📚 Differentiation: A Comprehensive Review for CIE A-Level | CIE A-Level 数学：微分考点精讲
Welcome to this comprehensive review of differentiation for CIE A-Level Mathematics (9709). Mastering differentiation is crucial, as it forms the foundation for many calculus problems, including curve sketching, optimisation, and kinematics. This article covers key concepts, rules, and common question types to help you excel in your exams.
欢迎阅读针对 CIE A-Level 数学 (9709) 的微分考点精讲。掌握微分至关重要，它是许多微积分问题（包括曲线草图、优化和运动学）的基础。本文涵盖核心概念、法则及常见题型，助你在考试中取得佳绩。
1. Basic Differentiation Rules | 基本求导法则
The derivative of a function f(x) is defined as the limit of the average rate of change. If y = f(x), then the derivative f'(x) = dy/dx = limh→0 (f(x+h) − f(x)) / h. Geometrically, it represents the gradient of the tangent to the curve at a point.
函数 f(x) 的导数定义为平均变化率的极限。若 y = f(x)，则导数 f'(x) = dy/dx = limh→0 (f(x+h) − f(x)) / h。从几何意义上看，它表示曲线上一点处切线的斜率。
The power rule is the most fundamental: d/dx (xⁿ) = n xⁿ⁻¹, where n is any real constant. For a constant function, the derivative is zero. The sum/difference rule and constant multiple rule allow us to differentiate term-by-term.
最基本的法则是幂法则：d/dx (xⁿ) = n xⁿ⁻¹，其中 n 为任意实常数。常数函数的导数为零。和差法则与常数倍法则让我们能够逐项求导。
d/dx (xⁿ) = n xⁿ⁻¹
d/dx (k) = 0
d/dx [f(x) ± g(x)] = f'(x) ± g'(x)
d/dx [c f(x)] = c f'(x)
For example, if y = 4x³ − 2x + 7, then dy/dx = 12x² − 2.
例如，若 y = 4x³ − 2x + 7，则 dy/dx = 12x² − 2。
2. Product Rule | 积的求导法则
When two functions are multiplied, we use the product rule. If y = u(x) v(x), then dy/dx = u’v + uv’. The rule states: derivative of the first times the second, plus first times derivative of the second.
当两个函数相乘时，我们使用积的求导法则。若 y = u(x) v(x)，则 dy/dx = u’v + uv’。即 “前导乘后，加上前乘后导”。
d/dx (u v) = u’ v + u v’
For instance, differentiate y = x² sin x. Let u = x², u’ = 2x; v = sin x, v’ = cos x. Then dy/dx = 2x sin x + x² cos x. Always present your answer in its simplest factorised form when possible.
例如，求 y = x² sin x 的导数。令 u = x², u’ = 2x; v = sin x, v’ = cos x。则 dy/dx = 2x sin x + x² cos x。如有可能，尽量将结果因式分解以简化表达。
3. Quotient Rule | 商的求导法则
For a function of the form y = u(x) / v(x), where v(x) ≠ 0, the quotient rule is used. The derivative is (u’v − uv’) / v². Remember the minus sign in the numerator – it is u’v subtract uv’, not the other way around.
形如 y = u(x) / v(x)（v(x) ≠ 0）的函数，需使用商的求导法则。导数为 (u’v − uv’) / v²。注意分子中是 u’v 减去 uv’，次序不可颠倒。
d/dx (u/v) = (u’v − u v’) / v²
Example: y = (eˣ) / (x+1). Here u = eˣ, u’ = eˣ; v = x+1, v’ = 1. So dy/dx = [eˣ (x+1) − eˣ·1] / (x+1)² = x eˣ / (x+1)².
例题：y = (eˣ) / (x+1)。此处 u = eˣ, u’ = eˣ; v = x+1, v’ = 1。则 dy/dx = [eˣ (x+1) − eˣ·1] / (x+1)² = x eˣ / (x+1)²。
4. Chain Rule | 链式法则
The chain rule handles composite functions. If y = f(g(x)), we set u = g(x) and then dy/dx = (dy/du) × (du/dx). This rule is essential for functions like sin(2x), e^(3x²) or ln(5x+1).
链式法则用于处理复合函数。若 y = f(g(x))，可设 u = g(x)，则 dy/dx = (dy/du) × (du/dx)。该法则对 sin(2x)、e^(3x²) 或 ln(5x+1) 等函数求导至关重要。
dy/dx = dy/du × du/dx
For y = (2x³ − 5)⁴, let u = 2x³ − 5, then y = u⁴. dy/du = 4u³, du/dx = 6x². Therefore dy/dx = 4(2x³ − 5)³ × 6x² = 24x²(2x³ − 5)³. With practice, you can apply the chain rule mentally without explicit substitution.
