Differential Calculus: Core Exam Topics for IB and OCR

📚 Differential Calculus: Core Exam Topics for IB and OCR | 微分考点精讲

Differential calculus forms the backbone of advanced mathematics in IB and OCR specifications. Understanding the derivative as a rate of change, mastering a range of differentiation techniques, and applying these skills to real-world problems are essential for exam success. This guide systematically covers the key topics you will encounter, from the limit definition through to parametric equations and optimisation, with clear explanations and practical examples.

微分学是 IB 与 OCR 数学课程的核心支柱。把导数理解为变化率、掌握多种微分技巧并应用到实际问题中，是考试取得高分的必备能力。本指南从极限定义出发，系统梳理参数方程、优化问题等关键考点，配合清晰解析与实用范例，助你稳扎稳打。

1. The Definition of the Derivative | 导数的定义

The derivative of a function f(x) at a point x = a is defined by the limit f'(a) = limₕ→₀ [f(a+h) − f(a)] / h, provided this limit exists. It represents the instantaneous rate of change of the function and the slope of the tangent line at that point.

函数 f(x) 在点 x = a 处的导数定义为极限 f'(a) = limₕ→₀ [f(a+h) − f(a)] / h（假设极限存在）。它表示函数在该点的瞬时变化率，也是切线的斜率。

For the function f(x) = x², the derivative from first principles gives f'(x) = 2x. This process reinforces the fundamental link between limits and differentiation.

以 f(x) = x² 为例，利用第一原理求导可得 f'(x) = 2x，这一过程强化了极限与微分之间的根本联系。

In IB exams, you may be asked to prove a derivative using the limit definition, while OCR often tests this concept in the context of gradient functions.

IB 考试可能要求用极限定义证明导数，而 OCR 常在梯度函数的情境中考查这一概念。

2. Basic Differentiation Rules | 基本求导法则

The power rule states that if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. Sum and constant multiple rules allow us to differentiate term by term: d/dx [c·f(x)] = c·f'(x) and d/dx [f(x) ± g(x)] = f'(x) ± g'(x).

幂法则指出，若 f(x) = xⁿ，则 f'(x) = nxⁿ⁻¹。常数倍法则与和差法则支持逐项求导：d/dx [c·f(x)] = c·f'(x)，且 d/dx [f(x) ± g(x)] = f'(x) ± g'(x)。

For example, y = 4x³ − 2x + 7 gives dy/dx = 12x² − 2. Memorising these elementary rules is the first step towards fluent differentiation.

例如 y = 4x³ − 2x + 7，求导得 dy/dx = 12x² − 2。熟记这些基本规则是流畅求导的第一步。

f(x)	f'(x)
k (constant)	0
xⁿ	nxⁿ⁻¹
k·u(x)	k·u'(x)
u ± v	u’ ± v’

3. The Chain Rule | 链式法则

When a function is composed of two functions, y = f(g(x)), the chain rule gives dy/dx = f'(g(x)) × g'(x), or in Leibniz notation, dy/dx = dy/du × du/dx where u = g(x).

当函数由两个函数复合而成，y = f(g(x))，链式法则表示为 dy/dx = f'(g(x)) × g'(x)，或用莱布尼兹记号写作 dy/dx = dy/du × du/dx，其中 u = g(x)。

To differentiate y = (3x² + 1)⁵, set u = 3x² + 1, giving dy/du = 5u⁴ and du/dx = 6x, hence dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴.

例如 y = (3x² + 1)⁵，设 u = 3x² + 1，则 dy/du = 5u⁴，du/dx = 6x，因此 dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴。

This rule is indispensable for differentiating powers of linear functions, trigonometric composites, and exponentials such as e^(kx).

链式法则在求线性函数的幂、三角复合函数以及 e^(kx) 等指数函数的导数时不可或缺。

4. Product and Quotient Rules | 乘积法则与商法则

For functions u(x) and v(x), the product rule is (uv)’ = u’v + uv’. The product rule is often recited as ‘derivative of the first times the second plus the first times derivative of the second’.

