Mastering Question 6: Differentiation & Applications | 掌握第6题：微分及其应用

📚 Mastering Question 6: Differentiation & Applications | 掌握第6题：微分及其应用

In the AQA A-Level Mathematics papers, Question 6 frequently tests your understanding of differentiation and its applications. This topic is a cornerstone of pure mathematics, and examiners love to see whether you can differentiate confidently, find stationary points, and interpret the second derivative correctly. This article breaks down everything you need to secure full marks on this type of question.

在 AQA A-Level 数学试卷中，第6题经常考查微分及其应用的理解。这一主题是纯数学的基石，考官们非常希望看到你能否自信地求导、求驻点，并正确解释二阶导数。本文将拆解你需要掌握的全部内容，助你在这类题目上拿到满分。

1. What to Expect in Question 6 | 第6题的常见题型

Question 6 in AQA A-Level Mathematics papers typically sits in the pure mathematics section. It often combines algebraic manipulation with differentiation skills. You may be asked to differentiate from first principles, apply the product, quotient or chain rule, or find and classify stationary points of a curve. The question usually carries between 6 and 9 marks, so method accuracy is essential.

AQA A-Level 数学试卷中的第6题通常位于纯数学部分。它常常将代数变形与微分技巧相结合。你可能会被要求用第一性原理求导、应用积法则、商法则或链式法则，或者求曲线的驻点并分类。该题通常分值为6到9分，因此方法的准确性至关重要。

To succeed, you need a systematic approach: read the question carefully, identify which rule applies, show all working steps, and check your final answer against the given conditions. Write down every derivative step — even if you can do it mentally — because AQA rewards method marks throughout.

要成功作答，你需要系统的方法：仔细审题，判断适用哪个法则，展示所有解题步骤，并根据题目条件检查最终答案。写下每一步求导过程——即使你能心算——因为 AQA 会根据过程给分。

2. Differentiation from First Principles | 用第一性原理求导

The definition of the derivative from first principles is based on the gradient of a chord as the interval tends to zero. For a function f(x), the derivative is given by:

第一性原理求导基于弦的斜率在区间趋于零时的极限。对于函数 f(x)，导数定义为：

f'(x) = lim(h→0) [f(x+h) − f(x)] / h

In AQA Question 6, you may be asked to prove that the derivative of a simple quadratic or cubic function is what you expect. For example, given f(x) = x² + 3x, you would substitute x + h into the function, subtract f(x), divide by h, and then take the limit as h approaches 0.

在 AQA 第6题中，你可能会被要求证明一个简单的二次或三次函数的导数正如你所料。例如，给定 f(x) = x² + 3x，你需要将 x + h 代入函数，减去 f(x)，除以 h，然后取 h 趋于0时的极限。

Step 1: Write f(x + h) and expand fully. | 第一步：写出 f(x + h) 并完整展开。
Step 2: Subtract f(x) and simplify. | 第二步：减去 f(x) 并化简。
Step 3: Divide by h. | 第三步：除以 h。
Step 4: Take the limit as h → 0. | 第四步：取 h → 0 时的极限。

Remember to state that h ≠ 0 when you cancel, but then let h tend to 0. This shows examiners that you understand the underlying limit concept.

记住在约分时说明 h ≠ 0，然后再令 h 趋于0。这向考官表明你理解背后的极限概念。

3. The Power Rule & Standard Derivatives | 幂法则与基本导数

The power rule is the most frequently used differentiation technique in AQA Question 6. For any function of the form y = axⁿ, the derivative is given by dy/dx = naxⁿ⁻¹. You must be confident applying this to negative and fractional powers as well.

幂法则是 AQA 第6题中最常用的求导技巧。对于任何形如 y = axⁿ 的函数，导数为 dy/dx = naxⁿ⁻¹。你必须能熟练地将其应用于负指数和分数指数。

Before differentiating, you should rewrite expressions to isolate powers of x. For example, 1/x should be rewritten as x⁻¹, √x as x^(1/2), and 1/x² as x⁻². This preparatory step is worth one method mark on its own.

