📚 Differentiation in IB Mathematics: Concepts and Techniques | IB数学微分：概念与技巧
Differentiation is a cornerstone of IB Mathematics, essential for both Analysis and Approaches (AA) and Applications and Interpretation (AI) at Standard and Higher Level. Mastering differentiation allows you to analyse rates of change, find slopes of curves, and solve real-world optimisation problems. This article covers the fundamental concepts and key techniques you need to succeed in your IB exams, from first principles to advanced applications.
微分是 IB 数学的基石，对于分析与方法 (AA) 以及应用与解释 (AI) 的标准级和高级都至关重要。掌握微分能让你分析变化率、求曲线斜率，并解决现实世界中的优化问题。本文涵盖了 IB 考试所需的基本概念和关键技巧，从第一原理到高级应用，助你顺利通关。
1. Understanding Limits | 理解极限
Limits form the foundation of differentiation. A limit describes the value a function f(x) approaches as x gets arbitrarily close to a specific point a. In IB, you will encounter limits when defining the derivative and evaluating behaviour near vertical asymptotes. The notation is lim x→a f(x) = L, meaning that as x tends to a, f(x) tends to the limit L.
极限是微分的基石。极限描述的是当 x 无限趋近于某一特定点 a 时，函数 f(x) 所趋向的值。在 IB 课程中，你会在定义导数以及分析函数在垂直渐近线附近的行为时遇到极限。其记号为 lim x→a f(x) = L，即当 x 趋近于 a 时，f(x) 趋近于极限 L。
is 4. This process of simplifying before substituting is a common strategy in IB problems.
例如，考虑函数 f(x) = (x² - 4)/(x - 2)。直接代入得到 0/0，但将分子因式分解为 (x-2)(x+2) 并约分后，x→2 时的极限为 4。这种先化简再代入的方法在 IB 问题中十分常见。
2. The Derivative as a Rate of Change | 导数作为变化率
to the curve y = f(x) at that point. If s(t) is a displacement function, the derivative s'(t) gives the velocity; if v(t) is velocity, v'(t) is acceleration. This interpretation is central to contextual problems in IB exams.
函数 f 在点 x 处的导数是 f 关于 x 的瞬时变化率。从几何角度看，它表示曲线 y = f(x) 在该点处切线的斜率。若 s(t) 为位移函数，导数 s'(t) 即为速度；若 v(t) 为速度，v'(t) 即为加速度。这种解释是 IB 考试中情境问题的核心。
When a question asks "Find the rate at which the volume is increasing," you are being asked to differentiate the volume function with respect to time. Being able to translate physical situations into derivative expressions is a must for both AA and AI papers.
当题目问到“求体积增大的速率”时，实际上就是要你对体积函数关于时间求导。能够将实际情景转化为导数表达式是 AA 和 AI 试卷的必备技能。
3. Differentiation from First Principles | 从第一原理求导
The formal definition of the derivative is based on limits:
f'(x) = limh→0 (f(x + h) - f(x)) / h
导数的正式定义基于极限：
f'(x) = limh→0 (f(x + h) - f(x)) / h
taking the limit as h→0 yields 2x. This process reinforces why differentiation rules work.
IB 学生，尤其是 AA 高级水平的学生，需要能够用这一定义推导出基本导数，比如 x² 或 1/x 的导数。对于 f(x) = x²，差商为 ((x+h)² - x²)/h = (2xh + h²)/h = 2x + h，然后取 h→0 时的极限，得到 2x。这一过程强化了对微分规则原理的理解。
4. Basic Differentiation Rules | 基本求导法则
tiate polynomial and rational functions efficiently. The key rule is: if f(x) = xⁿ, then f' ts. You must be comfortable rewriting expressions using index laws before differentiating.
