Mastering Differentiation for IB & Edexcel | 微分考点精讲

📚 Mastering Differentiation for IB & Edexcel | 微分考点精讲

Differentiation is a cornerstone of calculus, forming the foundation for understanding rates of change, slopes of curves, and optimization. Whether you are following the IB Mathematics Analysis & Approaches syllabus or the Edexcel A Level Mathematics specification, mastering differentiation is essential. This article breaks down each key topic, from the limit definition to advanced applications, providing clear explanations, examples, and exam-focused tips to help you build confidence and achieve top marks.

微分是微积分的基石，为理解变化率、曲线斜率和最优化问题奠定了基础。无论你学习的是 IB 数学分析与方法课程，还是 Edexcel A Level 数学大纲，掌握微分都是至关重要的。本文将逐一拆解各个核心知识点，从极限定义到高阶应用，提供清晰的解释、例题和应试技巧，帮助你建立信心，斩获高分。

1. Definition and First Principles | 导数定义与第一原理

The derivative of a function f(x) at a point x = a is defined as the limit of the difference quotient as h approaches zero: f'(a) = lim(h→0) [f(a+h) – f(a)] / h. This expression represents the instantaneous rate of change of f(x) at x = a, or geometrically, the slope of the tangent line to the curve y = f(x) at that point. Understanding this definition is vital, as it underpins all differentiation rules and may be tested directly in IB Paper 1 and Edexcel Pure Mathematics exams.

函数 f(x) 在点 x = a 处的导数定义为差商的极限，当 h 趋近于 0 时：f'(a) = lim(h→0) [f(a+h) – f(a)] / h。此表达式表示 f(x) 在 x = a 处的瞬时变化率，几何意义上即曲线 y = f(x) 在该点处切线的斜率。理解这一定义至关重要，因为它是一切微分法则的基础，并且在 IB 试卷一和 Edexcel 纯数学考试中可能直接考查。

To differentiate from first principles, substitute the function into the limit expression, expand, simplify, and then let h tend to zero. For example, for f(x) = x², the difference quotient becomes [(x+h)² – x²]/h = (2xh + h²)/h = 2x + h, and as h → 0, we get f'(x) = 2x. The same process applies to other simple functions like kx, x³, and 1/x, reinforcing the conceptual link between limits and derivatives.

要用第一原理求导，需将函数代入极限表达式，展开、化简，然后令 h 趋近于零。例如，对于 f(x) = x²，差商变为 [(x+h)² – x²]/h = (2xh + h²)/h = 2x + h，当 h → 0 时，得到 f'(x) = 2x。同样的过程适用于 kx、x³ 和 1/x 等简单函数，巩固了极限与导数之间的概念联系。

f'(x) = lim(h→0) [f(x+h) – f(x)] / h

2. Basic Differentiation Rules | 基本微分法则

Once the limit definition is understood, we can use standard rules to differentiate polynomials and power functions efficiently. The power rule states that if f(x) = xⁿ, then f'(x) = n xⁿ⁻¹. This holds for any real constant n, including negative and fractional exponents. For a constant c, the derivative is zero because a horizontal line has zero slope. The constant multiple rule allows us to pull out coefficients: d/dx [c f(x)] = c f'(x). The sum rule says the derivative of a sum is the sum of the derivatives.

在理解了极限定义之后，我们可以使用标准法则高效地对多项式和幂函数求导。幂法则指出，若 f(x) = xⁿ，则 f'(x) = n xⁿ⁻¹。这适用于任意实常数 n，包括负指数和分数指数。对于常数 c，其导数为零，因为水平线的斜率为零。常数倍法则允许我们提取系数：d/dx [c f(x)] = c f'(x)。和法则表明，和的导数等于导数之和。

Combining these rules, we can differentiate any polynomial term by term. For instance, if f(x) = 4x³ – 5x² + 2x – 7, then f'(x) = 12x² – 10x + 2. Rational and radical expressions should first be rewritten with exponents: √x becomes x^(1/2), and 1/x² becomes x⁻². This technique appears frequently in both IB and Edexcel exams, where simplification before differentiating saves time and reduces errors.