在求导之前，你应该改写表达式以分离 x 的幂。例如，1/x 应改写为 x⁻¹，√x 应改写为 x^(1/2)，1/x² 应改写为 x⁻²。这一准备步骤本身就值一分方法分。

You should also memorise the standard derivatives: the derivative of eˣ is eˣ, the derivative of sin x is cos x, and the derivative of cos x is −sin x. For AQA, you must know these without a formula book.

你还应该记住标准导数：eˣ 的导数是 eˣ，sin x 的导数是 cos x，cos x 的导数是 −sin x。对于 AQA，你必须无需公式手册就知道这些。

4. Product, Quotient and Chain Rules | 积法则、商法则与链式法则

When a function is a product of two simpler functions, you apply the product rule. If y = uv, then dy/dx = u·dv/dx + v·du/dx. When a function is a quotient, y = u/v, then dy/dx = (v·du/dx − u·dv/dx) / v². The chain rule handles composite functions: dy/dx = dy/du × du/dx.

当函数是两个较简单函数的乘积时，使用积法则。若 y = uv，则 dy/dx = u·dv/dx + v·du/dx。当函数是商时，y = u/v，则 dy/dx = (v·du/dx − u·dv/dx) / v²。链式法则处理复合函数：dy/dx = dy/du × du/dx。

In Question 6, you may need to combine these rules. A typical AQA question might ask you to differentiate y = x²·sin(3x). Here you need the product rule combined with the chain rule for sin(3x). The key to scoring full marks is to clearly label u, v, and their derivatives before substituting into the formula.

在第6题中，你可能需要结合这些法则。AQA 的典型题目可能要求你求 y = x²·sin(3x) 的导数。这里你需要积法则结合链式法则来处理 sin(3x)。拿满分的关键是先清楚地标出 u、v 及其导数，然后再代入公式。

Let u = x², so du/dx = 2x. | 令 u = x²，则 du/dx = 2x。
Let v = sin(3x), so dv/dx = 3cos(3x) using the chain rule. | 令 v = sin(3x)，则 dv/dx = 3cos(3x)（用链式法则）。
Then dy/dx = x²·3cos(3x) + sin(3x)·2x. | 那么 dy/dx = x²·3cos(3x) + sin(3x)·2x。

Always simplify your final answer by factoring out common terms — examiners prefer neat, factorised results.

始终通过提取公因式来化简最终答案——考官更喜欢整洁、因式分解后的结果。

5. Stationary Points: Finding | 驻点的求解

A stationary point occurs where the gradient of the curve is zero, that is, where dy/dx = 0. In Question 6, you may be given a cubic function such as y = 2x³ − 9x² + 12x + 4 and asked to find the coordinates of its stationary points.

驻点是曲线梯度为零的点，即 dy/dx = 0 处。在第6题中，你可能会得到一个三次函数，如 y = 2x³ − 9x² + 12x + 4，并要求求出其驻点坐标。

Step 1: Differentiate to get dy/dx = 6x² − 18x + 12. Step 2: Set this equal to zero: 6x² − 18x + 12 = 0. Step 3: Divide through by 6: x² − 3x + 2 = 0. Step 4: Factorise: (x − 1)(x − 2) = 0, giving x = 1 and x = 2. Step 5: Substitute these x-values back into the original equation to find the y-coordinates.

第一步：求导得 dy/dx = 6x² − 18x + 12。第二步：令其为零：6x² − 18x + 12 = 0。第三步：同除以6：x² − 3x + 2 = 0。第四步：因式分解：(x − 1)(x − 2) = 0，得到 x = 1 和 x = 2。第五步：将这些 x 值代回原方程，求出 y 坐标。

A common error is forgetting to substitute the x-values into the original function rather than the derivative. The y-coordinates for this example are y(1) = 2 − 9 + 12 + 4 = 9 and y(2) = 16 − 36 + 24 + 4 = 8. So the stationary points are (1, 9) and (2, 8).

