Second Derivatives: Concept and Applications | 二阶导数的概念与应用

📚 Second Derivatives: Concept and Applications | 二阶导数的概念与应用

The second derivative is one of the most powerful tools in A-Level calculus. While the first derivative tells us how a function is changing, the second derivative tells us how that change itself is changing — unlocking deeper information about curvature, acceleration, and the nature of stationary points.

二阶导数是A-Level微积分中最强大的工具之一。一阶导数告诉我们函数如何变化，而二阶导数则告诉我们这种变化本身如何变化——为我们揭示了曲线的弯曲程度、加速度以及驻点的性质等更深层的信息。

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

The second derivative is simply the derivative of the derivative. If y = f(x) is a differentiable function, then its first derivative f'(x) measures the rate of change of y with respect to x. Differentiating f'(x) again gives the second derivative, denoted f”(x).

二阶导数就是一阶导数的导数。如果 y = f(x) 是一个可微函数，那么它的一阶导数 f'(x) 表示 y 关于 x 的变化率。对 f'(x) 再次求导，就得到二阶导数，记作 f”(x)。

f”(x) = d/dx [f'(x)] = d²y/dx²

In Leibniz notation, if y is a function of x, the second derivative is written as d²y/dx². Note carefully the placement of the superscript 2: it appears in the numerator as d², and in the denominator as dx². This notation is not a fraction in the algebraic sense, but rather a conventional symbol representing the second derivative.

在莱布尼茨记号中，如果 y 是 x 的函数，二阶导数写作 d²y/dx²。请注意右上角数字 2 的位置：分子中写作 d²，分母中写作 dx²。这个记号并非代数意义上的分数，而是表示二阶导数的约定符号。

2. Common Notation | 常见记法

Depending on the textbook or examination board, you may encounter several different notations for the second derivative. For Edexcel A-Level Mathematics, you are expected to recognise and use both Lagrange and Leibniz notations fluently.

根据教材或考试局的不同，你会遇到二阶导数的多种记法。对于Edexcel A-Level数学，你需要熟练识别和使用拉格朗日记号和莱布尼茨记号。

Lagrange notation: f”(x) or y” — read as “f double prime of x”
Leibniz notation: d²y/dx² — read as “dee two y by dee x squared”
Operator notation: D²f or d²/dx² [f(x)]

拉格朗日记法：f”(x) 或 y” — 读作 “f 二阶导”
莱布尼茨记法：d²y/dx² — 读作 “d²y 比 dx²”
算子记法：D²f 或 d²/dx² [f(x)]

3. Worked Example: Computing Second Derivatives | 例题：计算二阶导数

Let us compute the second derivative of a polynomial function step by step.

让我们逐步计算一个多项式函数的二阶导数。

Example 1: Find f”(x) for f(x) = 3x⁴ − 5x³ + 2x² − 7x + 1.

例1：求 f(x) = 3x⁴ − 5x³ + 2x² − 7x + 1 的 f”(x)。

Solution: First differentiate once:

解：先求一阶导数：

f'(x) = 12x³ − 15x² + 4x − 7

Then differentiate again:

再求一次导：

f”(x) = 36x² − 30x + 4

Example 2: Find d²y/dx² for y = e²ˣsin x.

例2：求 y = e²ˣsin x 的 d²y/dx²。

Solution: Use the product rule first:

解：先用乘积法则：

dy/dx = 2e²ˣsin x + e²ˣcos x = e²ˣ(2sin x + cos x)

Differentiate again using the product rule on each term:

再对每一项用乘积法则求导：

d²y/dx² = 2e²ˣ(2sin x + cos x) + e²ˣ(2cos x − sin x)

= e²ˣ(4sin x + 2cos x + 2cos x − sin x) = e²ˣ(3sin x + 4cos x)

4. Geometric Meaning: Concavity | 几何意义：凹凸性

The first derivative f'(x) gives the slope of the tangent line at each point. The second derivative f”(x) describes how that slope is changing — which directly determines whether the curve bends upward or downward.