结合这些法则，我们可以逐项对任意多项式进行求导。例如，若 f(x) = 4x³ – 5x² + 2x – 7，则 f'(x) = 12x² – 10x + 2。有理式和根式应首先改写为指数形式：√x 变为 x^(1/2)，1/x² 变为 x⁻²。这一技巧在 IB 和 Edexcel 考试中频繁出现，求导前先化简可以节省时间并减少错误。

d/dx (xⁿ) = n xⁿ⁻¹ d/dx (c) = 0 d/dx [f(x) ± g(x)] = f'(x) ± g'(x)

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

When two functions are multiplied together, the derivative is not simply the product of their derivatives. The product rule must be applied: if y = u v, where u and v are functions of x, then dy/dx = u’ v + u v’. Memorising this as ‘first times derivative of second, plus second times derivative of first’ helps build fluency. For example, to differentiate y = x² sin x, let u = x², v = sin x, then dy/dx = 2x sin x + x² cos x.

当两个函数相乘时，导数并非简单地等于它们导数的乘积。必须使用乘积法则：若 y = u v，其中 u 和 v 均为关于 x 的函数，则 dy/dx = u’ v + u v’。将其记为“第一项乘以第二项的导数，加上第二项乘以第一项的导数”有助于熟练运用。例如，求 y = x² sin x 的导数，设 u = x²，v = sin x，则 dy/dx = 2x sin x + x² cos x。

The quotient rule handles division: if y = u/v, then dy/dx = (u’ v – u v’) / v². The order in the numerator is crucial—derivative of the top times the bottom, minus top times derivative of the bottom, all over bottom squared. Consider y = (3x + 1)/(x² – 2). Here, u = 3x + 1, v = x² – 2, so u’ = 3, v’ = 2x. Then dy/dx = [3(x² – 2) – (3x + 1)(2x)] / (x² – 2)², which simplifies to (-3x² – 2x – 6)/(x² – 2)². Both IB and Edexcel mark schemes reward correct application and simplification.

商法则处理除法：若 y = u/v，则 dy/dx = (u’ v – u v’) / v²。分子中的顺序至关重要——分子的导数乘以分母，减去分子乘以分母的导数，整体除以分母的平方。考虑 y = (3x + 1)/(x² – 2)，此处 u = 3x + 1，v = x² – 2，因此 u’ = 3，v’ = 2x。于是 dy/dx = [3(x² – 2) – (3x + 1)(2x)] / (x² – 2)²，化简后得 (-3x² – 2x – 6)/(x² – 2)²。IB 和 Edexcel 的评分标准都奖励正确的应用和化简。

Product Rule: d/dx (u v) = u’ v + u v’ Quotient Rule: d/dx (u/v) = (u’ v – u v’) / v²

4. Chain Rule | 链式法则

The chain rule is used to differentiate composite functions—functions of a function. If y = f(g(x)), then dy/dx = f'(g(x)) × g'(x). In Leibniz notation, if y = f(u) and u = g(x), then dy/dx = dy/du × du/dx. This rule is essential for a vast range of functions, including powers of linear expressions like (3x + 5)⁴, exponential functions such as e^(2x), and trigonometric functions like sin(5x). Mastering the chain rule opens the door to implicit and parametric differentiation.

链式法则用于对复合函数（即函数的函数）求导。若 y = f(g(x))，则 dy/dx = f'(g(x)) × g'(x)。用莱布尼茨记号表示，若 y = f(u) 且 u = g(x)，则 dy/dx = dy/du × du/dx。这一法则对大量函数类型至关重要，包括形如 (3x + 5)⁴ 的线性表达式的幂、指数函数如 e^(2x)，以及三角函数如 sin(5x)。熟练掌握链式法则将为进一步学习隐函数微分和参数微分打开大门。

A systematic approach helps: identify the outer and inner functions. For y = (2x³ – 7)⁵, the outer function is u⁵ and the inner is u = 2x³ – 7. Differentiate the outer with respect to u (5u⁴), then multiply by the derivative of the inner (6x²), giving dy/dx = 5(2x³ – 7)⁴ × 6x² = 30x² (2x³ – 7)⁴. In exams, failing to multiply by the derivative of the inner function is a common mistake, so always check the ‘inside’ derivative. For functions like ln(f(x)) and e^(f(x)), the chain rule yields f'(x)/f(x) and f'(x) e^(f(x)), respectively.

