📚 A-Level AQA Maths: Differentiation Essentials | A-Level AQA 数学：微分考点精讲
Differentiation is one of the cornerstones of A-Level Mathematics. This guide covers all key techniques and applications you need for the AQA specification, from basic power rules to implicit differentiation, parametric equations, optimisation, and connected rates of change. Work through the concepts step by step, and always link algebraic manipulation to graphical understanding.
微分是 A-Level 数学的核心基石之一。本指南涵盖 AQA 考试大纲要求的所有关键技巧与应用，从基本幂函数求导到隐函数、参数方程、最优化和相关变化率。请一步步掌握这些概念，并始终将代数操作与图形意义联系起来。
1. Basic Rules of Differentiation | 基本求导法则
The derivative of a constant function is zero. For any real power n, the derivative of xⁿ is nxⁿ⁻¹. If a function is multiplied by a constant, the constant factor stays unchanged. The derivative of a sum or difference is the sum or difference of the individual derivatives.
常数函数的导数为零。对于任意实数指数 n，xⁿ 的导数为 nxⁿ⁻¹。若函数乘以常数，该常数因子保持不变。和或差的导数等于各项导数的和或差。
d/dx (c) = 0, d/dx (xⁿ) = nxⁿ⁻¹, d/dx (cf(x)) = c f'(x), d/dx (f(x) ± g(x)) = f'(x) ± g'(x)
These rules allow you to differentiate any polynomial term by term. For example, the derivative of 4x³ − 2x + 7 is 12x² − 2.
这些法则使你可以逐项求导任何多项式。例如，4x³ − 2x + 7 的导数为 12x² − 2。
2. Chain Rule | 链式法则
When a function is composed of two functions, say y = f(u) and u = g(x), the derivative dy/dx is found by multiplying dy/du by du/dx. This is essential for handling powers of linear expressions, trigonometric functions of multiples, and exponentials of functions.
当函数为复合函数，如 y = f(u) 且 u = g(x)，则导数为 dy/dx = (dy/du) × (du/dx)。这是处理线性表达式的幂、倍角的三角函数以及函数的指数等方法的核心。
dy/dx = (dy/du) × (du/dx)
For example, to differentiate y = (3x² + 5)⁴, set u = 3x² + 5. Then y = u⁴ so dy/du = 4u³, and du/dx = 6x. Multiplying gives dy/dx = 24x (3x² + 5)³.
例如，求导 y = (3x² + 5)⁴，令 u = 3x² + 5，则 y = u⁴，dy/du = 4u³，du/dx = 6x，相乘得 dy/dx = 24x (3x² + 5)³。
3. Product Rule | 积法则
If y = u(x) v(x), then the derivative is given by the first function times the derivative of the second plus the second function times the derivative of the first. This rule is critical when differentiating expressions like x² sin x or eˣ ln x.
若 y = u(x) v(x)，则导数等于第一函数乘以第二函数的导数加上第二函数乘以第一函数的导数。在求导如 x² sin x 或 eˣ ln x 时，该法则极为关键。
d/dx (u v) = u’v + u v’
Consider y = x³ eˣ. Here u = x³, v = eˣ. Then u’ = 3x², v’ = eˣ. Applying the product rule gives dy/dx = 3x² eˣ + x³ eˣ = eˣ x² (3 + x).
考虑 y = x³ eˣ，令 u = x³，v = eˣ，则 u’ = 3x²，v’ = eˣ。应用积法则得 dy/dx = 3x² eˣ + x³ eˣ = eˣ x² (3 + x)。
4. Quotient Rule | 商法则
For y = u/v, the derivative is (v u’ − u v’) / v². Make sure the denominator is never zero where differentiation is applied. This rule handles rational functions and expressions like tan x when written as sin x/cos x.
对于 y = u/v，导数为 (v u’ − u v’) / v²。确保在求导点上分母不为零。该法则处理有理函数，也可用于将 tan x 表示为 sin x/cos x 时的求导。
d/dx (u/v) = (v u’ − u v’) / v²
Example: y = (2x + 1) / (x² + 3). Then u = 2x+1, v = x²+3, u’ = 2, v’ = 2x. The derivative is [ (x²+3)×2 − (2x+1)×2x ] / (x²+3)² = (2x²+6 − 4x² − 2x) / (x²+3)² = (−2x² − 2x + 6) / (x²+3)².