一阶导数 f'(x) 给出各点切线的斜率。二阶导数 f”(x) 描述的是斜率如何变化——这直接决定了曲线是向上弯曲还是向下弯曲。

If f”(x) > 0 on an interval, the slope is increasing, and the curve is said to be concave up (shaped like a cup ∪). If f”(x) < 0, the slope is decreasing, and the curve is concave down (shaped like a cap ∩).

如果在某区间上 f”(x) > 0，斜率在增大，称曲线为凹向上（形如 ∪）。如果 f”(x) < 0，斜率在减小，称曲线为凹向下（形如 ∩）。

Sign of f”(x) \| f”(x) 的符号	Curvature \| 弯曲方向	Shape \| 形状
f”(x) > 0	Concave up \| 凹向上	∪
f”(x) < 0	Concave down \| 凹向下	∩

A helpful mnemonic: when f”(x) > 0, the curve holds water like a cup (∪); when f”(x) < 0, it sheds water like a cap (∩).

一个有用的记忆方法：当 f”(x) > 0 时，曲线像杯子一样能盛水（∪）；当 f”(x) < 0 时，曲线像帽子一样不盛水（∩）。

5. Classifying Stationary Points | 驻点的分类

At a stationary point, f'(x) = 0. The sign of the second derivative at that point tells us what type of stationary point we have. This is known as the second derivative test.

在驻点处，f'(x) = 0。二阶导数在该点的符号告诉我们驻点的类型。这称为二阶导数判别法。

If f”(x) > 0 at the stationary point, the point is a local minimum (curve is concave up).
If f”(x) < 0 at the stationary point, the point is a local maximum (curve is concave down).
If f”(x) = 0 at the stationary point, the test is inconclusive — the point could be a maximum, minimum, or a point of inflection.

若驻点处 f”(x) > 0，该点为局部极小值（曲线凹向上）。
若驻点处 f”(x) < 0，该点为局部极大值（曲线凹向下）。
若驻点处 f”(x) = 0，判别法失效——该点可能是极大值、极小值或拐点。

6. Worked Example: Second Derivative Test | 例题：二阶导数判别法

Example: Find and classify the stationary points of f(x) = x³ − 6x² + 9x + 1.

例：求 f(x) = x³ − 6x² + 9x + 1 的驻点并判断其类型。

Solution: First, find the stationary points by setting f'(x) = 0.

解：首先令 f'(x) = 0，求驻点。

f'(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3) = 0

So x = 1 and x = 3 are the x-coordinates of the stationary points. Now compute the second derivative:

因此 x = 1 和 x = 3 是驻点的横坐标。现在计算二阶导数：

f”(x) = 6x − 12

Evaluate the second derivative at each stationary point:

在每个驻点处求二阶导数的值：

At x = 1: f”(1) = 6 − 12 = −6 < 0 → local maximum

At x = 3: f”(3) = 18 − 12 = 6 > 0 → local minimum

The corresponding y-values are f(1) = 1 − 6 + 9 + 1 = 5 and f(3) = 27 − 54 + 27 + 1 = 1. Hence (1, 5) is a local maximum and (3, 1) is a local minimum.

对应的 y 值为 f(1) = 1 − 6 + 9 + 1 = 5，f(3) = 27 − 54 + 27 + 1 = 1。因此 (1, 5) 是局部极大值点，(3, 1) 是局部极小值点。

7. Points of Inflection | 拐点

A point of inflection is a point where the curve changes its concavity — from concave up to concave down, or vice versa. At such a point, the second derivative must be zero, though not every point where f”(x) = 0 is necessarily a point of inflection.

拐点是曲线凹凸性发生改变的点——从凹向上变为凹向下，或反之。在拐点处，二阶导数必须为零，但并非所有 f”(x) = 0 的点都是拐点。

To confirm a point of inflection at x = a, we must check that the sign of f”(x) changes as x passes through a. Equivalently, if f”(a) = 0 and f”(x) changes sign across x = a, then (a, f(a)) is a point of inflection.

要确认 x = a 处是拐点，必须检验 f”(x) 的符号在 x 经过 a 时是否发生改变。等价地说，如果 f”(a) = 0 且 f”(x) 在 x = a 两侧变号，那么 (a, f(a)) 是拐点。

Example: For f(x) = x³, we have f”(x) = 6x. At x = 0, f”(0) = 0. Since f”(x) < 0 for x < 0 and f''(x) > 0 for x > 0, the sign changes, so (0, 0) is a point of inflection.

