📚 Applications of Differentiation: Tangents, Normals & Stationary Points | 微分的应用：切线、法线与驻点
4)-152, which appears frequently in Edexcel A-Level Mathematics practice sets. We take a deep dive into the applications of differentiation, focusing on tangents, normals, stationary points, and related exam-style problems. By mastering these concepts, you will be well equipped to tackle both pure and applied contexts in your exams.
本复习指南的灵感来源于一道经典例题 pdfjoiner_(4)-152，此类题目在 Edexcel A-Level 数学练习中频繁出现。我们将深入探讨微分的应用，重点涵盖切线、法线、驻点以及相关的考试题型。掌握这些概念后，你将能从容应对考试中纯数及应用情境的各种问题。
1. Review of Basic Differentiation Rules | 基本微分规则回顾
Before exploring applications, you must be completely comfortable with fundamental derivatives. The power rule states d/dx (xn) = n xn-1. For exponentials, d/dx (ex) = ex tan x) = sec² x. The sum, product, quotient, and chain rules allow you to handle combinations of these functions.
在探讨应用之前，你必须对基本求导法则感到得心应手。幂函数法则为 d/dx (xn) = n xn-1。指数函数的导数是它本身，而自然对数的导数是 1/x。三角函数的求导结果如下：d/dx (sin x) = cos x，d/dx (cos x) = -sin x，d/dx (tan x) = sec² x。运用和法则、乘积法则、商法则以及链式法则，你就能处理这些函数的各种组合。
- Power rule: d/dx (x⁵) = 5x⁴
- Exponential: d/dx (e²ˣ) = 2e²ˣ (using the chain rule)
- Product rule: d/dx (x sin x) = sin x + x cos x
- 幂函数法则： d/dx (x⁵) = 5x⁴
- 指数函数： d/dx (e²ˣ) = 2e²ˣ（运用链式法则）
- 乘积法则： d/dx (x sin x) = sin x + x cos x
2. Finding the Equation of a Tangent | 求切线方程
= f(x) at a point (x₁, y₁) is a straight line that just touches the curve. Its gradient mt is the value of the derivative f '(x₁). You then use the point–gradient form: y − y₁ = mt answer into the form ax + by + c = 0 or y = mx + c, as specified in Edexcel mark schemes.
曲线 y = f(x) 在点 (x₁, y₁) 处的切线是一条刚好接触曲线的直线，其斜率 mt 就是导数 f '(x₁) 的值。然后利用点斜式 y − y₁ = mt (x − x₁)。最后一定要按照 Edexcel 评分标准的要求，将答案化简为 ax + by + c = 0 或 y = mx + c 的形式。
Example: Find the tangent to y = x³ − 4x at x = 2. Here y = 0, f '(x) = 3x² − 4, so m = 8. Tangent: y = 8(x − 2), i.e. y = 8x − 16.
示例：求 y = x³ − 4x 在 x = 2 处的切线。此时 y = 0，f '(x) = 3x² − 4，故 m = 8。切线方程为 y = 8(x − 2)，即 y = 8x − 16。
3. Finding the Equation of a Normal | 求法线方程
The normal to a curve at a point is perpendicular to the tangent at that point. Therefore its gradient mn satisfies mn = −1 / mt, provided mt ≠ 0. If the tangent is horizontal (mt = x₁. Use the same point-gradient form with the normal gradient and simplify your answer.
曲线在某点的法线垂直于该点处的切线。因此，法线的斜率 mn 满足 mn = −1 / mt（前提是 mt ≠ 0）。如果切线是水平的（mt = 0），那么法线是垂直的，其方程为 x = x₁。将法线斜率代入点斜式，并化简方程即可。
For the previous example where mt ormal gradient is −1/8. Hence the normal at (2,0) is y − 0 = −⅛(x − 2), or x + 8y − 2 = 0.
