Worked Example 7.4.2: Differentiating a Polynomial and Finding a Tangent | 例题 7.4.2：多项式求导与切线方程

📚 Worked Example 7.4.2: Differentiating a Polynomial and Finding a Tangent | 例题 7.4.2：多项式求导与切线方程

In this AQA A-Level Mathematics worked example, we differentiate a polynomial function and use the derivative to find the equation of a tangent at a given point. The problem combines the power rule with coordinate geometry, and it is typical of the early pure mathematics content in AS and A-Level courses.

在这个 AQA A-Level 数学例题中，我们对一个多项式函数进行求导，并利用导数求出给定点处的切线方程。该问题将幂法则与解析几何结合起来，是 AS 和 A-Level 课程早期纯数学部分的典型内容。

This walkthrough follows the style of AQA Example 7.4.2, where learners are expected to work through a multi-step problem: differentiate, evaluate, locate the point, and then construct the equation of a straight line.

本讲解采用 AQA 例题 7.4.2 的风格，要求学习者完成一个多步骤问题：求导、代入求值、确定点的坐标，然后建立直线方程。

1. Context and AQA Specification | 背景与 AQA 考纲定位

This type of question sits within the AQA Pure Mathematics strand, normally covered in the first year of A-Level study or in the AS specification. It assesses basic differentiation rules, especially the power rule, and the geometric meaning of the derivative as the gradient of a tangent.

这类题目属于 AQA 纯数学部分，通常在 A-Level 第一年或 AS 阶段学习。它考查基本求导法则，特别是幂法则，以及导数作为切线斜率的几何意义。

The example is labelled Example 7.4.2, following a section on differentiating polynomial functions. You are expected to apply the power rule term by term and then interpret the derivative as a gradient at a specific point.

该例题编号为 7.4.2，位于多项式函数求导一节之后。你需要逐项使用幂法则求导，然后将导数理解为某一点处的斜率。

In AQA examinations, questions like this often appear in the non-calculator paper, so accurate arithmetic and clear working are essential. The method marks are awarded for showing the differentiation step, the substitution, and the use of the point-gradient formula.

在 AQA 考试中，这类题目经常出现在非计算器试卷中，因此精确的算术和清晰的推导过程非常重要。方法分通常分配给求导步骤、代入过程以及点斜式公式的使用。

2. The Problem Statement | 题目陈述

Given the function f(x) = 3x⁴ − 5x² + 2x − 7, find f'(x) and hence find the equation of the tangent to the curve at x = 2.

已知函数 f(x) = 3x⁴ − 5x² + 2x − 7，求 f'(x)，并由此求出曲线在 x = 2 处的切线方程。

dinate from the original function, and then write the tangent line in the form y = mx + c.

我们需要先写出简化后的导数，计算 x = 2 处的导数值以得到斜率，从原函数中求出对应的 y 坐标，然后将切线方程写成 y = mx + c 的形式。

Careful reading is important: the question asks for the tangent, not the normal. Using the wrong line would change the gradient relationship and lose all final accuracy marks.

仔细审题非常重要：题目要求的是切线，而不是法线。如果用了错误的直线类型，斜率关系就会出错，最终答案分也会全部丢失。

3. Key Concepts: Power Rule and Tangent Equation | 核心概念：幂法则与切线方程

ants differentiate to zero because their rate of change is zero. This rule must be applied separately to each term of a polynomial.

对于任意实数常量 n，幂法则指出 d/dx (xⁿ) = n xⁿ⁻¹。常数的导数为零，因为它的变化率为零。该法则必须分别应用于多项式的每一项。

The tangent to a curve at a point (x₁, y₁) has gradient m = f' point-gradient form as y − y₁ = m(x − x₁), which is then usually simplified to y = mx + c.

曲线在点 (x₁, y₁) 处的切线斜率为 m = f'(x₁)。其方程可以用点斜式写成 y − y₁ = m(x − x₁)，然后通常化简为 y = mx + c。

• Differentiate term by term using the power rule | 使用幂法则逐项求导
• Substitute the x-value into f'(x) to get the numerical gradient | 将 x 值代入 f'(x) 得到数值斜率
• Use the original function f(x) to find the y-coordinate | 使用原函数 f(x) 求 y 坐标
• Simplify the straight-line equation into the requested form | 将直线方程化简为题目要求的形式

4. Step 1: Differentiate Term by Term | 步骤一：逐项求导

. Apply the power rule to each non-constant term. The derivative of 3x⁴ is 3 × 4x³ = 12x³.

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从 f(x) = 3x⁴ − 5x² + 2x − 7 开始，对每一个非常数项应用幂法则。3x⁴ 的导数是 3 × 4x³ = 12x³。

of 2x is simply 2, because the power of x is 1 and 1 × 2x⁰ = 2. The derivative of −7 is 0.

−5x² 的导数是 −5 × 2x = −10x。2x 的导数就是 2，因为 x 的幂次为 1，而 1 × 2x⁰ = 2。−7 的导数是 0。

Therefore, the derivative is f'(x) = 12x³ − 10x + 2.

因此，导数为 f'(x) = 12x³ − 10x + 2。

Writing the derivative clearly and in descending powers of x helps with the next substitution step and reduces the chance of missing a term.

将导数按 x 的降幂清晰书写，有助于下一步代入并减少漏项的可能性。

5. Step 2: Simplify and Check the Derivative | 步骤二：化简并检查导数

The simplified derivative is f' r combining is needed because the terms have different powers of x: x³, x, and a constant.

化简后的导数为 f'(x) = 12x³ − 10x + 2。由于各项分别为 x³、x 和常数项，幂次不同，因此无需进一步合并。

A quick check shows that the degree has dropped from 4 to 3, and the original leading coefficient 3 has been multiplied by 4 to give 12. This is consistent with the power rule.

快速检查可以发现，多项式的次数从 4 降到 3，原首项系数 3 乘以 4 得到 12。这与幂法则一致。

Also note that every original term has been accounted for: the quartic term becomes cubic, the quadratic term becomes linear, the linear term becomes constant, and the constant disappears.

还要注意每一项都得到了处理：四次项变成三次项，二次项变成一次项，一次项变成常数，常数项消失。

6. Step 3: Evaluate the Gradient at x = 2 | 步骤三：求 x = 2 处的斜率

Substitute x = 2 into f'(x) to find the gradient of the tangent at that point.

将 x = 2 代入 f'(x)，求出该点处切线的斜率。

f'(2) = 12(2)³ − 10(2) + 2

f'(2) = 12(8) − 20 + 2 = 96 − 20 + 2 = 78

x = 2 is 78. This large positive value indicates a steeply increasing curve at that point.

在 x = 2 处，切线的斜率为 78。这个较大的正值说明曲线在该点附近急剧上升。

re multiplying: 2³ = 8, not 6 or 12. This is a common arithmetic slip under exam pressure.

计算时一定要先算幂再乘：2³ = 8，而不是 6 或 12。这是在考试压力下常见的算术错误。

7. Step 4: Find the Point on the Curve | 步骤四：求曲线上的点

nt where x = 2. Substitute x = 2 into the original function f(x), not into the derivative.

我们需要 x =

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