例如：y = (2x + 1) / (x² + 3)，u = 2x+1, v = x²+3, u’ = 2, v’ = 2x。导数为 [ (x²+3)×2 − (2x+1)×2x ] / (x²+3)² = (2x²+6 − 4x² − 2x) / (x²+3)² = (−2x² − 2x + 6) / (x²+3)²。
5. Exponential and Logarithmic Functions | 指数与对数函数求导
The derivative of eˣ is eˣ itself. For a general exponential aˣ (a > 0), the derivative is aˣ ln a. The natural logarithm differentiates as d/dx (ln x) = 1/x for x > 0. When the argument is a function, use the chain rule: d/dx (eᶠ⁽ˣ⁾) = f'(x) eᶠ⁽ˣ⁾ and d/dx (ln f(x)) = f'(x)/f(x).
eˣ 的导数就是 eˣ 本身。对于一般的指数函数 aˣ (a > 0)，导数为 aˣ ln a。自然对数的导数为 d/dx (ln x) = 1/x，要求 x > 0。当自变量是函数时，用链式法则：d/dx (eᶠ⁽ˣ⁾) = f'(x) eᶠ⁽ˣ⁾，d/dx (ln f(x)) = f'(x)/f(x)。
d/dx (eˣ) = eˣ, d/dx (aˣ) = aˣ ln a, d/dx (ln x) = 1/x
These forms appear frequently in growth and decay problems, as well as in further calculus like integration.
这些形式频繁出现在增长与衰减问题以及积分等高阶微积分中。
6. Trigonometric Functions | 三角函数求导
All trigonometric derivatives must be learned. The derivatives of sin x and cos x are cyclical: d/dx (sin x) = cos x; d/dx (cos x) = −sin x. From these, the derivative of tan x is sec² x. You may also need cosec, sec, and cot in AQA, which follow from the chain rule or quotient rule.
必须熟记所有三角函数的导数。sin x 和 cos x 的导数具有循环性：d/dx (sin x) = cos x；d/dx (cos x) = −sin x。由此可得 tan x 的导数为 sec² x。AQA 考试中也可能涉及 cosec、sec 和 cot，可通过链式法则或商法则推导。
d/dx (sin x) = cos x, d/dx (cos x) = −sin x, d/dx (tan x) = sec² x
For compound angles, always combine with the chain rule. Example: d/dx (sin 2x) = 2 cos 2x, and d/dx (cos² x) = 2 cos x (−sin x) = −sin 2x.
对于复合角，务必结合链式法则。例如：d/dx (sin 2x) = 2 cos 2x，d/dx (cos² x) = 2 cos x (−sin x) = −sin 2x。
7. Implicit Differentiation | 隐函数求导
When an equation defines y implicitly in terms of x, differentiate both sides with respect to x, treating y as a function of x. This means every time you differentiate a y-term, multiply by dy/dx. This method is essential for curves like circles, ellipses, and for finding gradients without explicit y = f(x).
当方程以隐式定义 y 与 x 的关系时，对等式两边关于 x 求导，并将 y 视为 x 的函数。这意味着每次对含有 y 的项求导时，需乘以 dy/dx。该方法对于圆、椭圆等曲线以及不需要显式 y = f(x) 求梯度的情况至关重要。
d/dx (yⁿ) = n yⁿ⁻¹ dy/dx
For the circle x² + y² = 25, differentiate: 2x + 2y dy/dx = 0 ⇒ dy/dx = −x/y. To find the tangent at (3,4), the gradient is −3/4.
对于圆 x² + y² = 25，求导：2x + 2y dy/dx = 0 ⇒ dy/dx = −x/y。在点 (3,4) 处的切线斜率为 −3/4。
8. Parametric Differentiation | 参数方程求导
When x and y are both given in terms of a third variable, typically t, differentiation is performed by dividing the derivative of y with respect to t by the derivative of x with respect to t. The second derivative d²y/dx² is obtained by differentiating dy/dx with respect to t and then dividing by dx/dt.
当 x 和 y 都以第三个变量（通常为 t）给出时，求导可通过将 y 关于 t 的导数除以 x 关于 t 的导数得到。二阶导数 d²y/dx² 则需先对 dy/dx 关于 t 求导，然后再除以 dx/dt。
dy/dx = (dy/dt) / (dx/dt), d²y/dx² = d/dt (dy/dx) / (dx/dt)
Given x = t² + 1, y = t³ − t, then dx/dt = 2t, dy/dt = 3t² − 1. Hence dy/dx = (3t² − 1)/(2t). At t = 1, the gradient is 1.
