变成零，表达式就没有定义。相反，我们需要对分子进行代数变形，以消去分母中的 h。
4. Step 2: Substitute and Expand | 第2步：代入并展开
Expand (x+h)² using the binomial expansion or standard identity. The correct expansion is x² + 2xh + h². Remember that (x+h)² means (x+h)(x+h), so it has three terms, not two. A very common error is to write x² + h², omitting the middle term 2xh.
利用二项式展开或标准恒等式展开 (x+h)²。正确的展开式是 x² + 2xh + h²。请记住 (x+h)² 表示 (x+h)(x+h)，所以它有三项，而不是两项。一个非常常见的错误是写成 x² + h²，遗漏了中间项 2xh。
(x+h)² = x² + 2xh + h²
Now substitute this expansion into the numerator of the difference quotient. The numerator becomes (x² + 2xh + h²) − x². You should write this out fully before simplifying to make your working clear for the examiner.
现在将这个展开式代入差商的分子中。分子变为 (x² + 2xh + h²) − x²。在化简之前，你应当完整地写出这一步骤，以便考官看清楚你的解题过程。
[ x² + 2xh + h² − x² ] / h
The two x² terms cancel exactly. This cancellation is vital because it removes the constant term and leaves only terms that contain h. After cancellation, the numerator is 2xh + h².
两个 x² 项恰好相消。这种相消非常关键，因为它消去了常数项，只留下含 h 的项。相消后，分子为 2xh + h²。
5. Step 3: Simplify the Numerator | 第3步：化简分子
After cancelling x², the numerator becomes 2xh + h². Both terms have a common factor of h. Factoring out h gives h(2x + h). This step is crucial because it allows us to cancel the h in the denominator.
约去 x² 后，分子变为 2xh + h²。这两项都有公因子 h。提取公因子 h 后得到 h(2x + h)。这一步非常关键，因为它使我们能够约去分母中的 h。
2xh + h² = h(2x + h)
The difference quotient can now be written as h(2x + h) divided by h. Since h approaches zero but is not equal to zero, we may divide the numerator and denominator by h. This operation is valid because in the limiting process h is never exactly zero, so we are not dividing by zero.
差商现在可以写成 h(2x + h) 除以 h。由于 h 趋于零但又不等于零，我们可以将分子和分母同时除以 h。这一操作是合理的，因为在极限过程中 h 永远不会真正等于零，因此我们并不是在除以零。
[ h(2x + h) ] / h
At this stage, some students try to substitute h=0 before cancelling,
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