例：对于 f(x) = x³，有 f”(x) = 6x。在 x = 0 处，f”(0) = 0。由于当 x < 0 时 f''(x) < 0，当 x > 0 时 f”(x) > 0，符号发生了改变，因此 (0, 0) 是拐点。

8. Distinction: Stationary vs Non-Stationary Inflection | 驻点拐点与非驻点拐点的区别

An important distinction must be made between stationary points of inflection and non-stationary points of inflection.

必须区分驻点型拐点与非驻点型拐点。

A stationary point of inflection occurs when f'(x) = 0 and f”(x) = 0 (with a sign change in f”(x)). For example, f(x) = x³ has a stationary point of inflection at x = 0, because both f'(0) = 0 and f”(0) = 0.

驻点型拐点出现在 f'(x) = 0 且 f”(x) = 0（且 f”(x) 变号）时。例如，f(x) = x³ 在 x = 0 处有驻点型拐点，因为 f'(0) = 0 且 f”(0) = 0。

A non-stationary point of inflection occurs when f”(x) = 0 with a sign change, but f'(x) ≠ 0. For example, f(x) = x³ − 3x has f”(x) = 6x. At x = 0, f”(0) = 0 and the sign of f” changes, so (0, 0) is an inflection point. However, f'(0) = −3 ≠ 0, so it is not a stationary point.

非驻点型拐点出现在 f”(x) = 0 且变号，但 f'(x) ≠ 0 时。例如，f(x) = x³ − 3x 的 f”(x) = 6x。在 x = 0 处，f”(0) = 0 且 f” 变号，所以 (0, 0) 是拐点。但 f'(0) = −3 ≠ 0，因此它不是驻点。

9. Kinematic Application: Acceleration | 运动学应用：加速度

In kinematics, if s(t) represents displacement as a function of time, then the first derivative ds/dt represents velocity, and the second derivative d²s/dt² represents acceleration.

在运动学中，如果 s(t) 表示位移关于时间的函数，那么一阶导数 ds/dt 表示速度，二阶导数 d²s/dt² 表示加速度。

v(t) = ds/dt, a(t) = dv/dt = d²s/dt²

Example: A particle moves along a straight line such that its displacement at time t seconds is s(t) = t³ − 6t² + 9t metres. Find the acceleration when t = 2.

例：一个质点沿直线运动，其 t 秒时的位移为 s(t) = t³ − 6t² + 9t 米。求 t = 2 时的加速度。

Velocity: v(t) = ds/dt = 3t² − 12t + 9.

速度：v(t) = ds/dt = 3t² − 12t + 9。

Acceleration: a(t) = dv/dt = 6t − 12.

加速度：a(t) = dv/dt = 6t − 12。

At t = 2: a(2) = 12 − 12 = 0 m/s².

当 t = 2 时：a(2) = 12 − 12 = 0 m/s²。

The particle is instantaneously not accelerating at t = 2, even though its velocity is v(2) = 3(4) − 24 + 9 = −3 m/s, meaning it is still moving in the negative direction.

在 t = 2 时，质点瞬时加速度为零，但其速度为 v(2) = 3(4) − 24 + 9 = −3 m/s，这意味着它仍然沿负方向运动。

10. Optimisation Problems | 优化问题

The second derivative test is particularly useful in optimisation problems. When you find a stationary point using f'(x) = 0, the second derivative provides a quick and reliable way to determine whether you have found a maximum or a minimum without constructing a sign table.

二阶导数判别法在优化问题中尤其有用。当你通过 f'(x) = 0 找到驻点后，二阶导数提供了一种快速且可靠的方法来判断你找到的是极大值还是极小值，而无需构造符号表。

Example: A farmer wants to fence a rectangular enclosure against a long straight wall. He has 200 metres of fencing. Find the maximum area he can enclose.

例：一位农夫想用一面长直墙围一个矩形围栏。他有 200 米长的栅栏。求他能围出的最大面积。

Let the width perpendicular to the wall be x metres, and the length parallel to the wall be y metres. Since the wall forms one side, the fencing satisfies 2x + y = 200, so y = 200 − 2x.

