Differentiation for GCSE AQA Maths | GCSE AQA 数学：微分考点精讲

📚 Differentiation for GCSE AQA Maths | GCSE AQA 数学：微分考点精讲

Differentiation is one of the most important topics in the GCSE AQA Mathematics course. It equips you with the tools to analyse how a function changes, to find the exact gradient of a curve at any point, and to locate and classify turning points such as maxima and minima. In this article, we will walk through the key concepts and techniques of differentiation, with clear explanations, step-by-step methods, and practical tips that are directly aligned with the AQA specification. Whether you are preparing for your mocks or final exams, this guide will help you build confidence and master the topic.

微分是 GCSE AQA 数学课程中最重要的课题之一。它让你掌握分析函数变化的方法，能够求出曲线上任意点的准确梯度，还能找出并判断极大值和极小值等驻点。本文将带你梳理微分的关键概念和技巧，通过清晰的解释、循序渐进的方法和贴近 AQA 考试大纲的实用建议，帮助你建立信心并掌握这一考点，无论你是在准备模拟考试还是正式大考，都会从中受益。

1. What is Differentiation? | 什么是微分？

Differentiation is a branch of calculus that allows us to find the instantaneous rate of change of a quantity. In graphical terms, it gives us a formula for the gradient of a curve at any point. For a straight line, the gradient is constant, but for a curve, the gradient changes continuously. By differentiating a function, we produce another function, called the derivative, which can tell us the gradient of the original function at any chosen x-value.

微分是微积分的一个分支，用来求一个量的瞬时变化率。从图像上看，它为我们提供了曲线上任意一点梯度的计算公式。对于直线来说，梯度是恒定的，但对于曲线，梯度是不断变化的。通过对一个函数进行微分，我们得到另一个函数，称为导数，它可以告诉我们原函数在任意 x 值处的梯度。

2. The Derivative and Rate of Change | 导数与变化率

If we have a function y = f(x), its derivative is written as dy/dx (pronounced “dee-y by dee-x”) or f'(x). The derivative represents the gradient of the tangent to the curve y = f(x) at a given point. It can also be interpreted as the rate at which y changes with respect to x. For example, if y represents distance and x represents time, dy/dx gives the speed.

如果我们有一个函数 y = f(x)，它的导数写作 dy/dx（读作“dee-y by dee-x”）或 f'(x)。导数表示曲线 y = f(x) 在给定点处切线的梯度。它也可以解释为 y 相对于 x 的变化率。例如，如果 y 表示距离，x 表示时间，那么 dy/dx 就给出速度。

In GCSE AQA, you will mostly work with polynomial functions, and you need to know how to find the derivative from the equation of the curve. Once you have dy/dx, you can substitute an x-coordinate to find the gradient of the curve at that point.

在 GCSE AQA 考试中，你主要处理多项式函数，需要知道如何从曲线方程中求出导数。一旦得到 dy/dx，就可以代入 x 坐标，求出曲线在该点的梯度。

3. Differentiating Powers of x | 幂函数的微分

The most fundamental rule for differentiation is the power rule. If y = xⁿ, then dy/dx = n xⁿ⁻¹. You simply multiply by the original power and then reduce the power by 1. For example, if y = x⁵, then dy/dx = 5x⁴. This rule works for any real power n, but at GCSE, n will typically be a positive integer, zero, or a negative integer.

微分最基本的法则是幂函数法则。如果 y = xⁿ，那么 dy/dx = n xⁿ⁻¹。你只需要乘上原来的指数，再把指数减 1。例如，如果 y = x⁵，那么 dy/dx = 5x⁴。这个法则对任何实数指数 n 都适用，但在 GCSE 中，n 通常是正整数、零或负整数。

For a constant term alone, such as y = 7, the derivative is zero because a horizontal line has zero gradient. In terms of the power rule, you can think of a constant as 7x⁰, and differentiating gives 0·7x⁻¹ = 0.

对于单独的常数项，例如 y = 7，它的导数为零，因为水平线的梯度为零。从幂函数法则的角度看，你可以把常数看作 7x⁰，微分后得到 0·7x⁻¹ = 0。

4. The Sum/Difference Rule | 和差法则

When a function is made up of several terms added or subtracted, you can differentiate each term separately and then combine the results. This is known as the sum/difference rule. For instance, if y = 3x⁴ + 2x² – 5x + 8, then:

当一个函数由几个项相加或相减组成时，你可以分别对每一项微分，然后把结果合并起来。这称为和差法则。例如，如果 y = 3x⁴ + 2x² – 5x + 8，那么：

dy/dx = 3·4x³ + 2·2x¹ – 5·1x⁰ + 0 = 12x³ + 4x – 5

You simply apply the power rule to each term, remembering that a constant term differentiates to zero. The coefficients remain attached to their respective differentiated powers.

你只需对每一项应用幂函数法则，记住常数项微分后为零。系数会保留在它们各自微分后的幂函数前面。

At GCSE level, you should be able to differentiate expressions like axⁿ + bxᵐ + c quickly and accurately, without making mistakes with the powers or the signs.