理解原理后，你就可以运用幂法则和其他速成方法来高效地对多项式和有理函数求导。核心法则是：若 f(x) = xⁿ，则 f'(x) = n·xⁿ⁻¹。这适用于任何实常数 n，包括负指数和分数指数。你必须能够熟练地在求导前用指数法则改写表达式。
A table of common derivatives is invaluable for quick reference:
| f(x) | f'(x) |
| c (constant) | 0 |
| x | 1 |
| x² | 2x |
| xⁿ | n xⁿ⁻¹ |
| √x = x½ | (1/2)x-½ = 1/(2√x) |
| 1/x = x⁻¹ | -x⁻² = -1/x² |
常用导数速查表对快速参考非常有用。练习将 1/√x 写成 x-½，将 ³√x 写成 x⅓，然后轻松运用幂法则求导。
5. The Chain Rule | 链式法则
/du · du/dx. In Leibniz notation, it appears as a simple product of derivatives, which makes it easy to remember and apply. IB questions frequently involve nested functions like sin(2x) or ex².
链式法则用于对复合函数求导，即形如 f(g(x)) 的函数。它的表述是：若 y = f(u) 且 u = g(x)，则 dy/dx = dy/du · du/dx。在莱布尼茨记号下，它表现为导数的简单乘积，易于记忆和应用。IB 考题经常涉及嵌套函数，如 sin(2x) 或 ex²。
For example, to differentiate h(x) = (3x² + 1)⁵, set u = 3x² + 1, giving h' ivative of the inner function. This step is the most common source of marks lost in exams.
例如，对 h(x) = (3x² + 1)⁵ 求导，设 u = 3x² + 1，则 h'(x) = 5u⁴ · 6x = 30x(3x²+1)⁴。始终记得乘上内层函数的导数。这一步是考试中最常见的失分点。
6. Product and Quotient Rules | 乘积法则与商法则
When differentiating a product of two functions, use the product rule: (uv)' = u'v + uv'. For a quotient, the rule is (u/v)' = (u'v - uv' ns. It is recommended to quote the rule first before substituting, to show working clearly.
对两个函数的乘积求导时，使用乘积法则：(uv)' = u'v + uv'。对于商，法则为 (u/v)' = (u'v - uv')/v²。这些是 IB 微积分中必不可少的，特别是在处理有理函数或代数函数与三角函数的乘积时。建议先写出法则公式，再代入，以便清晰地展示解题过程。
he final answer. For example, for y = (x² + 1)/(cos x), let u = x²+1 and v = cos x; then y' = (2x cos x + (x²+1) sin x) / cos² x.
一个常见陷阱是忘记商法则中的负号，或错误识别 u 和 v。应明确写出，并分别求导后再组合最终答案。例如，对于 y = (x² + 1)/(cos x)，令 u = x²+1，v = cos x；则 y' = (2x cos x + (x²+1) sin x) / cos² x。
7. Differentiation of Trigonometric Functions | 三角函数的微分
The derivatives of the six basic trigonometric functions must be memorised, but they reduce to a few key pairs. The most important are:
d/dx (sin x) = cos x, d/dx (cos x) = -sin x
六个基本三角函数的导数必须记熟，但它们可以归结为几个关键对。最重要的是：
d/dx (sin x) = cos x, d/dx (cos x) = -sin x
sin² x or tan(3x). Use the chain rule: the derivative of sin² x is 2 sin x cos x = sin 2x.
由此，借助商法则，可以推导出 tan x, cot x, sec x 和 csc x 的导数。在 IB 中，尤其是 AA 高级水平，可能会要求你对 sin² x 或 tan(3x) 进行求导。运用链式法则：sin² x 的导数为 2 sin x cos x = sin 2x。
Be careful with the argument inside: the derivative of sin(2x+1) is 2 cos(2x+1). Always multiply by the derivative of the inner linear function.
注意内部的参数：sin(2x+1) 的导数为 2 cos(2x+1)。始终要乘上内部线性函数的导数。
8. Differentiation of Exponential and Logarithmic Functions | 指数与对数函数的微分
The exponential function ex is unique because its derivative is itself: d/dx (ex) = ex. More generally, for ax, the derivative is ax esults are fundamental in modelling growth and decay, which feature prominently in IB exam papers.
指数函数 ex 的特殊之处在于其导数等于它自身：d/dx (ex) = ex。更一般地，对于 ax，导数为 ax ln a。自然对数函数的导数为 d/dx (ln x) = 1/x (x>0)。这些结果是增长和衰减建模的基础，常出现在 IB 试卷中。
When the exponent or argument is a function of x, always apply the chain rule. For instance, the derivative of e2x+3 is 2 e2x+3 /(5x)·5 = 1/x. Students often forget the factor from the derivative of the inner function.