一个常见错误是忘记将 x 值代入原函数而不是导数。本例的 y 坐标为 y(1) = 2 − 9 + 12 + 4 = 9，y(2) = 16 − 36 + 24 + 4 = 8。所以驻点为 (1, 9) 和 (2, 8)。

6. Stationary Points: Classifying | 驻点的分类

Once you have found stationary points, you must classify them as maximum, minimum, or point of inflection. The shape of a maximum point resembles the peak of a hill, while a minimum resembles the bottom of a valley. A point of inflection is where the curvature changes sign.

找到驻点后，你必须将它们分类为极大点、极小点或拐点。极大点形状像山峰，极小点像谷底。拐点是曲率改变符号的点。

There are two acceptable methods for classification in AQA. The first is the gradient table method: choose an x-value just before and just after each stationary point, substitute into dy/dx, and observe whether the gradient changes from positive to negative (maximum), negative to positive (minimum), or stays the same sign (point of inflection).

AQA 中有两种可接受的分类方法。第一种是梯度表法：在每个驻点的前后各选一个 x 值，代入 dy/dx，观察梯度是否从正变负（极大）、从负变正（极小），或符号不变（拐点）。

x	1	1.5	2	2.5
dy/dx	0	−1.5	0	+1.5

The gradient goes from positive to zero to negative at x = 1, so (1, 9) is a local maximum. The gradient goes from negative to zero to positive at x = 2, so (2, 8) is a local minimum.

在 x = 1 处梯度从正变零再变负，因此 (1, 9) 是局部极大值。在 x = 2 处梯度从负变零再变正，因此 (2, 8) 是局部极小值。

7. The Second Derivative Test | 二阶导数判定法

The second derivative, denoted f”(x) or d²y/dx², provides a quicker way to classify stationary points. For the function y = 2x³ − 9x² + 12x + 4, we already found dy/dx = 6x² − 18x + 12. Differentiating again gives d²y/dx² = 12x − 18.

二阶导数，记为 f”(x) 或 d²y/dx²，提供了一种更快的驻点分类方法。对于函数 y = 2x³ − 9x² + 12x + 4，我们已经求出 dy/dx = 6x² − 18x + 12。再次求导得 d²y/dx² = 12x − 18。

Substitute x = 1 into the second derivative: 12(1) − 18 = −6. Since d²y/dx² < 0, the point (1, 9) is a maximum. Substitute x = 2: 12(2) − 18 = +6. Since d²y/dx² > 0, the point (2, 8) is a minimum.

将 x = 1 代入二阶导数：12(1) − 18 = −6。由于 d²y/dx² < 0，点 (1, 9) 是极大值。代入 x = 2：12(2) − 18 = +6。由于 d²y/dx² > 0，点 (2, 8) 是极小值。

Be careful: if the second derivative equals zero at a stationary point, the test is inconclusive. In such cases you must fall back on the gradient table method or examine the sign of the first derivative on both sides.

注意：如果二阶导数在驻点处为零，则此判定法失效。此时你必须使用梯度表法，或检查一阶导数两侧的符号。

8. An Exam-Style Worked Example | 真题风格例题

Let us work through a complete Question 6 style problem from start to finish. The curve C has equation y = x³ − 6x² + 9x + 3. Find the coordinates of the stationary points and determine their nature.

让我们从头到尾完成一道第6题风格的完整题目。曲线 C 的方程为 y = x³ − 6x² + 9x + 3。求驻点坐标并判断其性质。

First, differentiate: dy/dx = 3x² − 12x + 9. Set dy/dx = 0: 3x² − 12x + 9 = 0. Divide by 3: x² − 4x + 3 = 0. Factorise: (x − 1)(x − 3) = 0, so x = 1 or x = 3.

首先求导：dy/dx = 3x² − 12x + 9。令 dy/dx = 0：3x² − 12x + 9 = 0。除以3：x² − 4x + 3 = 0。因式分解：(x − 1)(x − 3) = 0，所以 x = 1 或 x = 3。

Substitute back to find y: for x = 1, y = 1 − 6 + 9 + 3 = 7. For x = 3, y = 27 − 54 + 27 + 3 = 3. The stationary points are (1, 7) and (3, 3).

