📚 Using Differentiation | 微分应用
Differentiation is a cornerstone of A-Level Mathematics, allowing us to analyse rates of change, gradients of curves, and the behaviour of functions. In this article, we explore how to apply differentiation techniques to solve a variety of problems, from finding tangents and normals to optimising real-world models. Mastering these applications is essential for success in the Edexcel A-Level exam.
微分是A-Level数学的基石，它能让我们分析变化率、曲线梯度以及函数的行为。本文将探讨如何运用微分技巧解决各种问题，从求切线和法线方程到优化实际模型。掌握这些应用对于在Edexcel A-Level考试中取得成功至关重要。
1. Review of Basic Differentiation Rules | 基本微分法则回顾
Before applying differentiation, you must be fluent with standard derivatives. For any polynomial term axⁿ, the derivative is naxⁿ⁻¹. This rule extends to sums and constant multiples. You also need the derivatives of trigonometric, exponential, and logarithmic functions. The table below summarises the essential results for Edexcel A-Level Mathematics.
在应用微分之前，你必须熟练掌握标准导数。对于任意多项式项 axⁿ，其导数为 naxⁿ⁻¹。该法则同样适用于和与常数倍。你还需要三角函数、指数函数和对数函数的导数。下表总结了Edexcel A-Level数学中必备的结果。
| Function f(x) | Derivative f'(x) |
|---|---|
| xⁿ | n xⁿ⁻¹ |
| eˣ | eˣ |
| aˣ (a > 0) | aˣ ln a |
| ln x | 1/x |
| sin x | cos x |
| cos x | -sin x |
| tan x | sec² x |
You should also be comfortable with the chain rule, product rule, and quotient rule, as they are frequently needed when differentiating more complicated expressions. For example, if y = (2x + 1)⁵, use the chain rule: let u = 2x + 1, then dy/dx = 5u⁴ × 2 = 10(2x + 1)⁴.
你还应该熟练掌握链式法则、乘积法则和商法则，因为在求导更复杂的表达式时经常用到它们。例如，若 y = (2x + 1)⁵，使用链式法则：令 u = 2x + 1，则 dy/dx = 5u⁴ × 2 = 10(2x + 1)⁴。
2. Finding Equations of Tangents | 求切线方程
The derivative of a curve y = f(x) at a point x = a gives the gradient of the tangent at that point. Once you know the gradient m = f'(a) and the coordinates of the point (a, f(a)), you can write the equation of the tangent using the point-slope form: y − f(a) = m(x − a). This is one of the most direct applications of differentiation.
曲线 y = f(x) 在点 x = a 处的导数给出了该点切线的斜率。一旦知道斜率 m = f'(a) 以及点的坐标 (a, f(a))，就可以使用点斜式写出切线方程：y − f(a) = m(x − a)。这是微分最直接的应用之一。
Example: Find the equation of the tangent to the curve y = x³ − 4x at the point where x = 2.
First, compute y(2) = 2³ − 4×2 = 8 − 8 = 0, so the point is (2, 0). Next, differentiate: f'(x) = 3x² − 4. Evaluate at x = 2: f'(2) = 3×4 − 4 = 8. The tangent gradient is 8. Therefore, the tangent equation is y − 0 = 8(x − 2), which simplifies to y = 8x − 16.
示例：求曲线 y = x³ − 4x 在 x = 2 处的切线方程。
首先计算 y(2) = 2³ − 4×2 = 8 − 8 = 0，因此点为 (2, 0)。接着求导：f'(x) = 3x² − 4。代入 x = 2：f'(2) = 3×4 − 4 = 8。切线斜率为 8。因此，切线方程为 y − 0 = 8(x − 2)，化简得 y = 8x − 16。
3. Finding Equations of Normals | 求法线方程
The normal to a curve at a given point is the line perpendicular to the tangent at that point. If the gradient of the tangent is m = f'(a), then the normal gradient is −1/m, provided m ≠ 0. The equation of the normal then follows the same point-slope form: y − f(a) = (−1/m)(x − a). When m = 0, the tangent is horizontal, and the normal is vertical, giving an equation of the form x = a.
曲线在给定点处的法线是垂直于该点切线的直线。如果切线的斜率为 m = f'(a)，那么法线斜率为 −1/m，前提是 m ≠ 0。法线方程同样遵循点斜式：y − f(a) = (−1/m)(x − a)。当 m = 0 时，切线为水平线，法线为竖直线，此时方程为 x = a。
Using the previous example with m = 8 at (2, 0), the normal gradient is −1/8. The equation of the normal is y − 0 = −⅛(x − 2), or y = −⅛x + ¼. In exams, always check whether the final answer should be expressed in a simplified integer or fractional form.