系统化的方法很有帮助：识别外层函数和内层函数。对 y = (2x³ – 7)⁵ 而言，外层函数是 u⁵，内层函数是 u = 2x³ – 7。将外层对 u 求导得 5u⁴，然后乘以内层函数的导数 6x²，得到 dy/dx = 5(2x³ – 7)⁴ × 6x² = 30x² (2x³ – 7)⁴。考试中，忘记乘以内层函数的导数是常见错误，因此务必检查“内部”的导数。对于形如 ln(f(x)) 和 e^(f(x)) 的函数，链式法则分别给出 f'(x)/f(x) 和 f'(x) e^(f(x))。

dy/dx = dy/du × du/dx

5. Derivatives of Exponential and Logarithmic Functions | 指数与对数函数的导数

Exponential and logarithmic functions appear extensively in growth and decay models, making their derivatives critical. The derivative of eˣ is simply eˣ. Because of its unique property, eˣ is often called the natural exponential function. When the exponent is a function of x, apply the chain rule: d/dx (e^(ax+b)) = a e^(ax+b). For a general base a > 0, the derivative of aˣ is aˣ ln a, though this is less frequent in IB and Edexcel than eˣ.

指数函数和对数函数广泛出现在增长与衰减模型中，因此它们的导数至关重要。eˣ 的导数就是 eˣ 本身。由于其独特的性质，eˣ 常被称为自然指数函数。当指数为 x 的函数时，应用链式法则：d/dx (e^(ax+b)) = a e^(ax+b)。对于一般的底数 a > 0，aˣ 的导数为 aˣ ln a，但在 IB 和 Edexcel 考试中出现频率低于 eˣ。

The derivative of the natural logarithm ln x is 1/x, for x > 0. For ln(f(x)), the derivative is f'(x)/f(x). For instance, differentiate y = ln(3x² + 1); f(x) = 3x² + 1, f'(x) = 6x, so dy/dx = 6x/(3x² + 1). In IB, you may also see questions combining exponentials with logarithms in one expression, requiring product or quotient rules alongside chain rule. Practising a mix of these builds the necessary fluency.

自然对数 ln x 的导数是 1/x，其中 x > 0。对于 ln(f(x))，其导数为 f'(x)/f(x)。例如，求 y = ln(3x² + 1) 的导数：f(x) = 3x² + 1，f'(x) = 6x，因此 dy/dx = 6x/(3x² + 1)。在 IB 考试中，你可能会遇到在同一表达式中将指数与对数相结合的题目，这需要同时运用乘积法则或商法则以及链式法则。多加练习这类混合题型有助于培养必要的熟练度。

d/dx (eˣ) = eˣ d/dx (e^(f(x))) = f'(x) e^(f(x)) d/dx (ln x) = 1/x d/dx (ln f(x)) = f'(x)/f(x)

6. Derivatives of Trigonometric Functions | 三角函数的导数

The derivatives of the six standard trigonometric functions must be memorised. The most fundamental are: d/dx (sin x) = cos x, and d/dx (cos x) = -sin x. From these, the derivatives of tan x, sec x, csc x, and cot x can be derived using the quotient rule. For example, tan x = sin x / cos x, and d/dx (tan x) = sec² x. The derivative of sec x is sec x tan x, and csc x gives -csc x cot x. Note the symmetry and sign patterns—they are frequently tested.

六个标准三角函数的导数必须记忆。最基本的是：d/dx (sin x) = cos x，d/dx (cos x) = -sin x。由此，可以使用商法则推导出 tan x、sec x、csc x 和 cot x 的导数。例如，tan x = sin x / cos x，其导数 d/dx (tan x) = sec² x。sec x 的导数是 sec x tan x，csc x 给出 -csc x cot x。注意其中的对称性和符号规律——这些都是常考内容。

When the angle is a function of x, the chain rule applies: d/dx (sin(ax + b)) = a cos(ax + b). For composite trig functions like sin² x or cos³ x, rewrite as (sin x)² and apply the chain rule: 2 sin x cos x = sin 2x. Both IB and Edexcel questions often embed trig derivatives within product, quotient, or implicit contexts, so practice recognising these patterns quickly.

