Using Second Derivatives | 二阶导数的应用

📚 Using Second Derivatives | 二阶导数的应用

In A-Level Mathematics, the second derivative of a function tells us how the gradient itself is changing. It is the key tool for classifying stationary points, understanding concavity, and locating points of inflection. This article explains the main ideas and techniques required for Edexcel A-Level Maths, with worked reasoning and exam-style guidance.

在 A-Level 数学中，函数的二阶导数告诉我们梯度本身是如何变化的。它是判断驻点类型、理解凹凸性以及寻找拐点的关键工具。本文讲解 Edexcel A-Level 数学所需的主要概念与技巧，并配有推理过程和考试风格的指导。

1. Definition of the Second Derivative | 二阶导数的定义

If y = f(x), then the first derivative dy/dx represents the gradient of the curve. The second derivative is obtained by differentiating dy/dx with respect to x, so it is written as d²y/dx² or f”(x). It measures the rate at which the gradient is changing.

如果 y = f(x)，那么一阶导数 dy/dx 表示曲线的斜率。二阶导数是将 dy/dx 再对 x 求导，因此写作 d²y/dx² 或 f”(x)。它衡量斜率变化的速率。

For example, if y = x³ − 3x² + 2, then dy/dx = 3x² − 6x and d²y/dx² = 6x − 6. The second derivative is itself a function of x, so its value may change at different points on the curve.

例如，若 y = x³ − 3x² + 2，则 dy/dx = 3x² − 6x，且 d²y/dx² = 6x − 6。二阶导数本身也是 x 的函数，因此在曲线上不同点的取值可能不同。

2. Stationary Points and the Second Derivative Test | 驻点与二阶导数判别法

A stationary point occurs where dy/dx = 0. These points can be local maxima, local minima, or stationary points of inflection. The second derivative test uses the sign of d²y/dx² at the stationary point to classify it.

驻点出现在 dy/dx = 0 的位置。这些点可能是局部极大值、局部极小值或驻点型拐点。二阶导数判别法利用驻点处 d²y/dx² 的正负来判断其类型。

If d²y/dx² < 0 at a stationary point, the gradient is decreasing through zero, so the point is a local maximum. If d²y/dx² > 0, the gradient is increasing through zero, so the point is a local minimum.

若在驻点处 d²y/dx² < 0，则斜率经过零时在减小，因此该点是局部极大值。若 d²y/dx² > 0，则斜率经过零时在增大，因此该点是局部极小值。

This test is usually quicker than checking the sign of dy/dx on either side of the stationary point, but it can fail when d²y/dx² = 0.

这种方法通常比检查驻点两侧 dy/dx 的符号更快，但当 d²y/dx² = 0 时，判别法可能失效。

3. Concavity and Convexity | 凹性与凸性

The sign of the second derivative also tells us about the shape of a curve. If d²y/dx² > 0, the gradient is increasing and the curve is said to be convex, sometimes described as ‘happy’ or opening upwards. If d²y/dx² < 0, the gradient is decreasing and the curve is concave, sometimes described as ‘sad’ or opening downwards.

二阶导数的正负也告诉我们曲线的形状。若 d²y/dx² > 0，则斜率在增大，曲线称为凸的，有时被描述为“笑脸”或向上开口。若 d²y/dx² < 0，则斜率在减小，曲线称为凹的，有时被描述为“哭脸”或向下开口。

It is important to be precise with the words ‘concave’ and ‘convex’, as different exam boards and countries may use opposite conventions. For Edexcel A-Level, the practical test is simpler: positive second derivative means the curve bends upwards, negative means it bends downwards.

准确使用“凹”和“凸”这两个词很重要，因为不同的考试局和国家可能使用相反的约定。对 Edexcel A-Level 而言，实用的判断更简单：二阶导数为正表示曲线向上弯曲，为负表示向下弯曲。

4. Points of Inflection | 拐点

A point of inflection is where the curve changes concavity, from concave to convex or from convex to concave. At a point of inflection, the second derivative is usually zero, but d²y/dx² = 0 alone is not enough to guarantee an inflection.

拐点是曲线改变凹性的位置，即从凹变为凸或从凸变为凹。在拐点处，二阶导数通常为零，但 d²y/dx² = 0 本身并不足以保证一定是拐点。

To confirm a point of inflection, you must show that d²y/dx² changes sign on either side of the point. A stationary point of inflection is a point where both dy/dx = 0 and d²y/dx² = 0, and the second derivative changes sign.

要确认拐点，必须证明在该点两侧 d²y/dx² 改变符号。驻点型拐点是指 dy/dx = 0 且 d²y/dx² = 0，同时二阶导数改变符号的点。

For example, y = x³ has dy/dx = 3x² and d²y/dx² = 6x. At x = 0, both derivatives are zero, and d²y/dx² changes from negative to positive, so the origin is a stationary point of inflection.

例如，y = x³ 有 dy/dx = 3x²，d²y/dx² = 6x。在 x = 0 处，两个导数都为零，且 d²y/dx² 从负变为正，因此原点是驻点型拐点。

5. Classifying Stationary Points When f”(x) = 0 | 当 f”(x) = 0 时驻点的分类

If the second derivative test gives d²y/dx² = 0, the stationary point could be a maximum, a minimum, or a point of inflection. In this case, you should examine the sign of dy/dx just to the left and right of the point.

