Finding the Gradient of a Tangent to a Curve at a Point | IGCSE数学：曲线上一点处的切线斜率

📚 Finding the Gradient of a Tangent to a Curve at a Point | IGCSE数学：曲线上一点处的切线斜率

In IGCSE Mathematics, one of the most powerful ideas is finding how steep a curve is at a particular point. The gradient of a tangent to a curve at a point tells us the exact rate at which the curve is changing at that instant. This concept is the bridge between simple coordinate geometry and the beginnings of calculus.

在 IGCSE 数学中，一个非常重要的概念是求曲线在某一点处的陡峭程度。曲线上一点处的切线斜率告诉我们曲线在该瞬间变化的精确速率。这个概念将简单的坐标几何与微积分的入门知识联系了起来。

1. Why the Tangent Gradient Matters | 切线斜率为何重要

The gradient of a straight line is constant, but a curve has a different steepness at every point. If you draw a straight line that just touches the curve at one point, that line is called a tangent. The gradient of this tangent is the gradient of the curve at that exact point.

一条直线的斜率是恒定的，但曲线上的每一点都有不同的陡峭程度。如果你画一条直线，它只在一点恰好接触曲线，这条直线就称为切线。这条切线的斜率就是曲线在该点处的斜率。

Knowing this gradient tells you how fast a quantity is changing. For example, if a curve shows the distance travelled by a car over time, the gradient at any time gives the speed of the car at that exact moment. This is why tangent gradients are essential in physics, economics and many other fields.

知道了这个斜率，你就能知道某个量变化得多快。例如，如果一条曲线表示汽车行驶距离随时间的变化，那么任意时刻曲线的切线斜率就是该时刻汽车的瞬时速度。这就是为什么切线斜率在物理、经济等许多领域都至关重要。

2. Secant vs Tangent | 割线与切线

A secant line passes through two distinct points on a curve. A tangent line touches the curve at only one point, at least locally. As the two points of a secant move closer and closer together, the secant approaches the tangent.

割线通过曲线上的两个不同点。切线只在一点接触曲线，至少在局部是这样。当割线的两个点越来越靠近时，割线就会趋近于切线。

Consider the curve y = x². A line through (1,1) and (2,4) is a secant. If you move the second point closer to (1,1), the secant becomes a better approximation of the tangent at (1,1).

以曲线 y = x² 为例，经过 (1,1) 和 (2,4) 的直线是一条割线。如果将第二个点向 (1,1) 移动，割线就会越来越接近在 (1,1) 处的切线。

Secant: gradient is an average rate of change over an interval.
Tangent: gradient is the instantaneous rate of change at one point.

割线：斜率是一个区间内的平均变化率。
切线：斜率是某一点处的瞬时变化率。

3. Estimating Gradient from a Graph | 从图像估计斜率

Before learning differentiation, you can find an approximate tangent gradient by drawing. At the point of interest, draw a tangent line carefully. Then choose two points on this tangent line with easy coordinates, for example (x₁, y₁) and (x₂, y₂). The gradient is:

在学习微分之前，你可以通过画图来估算切线斜率。在感兴趣的点处，仔细画出一条切线。然后在这条直线上选择两个坐标容易读出的点，例如 (x₁, y₁) 和 (x₂, y₂)。斜率就是：

m = (y₂ − y₁) ÷ (x₂ − x₁)

For the curve y = x² at the point (1,1), a tangent drawn carefully will pass through (1,1) and approximately (2,3). The gradient is (3 − 1) ÷ (2 − 1) = 2.

对于曲线 y = x² 在点 (1,1) 处，仔细画出切线后，可以发现它约经过 (1,1) 和 (2,3)。因此斜率为 (3 − 1) ÷ (2 − 1) = 2。

The accuracy of this method depends on your drawing skill. In an exam, always use a sharp pencil and a ruler, and choose points far apart on the tangent to reduce error.

这种方法的结果是否准确取决于你的作图技巧。在考试中，一定要用削尖的铅笔和直尺，并且在切线上选择相距较远的点来减小误差。

4. Introduction to Differentiation | 微分入门

Differentiation is a mathematical method that gives the exact gradient of the tangent at any point on a curve. The derivative of y with respect to x is written as dy/dx and is called the gradient function.