当角度为 x 的函数时，需应用链式法则：d/dx (sin(ax + b)) = a cos(ax + b)。对于 sin² x 或 cos³ x 这类复合三角函数，可改写为 (sin x)² 并应用链式法则：2 sin x cos x = sin 2x。IB 和 Edexcel 的试题常将三角函数的导数嵌入乘积、商或隐函数的情境中，因此练习快速识别这些模式至关重要。

Function f(x)	Derivative f'(x)
sin x	cos x
cos x	-sin x
tan x	sec² x
sec x	sec x tan x
csc x	-csc x cot x
cot x	-csc² x

7. Implicit Differentiation | 隐函数微分

When y is not given explicitly as a function of x, but the relationship is defined by an equation involving both x and y, we use implicit differentiation. The technique involves differentiating both sides of the equation with respect to x, treating y as a function of x and applying the chain rule to any term containing y. This means d/dx (y) = dy/dx, and d/dx (y²) = 2y dy/dx. After differentiating, the resulting equation is solved for dy/dx.

当 y 没有明确表示成 x 的函数，但关系由同时包含 x 和 y 的方程定义时，我们使用隐函数微分。该方法涉及将方程两边同时对 x 求导，将 y 视作 x 的函数，并对任何含 y 的项应用链式法则。这意味着 d/dx (y) = dy/dx，而 d/dx (y²) = 2y dy/dx。求导后，解出所得方程中的 dy/dx 即可。

A typical IB or Edexcel question might ask for the gradient of a curve defined implicitly, say x² + y² = 25. Differentiating: 2x + 2y dy/dx = 0 → dy/dx = -x/y. Then the slope at (3,4) is -3/4. Implicit differentiation is also essential when the equation involves products of x and y, like x y + y² = 10. Here, d/dx (x y) requires the product rule: y + x dy/dx. Practice with a variety of equations to become comfortable gathering all dy/dx terms on one side.

典型的 IB 或 Edexcel 题目可能会求隐函数定义曲线的梯度，例如 x² + y² = 25。求导：2x + 2y dy/dx = 0 → dy/dx = -x/y。那么在点 (3,4) 处的斜率为 -3/4。当方程包含 x 与 y 的乘积时，如 x y + y² = 10，隐函数微分同样必不可少。此时 d/dx (x y) 需要使用乘积法则：y + x dy/dx。通过练习各类方程，熟悉将所有 dy/dx 项移到一边的做法。

Differentiate both sides w.r.t. x: d/dx (yⁿ) = n yⁿ⁻¹ dy/dx

8. Parametric Differentiation | 参数方程微分

In parametric equations, both x and y are expressed in terms of a third variable, usually t. To find dy/dx, we differentiate y and x with respect to t and use the formula dy/dx = (dy/dt) / (dx/dt). The second derivative d²y/dx² is found by differentiating dy/dx with respect to t and dividing by dx/dt: d²y/dx² = d/dt (dy/dx) / (dx/dt). This is a common topic in Edexcel’s Pure Mathematics and IB’s Analysis & Approaches HL papers, often followed by finding equations of tangents or stationary points.

在参数方程中，x 和 y 都用第三个变量（通常为 t）来表示。为求 dy/dx，我们将 y 和 x 分别对 t 求导，并使用公式 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：d²y/dx² = d/dt (dy/dx) / (dx/dt)。这是 Edexcel 纯数学和 IB 分析与方法 HL 试卷中的常见主题，通常紧接着要求求切线方程或驻点。

For example, given x = t² + 1, y = 2t³ – t, find dy/dx and d²y/dx². Here, dx/dt = 2t, dy/dt = 6t² – 1, so dy/dx = (6t² – 1)/(2t). To find the second derivative, differentiate dy/dx with respect to t using the quotient rule, then divide by dx/dt (2t). Simplification and substitution of specific t values to find tangent gradients or determine the nature of stationary points are standard exam tasks. Always check that dx/dt is not zero when using these formulas.

例如，给定 x = t² + 1，y = 2t³ – t，求 dy/dx 和 d²y/dx²。此处 dx/dt = 2t，dy/dt = 6t² – 1，因此 dy/dx = (6t² – 1)/(2t)。为求二阶导数，先用商法则将 dy/dx 对 t 求导，再除以 dx/dt（即 2t）。化简并代入特定 t 值以求切线梯度或判定驻点性质，是考试的典型任务。使用这些公式时，务必检查 dx/dt 是否不为零。

dy/dx = (dy/dt) / (dx/dt) d²y/dx² = d/dt (dy/dx) / (dx/dt)

9. Higher-Order Derivatives | 高阶导数

The derivative of a derivative is called the second derivative. If y = f(x), then d²y/dx² = f”(x) is the rate of change of the gradient. Higher-order derivatives continue this pattern: f”'(x) is the third derivative, and f⁽ⁿ⁾(x) the nth derivative. Second derivatives are crucial for classifying stationary points (concavity) and for problems involving acceleration as the derivative of velocity. Understanding the notation dy/dx, d²y/dx², f'(x), and f”(x) is essential.