沿用上例，在 (2, 0) 处 m = 8，法线斜率为 −1/8。法线方程为 y − 0 = −⅛(x − 2)，即 y = −⅛x + ¼。在考试中，始终检查最终答案是否应以最简整数或分数形式表示。
4. Stationary Points and How to Find Them | 驻点及其求法
A stationary point occurs where the derivative f'(x) = 0. Geometrically, this is where the tangent is horizontal. There are three types of stationary points: local maximum, local minimum, and point of inflection (which can be stationary or non‑stationary). To locate stationary points, set f'(x) = 0 and solve for x, then substitute back into y = f(x) to find the corresponding y‑coordinates.
驻点出现在导数 f'(x) = 0 的位置。从几何上看，这里是切线为水平的地方。驻点有三种类型：局部极大值、局部极小值以及拐点（可以是驻点拐点或非驻点拐点）。要寻找驻点，令 f'(x) = 0 解出 x，然后代回 y = f(x) 求相应的 y 坐标。
Example: Consider the function f(x) = x³ − 3x + 2. The derivative is f'(x) = 3x² − 3. Setting f'(x) = 0 gives 3x² − 3 = 0 ⇒ x² = 1 ⇒ x = 1 or x = −1. The corresponding points are (1, 1³−3×1+2 = 0) and (−1, (−1)³−3(−1)+2 = 4). So the stationary points are (1, 0) and (−1, 4).
示例：考虑函数 f(x) = x³ − 3x + 2。其导数为 f'(x) = 3x² − 3。令 f'(x) = 0，得 3x² − 3 = 0 ⇒ x² = 1 ⇒ x = 1 或 x = −1。对应点为 (1, 3−3+2 = 0) 和 (−1, (−1)+3+2 = 4)。因此驻点为 (1, 0) 和 (−1, 4)。
5. Classifying Stationary Points Using the Second Derivative | 用二阶导数对驻点分类
The second derivative f”(x) helps determine the nature of a stationary point. If f”(a) > 0, the function is concave up at x = a, indicating a local minimum. If f”(a) < 0, the curve is concave down, so the point is a local maximum.
二阶导数 f”(x) 有助于确定驻点的性质。若 f”(a) > 0，函数在 x = a 处凹向上，表明为局部极小值。若 f”(a) < 0，曲线凹向下，因此该点为局部极大值。
Continuing with f(x) = x³ − 3x + 2, the second derivative is f”(x) = 6x. At x = 1: f”(1) = 6 > 0, so (1, 0) is a local minimum. At x = −1: f”(−1) = −6 < 0, so (−1, 4) is a local maximum. If f''(a) = 0, the test is inconclusive; you need to examine the sign of f'(x) on either side of the stationary point (the first derivative test).
继续以 f(x) = x³ − 3x + 2 为例，二阶导数为 f”(x) = 6x。在 x = 1 处：f”(1) = 6 > 0，因此 (1, 0) 为局部极小值。在 x = −1 处：f”(−1) = −6 < 0，因此 (−1, 4) 为局部极大值。若 f''(a) = 0，则该检验无法得出结论；你需要检查 x = a 两侧 f'(x) 的符号变化（即第一导数检验法）。
6. Concavity and Points of Inflection | 凹凸性与拐点
Concavity describes whether a curve bends upwards (convex, f”(x) > 0) or downwards (concave, f”(x) < 0). A point of inflection occurs where the curve changes concavity, i.e., where f''(x) changes sign. It is important to note that at a point of inflection, f''(x) = 0 or f''(x) is undefined, but f''(x) = 0 alone is not sufficient — you must verify the sign change. A stationary point of inflection occurs when f'(x) = 0 and f''(x) changes sign.
凹凸性描述了曲线是向上弯曲（凸的，f”(x) > 0）还是向下弯曲（凹的，f”(x) < 0）。拐点出现在曲线凹凸性改变的位置，即 f''(x) 改变符号处。需要注意的是，在拐点处 f''(x) = 0 或 f''(x) 无定义，但仅凭 f''(x) = 0 并不充分——你必须验证符号发生变化。当 f'(x) = 0 且 f''(x) 改变符号时，该点为驻点拐点。
Example: f(x) = x³ has f'(x) = 3x², f”(x) = 6x. At x = 0, f'(0) = 0 and f”(0) = 0. However, for x < 0, f''(x) < 0 (concave down), and for x > 0, f”(x) > 0 (convex up). The concavity changes, so (0, 0) is a stationary point of inflection. This is a classic illustration of a horizontal inflection.