对于函数 u(x) 与 v(x)，乘积法则为 (uv)’ = u’v + uv’。可记忆为“第一者的导数乘以第二者，加上第一者乘以第二者的导数”。

The quotient rule states (u/v)’ = (u’v − uv’) / v², provided v ≠ 0. A helpful mnemonic is ‘low d-high minus high d-low over low squared’.

商法则为 (u/v)’ = (u’v − uv’) / v²（v ≠ 0）。口诀“分母微分子减分子微分母，除分母平方”便于记忆。

Given y = x² sin x, the product rule yields dy/dx = 2x sin x + x² cos x. For y = (x² + 1)/(x − 2), the quotient rule gives dy/dx = [2x(x − 2) − (x² + 1)] / (x − 2)² = (x² − 4x − 1)/(x − 2)².

若 y = x² sin x，乘积法则得 dy/dx = 2x sin x + x² cos x。对于 y = (x² + 1)/(x − 2)，商法则得 dy/dx = [2x(x − 2) − (x² + 1)]/(x − 2)² = (x² − 4x − 1)/(x − 2)²。

5. Implicit Differentiation | 隐函数微分

When y is not explicitly expressed as a function of x, we differentiate both sides of the equation with respect to x, treating y as an implicit function of x and applying the chain rule to terms involving y: d/dx (yⁿ) = nyⁿ⁻¹ dy/dx.

当 y 不是 x 的显函数时，我们将方程两边对 x 求导，将 y 视为 x 的隐函数，并对含 y 的项应用链式法则：d/dx (yⁿ) = nyⁿ⁻¹ dy/dx。

For the circle x² + y² = 25, differentiating gives 2x + 2y dy/dx = 0, so dy/dx = −x/y. This technique is vital for finding gradients of curves defined by relations like xy² + y = x³.

对于圆 x² + y² = 25，求导得 2x + 2y dy/dx = 0，因此 dy/dx = −x/y。这一技巧在求诸如 xy² + y = x³ 等关系确定的曲线斜率时至关重要。

IB and OCR both test implicit differentiation, often combined with finding equations of tangents and normals at given points.

IB 与 OCR 均考查隐函数微分，常结合求某点处的切线与法线方程。

6. Parametric Differentiation | 参数方程微分

When x and y are given in terms of a parameter t, x = f(t), y = g(t), the derivative dy/dx is obtained by dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0.

当 x 与 y 以参数 t 给出，x = f(t)，y = g(t)，导数 dy/dx 由 dy/dx = (dy/dt) / (dx/dt) 求得，只要 dx/dt ≠ 0。

For x = t² and y = 2t + 1, we have dx/dt = 2t, dy/dt = 2, so dy/dx = 2/(2t) = 1/t. This method extends to second derivatives: d²y/dx² = d(dy/dx)/dt ÷ dx/dt.

若 x = t²，y = 2t + 1，则 dx/dt = 2t，dy/dt = 2，故 dy/dx = 2/(2t) = 1/t。该方法可扩展到二阶导数：d²y/dx² = d(dy/dx)/dt ÷ dx/dt。

Parametric differentiation is particularly common in mechanics and curve-sketching questions across both IB and OCR syllabuses.

参数微分在力学与曲线描绘题中极为常见，同时覆盖 IB 与 OCR 考纲。

7. Higher-Order Derivatives | 高阶导数

The second derivative, f”(x) or d²y/dx², is the derivative of the first derivative. It describes the rate of change of the gradient and is used to determine concavity and points of inflection.

二阶导数 f”(x) 或 d²y/dx² 是一阶导数的导数，描述斜率的变化率，用于判断凹凸性与拐点。

If y = x³ − 3x² + 2x, then y’ = 3x² − 6x + 2 and y” = 6x − 6. Setting y” = 0 gives x = 1, a candidate for a point of inflection. Sign changes in y” confirm the concavity changes.