设垂直于墙的边宽为 x 米，平行于墙的边长为 y 米。由于墙作为一边，栅栏满足 2x + y = 200，因此 y = 200 − 2x。

The area is A = xy = x(200 − 2x) = 200x − 2x².

面积为 A = xy = x(200 − 2x) = 200x − 2x²。

Differentiate: dA/dx = 200 − 4x. Setting this to zero gives x = 50. The second derivative is d²A/dx² = −4 < 0, confirming a maximum. Thus y = 200 − 100 = 100, and the maximum area is A = 50 × 100 = 5000 m².

求导：dA/dx = 200 − 4x。令其为零得 x = 50。二阶导数为 d²A/dx² = −4 < 0，确认为极大值。因此 y = 200 − 100 = 100，最大面积为 A = 50 × 100 = 5000 m²。

11. When f”(x) = 0: What Else to Check | 当 f”(x) = 0 时：还需检验什么

There are situations where the second derivative test fails because f”(x) = 0 at the stationary point. In these cases, you must fall back on the first derivative test, examining the sign of f'(x) on either side of the stationary point.

在某些情况下，由于二阶导数在驻点处为零，判别法失效。此时你必须使用一阶导数判别法，即考察驻点两侧 f'(x) 的符号。

Example: Consider f(x) = x⁴. We have f'(x) = 4x³, so x = 0 is a stationary point. But f”(x) = 12x², and f”(0) = 0. The second derivative test is inconclusive. Checking the first derivative: f'(x) < 0 for x < 0 and f'(x) > 0 for x > 0. Since the sign of f’ changes from negative to positive, x = 0 is a local minimum.

例：考虑 f(x) = x⁴。有 f'(x) = 4x³，所以 x = 0 是驻点。但 f”(x) = 12x²，且 f”(0) = 0。二阶导数判别法失效。检验一阶导数：当 x < 0 时 f'(x) < 0，当 x > 0 时 f'(x) > 0。由于 f’ 的符号从负变正，x = 0 是局部极小值。

Always remember: f”(x) = 0 alone does not tell you the nature of a stationary point. Always verify with a sign change test when the second derivative is zero.

始终记住：仅凭 f”(x) = 0 不能判断驻点的性质。当二阶导数为零时，务必用变号检验来确认。

12. Summary and Exam Tips | 总结与考试要点

The second derivative is an indispensable tool in A-Level calculus. It connects the algebraic operation of repeated differentiation with the geometric notions of concavity and inflection, as well as physical notions such as acceleration.

二阶导数是A-Level微积分中不可或缺的工具。它将重复求导这一代数操作与凹凸性和拐点的几何概念，以及加速度等物理概念联系起来。

Always write f”(x) = d/dx[f'(x)] and remember the correct Leibniz notation d²y/dx².
For stationary point classification, evaluate f”(x) at the stationary point. If f” > 0, minimum; if f” < 0, maximum.
If f” = 0 at a stationary point, use the first derivative test instead.
Points of inflection satisfy f”(x) = 0 and require a sign change in f”(x).
In kinematics, acceleration is the second derivative of displacement: a = d²s/dt².
In optimisation, use the second derivative test to confirm the nature of turning points quickly.

始终写出 f”(x) = d/dx[f'(x)]，并记住莱布尼茨记号 d²y/dx² 的写法。
对驻点分类时，在驻点处计算 f”(x)。若 f” > 0，为极小值；若 f” < 0，为极大值。
若驻点处 f” = 0，改用一阶导数判别法。
拐点满足 f”(x) = 0 且 f”(x) 必须变号。
在运动学中，加速度是位移的二阶导数：a = d²s/dt²。
在优化问题中，用二阶导数判别法快速确认真值点的性质。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Sign of f”(x) \| f”(x) 的符号	Curvature \| 弯曲方向	Shape \| 形状
f”(x) > 0	Concave up \| 凹向上	∪
f”(x) < 0	Concave down \| 凹向下	∩

Second Derivatives: Concept and Applications | 二阶导数的概念与应用