例如 y = (2x³ − 5)⁴，设 u = 2x³ − 5，则 y = u⁴。dy/du = 4u³，du/dx = 6x²。故 dy/dx = 4(2x³ − 5)³ × 6x² = 24x²(2x³ − 5)³。熟练后可不必写出代换，直接使用链式法则。
5. Implicit Differentiation | 隐函数微分
When y is not given explicitly as a function of x, differentiate both sides of the equation with respect to x, treating y as a function of x. For each y-term, apply the chain rule and multiply by dy/dx.
当 y 未明确表示为 x 的函数时，可对方程两边关于 x 求导，并将 y 视为 x 的函数。对含有 y 的项，需使用链式法则并乘以 dy/dx。
d/dx (y²) = 2y · dy/dx
Consider the circle x² + y² = 25. Differentiating: 2x + 2y(dy/dx) = 0 ⇒ dy/dx = −x/y. Implicit differentiation often appears in tangent/normal problems where you first find the gradient at a given point.
考虑圆的方程 x² + y² = 25。求导得：2x + 2y(dy/dx) = 0 ⇒ dy/dx = −x/y。隐函数微分常出现在切线/法线问题中，需要先求出给定点处的梯度。
6. Parametric Differentiation | 参数函数微分
If x = f(t) and y = g(t), the derivative dy/dx is found by (dy/dt) / (dx/dt). The second derivative d²y/dx² can be obtained by differentiating dy/dx with respect to t and dividing by dx/dt.
若 x = f(t) 且 y = g(t)，则导数 dy/dx = (dy/dt) / (dx/dt)。二阶导数 d²y/dx² 可通过将 dy/dx 对 t 求导，再除以 dx/dt 得到。
dy/dx = (dy/dt) / (dx/dt)
Example: x = t² + 1, y = t³ − 3t. Then dx/dt = 2t, dy/dt = 3t² − 3, so dy/dx = (3t² − 3) / (2t) = (3/2)(t − 1/t). At the point where t = 2, dy/dx = (3/2)(2 − 0.5) = 2.25.
例题：x = t² + 1, y = t³ − 3t。则 dx/dt = 2t, dy/dt = 3t² − 3，故 dy/dx = (3t² − 3) / (2t) = (3/2)(t − 1/t)。在 t = 2 处，dy/dx = (3/2)(2 − 0.5) = 2.25。
7. Second Derivative and Concavity | 二阶导数与凹凸性
The second derivative, written as d²y/dx² or f”(x), is the derivative of the first derivative. It measures the rate of change of the gradient, revealing the concavity of the curve: if f”(x) > 0, the curve is concave upwards; if f”(x) < 0, it is concave downwards.
二阶导数记作 d²y/dx² 或 f”(x)，是一阶导数的导数。它衡量斜率的变化率，揭示曲线的凹凸性：若 f”(x) > 0，曲线向上凹；若 f”(x) < 0，曲线向下凹。
d²y/dx² = d/dx (dy/dx)
For y = x⁴ − 2x³, dy/dx = 4x³ − 6x², and d²y/dx² = 12x² − 12x = 12x(x − 1). Setting the second derivative to zero helps locate points of inflection, where concavity changes.
对于 y = x⁴ − 2x³，dy/dx = 4x³ − 6x²，d²y/dx² = 12x² − 12x = 12x(x − 1)。令二阶导数为零有助于找出凹凸性发生变化的拐点。
8. Tangents and Normals | 切线与法线
At a point (x₁, y₁) on a curve, the gradient of the tangent is m = dy/dx evaluated at that point. The equation of the tangent is y − y₁ = m(x − x₁). The normal is perpendicular to the tangent, so its gradient is −1/m, provided m ≠ 0.
在曲线上一点 (x₁, y₁) 处，切线的斜率 m 为在该点求出的 dy/dx 值。切线方程为 y − y₁ = m(x − x₁)。法线垂直于切线，故其斜率为 −1/m（当 m ≠ 0 时）。
Tangent: y − y₁ = m(x − x₁)
Normal: y − y₁ = (−1/m)(x − x₁)
Example: Find the tangent to y = x³ at x = 2. y₁ = 8, dy/dx = 3x² ⇒ m = 12. Tangent: y − 8 = 12(x − 2) ⇒ y = 12x − 16. Normal: gradient = −1/12, equation: y − 8 = −1/12 (x − 2).