若 y = x³ − 3x² + 2x，则 y’ = 3x² − 6x + 2，y” = 6x − 6。令 y” = 0 得 x = 1，可能是拐点；y” 变号即可确认凹凸性改变。

In kinematics, the second derivative of displacement with respect to time gives acceleration. IB and OCR frequently link second derivatives to geometrical interpretations of graphs.

在运动学中，位移对时间的二阶导数是加速度。IB 和 OCR 常将二阶导数与图像的几何解释相联。

8. Stationary Points and Points of Inflection | 驻点与拐点

Stationary points occur where dy/dx = 0. Their nature is classified using the second derivative test: if y” > 0, it is a local minimum; if y” < 0, a local maximum; if y'' = 0, further investigation is required.

驻点出现在 dy/dx = 0 处。其性质可通过二阶导数检验：若 y” > 0 则为局部极小值点；若 y” < 0 则为局部极大值点；若 y'' = 0 则需进一步分析。

A point of inflection is where the concavity changes sign (y” changes sign) and the tangent often crosses the curve. Not every point with y” = 0 is an inflection; a sign test is necessary.

拐点是凹凸性改变（y” 变号）的位置，通常切线穿过曲线。y” = 0 的点未必是拐点，必须进行符号检验。

For y = x⁴ − 4x³, y’ = 4x³ − 12x² = 4x²(x − 3), giving stationary points at x = 0 and x = 3. y” = 12x² − 24x; at x = 3, y” = 36 > 0 (minimum); at x = 0 the second derivative test fails, but analysis reveals a point of inflection.

例如 y = x⁴ − 4x³，y’ = 4x³ − 12x² = 4x²(x − 3)，驻点位于 x = 0 与 x = 3。y” = 12x² − 24x；x = 3 时 y” = 36 > 0（极小值）；x = 0 时二阶导数检验失效，但分析可知是拐点。

y’ sign change	Nature
+ to −	Local maximum
− to +	Local minimum
No sign change	Point of inflection

9. Optimization Problems | 优化问题

Optimization involves finding the maximum or minimum value of a quantity by expressing it as a function of one variable, differentiating, and setting the derivative equal to zero. The second derivative or a sign test confirms the nature of the extremum.

优化问题通过将待求极值的量表示为单一变量的函数，求导并令导数为零来寻找最大值或最小值。再利用二阶导数或符号检验确认极值性质。

A typical problem: a rectangular box with a square base and volume 500 cm³ is to be constructed. Find the dimensions that minimise the surface area. Let base side = x, height = h, then V = x²h = 500 ⇒ h = 500/x². Surface area S = 2x² + 4xh = 2x² + 2000/x. Differentiate: S’ = 4x − 2000/x² = 0 ⇒ x = ³√500 ≈ 7.94 cm. S” = 4 + 4000/x³ > 0 confirms a minimum.

典型例题：要制造一个底面为正方形、容积为 500 cm³ 的长方体盒子，求使表面积最小的尺寸。设底边长为 x，高为 h，则 V = x²h = 500 ⇒ h = 500/x²。表面积 S = 2x² + 4xh = 2x² + 2000/x。求导：S’ = 4x − 2000/x² = 0 ⇒ x = ³√500 ≈ 7.94 cm。S” = 4 + 4000/x³ > 0 确认为极小值。

Both IB and OCR exam papers frequently include contextual optimisation tasks, from minimising cost to maximising area or volume.

IB 与 OCR 试卷中常出现实际情境的优化题，如成本最小化、面积或体积最大化。

10. Related Rates of Change | 相关变化率

In related rates problems, two or more quantities that change with time are linked by an equation. Differentiation with respect to time using the chain rule connects their rates of change, e.g., dV/dt = dV/dr × dr/dt.