在 GCSE 阶段，你应该能够快速、准确地微分形如 axⁿ + bxᵐ + c 的表达式，避免在指数或符号上出错。

5. Finding the Gradient of a Curve | 求曲线的梯度

One of the main uses of differentiation is to find the gradient of a curve at a specific point. After obtaining the derivative dy/dx, substitute the x-coordinate of the point into the derivative. The resulting number is the gradient of the tangent at that point. For example, to find the gradient of y = x³ – 2x at x = 3, first find dy/dx = 3x² – 2, then substitute x = 3: 3(3)² – 2 = 25. So the gradient is 25.

微分的一个主要用途是求曲线在某一点处的梯度。在求得导数 dy/dx 之后，把该点的 x 坐标代入导数中，得到的数值就是该点切线的梯度。例如，要求曲线 y = x³ – 2x 在 x = 3 处的梯度，先求出 dy/dx = 3x² – 2，然后代入 x = 3：3(3)² – 2 = 25。因此梯度为 25。

This technique is very common in AQA exam questions, often appearing in contexts where you need to compare steepness or determine whether a function is increasing or decreasing at that point. If dy/dx > 0, the curve is increasing; if dy/dx < 0, it is decreasing.

这个技巧在 AQA 考试题中非常常见，通常会出现在需要比较陡峭程度或判断函数在该点是递增还是递减的情境中。如果 dy/dx > 0，曲线递增；如果 dy/dx < 0，曲线递减。

6. Equation of a Tangent | 切线方程

A tangent to a curve at a given point is a straight line that touches the curve at that point and has the same gradient as the curve. To find the equation of a tangent, follow these steps:

曲线在给定点处的切线是一条在该点与曲线相切、且梯度与曲线相同的直线。求切线方程的步骤如下：

Differentiate to get dy/dx, then substitute the x-coordinate to find the gradient m.
微分得到 dy/dx，然后代入 x 坐标求出梯度 m。
Using the original function, find the y-coordinate of the point if not already given.
如果 y 坐标没有直接给出，用原函数求出该点的 y 坐标。
Use the straight line formula y – y₁ = m(x – x₁) with (x₁, y₁) as the point of contact.
使用直线公式 y – y₁ = m(x – x₁)，其中 (x₁, y₁) 为切点坐标。
Rearrange to the required form, often y = mx + c or ax + by + c = 0.
将结果整理成题目要求的形式，通常是 y = mx + c 或 ax + by + c = 0。

For example, for the curve y = x² – 2x at x = 3, dy/dx = 2x – 2, so m = 4. The y-coordinate is 3² – 2·3 = 3. Equation: y – 3 = 4(x – 3), giving y = 4x – 9.

例如，对于曲线 y = x² – 2x 在 x = 3 处，dy/dx = 2x – 2，因此 m = 4。y 坐标为 3² – 2·3 = 3。切线方程为 y – 3 = 4(x – 3)，整理得 y = 4x – 9。

7. Equation of a Normal | 法线方程

The normal to a curve at a point is the line perpendicular to the tangent at that point. Its gradient is the negative reciprocal of the tangent’s gradient. If the gradient of the tangent is m, then the gradient of the normal is -1/m.

曲线在一点处的法线是垂直于该点切线的直线。它的梯度是切线梯度的负倒数。如果切线的梯度为 m，那么法线的梯度就是 -1/m。

To find the equation of the normal, first find the gradient of the tangent as before, then take the negative reciprocal, and finally use the same point with the line equation. Be careful with fractions and signs. For example, if the tangent gradient is 4, the normal gradient is -1/4.

要求法线方程，首先像之前一样求出切线梯度，然后取其负倒数，最后使用同一个点和直线方程。要小心分数和符号。例如，如果切线梯度为 4，法线梯度就是 -1/4。

AQA questions often ask for the normal in a particular form, so always read the question carefully. A common mistake is forgetting to change the sign or not inverting the fraction properly.

AQA 试题经常要求以特定形式给出法线方程，因此务必仔细读题。一个常见错误是忘记变号或没有正确地取倒数。

8. Turning Points – Maximum and Minimum | 驻点 – 极大值与极小值

Turning points are points on a curve where the gradient changes sign, i.e. where the curve goes from increasing to decreasing (maximum) or from decreasing to increasing (minimum). At a turning point, the gradient is zero, so dy/dx = 0.

驻点是曲线上梯度改变符号的点，也就是曲线从递增变为递减（极大值点）或从递减变为递增（极小值点）的地方。在驻点处，梯度为零，因此 dy/dx = 0。

To find turning points, solve the equation dy/dx = 0 to find the x-coordinates. Substitute these x-values back into the original function y = f(x) to get the corresponding y-coordinates. You can then classify the turning points using one of two methods: the second derivative test or examining the sign of dy/dx on either side.