例题：求曲线 y = x³ 在 x = 2 处的切线。y₁ = 8，dy/dx = 3x² ⇒ m = 12。切线：y − 8 = 12(x − 2) ⇒ y = 12x − 16。法线斜率为 −1/12，方程为 y − 8 = −1/12 (x − 2)。
9. Increasing and Decreasing Functions | 增函数与减函数
A function f(x) is increasing on an interval if f'(x) > 0, and decreasing if f'(x) < 0. To determine these intervals, find the derivative, solve f'(x) = 0 for critical values, and test the sign of f'(x) in each sub-interval.
若在某一区间内 f'(x) > 0，则函数递增；若 f'(x) < 0，则函数递减。为确定这些区间，需求出导数，解方程 f'(x) = 0 找到临界值，再在每个子区间内测试 f'(x) 的正负号。
For f(x) = x³ − 3x + 2, f'(x) = 3x² − 3 = 3(x−1)(x+1). Critical points at x = −1, 1. Test intervals: x < −1 ⇒ f'(x) > 0 (increasing); −1 < x < 1 ⇒ f'(x) < 0 (decreasing); x > 1 ⇒ f'(x) > 0 (increasing). These intervals are essential for sketching graphs.
对于 f(x) = x³ − 3x + 2，f'(x) = 3x² − 3 = 3(x−1)(x+1)，临界点为 x = −1 和 1。测试区间：x < −1 时 f'(x) > 0（递增）；−1 < x < 1 时 f'(x) < 0（递减）；x > 1 时 f'(x) > 0（递增）。这些区间对绘制草图至关重要。
10. Stationary Points and Curve Sketching | 驻点与曲线作图
Stationary points occur where f'(x) = 0. To classify them, use the second derivative test: if f”(x) > 0, it is a minimum point; if f”(x) < 0, it is a maximum point. If f''(x) = 0, check the sign of f'(x) on either side to identify a possible point of inflection.
驻点出现在 f'(x) = 0 处。用二阶导数判别其类型：若 f”(x) > 0，为极小值点；若 f”(x) < 0，为极大值点。若 f''(x) = 0，则需检查该点两侧 f'(x) 的符号，判断是否为拐点。
For f(x) = x³ − 3x, f'(x) = 3(x² − 1), stationary at x = ±1. f”(x) = 6x. At x = 1, f”(1) = 6 > 0 ⇒ minimum; at x = −1, f”(−1) = −6 < 0 ⇒ maximum. Then sketch the curve by combining intercepts, stationary points and the sign of f'(x).
以 f(x) = x³ − 3x 为例，f'(x) = 3(x² − 1)，驻点在 x = ±1。f”(x) = 6x。x = 1 处 f”(1) = 6 > 0，为极小值；x = −1 处 f”(−1) = −6 < 0，为极大值。结合截距、驻点及 f'(x) 的正负即可绘制曲线草图。
11. Related Rates of Change | 相关变化率
In many practical problems, two or more variables change with time, and their rates of change are linked. The chain rule gives dy/dt = dy/dx × dx/dt. This allows us to find an unknown rate from a known rate and a geometric relationship.
在许多实际问题中，两个或多个变量随时间变化，其变化率相互关联。通过链式法则 dy/dt = dy/dx × dx/dt，可由已知变化率及几何关系求出未知变化率。
dy/dt = dy/dx · dx/dt
Example: A spherical balloon is inflated such that its volume increases at 100 cm³/s. Find the rate of increase of the radius when r = 5 cm. Volume V = 4/3 π r³, so dV/dr = 4π r². Given dV/dt = 100, we have dV/dt = (dV/dr)(dr/dt) ⇒ 100 = 4π (5)² × dr/dt ⇒ dr/dt = 1/π cm/s.
例题：给球形气球充气，体积以 100 cm³/s 的速率增加，求当 r = 5 cm 时半径的增加速率。体积 V = 4/3 π r³，故 dV/dr = 4π r²。已知 dV/dt = 100，由 dV/dt = (dV/dr)(dr/dt) 得 100 = 4π (5)² × dr/dt ⇒ dr/dt = 1/π cm/s。
Always write down the given rate and the required rate, identify the connecting equation, differentiate with respect to time, and substitute known values. Check that units are consistent.
解题步骤：写出已知变化率和待求变化率，确定关联方程，然后对时间求导，代入已知值。注意单位的一致性。
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