📚 Differentiation Mastery for IB AQA Mathematics | IB AQA 数学：微分 考点精讲
Differentiation is a cornerstone of calculus, essential for analysing rates of change, slopes of curves, and optimisation problems. In the IB AQA Mathematics syllabus, you are expected to master a range of differentiation techniques, from basic rules to advanced applications. This guide breaks down every key topic, provides step-by-step explanations, and aligns closely with exam-style questions to ensure you are fully prepared.
微分是微积分的基础，对于分析变化率、曲线斜率和最优化问题至关重要。在 IB AQA 数学大纲中，你需要掌握从基本规则到高级应用的一系列微分技巧。本指南将逐一解析每个关键知识点，提供循序渐进的解释，并紧密结合考试题型，确保你做好充分准备。
1. Definition and First Principles | 定义与第一性原理
The derivative of a function f(x) at a point x is defined as the limit of the average rate of change as the interval approaches zero: f'(x) = limₕ→₀ [f(x+h) − f(x)] / h. This is known as differentiation from first principles. Although you will rarely use this definition directly in exam calculations, understanding it strengthens your grasp of what a derivative truly represents: the instantaneous rate of change or the gradient of the tangent line.
函数 f(x) 在某一点 x 的导数定义为当区间趋近于零时平均变化率的极限：f'(x) = limₕ→₀ [f(x+h) − f(x)] / h。这被称为从第一性原理求导。尽管考试中极少直接使用定义计算，但理解其含义能帮助你真正掌握导数的本质：瞬时变化率或切线斜率。
In the AQA formula booklet, you will not find the limit definition, but you may be asked to prove the derivative of simple functions like x² or 1/x using first principles. For x², substitute into the limit: [(x+h)² − x²]/h = (2xh + h²)/h = 2x + h, which approaches 2x as h → 0. Similarly, for sin x, the first-principles approach relies on small-angle approximations and the addition formula.
在 AQA 公式手册中，不会给出极限定义，但可能会要求你用第一性原理证明简单函数（如 x² 或 1/x）的导数。以 x² 为例，代入极限：[(x+h)² − x²]/h = (2xh + h²)/h = 2x + h，当 h → 0 时趋于 2x。类似地，对于 sin x，第一性原理推导需要用到小角近似和加法公式。
2. Power Rule and Basic Derivatives | 幂法则与基本导数
The most frequently used rule is the power rule: if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. This works for any real constant n, including negative and fractional indices. For example, the derivative of √x is (1/2)x⁻½, and the derivative of 1/x² is −2x⁻³. Remember to rewrite roots and reciprocals as powers before differentiating.
最常用的法则是幂法则：若 f(x) = xⁿ，则 f'(x) = nxⁿ⁻¹。这适用于任何实常数 n，包括负指数和分数指数。例如，√x 的导数为 (1/2)x⁻½，1/x² 的导数为 −2x⁻³。务必先将根式和倒数改写为幂的形式再求导。
You must also memorise the derivatives of fundamental functions that appear in the AQA booklet: d/dx (eˣ) = eˣ, d/dx (ln x) = 1/x, d/dx (sin x) = cos x, d/dx (cos x) = −sin x, and d/dx (tan x) = sec² x. Constants vanish: d/dx (c) = 0, and constant multiples factor out: d/dx [c·f(x)] = c·f'(x).
你还需要熟记 AQA 公式手册中基本函数的导数：d/dx (eˣ) = eˣ，d/dx (ln x) = 1/x，d/dx (sin x) = cos x，d/dx (cos x) = −sin x，d/dx (tan x) = sec² x。常数的导数为零：d/dx (c) = 0，常数倍可提取：d/dx [c·f(x)] = c·f'(x)。
When differentiating sums and differences, simply differentiate term by term: d/dx (3x² + 5x − 7) = 6x + 5.
求导和差的函数时，逐项求导即可：d/dx (3x² + 5x − 7) = 6x + 5。
3. Exponential and Logarithmic Differentiation | 指数与对数微分
Functions involving eˣ and ln x are common in IB exam papers. The natural exponential function is its own derivative, making it straightforward: d/dx (eᵏˣ) = k eᵏˣ. For ln x, remember that it is only defined for x > 0, and its derivative is 1/x. When the argument is a linear function, use the chain rule in one step: d/dx [ln(ax+b)] = a/(ax+b).
涉及 eˣ 和 ln x 的函数在 IB 试卷中十分常见。自然指数函数的导数就是其本身，非常简单：d/dx (eᵏˣ) = k eᵏˣ。对于 ln x，需注意其定义域为 x > 0，导数为 1/x。当参数为线性函数时，可直接使用链式法则一步到位：d/dx [ln(ax+b)] = a/(ax+b)。
For general exponential functions aˣ, rewrite using base e: aˣ = e^(x ln a), then differentiate to obtain aˣ ln a. This technique is especially useful when differentiating functions like xˣ, where both base and exponent vary — you take natural logarithms of both sides and use implicit differentiation.
