Applications of Differentiation | 微分的应用

📚 Applications of Differentiation | 微分的应用

Differentiation is one of the most powerful tools in A-Level Mathematics. It allows us to analyse how quantities change, find maximum and minimum values, and solve real-world optimisation problems. This article covers the key applications of differentiation required for AQA A-Level Mathematics, with worked examples and common pitfalls to avoid.

微分是A-Level数学中最强大的工具之一。它使我们能够分析量的变化方式，找到最大值和最小值，并解决现实中的优化问题。本文涵盖AQA A-Level数学要求的微分核心应用，包含典型例题和常见易错点。

1. The Gradient Function | 梯度函数

The derivative f'(x) represents the gradient of the tangent to the curve y = f(x) at any point x. For a function f(x) = xⁿ, the derivative is given by f'(x) = nxⁿ⁻¹. This rule, known as the power rule, is the foundation of all differentiation in this course.

导数 f'(x) 表示曲线 y = f(x) 在任意点 x 处切线的斜率。对于函数 f(x) = xⁿ，其导数为 f'(x) = nxⁿ⁻¹。这条法则称为幂法则，是本课程中所有微分的基础。

For example, if f(x) = 3x⁴ – 2x² + 5x – 7, then differentiating term by term gives f'(x) = 12x³ – 4x + 5. Remember that constant terms differentiate to zero, since a constant function has zero gradient everywhere.

例如，若 f(x) = 3x⁴ – 2x² + 5x – 7，则逐项微分得 f'(x) = 12x³ – 4x + 5。请记住常数项微分为零，因为常数函数处处斜率为零。

The derivative of a sum is the sum of the derivatives: d/dx [f(x) + g(x)] = f'(x) + g'(x)
The derivative of k·f(x) is k·f'(x), where k is a constant
Write all terms in the form axⁿ before differentiating (e.g., 1/x² = x⁻²)

和的导数等于导数的和：d/dx [f(x) + g(x)] = f'(x) + g'(x)
k·f(x) 的导数为 k·f'(x)，其中 k 为常数
微分前将所有项写成 axⁿ 的形式（例如 1/x² = x⁻²）

2. Tangents and Normals | 切线与法线

At a point (x₁, y₁) on a curve y = f(x), the gradient of the tangent is f'(x₁). The tangent line has equation y – y₁ = f'(x₁)(x – x₁). The normal is perpendicular to the tangent, so its gradient is the negative reciprocal: -1/f'(x₁), provided f'(x₁) ≠ 0.

在曲线 y = f(x) 上的点 (x₁, y₁) 处，切线的斜率为 f'(x₁)。切线方程为 y – y₁ = f'(x₁)(x – x₁)。法线垂直于切线，因此其斜率为负倒数：-1/f'(x₁)，前提是 f'(x₁) ≠ 0。

For example, consider f(x) = x² – 4x + 3 at x = 2. We have f(2) = 4 – 8 + 3 = -1 and f'(x) = 2x – 4, so f'(2) = 0. The tangent is horizontal: y = -1. The normal is then vertical: x = 2.

例如，考虑 f(x) = x² – 4x + 3 在 x = 2 处。我们有 f(2) = 4 – 8 + 3 = -1 且 f'(x) = 2x – 4，所以 f'(2) = 0。切线为水平线：y = -1。此时法线为竖直线：x = 2。

Tangent: y – y₁ = f'(x₁)(x – x₁) | Normal: y – y₁ = -1/f'(x₁) · (x – x₁)

3. Stationary Points | 驻点

A stationary point occurs where f'(x) = 0. At such a point, the curve is momentarily flat, and the tangent is horizontal. Stationary points are classified as local maxima, local minima, or stationary points of inflection.

驻点出现在 f'(x) = 0 处。在驻点上，曲线暂时是平的，切线为水平。驻点分为局部最大值、局部最小值和驻点拐点三类。

To find stationary points: (1) differentiate f(x) to get f'(x); (2) solve f'(x) = 0 for x; (3) substitute each x back into f(x) to find the corresponding y-coordinate.

