Example 4.2.3: Differentiating to Find Tangents and Normals | 示例4.2.3：利用微分求切线与法线

📚 Example 4.2.3: Differentiating to Find Tangents and Normals | 示例4.2.3：利用微分求切线与法线

Differentiation is one of the cornerstones of A-Level Mathematics, and the ability to find tangents and normals to a curve is tested in almost every AQA paper. In this worked example, we take a typical examination question: we differentiate a cubic function, evaluate its gradient at a given point, and then construct the equations of both the tangent and the normal. Every step is explained in full so that you can replicate the process confidently in your own work.

微分是 A-Level 数学的基石之一，而求曲线上某点处的切线与法线几乎是每一份 AQA 试卷都会考查的内容。在本例题中，我们选取一道典型的考试题目：对一个三次函数求导，计算它在给定点处的斜率，然后写出该点处切线与法线的方程。我们将逐步详细讲解，使你能够在考试中自信地复现这一完整过程。

1. The Problem | 题目

We begin by stating the problem in the exact form in which it might appear on your AQA examination paper. This question combines the techniques of differentiation with the coordinate geometry of straight lines.

我们先把题目完整地写出来，形式与 AQA 试卷真题一致。本题把微分技巧与直线坐标几何结合到了一起。

The curve C has equation:

曲线 C 的方程为：

y = x³ − 4x + 1

(a) Find dy/dx. (b) Find the gradient of the curve C at the point P where x = 2. (c) Find the equation of the tangent to C at P. (d) Find the equation of the normal to C at P.

(a) 求 dy/dx；(b) 求曲线 C 在点 P（x = 2）处的斜率；(c) 求 C 在点 P 处的切线方程；(d) 求 C 在点 P 处的法线方程。

Notice that part (a) is a pure differentiation question, parts (b) and (c) require us to interpret the derivative as a gradient, and part (d) uses the key fact that the normal is perpendicular to the tangent.

请注意：第 (a) 问是纯粹的求导问题；第 (b)(c) 问要求我们把导数解释为斜率；第 (d) 问则要用到”法线与切线互相垂直”这一关键性质。

2. Recap: Differentiation Rules | 复习：微分法则

Before we differentiate, let us recall the three rules that we will use. These rules hold for all real values of n and are given on the AQA formula sheet, but you should know how to apply them without hesitation.

在正式求导之前，我们先复习将要用到的三条法则。这些法则对一切实数 n 均成立，虽然 AQA 公式表中会给出，但你应当能够不加犹豫地运用它们。

The power rule states that for y = xⁿ, we have:

幂法则指出：若 y = xⁿ，则：

dy/dx = nxⁿ⁻¹

The derivative of a constant is zero: if y = k, then dy/dx = 0. The derivative of kx is simply k. Finally, the sum (and difference) rule tells us that we differentiate term by term: d/dx (u + v) = du/dx + dv/dx.

常数的导数为零：若 y = k，则 dy/dx = 0。kx 的导数就是 k。最后，和（差）法则告诉我们，可以对式子逐项求导：d/dx (u + v) = du/dx + dv/dx。

In our curve, y = x³ − 4x + 1, we must differentiate each of the three terms separately and then combine the results.

对于我们的曲线 y = x³ − 4x + 1，我们需要分别对三个项求导，然后把结果合并起来。

3. Step 1: Find the Gradient Function | 第一步：求导函数

We now apply the power rule to each term of y = x³ − 4x + 1. The first term x³ gives 3x². The second term −4x is a linear term, so its derivative is simply the coefficient −4. The third term, the constant 1, has derivative 0.

现在我们对 y = x³ − 4x + 1 逐项运用幂法则。第一项 x³ 的导数为 3x²。第二项 −4x 是一次项，它的导数就是其系数 −4。第三项是常数 1，其导数为 0。

Therefore, the gradient function of the curve is:

因此，曲线的导函数为：

dy/dx = 3x² − 4

This expression tells us the gradient of the curve at any value of x. For example, if we wanted the gradient at x = 1, we would compute 3(1)² − 4 = −1. In the next step, we substitute x = 2 as required by the question.