对于一般指数函数 aˣ，可利用换底公式改写：aˣ = e^(x ln a)，然后求导得 aˣ ln a。当底数和指数均为变量，如 xˣ 时，这种技巧尤为重要——你需要对等式两边取自然对数，再使用隐函数求导。
4. Trigonometric Differentiation | 三角函数微分
The derivatives of sine and cosine form a neat pattern: sin x → cos x → −sin x → −cos x → sin x. It is essential to work in radians, as the derivative formulas only hold when x is measured in radians. For tan x, you can derive the result using the quotient rule or simply remember d/dx (tan x) = sec² x.
正弦和余弦的导数形成一个简洁的循环：sin x → cos x → −sin x → −cos x → sin x。必须使用弧度制，因为只有 x 以弧度为单位时这些导数公式才成立。对于 tan x，你可以用商法则推导，或者直接记住 d/dx (tan x) = sec² x。
The reciprocal trigonometric functions also appear: d/dx (cot x) = −csc² x, d/dx (sec x) = sec x tan x, d/dx (csc x) = −csc x cot x. These can be derived but memorising them saves time. When differentiating sin(kx) or cos(kx+b), multiply by the derivative of the inner function: d/dx [sin(2x+π)] = 2 cos(2x+π).
倒数三角函数也会出现：d/dx (cot x) = −csc² x，d/dx (sec x) = sec x tan x，d/dx (csc x) = −csc x cot x。虽然可以推导，但记住它们能节省时间。当对 sin(kx) 或 cos(kx+b) 求导时，记得乘以内层函数的导数：d/dx [sin(2x+π)] = 2 cos(2x+π)。
5. The Chain Rule | 链式法则
The chain rule is one of the most powerful tools in differentiation. If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). In words: differentiate the outer function, leave the inner function untouched, then multiply by the derivative of the inner function. This rule is indispensable for composite functions like (3x²+1)⁵, sin(ln x), and e^(2x³).
链式法则是最强大的微分工具之一。若 y = f(g(x))，则 dy/dx = f'(g(x)) · g'(x)。换句话说：先对外层函数求导，保持内层函数不变，再乘以内层函数的导数。对于 (3x²+1)⁵、sin(ln x)、e^(2x³) 等复合函数，链式法则不可或缺。
Many students find it helpful to use an intermediate variable. Let u = g(x), then y = f(u), and dy/dx = (dy/du) × (du/dx). This method makes it easier to handle repeated applications. In IB exams, you will often chain multiple functions, e.g., differentiating cos³(4x) requires two applications of the chain rule.
许多学生发现引入中间变量更容易操作。令 u = g(x)，则 y = f(u)，且 dy/dx = (dy/du) × (du/dx)。这种方法便于处理多次链式法则。在 IB 考试中，经常需要嵌套多次链式法则，例如求导 cos³(4x) 就需要两次链式法则。
Always check for a chain rule: if a function is not a simple xⁿ, eˣ, sin x, etc., but contains a more complicated argument, the chain rule is needed. Functions like √(2x+5) should be rewritten as (2x+5)^(½) and then differentiated using the chain rule.
务必检查是否需要链式法则：如果函数不是简单的 xⁿ、eˣ、sin x 等形式，而是带有复杂参数的，就需要链式法则。例如 √(2x+5) 应改写为 (2x+5)^(½) 后再用链式法则求导。
6. Product and Quotient Rules | 乘积法则与商法则
When two functions are multiplied, the product rule applies: if y = u(x)·v(x), then dy/dx = u’ v + u v’. The order does not matter, but keeping it symmetrical helps avoid mistakes. A common exam question asks you to differentiate x² sin x or eˣ ln x, where both parts are non-constant.
当两个函数相乘时，需使用乘积法则：若 y = u(x)·v(x)，则 dy/dx = u’ v + u v’。顺序不重要，但保持对称有助于避免错误。考试常出现要求对 x² sin x 或 eˣ ln x 求导的题目，其中两部分均非常数。
When two functions form a fraction, the quotient rule is needed: if y = u(x)/v(x), then dy/dx = (u’ v − u v’) / v². A memory aid is ‘low d-high minus high d-low over low squared’. Take care with the minus sign—it is u’ v minus u v’, not the reverse. This rule is essential for rational functions like (3x+1)/(x²+2) or tan x = sin x / cos x.