对于上例中 mt = 8 的情况，法线斜率为 −1/8。因此，点 (2,0) 处的法线方程为 y − 0 = −⅛(x − 2)，即 x + 8y − 2 = 0。
4. Increasing and Decreasing Functions | 函数的增减性
A function f(x) is increasing on an interval if f '(x) > 0 for all x in that interval. It is decreasing if f '(x) < 0. At points where f ' ary point. Edexcel questions often ask you to find ranges of x for which the function is increasing or decreasing by solving inequalities involving the derivative.
如果在某个区间内对所有 x 均有 f '(x) > 0，则函数在该区间上单调递增。若 f '(x) < 0，则函数单调递减。当 f '(x) = 0 时，函数可能存在驻点。Edexcel 试题经常要求你通过求解含导数的不等式，找出函数递增或递减的 x 的范围。
For f(x) = 2x³ + 3x² − 36x + 5, compute f '(x) = 6x² + 6x − 36 = 6(x² + x − 6) = 6(x+3)(x−2). So f '(x) > 0 when x < −3 or x > 2 (increasing), and f '(x) < 0 when −3 < x < 2 (decreasing).
对于 f(x) = 2x³ + 3x² − 36x + 5，求得 f '(x) = 6x² + 6x − 36 = 6(x² + x − 6) = 6(x+3)(x−2)。因此，当 x < −3 或 x > 2 时 f '(x) > 0（递增），当 −3 < x < 2 时 f '(x) < 0（递减）。
5. Stationary Points: Introduction | 驻点简介
Stationary points occur where the derivative equals zero, i.e. f ' dient of the tangent is zero, meaning the tangent is horizontal. Stationary points come in three types: local maxima, local minima, and points of inflection with zero gradient. Identifying them is a core skill in A-Level calculus.
驻点出现在导数为零的地方，即 f '(x) = 0。在这些点上，切线的斜率为零，意味着切线是水平的。驻点分为三类：局部极大值点、局部极小值点以及斜率为零的拐点。识别驻点是 A-Level 微积分的一项核心技能。
For the cubic f(x) = x³ − 3x² + 4, we set f '(x) = 3x² − 6x = 3x(x − 2) = 0, giving stationary points at x = 0 and x = 2.
对于三次函数 f(x) = x³ − 3x² + 4，令 f '(x) = 3x² − 6x = 3x(x − 2) = 0，得驻点位于 x = 0 和 x = 2。
6. Classifying Stationary Points Using First Derivative Test | 用一阶导数判别法分类驻点
The first derivative test examines the sign of f '(x) on either side of the stationary point. If f '(x) changes from positive to negative, the point is a local maximum. If it changes from negative to positive, it is a local minimum. If the sign does not change, it could be a point of inflection. Always check values just to the left and right of the point.
一阶导数判别法通过检查驻点左右两侧 f '(x) 的符号来分类。若 f '(x) 由正变负，则该点为局部极大值点；若由负变正，则为局部极小值点；若符号不变，则该点可能是一个拐点。记得一定要检查该点紧邻左右两侧的值。
For x = 0 in the example above, f '(x) = 3x(x−2). Just left (x = −0.1): f '(−0.1) > 0; just right (x = 0.1): f '(0.1) < 0. So x = 0 is a maximum. At x = 2, f '(1.9) < 0 and f '(2.1) > 0, so it is a minimum.
上例中的 x = 0 处，f '(x) = 3x(x−2)。左侧 (x = −0.1): f '(−0.1) > 0；右侧 (x = 0.1): f '(0.1) < 0。故 x = 0 为极大值点。在 x = 2 处，f '(1.9) < 0 而 f '(2.1) > 0，故为极小值点。
7. The Second Derivative Test | 二阶导数判别法
When finding f '(x) is manageable, the second derivative test provides a quick classification. Compute f ''(x) and substitute the x‑coordinate of the stationary point. If f ''(x) < 0, the function is concave down, indicating a local maximum. If f ''(x) > 0, it is concave up, indicating a local minimum. If f ''(x) = 0, the test is inconclusive; use the first derivative test instead.