设 x = t² + 1, y = t³ − t，则 dx/dt = 2t, dy/dt = 3t² − 1，于是 dy/dx = (3t² − 1)/(2t)。当 t = 1 时，梯度为 1。
9. Second Derivative and Concavity | 二阶导数与凹凸性
The second derivative, written f”(x) or d²y/dx², measures the rate of change of the gradient. Its sign tells us about the curvature: f”(x) > 0 indicates the curve is convex (bending upwards), f”(x) < 0 indicates concave (bending downwards). Points where f''(x) = 0 and concavity changes are points of inflection.
二阶导数写作 f”(x) 或 d²y/dx²，度量梯度的变化率。其符号指示曲线的弯曲方向：f”(x) > 0 表示曲线下凸（向上弯），f”(x) < 0 表示上凸（向下弯）。f''(x) = 0 且凹凸性改变的点为拐点。
For y = x³ − 3x, the first derivative is 3x² − 3; the second is 6x. Setting second derivative to zero gives x = 0, where the curve changes from concave to convex.
对于 y = x³ − 3x，一阶导数为 3x² − 3；二阶导数为 6x。令二阶导数为零得 x = 0，在此处曲线由上凸变为下凸。
10. Tangents and Normals | 切线与法线
Once the derivative at a point is known, the equation of the tangent can be written using the point-slope form. The normal is perpendicular to the tangent; its gradient is the negative reciprocal of the tangent’s gradient. Accurate use of algebra is needed to form these linear equations.
一旦知道某点的导数，即可用点斜式写出切线方程。法线与切线垂直，其斜率为切线斜率的负倒数。需要准确代数运算来建立这些线性方程。
m_tangent = f'(a), m_normal = −1/f'(a) (if f'(a) ≠ 0)
For y = x² at x = 3, f'(3) = 6, so tangent gradient = 6, normal gradient = −1/6. The point is (3,9). Tangent: y − 9 = 6(x − 3). Normal: y − 9 = −1/6 (x − 3).
对于 y = x² 在 x=3 处，f'(3) = 6，切线斜率为 6，法线斜率为 −1/6。点为 (3,9)。切线：y − 9 = 6(x − 3)；法线：y − 9 = −1/6 (x − 3)。
11. Stationary Points and Curve Sketching | 驻点与曲线草图
Stationary points occur where f'(x) = 0. Their nature (maximum, minimum, or inflection) can be determined by the second derivative test or by checking the sign change of the first derivative. Combining stationary points with intercepts and asymptotes allows a reliable sketch of the function.
驻点出现在 f'(x) = 0 处。其性质（极大值、极小值或拐点）可通过二阶导数检验或检查一阶导数符号变化来判定。结合驻点、截距和渐近线可绘出可靠的函数草图。
Classify with second derivative: if f”(x) > 0, the point is a local minimum; if f”(x) < 0, it is a local maximum. If f''(x) = 0, further investigation is needed.
用二阶导数分类：若 f”(x) > 0，则为局部极小点；若 f”(x) < 0，则为局部极大点。若 f''(x) = 0，则需进一步检验。
12. Optimisation and Connected Rates of Change | 最优化与相关变化率
Optimisation problems involve finding the maximum or minimum of a quantity by expressing it as a function of one variable and then finding the stationary points. Connected rates of change use the chain rule in the form dA/dt = dA/dx × dx/dt, where A and x depend on time t. Such problems often involve geometry, physics, or economics.
最优化问题通过将目标量表达为单变量函数，再寻找驻点来求得最大值或最小值。相关变化率利用链式法则形式 dA/dt = dA/dx × dx/dt，其中 A 和 x 依赖于时间 t。这类问题常涉及几何、物理或经济背景。
dA/dt = (dA/dx) (dx/dt)
A typical example: a spherical balloon’s radius r increases at 2 cm/s. Find the rate of change of volume V when r = 5 cm. V = (4/3)πr³, so dV/dt = 4πr² (dr/dt). Substituting gives dV/dt = 4π(25)(2) = 200π cm³/s.
典型例子：球形气球半径 r 以 2 cm/s 增大。求 r = 5 cm 时体积 V 的变化率。V = (4/3)πr³，则 dV/dt = 4πr² (dr/dt)，代入得 dV/dt = 4π(25)(2) = 200π cm³/s。
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