📚 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 |
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
nt x = a gives the gradient of the tangent at that point. Once you know the gradient m = f' 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。
ample: 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' s 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
e line perpendicular to the tangent at that point. If the gradient of the tangent is m = f' e 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。
eck 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' nts: 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 ⇒ ³−3×1+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'' 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
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'' lection 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'' etch 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‑ press the quantity to be optimised as a function of one variable, using given constraints to eliminate other variables. (ii) Differentiate and set f' 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。
− x)², with 0 < x < 6. Expand: V = 4x(36 − 12x + x²) = 4x³ − 48x² + 144x. Differentiate: V' = 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'' 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
le 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。
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.
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
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.
s 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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