示例：f(x) = x³，f'(x) = 3x²，f”(x) = 6x。在 x = 0 处，f'(0) = 0，f”(0) = 0。但 x < 0 时 f''(x) < 0（凹向下），x > 0 时 f”(x) > 0（凸向上）。凹凸性发生改变，因此 (0, 0) 是一个驻点拐点。这是水平拐点的经典示例。
7. Curve Sketching Using Differentiation | 利用微分作函数图像
Differentiation provides powerful tools for sketching graphs. The systematic approach is: (1) Find intercepts by setting x = 0 and y = 0. (2) Identify any asymptotes or domain restrictions. (3) Compute f'(x) to locate stationary points and intervals of increase/decrease. (4) Compute f”(x) to determine concavity and inflection points. (5) Plot the key points and sketch the curve, ensuring it blends smoothly.
微分为绘制函数图像提供了强大的工具。系统性的方法是：(1) 设 x = 0 和 y = 0 求截距。(2) 识别任何渐近线或定义域限制。(3) 计算 f'(x) 找出驻点以及递增/递减区间。(4) 计算 f”(x) 确定凹凸性和拐点。(5) 标出关键点并勾勒曲线，确保平滑连接。
Consider a cubic y = x³ − 6x² + 9x. Intercepts: y(0)=0; x(x²−6x+9)=0 ⇒ x(x−3)²=0, so x = 0 and x = 3 (double root). f'(x) = 3x² − 12x + 9 = 3(x−1)(x−3). Stationary points at x=1 (y=4) and x=3 (y=0). f”(x) = 6x −12 = 6(x−2). For x<2, f''<0 (concave down); x>2, f”>0 (convex up). At x=2, y=2³−24+18=2, inflection point (2,2). The sketch reveals a local max at (1,4), a local min at (3,0), and a smooth inflection at (2,2).
考虑三次函数 y = x³ − 6x² + 9x。截距：y(0)=0；x(x²−6x+9)=0 ⇒ x(x−3)²=0，故 x = 0 和 x = 3（重根）。f'(x) = 3x² − 12x + 9 = 3(x−1)(x−3)。驻点在 x=1 (y=4) 和 x=3 (y=0)。f”(x) = 6x − 12 = 6(x−2)。当 x<2 时 f''<0（凹向下）；x>2 时 f”>0（凸向上）。在 x=2 处，y=8−24+18=2，拐点为 (2,2)。草图显示在 (1,4) 处有局部极大值，在 (3,0) 处有局部极小值，并在 (2,2) 处有一个平滑的拐点。
8. Optimisation Problems | 优化问题
Optimisation is the process of maximising or minimising a quantity, often in real‑world contexts. The steps are: (i) Express the quantity to be optimised as a function of one variable, using given constraints to eliminate other variables. (ii) Differentiate and set f'(x) = 0 to find critical points. (iii) Determine whether each critical point gives a maximum or minimum, using the second derivative test or by checking endpoints. (iv) Answer the question in context, verifying that the solution is feasible.
优化是最大化或最小化某个量的过程，常出现在实际背景中。步骤为：(i) 将要优化的量表示为单变量函数，利用给定约束消去其他变量。(ii) 求导并令 f'(x) = 0 以找到临界点。(iii) 使用二阶导数检验或检查端点，确定每个临界点对应于最大值还是最小值。(iv) 结合题意回答问题，验证解的合理性。
Classic example: An open box is made by cutting squares of side x cm from the corners of a 12 cm × 12 cm square sheet and folding up the sides. The volume V = x(12 − 2x)² = 4x(6 − x)², with 0 < x < 6. Expand: V = 4x(36 − 12x + x²) = 4x³ − 48x² + 144x. Differentiate: V' = 12x² − 96x + 144 = 12(x² − 8x + 12) = 12(x − 2)(x − 6). Stationary points at x = 2 and x = 6 (boundary). V'' = 24x − 96. At x=2, V'' = 48 − 96 = −48 < 0, so a maximum. The maximum volume is V(2) = 2 × (8)² = 128 cm³. Always check the domain boundaries; here x=6 gives V=0, so x=2 is indeed the optimal cut.