微分是一种数学方法，它可以精确地求出曲线上任意一点处切线的斜率。y 关于 x 的导数写作 dy/dx，称为导函数。

For a curve y = f(x), the derivative dy/dx gives a formula for the gradient at any x. To find the gradient at one specific point, you simply substitute the x-coordinate of that point into dy/dx.

对于曲线 y = f(x)，导数 dy/dx 给出了任意 x 值处斜率的公式。要求某一点处的斜率，你只需将该点的 x 坐标代入 dy/dx 即可。

5. The Power Rule | 幂法则

The most important rule for IGCSE is the power rule. If y = xⁿ, then the derivative is:

在 IGCSE 中最重要的规则就是幂法则。如果 y = xⁿ，那么导数为：

dy/dx = n xⁿ⁻¹

Multiply by the power, then reduce the power by one. For example:

先乘上指数，再将指数减一。例如：

y = x² → dy/dx = 2x
y = x³ → dy/dx = 3x²
y = x⁵ → dy/dx = 5x⁴

y = x² → dy/dx = 2x
y = x³ → dy/dx = 3x²
y = x⁵ → dy/dx = 5x⁴

This rule works for positive, negative and fractional powers. For IGCSE, you usually work with positive integer powers and simple linear expressions.

这条法则适用于正指数、负指数和分数指数。在 IGCSE 中，通常只需要处理正整数指数和简单的线性表达式。

6. Rules for Constants and Sums | 常数与求和规则

A constant term has a derivative of zero, because a constant does not change. For example, if y = 7, then dy/dx = 0. If a term has a coefficient, keep the coefficient and differentiate the power part separately.

常数项的导数为零，因为常数不会变化。例如，如果 y = 7，那么 dy/dx = 0。如果一个项带有系数，就保留系数，并分别对幂部分求导。

General rule: if y = k xⁿ, then dy/dx = k n xⁿ⁻¹. For example:

一般规则：如果 y = k xⁿ，那么 dy/dx = k n xⁿ⁻¹。例如：

y = 4x² → dy/dx = 8x
y = 3x³ → dy/dx = 9x²

y = 4x² → dy/dx = 8x
y = 3x³ → dy/dx = 9x²

When a curve is written as a sum of terms, differentiate each term separately. The derivative of a sum is the sum of the derivatives. For example, y = x² + 3x − 5 becomes dy/dx = 2x + 3.

当曲线写成若干项之和时，可以逐项求导。和函数的导数等于各项导数之和。例如，y = x² + 3x − 5 的导数为 dy/dx = 2x + 3。

7. Finding the Gradient at a Given Point | 求给定点处的斜率

Once you have the gradient function, finding the gradient at a specific point is straightforward. Substitute the x-coordinate of the point into dy/dx. The resulting number is the gradient of the tangent there.

一旦你有了导函数，求特定点处的斜率就非常简单了。将该点的 x 坐标代入 dy/dx，得到的数就是该点处切线的斜率。

Example 1: Find the gradient of y = x² + 3x − 4 at x = 2.

示例 1：求 y = x² + 3x − 4 在 x = 2 处的斜率。

First differentiate: dy/dx = 2x + 3. Then substitute x = 2: dy/dx = 2(2) + 3 = 7. So the tangent gradient at x = 2 is 7.

首先求导：dy/dx = 2x + 3。然后代入 x = 2：dy/dx = 2(2) + 3 = 7。所以在 x = 2 处切线斜率为 7。

Example 2: Find the gradient of y = x³ − 6x at x = 1.

示例 2：求 y = x³ − 6x 在 x = 1 处的斜率。

dy/dx = 3x² − 6. At x = 1: 3(1)² − 6 = 3 − 6 = −3. The gradient is −3, meaning the curve is sloping downwards at that point.

dy/dx = 3x² − 6。在 x = 1 处：3(1)² − 6 = 3 − 6 = −3。斜率为 −3，表示曲线在该点向下倾斜。

8. Equation of the Tangent | 切线方程

To find the equation of a tangent, you need two pieces of information: the gradient m at the point, and the coordinates of the point (x₁, y₁). Then use the point-slope form:

要求切线的方程，你需要两个信息：该点处的斜率 m，以及该点的坐标 (x₁, y₁)。然后使用点斜式：

y − y₁ = m (x − x₁)

Example: Find the equation of the tangent to y = x² at the point (3, 9).

