📚 Mastering First Principles Differentiation for AQA A-Level Maths | 掌握 AQA A-Level 数学第一原理求导
In AQA A-Level Mathematics Paper 1, a typical Question 1 often asks students to differentiate a simple polynomial from first principles, and then use the derivative to find the gradient of a tangent at a given point.
在 AQA A-Level 数学 Paper 1 中，典型的第一题通常要求学生从第一原理出发对简单多项式求导，然后利用导数求曲线上某一点处切线的斜率。
This topic appears regularly because it tests careful algebra, limit notation, and the connection between a derivative and the gradient of a curve.
这个主题经常出现，因为它考查细致的代数运算、极限记号，以及导数与曲线斜率之间的联系。
1. What Question 1 Usually Asks | 第一题通常考查什么
arts: first, differentiate a given function f(x) from first principles; second, evaluate f'(a) to give the gradient at x = a, or use that gradient to find an equation of a tangent.
AQA Paper 1 第一题通常是一道结构简短的两问：先用第一原理求给定函数 f(x) 的导数；再计算 f'(a) 给出 x = a 处的斜率，或者利用该斜率求切线方程。
Typical functions include x², 3x² − 5x, 2x² + 7x − 1, or similar quadratics where the expansion of (x + h)² is manageable and the cancellation of terms is clear.
常见函数包括 x²、3x² − 5x、2x² + 7x − 1 或类似的二次函数，因为 (x + h)² 的展开容易处理，项的抵消也很清晰。
Because this is usually the first question on the paper, examiners expect tidy, methodical working that follows the definition exactly.
由于这通常是试卷上的第一题，考官希望看到整洁、有条理的步骤，并且严格遵循定义。
2. Formal Definition of the Derivative | 导数的正式定义
ivative of f(x) at a general point x is defined as the limit of the difference quotient: f'(x) = limₕ→₀ [f(x + h) − f(x)] / h, provided this limit exists.
函数 f(x) 在一般点 x 处的导数定义为差商的极限：f'(x) = limₕ→₀ [f(x + h) − f(x)] / h，前提是该极限存在。
In an AQA first principles question you must quote this definition at the start; a method mark is often awarded for writing it down correctly with f(x + h) and f(x) in the numerator.
在 AQA 第一原理题中，你必须在开头写出这个定义；只要分子中的 f(x + h) 和 f(x) 书写正确，评分标准通常就会给出方法分。
f '(x) = limₕ→₀ (f(x + h) − f(x)) / h
3. Setting Up f(x + h) | 建立 f(x + h)
appears, using brackets around x + h. For example, if f(x) = x², then f(x + h) = (x + h)².
将 x + h 原封不动地代入函数中出现 x 的位置，并在 x + h 外加括号。例如，若 f(x) = x²，则 f(x + h) = (x + h)²。
ch as f(x) = 3x² − 5x + 2, because f(x + h) must be 3(x + h)² − 5(x + h) + 2, not 3x² + h.
当函数中含有系数或负项时，括号尤为重要。例如 f(x) = 3x² − 5x + 2，那么 f(x + h) 必须写成 3(x + h)² − 5(x + h) + 2，而不是 3x² + h。
Take your time with this substitution step: many AQA candidates lose accuracy here, and an incorrect f(x + h) ruins the entire derivative.
这一步代入要仔细：许多 AQA 考生就在这里丢分，如果 f(x + h) 写错，整个导数都会出错。
4. Expanding the Difference Quotient | 展开差商
After writing f(x + h) − f(x), expand all brackets and collect like terms. In a correctly set up quadratic example, the original f(x) terms always cancel, leaving only terms that contain h.
写出 f(x + h) − f(x) 之后，展开所有括号并合并同类项。在正确建立的二次函数例子中，原来的 f(x) 项总会抵消，只留下含 h 的项。
For f(x) = x², the numerator is (x + h)² − x² = x² + 2xh + h² − x² = 2xh + h².
对于 f(x) = x²，分子为 (x + h)² − x² = x² + 2xh + h² − x² = 2xh + h²。
step: 2xh + h² = h(2x + h), which allows the h in the denominator to be cancelled cleanly.
对分子进行因式分解是关键一步：2xh + h² = h(2x + h)，这样就能与分母中的 h 干净地约掉。
5. Dividing by h and Taking the Limit | 除以 h 并取极限
Divide the whole numerator by h, then let h tend to 0. Any term still containing h will vanish in the limit, leaving the derivative as a function of x only.
将整个分子除以 h，然后令 h 趋于 0。任何仍含有 h 的项都会在极限中消失，最终得到只含 x 的导数。
For f(x) = x², the quotient is (2xh + h²)/h = 2x + h. As h → 0, h disappears, so f'(x) = 2x.
对于 f(x) = x²，差商为 (2xh + h²)/h = 2x + h。当 h → 0 时，h 消失，因此 f'(x) = 2x。
Note that we divide by h before taking the limit because h is not exactly zero in the difference quotient; it is only allowed to approach zero.