经典示例：用边长为 12 cm 的正方形纸板，从四个角各剪去边长为 x cm 的小正方形，然后折起制成开口盒。体积 V = x(12 − 2x)² = 4x(6 − x)²，其中 0 < x < 6。展开：V = 4x(36 − 12x + x²) = 4x³ − 48x² + 144x。求导：V' = 12x² − 96x + 144 = 12(x² − 8x + 12) = 12(x − 2)(x − 6)。驻点为 x=2 和 x=6（边界）。V'' = 24x − 96。当 x=2 时，V'' = 48−96 = −48 < 0，为极大值。最大体积为 V(2) = 2 × 8² = 128 cm³。始终检查定义域边界；这里 x=6 时 V=0，因此 x=2 确实是最优剪裁尺寸。
9. Kinematics – Motion in a Straight Line | 直线运动学
In kinematics, differentiation links displacement (s), velocity (v), and acceleration (a) with respect to time t. If displacement s is given as a function of time, then velocity v = ds/dt, and acceleration a = dv/dt = d²s/dt². Problems often ask for times when a particle is at rest (v = 0), its maximum displacement, or its acceleration at specific instants.
在运动学中，微分将位移 (s)、速度 (v) 和加速度 (a) 与时间 t 联系起来。若位移 s 表示为时间的函数，则速度 v = ds/dt，加速度 a = dv/dt = d²s/dt²。常考问题包括粒子静止的时刻 (v = 0)、最大位移，或特定时刻的加速度。
Example: A particle moves along a line with displacement s(t) = t³ − 6t² + 9t metres, where t ≥ 0 seconds. Find the times when the particle is at rest and its acceleration at those times.
v(t) = ds/dt = 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t − 1)(t − 3). Set v = 0 ⇒ t = 1, t = 3. The particle is at rest at t = 1 s and t = 3 s.
a(t) = dv/dt = 6t − 12. At t = 1: a = 6 − 12 = −6 m/s² (decelerating). At t = 3: a = 18 − 12 = 6 m/s². This analysis helps describe the motion without needing a graph.
示例：一粒子沿直线运动，位移为 s(t) = t³ − 6t² + 9t 米，其中 t ≥ 0 秒。求粒子静止的时刻及该时刻的加速度。
v(t) = ds/dt = 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t − 1)(t − 3)。令 v = 0 ⇒ t = 1, t = 3。粒子在 t = 1 s 和 t = 3 s 时静止。
a(t) = dv/dt = 6t − 12。t = 1 时：a = 6 − 12 = −6 m/s²（减速）；t = 3 时：a = 18 − 12 = 6 m/s²。这种分析无需图像即可描述运动。
10. Related Rates of Change | 相关变化率
Related rates problems involve two or more varying quantities that are linked by an equation. The chain rule allows us to relate their rates of change with respect to time. If a quantity A depends on a variable r, and r changes over time, then dA/dt = (dA/dr) × (dr/dt). The key is to identify what is constant and what varies, then differentiate the connecting equation with respect to t.
相关变化率问题涉及两个或更多由方程关联的变量。链式法则使我们能够将它们关于时间的变化率联系起来。如果某个量 A 依赖于变量 r，而 r 随时间变化，则 dA/dt = (dA/dr) × (dr/dt)。关键是区分哪些量是常数、哪些在变化，然后对联系方程关于 t 求导。
Example: A spherical balloon is inflated so that its radius increases at a constant rate of 2 cm/s. Find the rate at which the volume is increasing when the radius is 5 cm.
Volume V = 4/3 π r³, so dV/dr = 4π r². Given dr/dt = 2 cm/s. By the chain rule, dV/dt = 4π r² × 2 = 8π r². At r = 5, dV/dt = 8π × 25 = 200π cm³/s. Always include units and state the final answer clearly. This technique is widely tested in applied problems.
示例：一个球形气球充气时，其半径以 2 cm/s 的恒定速率增加。求当半径为 5 cm 时体积增大的速率。
体积 V = 4/3 π r³，故 dV/dr = 4π r²。已知 dr/dt = 2 cm/s。由链式法则，dV/dt = 4π r² × 2 = 8π r²。当 r = 5 时，dV/dt = 8π × 25 = 200π cm³/s。务必包含单位并清晰地陈述最终答案。该技巧在应用题中被广泛考查。
In harder examples, you may need to use implicit differentiation or substitute a relation (like Pythagorean theorem) before differentiating. Always write down the known rates
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