求驻点的步骤：(1) 对 f(x) 求导得到 f'(x)；(2) 解方程 f'(x) = 0 求 x；(3) 将每个 x 代回 f(x) 求对应的 y 坐标。

Consider f(x) = x³ – 3x. Then f'(x) = 3x² – 3 = 3(x – 1)(x + 1). Setting f'(x) = 0 gives x = 1 or x = -1. Substituting: f(1) = -2 and f(-1) = 2. Thus the stationary points are (1, -2) and (-1, 2).

考虑 f(x) = x³ – 3x。则 f'(x) = 3x² – 3 = 3(x – 1)(x + 1)。令 f'(x) = 0 得 x = 1 或 x = -1。代回：f(1) = -2，f(-1) = 2。因此驻点为 (1, -2) 和 (-1, 2)。

4. Classifying Stationary Points | 驻点的分类

There are two main methods to determine the nature of a stationary point. The first derivative test examines the sign of f'(x) immediately to the left and right of the stationary point.

判断驻点性质有两种主要方法。一阶导数检验法考察驻点两侧邻近处 f'(x) 的符号。

If f'(x) changes from positive to negative, the point is a local maximum
If f'(x) changes from negative to positive, the point is a local minimum
If the sign does not change, the point is a stationary point of inflection

若 f'(x) 由正变负，则该点为局部最大值
若 f'(x) 由负变正，则该点为局部最小值
若符号不变，则该点为驻点拐点

For f(x) = x³ – 3x, we have f'(-2) = 3(4) – 3 = 9 > 0, f'(0) = -3 < 0, and f'(2) = 9 > 0. Hence (-1, 2) is a maximum and (1, -2) is a minimum.

对于 f(x) = x³ – 3x，f'(-2) = 3(4) – 3 = 9 > 0，f'(0) = -3 < 0，f'(2) = 9 > 0。因此 (-1, 2) 为最大值，(1, -2) 为最小值。

5. The Second Derivative Test | 二阶导数检验法

The second derivative f”(x) measures the rate of change of the gradient. At a stationary point where f'(x) = 0:

二阶导数 f”(x) 衡量梯度的变化率。在 f'(x) = 0 的驻点处：

If f”(x) > 0, the gradient is increasing, so the point is a local minimum
If f”(x) < 0, the gradient is decreasing, so the point is a local maximum
If f”(x) = 0, the test is inconclusive; use the first derivative test

若 f”(x) > 0，梯度递增，该点为局部最小值
若 f”(x) < 0，梯度递减，该点为局部最大值
若 f”(x) = 0，检验无效；应使用一阶导数检验法

For f(x) = x³ – 3x, f”(x) = 6x. At x = -1, f”(-1) = -6 < 0, confirming a maximum. At x = 1, f''(1) = 6 > 0, confirming a minimum. This method is generally quicker, but always check the condition f”(x) = 0 carefully.

对于 f(x) = x³ – 3x，f”(x) = 6x。在 x = -1 处，f”(-1) = -6 < 0，确认是最大值。在 x = 1 处，f''(1) = 6 > 0，确认是最小值。此法通常更快，但遇到 f”(x) = 0 时必须谨慎处理。

6. Increasing and Decreasing Functions | 增函数与减函数

A function is increasing on an interval if f'(x) ≥ 0 for all x in that interval, and decreasing if f'(x) ≤ 0. This is a direct consequence of the derivative representing the gradient, or rate of change.

若在某个区间内对所有 x 有 f'(x) ≥ 0，则函数在该区间内递增；若 f'(x) ≤ 0，则函数递减。这是导数表示斜率或变化率的直接推论。

Example: determine where f(x) = x³ – 12x + 5 is increasing. First, f'(x) = 3x² – 12 = 3(x – 2)(x + 2). A quadratic with a positive leading coefficient is positive outside the roots, so f'(x) > 0 when x < -2 or x > 2. Therefore f is increasing on (-∞, -2] and [2, ∞).

例如：确定 f(x) = x³ – 12x + 5 在何处递增。首先，f'(x) = 3x² – 12 = 3(x – 2)(x + 2)。首项系数为正的二次函数在两根之外为正，所以当 x < -2 或 x > 2 时 f'(x) > 0。因此 f 在 (-∞, -2] 和 [2, ∞) 上递增。

Always sketch the sign diagram of f'(x) to avoid mistakes when the derivative factorises into more than two linear factors.