注意，我们先除以 h 再取极限，因为在差商中 h 并不严格等于零，它只是被允许趋于零。
6. Worked Example with Coefficients | 含系数的完整示例
Let f(x) = 3x² − 5x + 2. First, find f(x + h) = 3(x + h)² − 5(x + h) + 2.
设 f(x) = 3x² − 5x + 2。首先求 f(x + h) = 3(x + h)² − 5(x + h) + 2。
Expand to get f(x + h) = 3x² + 6xh + 3h² − 5x − 5h + 2. Then subtract f(x):
展开得到 f(x + h) = 3x² + 6xh + 3h² − 5x − 5h + 2。然后减去 f(x)：
f(x + h) − f(x) = (3x² + 6xh + 3h² − 5x − 5h + 2) − (3x² − 5x + 2) = 6xh + 3h² − 5h.
f(x + h) − f(x) = (3x² + 6xh + 3h² − 5x − 5h + 2) − (3x² − 5x + 2) = 6xh + 3h² − 5h。
as h(6x + 3h − 5), divide by h to get 6x + 3h − 5, and take the limit as h → 0 to obtain f'(x) = 6x − 5.
将分子因式分解为 h(6x + 3h − 5)，除以 h 得到 6x + 3h − 5，再取 h → 0 的极限，得到 f'(x) = 6x − 5。
7. Evaluating the Gradient at a Point | 求某一点处的斜率
If part (b) asks for the gradient of the curve at x = 2, substitute x = 2 into f'(x). For f'(x) = 6x − 5, f'(2) = 12 − 5 = 7.
如果第 (b) 问要求曲线在 x = 2 处的斜率，就把 x = 2 代入 f'(x)。对于 f'(x) = 6x − 5，f'(2) = 12 − 5 = 7。
substitute x = 2 into the original f(x): f(2) = 3(4) − 10 + 2 = 4, so the point is (2, 4).
要求曲线上该点的完整坐标，就把 x = 2 代入原函数 f(x)：f(2) = 3(4) − 10 + 2 = 4，所以该点是 (2, 4)。
The value f'(2) = 7 is the gradient of the tangent to the curve y = 3x² − 5x + 2 at the point (2, 4).
f'(2) = 7 就是曲线 y = 3x² − 5x + 2 在点 (2, 4) 处切线的斜率。
8. Equation of a Tangent | 切线方程
Once you know a point (x₁, y₁) on a curve and the gradient m = f'(x₁), the tangent equation is y − y₁ = m(x − x₁).
一旦你知道曲线上一点 (x₁, y₁) 和斜率 m = f'(x₁)，切线方程就是 y − y₁ = m(x − x₁)。
e with (2, 4) and m = 7, the tangent is y − 4 = 7(x − 2), which simplifies to y = 7x − 10.
用上面的例子，点 (2, 4) 且 m = 7，切线为 y − 4 = 7(x − 2)，化简得 y = 7x − 10。
equired form, such as ax + by + c = 0, so rearrange carefully at the end: 7x − y − 10 = 0.
AQA 可能指定答案形式，例如 ax + by + c = 0，因此最后要仔细移项：7x − y − 10 = 0。
9. Normal Line and Negative Reciprocal | 法线与负倒数
The normal to a curve at a point is perpendicular to the tangent at that point, so its gradient is −1 / m, where m is the tangent gradient.
曲线上某一点的法线在该点与切线垂直，因此其斜率为 −1 / m，其中 m 是切线的斜率。
gradient is −1/7. The equation of the normal at (2, 4) is therefore y − 4 = (−1/7)(x − 2).
如果切线斜率为 7，法线斜率就是 −1/7。因此 (2, 4) 处法线方程为 y − 4 = (−1/7)(x − 2)。
ly through by 7 to avoid fractions: 7y − 28 = −x + 2, which rearranges to x + 7y − 30 = 0.
两边乘以 7 以消去分数：7y − 28 = −x + 2，整理得 x + 7y − 30 = 0。
10. Common AQA Mark Points | AQA 常见得分点
AQA mark schemes for first principles usually award marks for stating the definition, writing f(x + h) correctly, expanding and cancelling, dividing by h, and taking the limit.
AQA 第一原理题的评分标准通常会给以下步骤打分：写出定义、正确写出 f(x + h)、展开并抵消、除以 h、取极限。
M1: correct use of the limit definition with f(x + h) and f(x)
M1：正确使用含 f(x + h) 和 f(x) 的极限定义
A1: correct expansion of brackets and cancellation of original terms
A1：正确展开括号并抵消原项
A1: final derivative f'(x) with no h remaining
A1：最终导数 f'(x) 中不含 h
If a tangent is requested, there is usually an extra mark for substituting the correct x value and simplifying to a linear equation.
如果题目还要求切线方程，通常还会额外给分：代入正确的 x 值并化简为线性方程。
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