📚 Example 4.6.1: Differentiation from First Principles | 例4.6.1：从第一原理求导
Differentiation from first principles is the formal way to find the gradient of a curve at a single point. In AQA A Level Mathematics, this method is tested early in the calculus topics because it builds the limit definition of the derivative. Example 4.6.1 walks through the process for a simple power function, showing every algebraic step that leads to the derivative formula.
从第一原理求导是求曲线在单点处梯度的正式方法。在 AQA A Level 数学中，这一方法在微积分主题初期就会考查，因为它建立了导数的极限定义。例4.6.1 演示了对一个简单幂函数使用该过程的完整步骤，展示了推出导数公式的每一步代数运算。
1. What Is Differentiation from First Principles? | 什么是第一原理求导？
Differentiation from first principles uses the idea that the derivative is the limit of the average rate of change as the interval shrinks to zero. Instead of relying on memorised rules, you start from a secant line between two close points on a curve and then let the distance between those points tend to zero.
第一原理求导利用的思想是：导数是平均变化率在区间趋近于零时的极限。我们不依赖记忆中的求导法则，而是从曲线上两个接近点之间的割线出发，然后让这两点之间的距离趋近于零。
This definition is fundamental because it explains why the gradient of a tangent is not exactly the gradient of any secant, but the limiting value. In AQA exams, you may be asked to apply this definition to quadratic, cubic, or simple algebraic functions.
这一定义之所以根本，是因为它解释了为什么切线的梯度不是任何割线的梯度，而是一个极限值。在 AQA 考试中，你可能会被要求将这一定义应用于二次函数、三次函数或简单的代数函数。
2. The Gradient of a Chord and the Tangent | 割线与切线的梯度
s P(x, f(x)) and Q(x+h, f(x+h)). The gradient of the chord PQ is the change in y divided by the change in x. This gives the difference quotient [f(x+h) - f(x)] / h.
考虑曲线 y = f(x) 以及两点 P(x, f(x)) 和 Q(x+h, f(x+h))。弦 PQ 的梯度等于 y 的变化量除以 x 的变化量。这样就得到差分商 [f(x+h) - f(x)] / h。
As h becomes smaller, Q moves closer to P, and the chord becomes closer to the tangent at P. The tangent gradient is therefore the limit of the difference quotient as h approaches zero. This geometric idea is the heart of first principles differentiation.
当 h 变小时，Q 向 P 靠近，弦越来越接近 P 点处的切线。因此，切线梯度就是差分商在 h 趋近于零时的极限。这个几何思想正是第一原理求导的核心。
3. The Formal Limit Definition | 正式的极限定义
The derivative of f at a point x is defined by the following limit, provided the limit exists. The notation f'(x) is read as 'f dash of x' and represents the gradient function of the original curve.
函数 f 在点 x 处的导数由以下极限定义，前提是该极限存在。记号 f'(x) 读作 'f dash of x'，表示原曲线的梯度函数。
f'(x) = lim (h → 0) [f(x+h) - f(x)] / h
This formula is the starting point for all first principles calculations. You must be able to write it down accurately, substitute the given function, expand brackets, simplify, and then evaluate the limit by setting h equal to zero only after cancellation.
这个公式是所有第一原理计算的起点。你必须能够准确地写出它，代入给定的函数，展开括号，化简，并且只有在约分之后才能将 h 设为零来求极限。
4. Worked Example 4.6.1 Step 1: Setting Up the Difference Quotient | 例4.6.1 第一步：建立差分商
ing f(x+h) = (x+h)². Then form the difference quotient [f(x+h) - f(x)] / h as shown below.
在这个例子中，我们用第一原理对 f(x) = x² 求导。首先写出 f(x+h) = (x+h)²。然后如下所示建立差分商 [f(x+h) - f(x)] / h。
f'(x) = lim (h → 0) [(x+h)² - x²] / h
irectly, because that would make the denominator zero. Instead, expand the numerator completely and look for factors of h that can be cancelled with the denominator.
在这个阶段，不要直接将 h = 0 代入，因为那样会使分母为零。相反，应完全展开分子，并寻找可以与分母约去的 h 因子。
5. Worked Example 4.6.1 Step 2: Simplifying and Taking the Limit | 例4.6.1 第二步：化简并取极限
Expand (x+h)² to get x² + 2xh + h². Then subtract x², leaving 2xh + h² in the numerator. The full expression becomes (2xh + h²) / h.