当导数分解为两个以上线性因式时，务必画出 f'(x) 的符号图以避免出错。

7. Optimisation Problems | 优化问题

Optimisation uses differentiation to find the maximum or minimum value of a quantity subject to given constraints. The standard approach is to express the quantity as a single-variable function, then find stationary points and determine their nature.

优化问题利用微分在给定约束下求某个量的最大值或最小值。标准方法是将该量表示为单变量函数，然后求驻点并判断其性质。

Problem: A rectangular field is to be enclosed with 400 m of fencing. Find the maximum possible area. Let the length be x and width be y. Then 2x + 2y = 400, so y = 200 – x. The area is A = xy = x(200 – x) = 200x – x². Differentiating: A'(x) = 200 – 2x. Setting A'(x) = 0 gives x = 100. Since A”(x) = -2 < 0, this is a maximum. Hence y = 100 and A_max = 100 × 100 = 10000 m².

问题：用400米围栏围一个矩形场地，求最大面积。设长为 x，宽为 y。则 2x + 2y = 400，所以 y = 200 – x。面积为 A = xy = x(200 – x) = 200x – x²。求导：A'(x) = 200 – 2x。令 A'(x) = 0 得 x = 100。因为 A”(x) = -2 < 0，所以是最大值。因此 y = 100，A_max = 100 × 100 = 10000 m²。

A_max = 10000 m² when the rectangle is a 100 m × 100 m square

8. Modelling with Differentiation | 微分建模

Real-world problems often involve business, physics, or geometry. A classic AQA example is revenue optimisation. Suppose a company sells x units of a product, and the profit is given by P(x) = -0.5x² + 50x – 200 for 0 ≤ x ≤ 100. Find the production level that maximises profit.

现实问题常涉及商业、物理或几何。一个典型的AQA例题是收益优化。假设一家公司销售 x 件产品，利润由 P(x) = -0.5x² + 50x – 200 给出，其中 0 ≤ x ≤ 100。求使利润最大化的产量。

Differentiating: P'(x) = -x + 50. Setting P'(x) = 0 gives x = 50. Since P”(x) = -1 < 0, this is a maximum. The maximum profit is P(50) = -0.5(2500) + 2500 - 200 = -1250 + 2500 - 200 = 1050. Also check endpoints: P(0) = -200 and P(100) = -5000 + 5000 - 200 = -200, so x = 50 indeed gives the global maximum.

求导：P'(x) = -x + 50。令 P'(x) = 0 得 x = 50。因为 P”(x) = -1 < 0，所以是最大值。最大利润为 P(50) = -0.5(2500) + 2500 - 200 = -1250 + 2500 - 200 = 1050。还需检查端点：P(0) = -200，P(100) = -5000 + 5000 - 200 = -200，因此 x = 50 确实是全局最大值。

9. Rates of Change | 变化率

The derivative dy/dx represents the rate of change of y with respect to x. In applied contexts, this often involves related rates. The chain rule d/dx [f(g(x))] = f'(g(x)) · g'(x) is essential for these problems.

导数 dy/dx 表示 y 关于 x 的变化率。在应用情境中，这常涉及相关变化率问题。链式法则 d/dx [f(g(x))] = f'(g(x)) · g'(x) 是解决此类问题的关键。

Example: The radius r of a circular patch of oil is increasing at 0.5 cm/s. Find the rate at which the area A is increasing when r = 10 cm. Since A = πr², we have dA/dr = 2πr. Using dA/dt = dA/dr × dr/dt = 2πr × 0.5 = πr. At r = 10, dA/dt = 10π cm²/s.

例如：圆形油膜的半径 r 以 0.5 cm/s 的速度增大。求当 r = 10 cm 时面积 A 的增长率。由于 A = πr²，我们有 dA/dr = 2πr。利用 dA/dt = dA/dr × dr/dt = 2πr × 0.5 = πr。当 r = 10 时，dA/dt = 10π cm²/s。

Identify which rates are given and which rate is required
Write an equation linking the relevant variables
Differentiate with respect to time t using the chain rule

明确哪些变化率已知、哪些要求
写出关联各变量的方程
用链式法则对时间 t 求导

10. Curve Sketching | 曲线作图

Differentiation provides key information for sketching curves: stationary points, intervals of increase and decrease, and points of inflection. Combine this with the y-intercept (set x = 0) and roots (set f(x) = 0) to produce an accurate graph.