当 f '(x) 容易再次求导时，二阶导数判别法可以快速分类。计算 f ''(x)，并代入驻点的 x 坐标。若 f ''(x) < 0，则函数在该点向下凹，表明是局部极大值；若 f ''(x) > 0，则向上凹，表明是局部极小值。如果 f ''(x) = 0，此判别法失效，需改用一阶导数判别法。
In our cubic example, f ''(x) = 6x − 6. At x = 0, f ''(0) = −6 < 0 → maximum. At x = 2, f ''(2) = 6 > 0 → minimum. This confirms the first derivative test result efficiently.
在上述三次函数例子中，f ''(x) = 6x − 6。在 x = 0 处，f ''(0) = −6 < 0 → 极大值；在 x = 2 处，f ''(2) = 6 > 0 → 极小值。这高效地验证了一阶导数判别法的结果。
8. Points of Inflection | 拐点
up to concave down, or vice versa. At such a point, the tangent may cross the curve, and f ''(x) = 0 is a necessary condition, but not sufficient; there must be a sign change in f ''(x). Not every point with f ''(x) = 0 is an inflection—always check the sign of the second derivative on either side.
拐点是曲线凹凸性改变的地方：从上凹变为向下凹，或反之。在这样的点上，切线可能穿过曲线，且 f ''(x) = 0 是一个必要条件，但并不充分；必须有 f ''(x) 的符号变化。并非所有满足 f ''(x) = 0 的点都是拐点——一定要检查该点两侧二阶导数的符号。
Consider f(x) = x³ − 3x² + 3x. Then f ''(x) = 6x − 6, so f ''(1) = 0. Testing signs: for x < 1, f ''(x) < 0; for x > 1, f ''(x) > 0. Hence (1,1) is a point of inflection, even though f '(1) ≠ 0 (so it is a non-stationary inflection).
考虑 f(x) = x³ − 3x² + 3x，则 f ''(x) = 6x − 6，故 f ''(1) = 0。检验符号：当 x < 1 时 f ''(x) < 0；当 x > 1 时 f ''(x) > 0。因此 (1,1) 是一个拐点，尽管 f '(1) ≠ 0（即为非驻点拐点）。
9. Optimisation Problems | 优化问题
Differentiation is widely used to optimise quantities: maximise area, minimise surface area, maximise profit, or minimise cost. Start by expressing the quantity to be optimised as a function of a single variable using given constraints. Then find its derivative, locate stationary points, and classify them to determine the optimal value. Always check that the solution makes physical sense and falls within the domain.
微分广泛应用于优化问题中：最大化面积、最小化表面积、最大化利润或最小化成本。首先，利用给定的约束条件，将要优化的量表示为一个单变量的函数。然后求出它的导数，找到驻点，并进行分类以确定最优值。务必检查解在实际意义上合理且在定义域内。
a A = x(200 − 2x) = 200x − 2x². dA/dx = 200 − 4x = 0 → x = 50. Second derivative = −4 < 0, so maximum area = 50×100 = 5000 m².
示例：用 200 米围栏与一直墙围成一个矩形区域。最大化面积。设宽度为 x，长为 200 − 2x，则面积 A = x(200 − 2x) = 200x − 2x²。dA/dx = 200 − 4x = 0 → x = 50。二阶导数 = −4 < 0，故最大面积 = 50×100 = 5000 平方米。
10. Connected Rates of Change and Exam Strategies | 相关变化率与考试策略
× dx/dt. Identify the relationship between the variables, differentiate with respect to time, and substitute known rates. Always write units clearly.
在许多 Edexcel 试题中，两个或更多的量会随时间变化，而微分通过链式法则将它们的变化率联系起来：dy/dt = dy/dx × dx/dt。找出变量之间的关系，对时间求导，并代入已知变化率。记得清晰注明单位。
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