如果二阶导数判别法得到 d²y/dx² = 0，驻点可能是极大值、极小值或拐点。这时应当检查该点左侧和右侧 dy/dx 的符号。

If dy/dx changes from positive to negative, the point is a maximum. If it changes from negative to positive, the point is a minimum. If dy/dx has the same sign on both sides, it is a stationary point of inflection.

如果 dy/dx 从正变为负，该点是极大值。如果从负变为正，该点是极小值。如果 dy/dx 在两侧符号相同，则为驻点型拐点。

Example: y = x⁴ has dy/dx = 4x³ and d²y/dx² = 12x². At x = 0 both are zero. Since dy/dx changes from negative to positive, the point is a minimum.
示例：y = x⁴ 有 dy/dx = 4x³，d²y/dx² = 12x²。在 x = 0 处两者都为零。由于 dy/dx 从负变为正，该点是极小值。
Example: y = x³ has dy/dx = 3x² and d²y/dx² = 6x. Since dy/dx is positive on both sides of x = 0, it is a stationary point of inflection.
示例：y = x³ 有 dy/dx = 3x²，d²y/dx² = 6x。由于在 x = 0 两侧 dy/dx 都为正，因此是驻点型拐点。

6. Connecting First and Second Derivatives | 一阶导数与二阶导数的联系

It helps to think of dy/dx as the velocity of y with respect to x, and d²y/dx² as the acceleration of y with respect to x. A positive second derivative means the first derivative is increasing, just as positive acceleration means velocity is increasing.

把 dy/dx 理解为 y 对 x 的速度，把 d²y/dx² 理解为 y 对 x 的加速度会很有帮助。正的二阶导数表示一阶导数在增大，正如正的加速度表示速度在增大。

This analogy is especially useful in kinematics. If displacement is s, then ds/dt is velocity v, and d²s/dt² is acceleration a. Many Edexcel questions ask you to use the second derivative to find acceleration from a displacement-time expression.

这种类比在运动学中特别有用。如果位移为 s，则 ds/dt 是速度 v，而 d²s/dt² 是加速度 a。许多 Edexcel 题目要求你利用二阶导数从位移-时间表达式中求加速度。

For example, if s = t³ − 2t² + 5t, then v = ds/dt = 3t² − 4t + 5 and a = d²s/dt² = 6t − 4. Substituting a time value gives the acceleration at that instant.

例如，若 s = t³ − 2t² + 5t，则 v = ds/dt = 3t² − 4t + 5，a = d²s/dt² = 6t − 4。代入某一时刻即可求出该瞬时的加速度。

7. Worked Example: Finding and Classifying Stationary Points | 示例：求驻点并分类

Find the stationary points of y = 2x³ − 9x² + 12x − 4 and determine their nature.

求 y = 2x³ − 9x² + 12x − 4 的驻点，并判断其类型。

First differentiate: dy/dx = 6x² − 18x + 12. Set dy/dx = 0:

先求导：dy/dx = 6x² − 18x + 12。令 dy/dx = 0：

6x² − 18x + 12 = 0

Divide by 6: x² − 3x + 2 = 0, so (x − 1)(x − 2) = 0. Hence x = 1 or x = 2.

除以 6：x² − 3x + 2 = 0，因此 (x − 1)(x − 2) = 0，得到 x = 1 或 x = 2。

Now find the second derivative: d²y/dx² = 12x − 18.

再求二阶导数：d²y/dx² = 12x − 18。

At x = 1, d²y/dx² = 12 × 1 − 18 = −6. Since this is negative, the point is a local maximum. The y-coordinate is 2(1)³ − 9(1)² + 12(1) − 4 = 1, so the maximum is at (1, 1).

在 x = 1 处，d²y/dx² = 12 × 1 − 18 = −6。因为为负，该点是局部极大值。y 坐标为 2(1)³ − 9(1)² + 12(1) − 4 = 1，因此极大值在 (1, 1)。

At x = 2, d²y/dx² = 12 × 2 − 18 = 6. Since this is positive, the point is a local minimum. The y-coordinate is 2(8) − 9(4) + 24 − 4 = 16 − 36 + 24 − 4 = 0, so the minimum is at (2, 0).

在 x = 2 处，d²y/dx² = 12 × 2 − 18 = 6。因为为正，该点是局部极小值。y 坐标为 2(8) − 9(4) + 24 − 4 = 16 − 36 + 24 − 4 = 0，因此极小值在 (2, 0)。

8. Using the Second Derivative for Curve Sketching | 利用二阶导数绘制曲线草图

The second derivative helps you sketch curves more accurately by showing where the curve is concave or convex. Combined with stationary points, intercepts, and behaviour as x tends to ±∞, it gives a clear picture of the graph’s shape.