当指数或对数的自变量为 x 的函数时，务必运用链式法则。例如，e2x+3 的导数为 2 e2x+3，ln(5x) 的导数为 1/(5x)·5 = 1/x。学生常会遗忘内层函数的求导因子。
9. Implicit Differentiation | 隐函数微分
x, treating y as an implicit function of x and using the chain rule: d/dx (y²) = 2y dy/dx.
数学中许多关系是隐式定义的，比如圆的方程 x² + y² = r²。隐函数微分使你能在不显式解出 y 的情况下求出 dy/dx。方法是对方程两边关于 x 求导，将 y 视为 x 的隐函数并运用链式法则：d/dx (y²) = 2y dy/dx。
x dy/dx, then solve for dy/dx. This is standard AA HL material and occasionally appears in AI HL calculus problems.
例如，对于 x³ + y³ = 6xy，求导得 3x² + 3y² dy/dx = 6y + 6x dy/dx，然后解出 dy/dx。这是 AA 高级的标准内容，也偶尔出现在 AI 高级的微积分问题中。
10. Applications of Derivatives: Tangents and Normals | 导数的应用：切线与法线
t point. Thus, the equation of the tangent at (x₁, y₁) is y - y₁ = m (x - x₁), where m = f' combine differentiation with coordinate geometry and are frequent in both SL and HL exams.
函数在某点的导数给出了曲线在该点处切线的斜率。因此，点 (x₁, y₁) 处的切线方程为 y - y₁ = m (x - x₁)，其中 m = f'(x₁)。法线垂直于切线，所以其斜率为 -1/m（若 m ≠ 0）。此类问题将微分与坐标几何相结合，在 SL 和 HL 考试中都很常见。
= 3x² - 3, so m = 0. The tangent is horizontal: y = -2, and the normal is vertical: x = 1.
一道典型的 IB 题可能给出曲线 y = x³ - 3x，并要求求出 x=1 处的切线和法线方程。先求得 y = -2，然后 dy/dx = 3x² - 3，故 m = 0。切线为水平线 y = -2，法线为垂直线 x = 1。
11. Optimization Problems | 优化问题
Optimisation involves finding maximum or minimum values of a quantity, which often translates to setting the derivative equal to zero and solving for critical points. In IB, you will model a real-world situation, express the quantity to be optimised as a function of one variable, differentiate, and verify whether the critical point gives a maximum or minimum (using the second derivative or sign analysis).
优化问题涉及求某一量的最大值或最小值，这通常转化为令导数等于零并求解临界点。在 IB 中，你需要对现实情境建模，将被优化的量表示为一个变量的函数，求导，并验证临界点给出最大值还是最小值（使用二阶导数或符号分析法）。
For instance, to maximise the area of a rectangular enclosure with a fixed perimeter, or to minimise the surface area for a given volume as in packaging design. Clearly defining the domain and checking endpoints is crucial.
例如，在周长固定的情况下最大化矩形围场的面积，或在体积给定的情况下最小化表面积（如包装设计）。清晰地定义自变量范围并检查端点至关重要。
12. Related Rates | 相关变化率
Related rates problems deal with two or more variables that are changing with respect to time, linked by an equation. By differentiating the equation with respect to time t (often using the chain rule implicitly), you can find the rate of change of one variable given the rate of change of another. IB AI HL and AA HL both include questions on this topic.
相关变化率问题处理两个或更多随时间变化的变量，它们由一个方程相互联系。通过对方程关于时间 t 进行隐式求导（常使用链式法则），你可以在已知一个变量变化率的情况下求出另一个变量的变化率。IB AI 高级和 AA 高级都包含这一主题的问题。
ith respect to t gives 2x dx/dt + 2y dy/dt = 0, from which the desired rate can be solved.
一个经典例子：一把梯子靠在墙上滑下。已知梯子底端远离墙壁的速率，求顶端下降的速率。利用 x² + y² = L² 并对 t 求导，得 2x dx/dt + 2y dy/dt = 0，由此可解出所求速率。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图