这个表达式给出了曲线在任意 x 处的斜率。例如，若要求 x = 1 处的斜率，我们就计算 3(1)² − 4 = −1。下一步，我们按要求代入 x = 2。

It is worth noting that the notation f'(x) is often used instead of dy/dx. If we wrote f(x) = x³ − 4x + 1, then f'(x) = 3x² − 4. The two notations mean exactly the same thing.

值得注意的是，有时会用 f'(x) 来代替 dy/dx。若写 f(x) = x³ − 4x + 1，那么 f'(x) = 3x² − 4。这两种记号的含义完全相同。

4. Step 2: Gradient at x = 2 | 第二步：求 x = 2 处的斜率

To find the gradient of the curve at the point P where x = 2, we substitute x = 2 into the gradient function dy/dx = 3x² − 4.

为了求曲线在点 P（x = 2）处的斜率，我们把 x = 2 代入导函数 dy/dx = 3x² − 4。

At x = 2: dy/dx = 3(2)² − 4 = 12 − 4 = 8

So the gradient of the curve at P is 8. This number is also the gradient of the tangent line at P, because the tangent is, by definition, the line that touches the curve at that point and has the same gradient as the curve there.

因此曲线在点 P 处的斜率为 8。这个数值同时也是点 P 处切线的斜率，因为根据定义，切线就是与曲线在该点相切且斜率与之相同的直线。

We also need the y-coordinate of P. Substituting x = 2 into the original equation of the curve:

我们还需要点 P 的纵坐标。把 x = 2 代入曲线原方程：

y = 2³ − 4(2) + 1 = 8 − 8 + 1 = 1

Hence the point P has coordinates (2, 1). We now have everything we need to write down the equations of the tangent and the normal.

因此点 P 的坐标为 (2, 1)。现在我们已经具备写出切线与法线方程所需的全部条件。

5. Step 3: Equation of the Tangent | 第三步：切线方程

The tangent is a straight line passing through P(2, 1) with gradient m = 8. We use the point-gradient form of a straight line:

切线是过点 P(2, 1) 且斜率为 m = 8 的直线。我们使用直线的点斜式方程：

y − y₁ = m(x − x₁)

Substituting x₁ = 2, y₁ = 1 and m = 8 gives:

代入 x₁ = 2，y₁ = 1，m = 8，得：

y − 1 = 8(x − 2)

Expanding the bracket yields y − 1 = 8x − 16. Adding 1 to both sides gives the final equation of the tangent:

展开括号得 y − 1 = 8x − 16。两边同时加 1，得到切线的最终方程：

y = 8x − 15

Always check your answer by confirming that the point P lies on the line: 8(2) − 15 = 16 − 15 = 1, which matches the y-coordinate of P. This simple check can catch arithmetic errors.

务必检验答案：确认点 P 在这条直线上——8(2) − 15 = 16 − 15 = 1，恰好等于点 P 的纵坐标。这个简单的检验可以帮你发现运算错误。

6. Step 4: Equation of the Normal | 第四步：法线方程

The normal to a curve at a point is the straight line that passes through that point and is perpendicular to the tangent. There is a crucial relationship between the gradients of two perpendicular lines: if a line has gradient m, then any line perpendicular to it has gradient −1/m, provided m ≠ 0.

曲线在某点处的法线，就是过该点且与切线垂直的直线。两条垂直直线的斜率之间存在一个关键关系：若一条直线的斜率为 m，则与之垂直的直线的斜率为 −1/m（前提是 m ≠ 0）。

Since the tangent has gradient 8, the normal has gradient:

由于切线的斜率为 8，所以法线的斜率为：

m_normal = −1/8

Using the point-gradient form again, with P(2, 1) and m = −1/8:

再次使用点斜式，代入 P(2, 1) 和 m = −1/8：

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Example 4.2.3: Differentiating to Find Tangents and Normals | 示例4.2.3：利用微分求切线与法线