二阶导数通过显示曲线何处为凹或凸，帮助你更准确地绘制曲线草图。结合驻点、截距以及 x 趋于 ±∞ 时的行为，可以清晰地了解图像形状。

For a cubic, the second derivative is linear, so there is exactly one possible inflection point where d²y/dx² = 0. This point divides the curve into a concave section and a convex section.

对于三次函数，二阶导数是线性的，因此恰有一个可能的拐点，即 d²y/dx² = 0 处。该点将曲线分为凹段和凸段。

For example, y = x³ − 3x has d²y/dx² = 6x. The concavity changes at x = 0, and the curve is concave for x < 0 and convex for x > 0.

例如，y = x³ − 3x 有 d²y/dx² = 6x。凹性在 x = 0 处改变，曲线在 x < 0 时为凹，在 x > 0 时为凸。

9. Common Mistakes with Second Derivatives | 二阶导数的常见错误

A common mistake is to assume that d²y/dx² = 0 automatically means a point of inflection. You must check that the second derivative changes sign. For example, y = x⁴ has d²y/dx² = 12x², which is zero at x = 0, but the concavity does not change; the curve is convex on both sides.

一个常见错误是认为 d²y/dx² = 0 就自动意味着拐点。你必须检查二阶导数是否改变符号。例如，y = x⁴ 有 d²y/dx² = 12x²，在 x = 0 处为零，但凹性没有改变；曲线两侧都是凸的。

Another mistake is forgetting to find the y-coordinate after classifying a stationary point. A maximum or minimum is a point, so it needs both coordinates.

另一个错误是在判断驻点类型后忘记求 y 坐标。极大值或极小值是一个点，因此需要两个坐标。

Students also sometimes differentiate incorrectly when using the chain, product, or quotient rule first. The second derivative is only reliable if the first derivative is correct.

学生有时在先使用链式法则、乘积法则或商法则时求导错误。只有一阶导数正确，二阶导数才可靠。

10. Second Derivatives in Optimisation Problems | 二阶导数在最优化问题中的应用

Optimisation questions ask you to find maximum or minimum values, such as the largest area, smallest surface area, or maximum profit. After finding a stationary point, the second derivative test gives a quick justification for whether the value is a maximum or minimum.

最优化问题要求你找出最大或最小值，例如最大面积、最小表面积或最大利润。找到驻点后，二阶导数判别法可以快速证明该值是极大值还是极小值。

For example, if a volume V is modelled by V = 12x − x³, then dV/dx = 12 − 3x² and d²V/dx² = −6x. At x = 2, dV/dx = 0 and d²V/dx² = −12, so the volume is maximised at x = 2.

例如，若体积 V 由 V = 12x − x³ 建模，则 dV/dx = 12 − 3x²，d²V/dx² = −6x。在 x = 2 处，dV/dx = 0 且 d²V/dx² = −12，因此体积在 x = 2 处取得最大值。

In such questions, the second derivative test is often the neatest way to show that a stationary point is the required maximum or minimum, rather than testing signs on either side.

在这类题目中，二阶导数判别法通常是证明驻点为所需极大值或极小值的最简洁方法，而不必检查两侧符号。

11. Summary Table for the Second Derivative Test | 二阶导数判别法总结表

The following table summarises how the first and second derivatives are used together to classify stationary points.

下表总结了如何结合使用一阶导数和二阶导数来判断驻点类型。

Condition at stationary point	Conclusion
dy/dx = 0 and d²y/dx² < 0	Local maximum
dy/dx = 0 and d²y/dx² > 0	Local minimum
dy/dx = 0 and d²y/dx² = 0	Inconclusive; test dy/dx sign on either side
d²y/dx² changes sign	Point of inflection

In exam work, always show the value you calculate for d²y/dx² and write a conclusion such as ‘negative, so maximum’ or ‘positive, so minimum’.

在考试作答时，始终写出你计算出的 d²y/dx² 值，并写出结论，例如“为负，所以是极大值”或“为正，所以是极小值”。

12. Exam Tips and Final Reminders | 考试提示与最后提醒

When a question says ‘determine the nature of the stationary points’, you may use either the second derivative test or a sign table. If the question specifically asks for the second derivative test, you must use it.

当题目说“判断驻点的类型”时，你可以使用二阶导数判别法或符号表。如果题目明确要求使用二阶导数判别法，则必须使用该方法。

Always simplify your first derivative before differentiating again. This reduces errors and makes the second derivative easier to evaluate at a point.

在再次求导之前，一定要先化简一阶导数。这可以减少错误，并使二阶导数在某点的计算更容易。

Remember that a point of inflection is about concavity change, not necessarily a horizontal gradient. A non-stationary point of inflection can have dy/dx ≠ 0 while d²y/dx² = 0 and changes sign.

请记住，拐点关乎凹性的改变，而不一定是水平斜率。非驻点型拐点可以具有 dy/dx ≠ 0，同时 d²y/dx² = 0 且改变符号。

Mastering the second derivative will improve your curve sketching, optimisation, and kinematics work across the Edexcel A-Level specification.

掌握二阶导数将提升你在 Edexcel A-Level 课程中曲线草图、最优化和运动学方面的解题能力。

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Using Second Derivatives | 二阶导数的应用