Example 3.4.2: Differentiation from First Principles and Tangent Equation | 例题 3.4.2：第一性原理求导与切线方程

📚 Example 3.4.2: Differentiation from First Principles and Tangent Equation | 例题 3.4.2：第一性原理求导与切线方程

from first principles, then use the derivative to find the tangent to the curve at x = 2.

在这个 AQA A-Level 纯数学例题中，给定二次函数 f(x) = 3x² − 5x + 2。我们使用第一性原理对其求导，然后利用导数求曲线在 x = 2 处的切线。

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1. Understanding the Problem | 理解题意

This question combines two core AQA Pure Maths skills: differentiation from first principles and coordinate geometry of tangents. You must first find the gradient function f '(x) using the limit definition, then evaluate the derivative and the original function at the given x-value.

本题结合了 AQA 纯数学两大核心技能：第一性原理求导和切线的坐标几何。你必须先用极限定义求出梯度函数 f '(x)，再在给定 x 值处计算导数和原函数的值。

es the parabola at exactly one point and has the same gradient as the curve at that point.

函数 f(x) = 3x² − 5x + 2 是二次函数，因此其图像是一条抛物线。x = 2 处的切线与抛物线恰好相切于一点，并且在该点处与曲线具有相同的梯度。

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2. First Principles Formula | 第一性原理公式

The formal definition of the derivative is:

导数的正式定义如下：

f '(x) = lim (h → 0) [ f(x + h) − f(x) ] / h

In words, this is the gradient of the chord joining the point (x, f(x)) to a nearby point (x + h, f(x + h)), as the horizontal distance h shrinks to zero.

用文字来说，这是连接点 (x, f(x)) 与邻近点 (x + h, f(x + h)) 的弦的梯度，且水平距离 h 趋向于零。

AQA expects this definition to be stated accurately before any substitution, because the method marks depend on showing the limit process.

AQA 要求在代入之前准确写出该定义，因为方法分取决于展示极限过程。

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3. Expanding f(x + h) | 展开 f(x + h)

Replace every x in f(x) with x + h:

将 f(x) 中的每个 x 替换为 x + h：

f(x + h) = 3(x + h)² − 5(x + h) + 2

Use the square expansion (a + b)² = a² + 2ab + b² with a = x and b = h. This gives:

使用平方展开公式 (a + b)² = a² + 2ab + b²，其中 a = x，b = h。得到：

3(x + h)² = 3(x² + 2xh + h²) = 3x² + 6xh + 3h²

Then distribute the remaining coefficients carefully:

然后仔细分配剩余系数：

−5(x + h) = −5x − 5h

So the full expansion is:

因此完整展开式为：

f(x + h) = 3x² + 6xh + 3h² − 5x − 5h + 2

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4. Simplifying the Difference Quotient | 化简差商

tract f(x) from f(x + h). Since f(x) = 3x² − 5x + 2, the terms 3x², −5x and +2 cancel out:

用 f(x + h) 减去 f(x)。因为 f(x) = 3x² − 5x + 2，项 3x²、−5x 和 +2 相互抵消：

f(x + h) − f(x) = 6xh + 3h² − 5h

Now divide by h. Every remaining term contains a factor h, so the division is valid for h ≠ 0:

现在除以 h。每一项都含有因子 h，因此在 h ≠ 0 时除法成立：

[ f(x + h) − f(x) ] / h = (6xh + 3h² − 5h) / h = 6x + 3h − 5

Be careful with signs: subtracting −5x gives +5x, which cancels the original −5x. The remaining −5h comes from −5(x + h) after cancellation.

注意符号：减去 −5x 得到 +5x，与原来的 −5x 抵消。剩下的 −5h 来自抵消后的 −5(x + h)。

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5. Taking the Limit | 取极限

As h approaches zero, the chord gradient tends to the tangent gradient. Mathematically, the only term containing h, which is 3h, tends to 0:

当 h 趋向于零时，弦的梯度趋向于切线的梯度。从数学上看，唯一含 h 的项 3h 趋于 0：

f '(x) = lim (h → 0) (6x + 3h − 5) = 6x − 5

Therefore the gradient function is f '(x) = 6x − 5. This means the gradient of the curve at any

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