微分提供了作图的关健信息：驻点、增减区间和拐点。结合 y 截距（令 x = 0）和根（令 f(x) = 0），即可画出准确的图形。

For f(x) = x³ – 3x² + 2: f'(x) = 3x² – 6x = 3x(x – 2), so stationary points occur at x = 0 and x = 2. We have f(0) = 2 and f(2) = 8 – 12 + 2 = -2. The second derivative f”(x) = 6x – 6. At x = 0, f”(0) = -6 < 0 so (0, 2) is a maximum. At x = 2, f''(2) = 6 > 0 so (2, -2) is a minimum. The y-intercept is at (0, 2) and f(x) = 0 has one negative root and one positive root (test x = -1: f(-1) = -1 – 3 + 2 = -2; test x = 3: f(3) = 27 – 27 + 2 = 2, so a root lies between 2 and 3).

对于 f(x) = x³ – 3x² + 2：f'(x) = 3x² – 6x = 3x(x – 2)，因此驻点出现在 x = 0 和 x = 2 处。f(0) = 2，f(2) = 8 – 12 + 2 = -2。二阶导数 f”(x) = 6x – 6。在 x = 0 处，f”(0) = -6 < 0，所以 (0, 2) 是最大值。在 x = 2 处，f''(2) = 6 > 0，所以 (2, -2) 是最小值。y截距为 (0, 2)，f(x) = 0 有一个负根和一个正根（检验 x = -1：f(-1) = -1 – 3 + 2 = -2；检验 x = 3：f(3) = 27 – 27 + 2 = 2，因此根在 2 和 3 之间）。

11. Points of Inflection | 拐点

A point of inflection is where the curve changes from concave up (convex) to concave down, or vice versa. At a point of inflection, f”(x) = 0 (provided f” changes sign). Note that stationary points of inflection also have f'(x) = 0.

拐点是曲线从凹向上（凸）变为凹向下，或反之的点。在拐点处，f”(x) = 0（前提是 f” 变号）。注意驻点拐点同时满足 f'(x) = 0。

For f(x) = x³, f'(x) = 3x² and f”(x) = 6x. At x = 0, both f'(0) = 0 and f”(0) = 0. The second derivative changes from negative (x < 0) to positive (x > 0), so (0, 0) is a stationary point of inflection. The curve is concave down for x < 0 and concave up for x > 0.

对于 f(x) = x³，f'(x) = 3x²，f”(x) = 6x。在 x = 0 处，f'(0) = 0 且 f”(0) = 0。二阶导数从负（x < 0）变为正（x > 0），因此 (0, 0) 是驻点拐点。曲线在 x < 0 时凹向下，在 x > 0 时凹向上。

Condition for inflection: f”(x) = 0 and f” changes sign across x

12. Common Exam Pitfalls | 常见考试误区

Students frequently lose marks on differentiation questions due to avoidable errors. Here are the most common pitfalls to watch for in the AQA examination.

学生在微分题目中常因可避免的错误而失分。以下是AQA考试中最常见的易错点。

Pitfall / 误区	Correct Approach / 正确做法
Differentiating 1/x as 1 instead of x⁻¹	Rewrite as x⁻¹, giving derivative -x⁻²
Forgetting the constant when differentiating	Constant terms always differentiate to 0
Not checking the nature of stationary points	Always use f”(x) or the sign test to classify
Ignoring endpoints in optimisation problems	Compare stationary values with endpoint values
Using the product/quotient rules incorrectly	Recognise when f = u·v and apply the correct rule

Always read the question carefully: does it ask for the equation of the tangent, the gradient, or the coordinates of the stationary point? Write your final answer with appropriate units and full sentences in applied contexts.

务必仔细审题：题目问的是切线方程、斜率还是驻点坐标？在应用情境中，最终答案要带合适的单位并用完整句子作答。

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Applications of Differentiation | 微分的应用