示例：求曲线 y = x² 在点 (3, 9) 处的切线方程。

Differentiate: dy/dx = 2x. At x = 3, m = 6. Using y − 9 = 6(x − 3), we get y = 6x − 9. So the tangent equation is y = 6x − 9.

求导：dy/dx = 2x。在 x = 3 处，m = 6。代入 y − 9 = 6(x − 3)，得到 y = 6x − 9。因此切线方程为 y = 6x − 9。

Always simplify the equation into the form y = mx + c or ax + by + c = 0, depending on what the question asks.

记得把方程化简为 y = mx + c 或 ax + by + c = 0 的形式，具体取决于题目的要求。

9. Equation of the Normal | 法线方程

The normal is a line perpendicular to the tangent at the same point. If the tangent has gradient m, the normal has gradient −1/m, provided m ≠ 0.

法线是在同一点处与切线垂直的直线。如果切线的斜率为 m，那么法线的斜率为 −1/m，前提是 m ≠ 0。

Example: For the curve y = x² at (3, 9), the tangent gradient is 6. The normal gradient is −1/6. Using y − 9 = −1/6 (x − 3), the normal equation becomes y = −x/6 + 9.5 or 6y + x = 57.

示例：对于曲线 y = x² 在 (3, 9) 处，切线斜率为 6。法线斜率为 −1/6。代入 y − 9 = −1/6 (x − 3)，法线方程为 y = −x/6 + 9.5，或写成 6y + x = 57。

Notice that multiplying both sides by 6 gives a neat integer equation. In IGCSE questions, you may be asked for the normal equation, so remember the negative reciprocal rule.

注意两边同时乘以 6 可以得到整洁的整数方程。在 IGCSE 题目中，有时会要求求法线方程，所以务必记住负倒数规则。

10. Common Mistakes and Tips | 常见错误与小贴士

Students often make small errors when finding tangent gradients. Here are the most common ones and how to avoid them.

学生在求切线斜率时经常犯一些小错误。以下是最常见的错误以及如何避免它们。

Confusing secant with tangent: A secant passes through two points; a tangent touches one point.
Forgetting to differentiate: Do not use the original equation when substituting x; use dy/dx instead.
Wrong power rule: Always multiply by the power and then subtract one. x⁵ → 5x⁴, not 5x⁵.
Sign errors: Be careful with negative terms, e.g. y = x³ − 6x gives dy/dx = 3x² − 6.
Not simplifying: Expand and simplify the tangent equation fully.

混淆割线与切线：割线经过两个点，切线只接触一个点。
忘记求导：代入 x 时应使用 dy/dx，而不是原方程。
幂法则出错：一定要先乘指数，再减一。x⁵ → 5x⁴，而不是 5x⁵。
符号错误：注意负项，例如 y = x³ − 6x 的导数是 dy/dx = 3x² − 6。
没有化简：将切线方程展开并化简完整。

11. Practice Questions | 练习题

Try these questions yourself before checking the answers.

请先自己完成以下练习，再对照答案。

Q1. Find the gradient of the curve y = x³ − 4x at the point where x = 2.

练习题 1：求曲线 y = x³ − 4x 在 x = 2 处的斜率。

Q2. Find the equation of the tangent to y = 3x² + 1 at the point (1, 4).

练习题 2：求曲线 y = 3x² + 1 在点 (1, 4) 处的切线方程。

Q3. Find the gradient of the normal to y = 2x² − x at the point where x = 1.

练习题 3：求曲线 y = 2x² − x 在 x = 1 处的法线斜率。

Answers: Q1: dy/dx = 3x² − 4, so at x = 2, gradient = 3(4) − 4 = 8. Q2: dy/dx = 6x, so m = 6; tangent: y − 4 = 6(x − 1), giving y = 6x − 2. Q3: dy/dx = 4x − 1, so m = 3; normal gradient = −1/3.

答案：练习 1：dy/dx = 3x² − 4，因此 x = 2 时，斜率为 3(4) − 4 = 8。练习 2：dy/dx = 6x，因此 m = 6；切线：y − 4 = 6(x − 1)，即 y = 6x − 2。练习 3：dy/dx = 4x − 1，因此 m = 3；法线斜率为 −1/3。

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Finding the Gradient of a Tangent to a Curve at a Point | IGCSE数学：曲线上一点处的切线斜率