展开 (x+h)² 得到 x² + 2xh + h²。然后减去 x²，分子剩下 2xh + h²。整个表达式变为 (2xh + h²) / h。
f'(x) = lim (h → 0) (2xh + h²) / h
he limiting process, you can cancel the common factor h. This leaves lim (h → 0) (2x + h).
从分子中提取公因式 h，得到 h(2x + h) / h。由于在取极限的过程中 h 不为零，你可以约去公因式 h。这样就剩下 lim (h → 0) (2x + h)。
he derivative is therefore 2x, which matches the standard power rule result for f(x) = x².
现在令 h 趋近于零来取极限。项 2x 保持不变，而 h 变为零。因此导数为 2x，这与 f(x) = x² 的标准幂法则结果一致。
6. Extending the Method to f(x) = xⁿ | 将方法推广到 f(x) = xⁿ
s using the binomial expansion to write (x+h)ⁿ as a sum of terms, then cancelling the factor h.
同样的第一原理过程可以应用于 x 的更高次幂，例如 f(x) = x³ 或 f(x) = x⁴。关键步骤是使用二项式展开将 (x+h)ⁿ 写成多项之和，然后约去因子 h。
= (3x²h + 3xh² + h³) / h = 3x² + 3xh + h². Taking the limit as h tends to zero gives 3x².
对于 f(x) = x³，你会得到 h)³ - x³] / h = [x³ + 3x²h + 3xh² + h³ - x³] / h = (3x²h + 3xh² + h³) / h = 3x² + 3xh + h²。令 h 趋近于零取极限，结果为 3x²。
7. Connecting to the Power Rule | 与幂法则的联系
n shows that the derivative is n xⁿ⁻¹. In Example 4.6.1, n = 2, so the result is 2x¹ = 2x.
第一原理方法为幂法则提供了一个证明。对于 f(x) = xⁿ，其中 n 是正整数，反复展开表明导数为 n xⁿ⁻¹。在例4.6.1 中，n = 2，因此结果为 2x¹ = 2x。
Understanding this connection helps you check your first principles work. If your limit result does not match the power rule for simple polynomial functions, you have likely made an algebraic error in expanding or simplifying.
理解这种联系有助于检查你的第一原理运算。对于简单的多项式函数，如果你的极限结果与幂法则不一致，那么你很可能在展开或化简时出现了代数错误。
8. Common Mistakes and How to Avoid Them | 常见错误及如何避免
ined expression 0/0. Always expand and cancel the common factor h before taking the limit. Another common error is incorrectly expanding brackets such as (x+h)².
一个常见错误是过早代入 h = 0，从而产生未定义的表达式 0/0。务必在取极限之前展开并约去公因式 h。另一个常见错误是错误地展开诸如 (x+h)² 这样的括号。
+ 3(x+h)] - [x² + 3x], and each part must be handled carefully. Write every step to avoid sign errors.
学生有时会忘记正确减去 f(x)，尤其是当函数有多个项时。对于 f(x) = x² + 3x，分子为 [(x+h)² + 3(x+h)] - [x² + 3x]，每一项都必须小心处理。写出每一步以避免符号错误。
9. Exam Tips for AQA Mark Schemes | AQA 评分标准的考试技巧
h, and then clearly state the limit as h → 0. Missing the limit statement can cost a mark.
在 AQA 考试中，第一原理题通常占四到六分。你必须写出定义、代入函数、展开并化简、约去 h，然后清楚地写出 h → 0 时的极限。遗漏极限陈述可能会丢分。
Even if your final answer is correct, you must show the intermediate algebra to gain full marks. AQA mark schemes allocate method marks for the difference quotient and for the cancellation step. Practise writing the full solution in a logical order under timed conditions.
即使最终答案正确，你也必须展示中间的代数步骤才能获得满分。AQA 评分标准会为差分商和约分步骤分配方法分。练习在限时条件下按逻辑顺序写出完整解答。
10. Practice Questions and Summary | 练习题与总结
= 3x², and f(x) = x³ + 1. Check your final answers by comparing them with the power rule.
尝试以下练习题来巩固理解。用第一原理求导：f(x) = x² + 2x、f(x) = 3x² 和 f(x) = x³ + 1。将最终答案与幂法则比较来检查。
Differentiation from first principles is not just a formula to memorise; it is a concept that links average gradients to instantaneous gradients. Mastering Example 4.6.1 gives you the confidence to handle any first principles question in AQA A Level Mathematics.
第一原理求导不仅仅是要记忆的公式；它是一个将平均梯度与瞬时梯度联系起来的核心概念。掌握例4.6.1 能让你有信心应对 AQA A Level 数学中任何第一原理题目。
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