在相关变化率问题中，两个或多个随时间变化的量通过方程关联。利用链式法则对时间求导，可将它们的变化率联系起来，如 dV/dt = dV/dr × dr/dt。

If a spherical balloon is inflated and its radius increases at 2 cm/s, the rate of change of volume when r = 5 cm is found from V = (4/3)πr³, so dV/dt = 4πr² dr/dt = 4π(5)² × 2 = 200π cm³/s.

若球状气球充气时半径以 2 cm/s 增大，当 r = 5 cm 时体积的变化率可由 V = (4/3)πr³ 求得：dV/dt = 4πr² dr/dt = 4π(5)² × 2 = 200π cm³/s。

These questions demand careful interpretation of the problem statement and correct application of differentiation with respect to time, a skill valued in both IB and OCR assessments.

此类题目要求仔细解读题意并正确对时间求导，是 IB 与 OCR 评价中的重点技能。

11. Differentiation of Exponentials and Logarithms | 指数与对数微分

The derivative of eˣ is eˣ itself. For e^(kx), the chain rule gives ke^(kx). The derivative of ln x is 1/x (for x > 0). For ln(kx), the derivative is still 1/x, since ln(kx) = ln k + ln x.

eˣ 的导数就是 eˣ 自身。对于 e^(kx)，链式法则给出 ke^(kx)。ln x 的导数为 1/x（x > 0）。对于 ln(kx)，导数仍为 1/x，因为 ln(kx) = ln k + ln x。

For aˣ (a > 0), convert to e^(x ln a), so d/dx(aˣ) = aˣ ln a. Logarithmic differentiation is a powerful technique for differentiating forms like [f(x)]^(g(x)) by taking natural logs of both sides.

对于 aˣ（a > 0），可转化为 e^(x ln a)，故 d/dx(aˣ) = aˣ ln a。对数微分法是处理 [f(x)]^(g(x)) 型函数的强大工具，通过对两边取自然对数后进行求导。

IB Analysis & Approaches and OCR Mechanics specification both require fluency in these derivatives, often mixing them with the chain, product, and quotient rules.

IB 分析与方法和 OCR 力学考纲都要求熟练掌握这些导数，并常与链式法则、乘积法则、商法则混合考查。

12. Trigonometric Differentiation | 三角函数的微分

The fundamental trigonometric derivatives are: d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(tan x) = sec² x. All angles are in radians unless stated otherwise; this is essential for limit results to hold.

基本三角函数的导数为：d/dx(sin x) = cos x，d/dx(cos x) = −sin x，d/dx(tan x) = sec² x。除非特别说明，所有角度均以弧度为单位，这对极限成立至关重要。

With the chain rule, we can differentiate sin(kx): d/dx(sin(2x)) = 2cos(2x). For sec x = 1/cos x, use the quotient rule: d/dx(sec x) = sec x tan x. Derivatives of cot x, csc x appear in further mathematics contexts.

结合链式法则可求导 sin(kx)：d/dx(sin(2x)) = 2cos(2x)。对 sec x = 1/cos x 使用商法则：d/dx(sec x) = sec x tan x。cot x、csc x 的导数出现在进阶数学中。

Typical exam question: find the equation of the tangent to y = sin x at x = π/4. Gradient is cos(π/4) = √2/2, point is (π/4, √2/2), giving y − √2/2 = (√2/2)(x − π/4).

典型考题：求 y = sin x 在 x = π/4 处的切线方程。斜率为 cos(π/4) = √2/2，点为 (π/4, √2/2)，故切线为 y − √2/2 = (√2/2)(x − π/4)。

OCR’s Pure Mathematics and IB SL/HL both require proficiency in differentiating trigonometric functions, often within broader problem-solving settings.

OCR 纯数学与 IB SL/HL 均要求熟练掌握三角函数微分，通常嵌入更广泛的解题情境中。

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Differential Calculus: Core Exam Topics for IB and OCR | 微分 考点精讲