要求驻点，先解方程 dy/dx = 0 以求出 x 坐标，然后将这些 x 值代回原函数 y = f(x) 中，得到对应的 y 坐标。接着，你可以用两种方法之一来判断驻点类型：二阶导数检验法，或者检查 dy/dx 在驻点左右两侧的符号。

9. Using the Second Derivative | 使用二阶导数

The second derivative, written d²y/dx² or f”(x), is the derivative of the derivative. It tells us the rate of change of the gradient and helps classify turning points. The rule is:

二阶导数，写作 d²y/dx² 或 f”(x)，是导数的导数。它告诉我们梯度的变化率，有助于判断驻点类型。规则如下：

If d²y/dx² > 0 at a point where dy/dx = 0, the point is a minimum.
如果在 dy/dx = 0 的点处 d²y/dx² > 0，该点是极小值点。
If d²y/dx² < 0 at a point where dy/dx = 0, the point is a maximum.
如果在 dy/dx = 0 的点处 d²y/dx² < 0，该点是极大值点。
If d²y/dx² = 0, the test is inconclusive; you should use the first derivative sign test instead.
如果 d²y/dx² = 0，该检验法无法判断；这时应改用一阶导数符号检验法。

To find d²y/dx², simply differentiate dy/dx again. For example, if dy/dx = 3x² – 12, then d²y/dx² = 6x. At x = 2, d²y/dx² = 12 > 0, so it is a minimum.

要求 d²y/dx²，只需对 dy/dx 再微分一次。例如，如果 dy/dx = 3x² – 12，那么 d²y/dx² = 6x。在 x = 2 处，d²y/dx² = 12 > 0，因此是极小值点。

At GCSE, the second derivative test is the preferred method for classifying turning points, and you are expected to show this reasoning clearly.

在 GCSE 中，二阶导数检验法是判断驻点类型的首选方法，你需要在解答中清晰地展示这一推理过程。

10. Sketching Gradient Functions | 画导数函数草图

AQA may ask you to sketch the graph of the gradient function f'(x) given the graph of f(x). The key idea is that where the original curve is increasing, the gradient function is above the x-axis (positive), and where it is decreasing, f'(x) is below the x-axis (negative). At turning points of f(x), the gradient is zero, so f'(x) crosses the x-axis.

AQA 可能会要求你根据 f(x) 的图形，画出导数函数 f'(x) 的草图。核心思想是：原曲线递增的区域，导数函数在 x 轴上方（正值）；原曲线递减的区域，f'(x) 在 x 轴下方（负值）。在 f(x) 的驻点处，梯度为零，因此 f'(x) 会穿过 x 轴。

To sketch accurately, identify intervals of increase and decrease, note x-coordinates of stationary points, and consider the steepness. A steeper curve gives a larger magnitude of f'(x). This is a visual way to test understanding of the derivative concept.

要画得准确，你需要识别递增和递减区间，记录驻点的 x 坐标，并考虑陡峭程度。曲线越陡，f'(x) 的绝对值越大。这是检验对导数概念理解的一种直观方式。

Conversely, you may be given the graph of f'(x) and asked to identify where f(x) has maxima or minima, or where it is increasing. Remember that the x-intercepts of f'(x) give the stationary points of f(x).

反过来，你也可能看到 f'(x) 的图形，并被要求指出 f(x) 在哪里有极大值或极小值，或是递增区间。记住，f'(x) 与 x 轴的交点给出了 f(x) 的驻点。

11. Applications of Differentiation | 微分的应用

Beyond pure curve analysis, differentiation in GCSE AQA can appear in optimisation problems. For example, you might be asked to find the maximum area of a shape with a given perimeter, or to minimise surface area for a given volume. The technique is always the same: express the quantity to be maximised or minimised as a function of one variable, differentiate, set the derivative to zero, and then verify it is a maximum or minimum using the second derivative.

除了纯粹的曲线分析，GCSE AQA 的微分考题还会出现在优化问题中。比如，要求你在周长固定的情况下求最大面积，或者在体积给定时求最小表面积。解题技巧始终如一：将需要最大化或最小化的量表示为一个变量的函数，再进行微分，令导数为零，然后用二阶导数验证它是极大值还是极小值。

A typical exam question might read: “A farmer wants to make a rectangular enclosure using 60 m of fencing. One side is against a wall. Find the maximum area.” You would introduce variables, write an expression for area, eliminate one variable using the fencing constraint, differentiate area with respect to the remaining variable, solve dA/dx = 0, and then prove it is a maximum.

一道典型的考题可能是：“一个农民想用 60 米的围栏围成一个矩形场地，其中一边靠墙。求最大面积。”你需要引入变量，写出面积表达式，利用围栏长度约束消去一个变量，对剩余变量求面积的导数，解 dA/dx = 0，然后证明该点是极大值。

Always check that your answer makes sense in context, and clearly state the final maximum or minimum value.

一定要检查答案在真实情境中是否合理，并明确写出最终的最大值或最小值。

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Differentiation for GCSE AQA Maths | GCSE AQA 数学：微分 考点精讲