当两个函数构成分式时，需使用商法则：若 y = u(x)/v(x)，则 dy/dx = (u’ v − u v’) / v²。记忆口诀是“分母平方分之（分子导乘分母减分母导乘分子）”。注意减号——是 u’ v 减去 u v’，不可颠倒。对于 (3x+1)/(x²+2) 或 tan x = sin x / cos x 等有理函数，商法则是必须的。
Often it is wise to simplify before differentiating. For instance, (x²+x)/x can be simplified to x+1, making differentiation trivial. Similarly, use algebraic manipulation to avoid the quotient rule where possible—but be prepared to use it when necessary.
通常建议在求导前先化简。例如 (x²+x)/x 可以简化为 x+1，使求导变得简单。同样地，尽可能通过代数变形避免使用商法则——但在必要时要能熟练运用。
7. Implicit Differentiation | 隐函数微分
Not all curves are given as y = f(x). Sometimes you meet equations like x² + y² = 25 or xy + sin y = 3. Implicit differentiation allows you to find dy/dx without solving for y explicitly. Differentiate both sides with respect to x, treating y as a function of x. Whenever you differentiate a y-term, attach dy/dx (or y’) as required.
并非所有曲线都以 y = f(x) 的形式给出。有时你会遇到像 x² + y² = 25 或 xy + sin y = 3 这样的方程。隐函数微分使你在无需显式解出 y 的情况下求得 dy/dx。对 x 两边求导，将 y 视作 x 的函数。每次对含有 y 的项求导，都要相应地加上 dy/dx（或 y’）。
For x² + y² = 25, differentiating gives 2x + 2y dy/dx = 0, so dy/dx = −x/y. Notice the answer is in terms of both x and y—this is perfectly acceptable. For xy, use the product rule: d/dx (xy) = 1·y + x·dy/dx. For sin y, use the chain rule: d/dx (sin y) = cos y · dy/dx.
以 x² + y² = 25 为例，求导得 2x + 2y dy/dx = 0，故 dy/dx = −x/y。注意结果中同时出现了 x 和 y——这完全正确。对于 xy，使用乘积法则：d/dx (xy) = 1·y + x·dy/dx。对于 sin y，使用链式法则：d/dx (sin y) = cos y · dy/dx。
Implicit differentiation frequently appears in problems involving tangents and normals to curves defined implicitly. You may also need to find second derivatives implicitly by differentiating dy/dx again, treating dy/dx as another function of x.
隐函数求导常出现在涉及隐式定义曲线的切线和法线问题中。你可能还需要通过再次对 dy/dx 求导来隐式地求二阶导数，此时将 dy/dx 视作 x 的另一个函数处理。
8. Higher-Order Derivatives | 高阶导数
The first derivative f'(x) gives the gradient function; differentiating again yields the second derivative f”(x) or d²y/dx². The second derivative describes the rate of change of the gradient, which relates to concavity and points of inflection. A positive second derivative means the curve is concave up (convex), negative means concave down (concave).
一阶导数 f'(x) 给出斜率函数；再次求导得到二阶导数 f”(x) 或 d²y/dx²。二阶导数描述了斜率的变化率，与曲线的凹凸性和拐点有关。二阶导数为正表示曲线下凸（凹向上），为负表示上凸（凹向下）。
For kinematic problems, if displacement s is a function of time t, velocity v = ds/dt, and acceleration a = d²s/dt². These concepts are directly examined in the context of motion along a straight line. You must be able to interpret and calculate higher derivatives for polynomial, trigonometric, and exponential functions.
在运动学问题中，若位移 s 是时间 t 的函数，则速度 v = ds/dt，加速度 a = d²s/dt²。这些概念在直线运动问题中直接考查。你必须能够对多项式、三角函数和指数函数求出高阶导数并进行解释。
9. Tangents and Normals | 切线与法线
One of the most common applications of differentiation is finding the equation of a tangent or normal to a curve at a given point. The gradient of the tangent at (x₁, y₁) is m = f'(x₁). The equation of the tangent is then y − y₁ = m(x − x₁). The normal is perpendicular to the tangent, so its gradient is −1/m (provided m ≠ 0).
微分最常见的应用之一是求曲线在某一点的切线或法线方程。在 (x₁, y₁) 处的切线斜率为 m = f'(x₁)。切线方程为 y − y₁ = m(x − x₁)。法线与切线垂直，故其斜率为 −1/m（假设 m ≠ 0）。
Always substitute the x-coordinate into the original function to confirm the y-coordinate, then into the derivative to find the gradient. For example, find the tangent to y = x³ − 3x at x = 2: y(2) = 8 − 6 = 2, y’ = 3x² − 3, so gradient = 12 − 3 = 9. Equation: y − 2 = 9(x − 2).