代回求 y：当 x = 1，y = 1 − 6 + 9 + 3 = 7。当 x = 3，y = 27 − 54 + 27 + 3 = 3。驻点为 (1, 7) 和 (3, 3)。

Now find the second derivative: d²y/dx² = 6x − 12. At x = 1, d²y/dx² = 6(1) − 12 = −6 < 0, so (1, 7) is a local maximum. At x = 3, d²y/dx² = 6(3) − 12 = +6 > 0, so (3, 3) is a local minimum.

现在求二阶导数：d²y/dx² = 6x − 12。在 x = 1 处，d²y/dx² = 6(1) − 12 = −6 < 0，因此 (1, 7) 是局部极大值。在 x = 3 处，d²y/dx² = 6(3) − 12 = +6 > 0，因此 (3, 3) 是局部极小值。

9. Common Pitfalls & Examiner Tips | 常见误区与考官提示

Many students lose marks in Question 6 not because they cannot differentiate, but because of avoidable errors. The most frequent mistake is incorrectly expanding brackets when using the first principles method. Another common error is mixing up the product rule and quotient rule — remember the quotient rule has a minus sign in the numerator, not a plus sign.

许多学生在第6题失分并非因为不会求导，而是因为可避免的错误。最常见的错误是在使用第一性原理方法时展开括号出错。另一个常见错误是混淆积法则和商法则——记住商法则的分子中是减号，不是加号。

Always write dy/dx = 0 before solving for stationary points. | 求解驻点前务必写出 dy/dx = 0。
When differentiating a product, do not differentiate the whole expression at once. | 对乘积求导时，不要对整个表达式一次求导。
Check whether your stationary points are indeed stationary by substituting back into dy/dx. | 将结果代回 dy/dx，检查是否确实是驻点。
Use the second derivative test where possible — it is faster and clearer. | 尽可能使用二阶导数判定法——它更快更清晰。
If d²y/dx² = 0 at a stationary point, do not guess; use the gradient table. | 如果驻点处 d²y/dx² = 0，不要猜；使用梯度表。

Examiners also reward correct notation. Write f'(x) or dy/dx consistently, and avoid using the same symbol for different variables. A well-set-out solution with labelled steps will earn method marks even if your final answer has a small arithmetic slip.

考官也会对正确的符号给予奖励。统一使用 f'(x) 或 dy/dx，避免对不同的变量使用同一符号。一份步骤清晰、标注完整的解答即使最终答案有小的计算失误，也能获得方法分。

10. Practice Questions | 练习题目

To solidify your skills, try these three practice questions. Attempt each one fully before checking your working.

为巩固你的技能，尝试以下三道练习题。请先完整作答，再核对你的过程。

Question A: Differentiate y = (x² + 1)(x³ − 2) using the product rule. | 题目A：使用积法则求 y = (x² + 1)(x³ − 2) 的导数。

Question B: Find the stationary points of y = x³ − 3x² + 2 and classify each using the second derivative test. | 题目B：求 y = x³ − 3x² + 2 的驻点，并用二阶导数判定法分类。

Question C: Use first principles to differentiate f(x) = x² + 4x. | 题目C：用第一性原理求 f(x) = x² + 4x 的导数。

For Question B, the answer is: dy/dx = 3x² − 6x = 3x(x − 2), giving x = 0 and x = 2. The points are (0, 2) which is a maximum, and (2, −2) which is a minimum. For Question C, f'(x) = 2x + 4 using the limit definition.

对于题目B，答案为：dy/dx = 3x² − 6x = 3x(x − 2)，解得 x = 0 和 x = 2。点为 (0, 2) 是极大值，(2, −2) 是极小值。对于题目C，用极限定义得 f'(x) = 2x + 4。

By mastering differentiation and its applications, you will approach Question 6 of the AQA A-Level Mathematics paper with confidence. Remember: practise each rule separately, then combine them, and always present your working clearly. Consistent practice is the key to turning these marks from potential into guaranteed.

通过掌握微分及其应用，你将满怀信心地面对 AQA A-Level 数学试卷的第6题。记住：先单独练习每个法则，再组合使用，并且始终清晰地展示你的解题过程。持续练习是将这些分数从可能变为确定的关键。

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Mastering Question 6: Differentiation & Applications | 掌握第6题：微分及其应用