务必先将 x 坐标代入原函数确认 y 坐标，再代入导数求斜率。例如求 y = x³ − 3x 在 x = 2 处的切线：y(2) = 8 − 6 = 2，y’ = 3x² − 3，于是斜率 = 12 − 3 = 9。方程为：y − 2 = 9(x − 2)。
IB questions sometimes ask for the normal at a point where the tangent is horizontal (m = 0). In that case the normal is vertical, and its equation is simply x = x₁.
IB 试题有时会要求在切线水平（m = 0）的点求法线。此时法线为竖直线，方程就是 x = x₁。
10. Rates of Change and Optimisation | 变化率与最优化
Connected rates of change problems involve relating two or more varying quantities. If a variable y depends on x, and x depends on time t, then dy/dt = (dy/dx)·(dx/dt). This is a direct application of the chain rule. Classic examples include a ladder sliding down a wall, water pouring into a conical tank, or a spherical balloon being inflated.
相关变化率问题涉及将两个或多个变化量关联起来。若变量 y 依赖于 x，而 x 又依赖于时间 t，则 dy/dt = (dy/dx)·(dx/dt)。这是链式法则的直接应用。经典例子包括梯子沿墙下滑、水注入圆锥形容器或球形气球充气等场景。
Optimisation problems require finding maximum or minimum values of a function within a given context. Steps: write the quantity to be optimised as a function of one variable, often using a constraint to eliminate another variable; differentiate and set f'(x) = 0 to find stationary points; use the second derivative test or a sign diagram to classify the nature (maximum or minimum); and finally interpret the result in context.
最优化问题要求求出函数在给定情境下的最大值或最小值。步骤：将需优化的量表示为单变量函数，通常借助约束条件消去另一变量；求导并令 f'(x) = 0 寻找驻点；使用二阶导数判定或符号表判别驻点性质（极大或极小）；最后结合情境解释结果。
For IB AQA exams, optimisation frequently involves volumes, areas, or costs. Remember to check endpoints of the domain if the interval is closed. A typical problem: find the dimensions of a cylinder with given surface area that maximises volume.
在 IB AQA 考试中，最优化常涉及体积、面积或成本。如果定义域是闭区间，别忘了检查区间端点。一个典型问题是：给定表面积的圆柱体中，求使容积最大的尺寸。
11. Stationary Points and Curve Sketching | 驻点与曲线草图
Stationary points occur where f'(x) = 0. To determine their nature, use the second derivative: if f”(x) > 0, it is a local minimum; if f”(x) < 0, a local maximum; if f''(x) = 0, the test is inconclusive, and you must check the sign of f'(x) on either side. Points of inflection occur where the concavity changes, which often coincides with f''(x) = 0, but not always.
驻点出现在 f'(x) = 0 处。要判断其性质，可使用二阶导数：若 f”(x) > 0，则为局部极小值；若 f”(x) < 0，则为局部极大值；若 f''(x) = 0，判定失效，则需检查 f'(x) 在两侧的符号。拐点出现在凹性变化处，通常与 f''(x) = 0 吻合，但不绝对。
Curve sketching in calculus involves identifying intercepts, stationary points, asymptotes, and behaviour at infinity. Combine these with the first and second derivatives to produce an accurate sketch. Exam questions may ask you to sketch a gradient function f'(x) given the graph of f(x), or vice versa.
微积分中的曲线草图绘制包括标出截距、驻点、渐近线和无穷远处的趋势。结合一阶和二阶导数，即可画出准确的草图。试题可能要求你根据 f(x) 的图像画出导函数 f'(x) 的图像，或反过来。
12. Exam Strategy and Common Pitfalls | 考试策略与常见陷阱
When facing an unfamiliar function, simplify first: expand brackets, combine fractions, or take logs if it makes differentiation easier. Always consider the domain; for example, ln x requires x > 0, so solutions outside this range must be discarded. In optimisation, always verify that your answer is indeed a maximum (or minimum) and not a saddle point.
面对陌生函数时，先化简：展开括号、合并分式或在有助于求导时取对数。始终考虑定义域；例如 ln x 要求 x > 0，因此定义域外的解必须舍去。在最优化问题中，务必验证所得解确实是极大值（或极小值）而非鞍点。
A common mistake is forgetting to multiply by the derivative of the inner function when using the chain rule. Another is misapplying the quotient rule—remember it is u’v − uv’, not uv’ − u’v. Also ensure you work in radians for all calculus involving trigonometric functions. Finally, always show clear steps in the exam; even if your final answer is wrong, method marks are awarded.
常见错误包括：使用链式法则时忘记乘以内层函数的导数；错误应用商法则——记住是 u’v − uv’，而不是 uv’ − u’v；此外，所有涉及三角函数的微积分都必须使用弧度制。最后，考试中务必展示清晰的步骤；即使最终答案有误，过程分依然可得。
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