导数的导数称为二阶导数。若 y = f(x)，则 d²y/dx² = f”(x) 表示梯度的变化率。更高阶的导数延续这一模式：f”'(x) 为三阶导数，f⁽ⁿ⁾(x) 为 n 阶导数。二阶导数对于判别驻点类型（凹凸性）以及涉及加速度作为速度导数的问题至关重要。理解 dy/dx、d²y/dx²、f'(x) 和 f”(x) 这些记号是必不可少的。

In IB and Edexcel, you might be asked to find the nth derivative of a given function for small n, or to use the second derivative test. For polynomial functions, repeated differentiation reduces the degree until it becomes zero. For example, if f(x) = x⁴ + 2x³, then f'(x) = 4x³ + 6x², f”(x) = 12x² + 12x, f”'(x) = 24x + 12, and f⁽⁴⁾(x) = 24. Exponentials and trigonometric functions produce cyclic patterns: the derivatives of sin x cycle through sin x, cos x, -sin x, -cos x, and back.

在 IB 和 Edexcel 考试中，你可能会被要求求某函数的小 n 阶导数，或运用二阶导数检验法。对于多项式函数，逐次求导会降低次数直至为零。例如，若 f(x) = x⁴ + 2x³，则 f'(x) = 4x³ + 6x²，f”(x) = 12x² + 12x，f”'(x) = 24x + 12，f⁽⁴⁾(x) = 24。指数函数和三角函数则产生循环模式：sin x 的导数依次为 cos x、-sin x、-cos x，又回到 sin x。

Second derivative: d²y/dx² = d/dx (dy/dx) f”(x) = (f'(x))’

10. Applications: Tangents, Normals, and Stationary Points | 应用：切线、法线与驻点

One of the most immediate applications of differentiation is finding slopes of tangents and normals. The gradient of the tangent to y = f(x) at x = a is m = f'(a). The equation of the tangent is y – f(a) = m (x – a). The normal line is perpendicular to the tangent, so its gradient is -1/m, provided m ≠ 0. These lines appear often in coordinate geometry problems linked to differentiation.

微分最直接的应用之一是求切线和法线的斜率。y = f(x) 在 x = a 处切线的斜率为 m = f'(a)。切线方程为 y – f(a) = m (x – a)。法线垂直于切线，因此其斜率为 -1/m（假定 m ≠ 0）。这些直线经常出现在与微分相关的解析几何问题中。

Stationary points occur where f'(x) = 0. To determine their nature (maximum, minimum, or point of inflection), we can use the second derivative test: if f”(a) > 0, the point is a local minimum; if f”(a) < 0, it is a local maximum. If f''(a) = 0, the test is inconclusive, and we must examine the sign change of the first derivative around a. Questions often ask for the coordinates and nature of turning points on cubic or quartic curves. Being systematic—find derivative, set to zero, solve for x, find y, test using f'' or gradient sign—will secure full marks.

驻点出现在 f'(x) = 0 处。要确定其性质（极大值、极小值或拐点），可使用二阶导数检验法：若 f”(a) > 0，该点为局部极小值；若 f”(a) < 0，则为局部极大值。如果 f''(a) = 0，该检验无法确定，必须考察一阶导数在 a 点两侧的符号变化。考题常要求找出三次或四次曲线上转向点的坐标及性质。系统地操作——求导、令其为零、解出 x、求 y、用 f'' 或斜率符号检验——将确保拿到满分。

Tangent: y – y₁ = m (x – x₁) Normal: y – y₁ = -1/m (x – x₁) Stationary: f'(x) = 0

11. Optimization Problems | 优化问题

Optimization is about finding the maximum or minimum value of a quantity under given constraints—a favourite application in both IB and Edexcel, often set in real-world contexts. The approach is to express the quantity to be optimized as a function of a single variable, using the given constraints to eliminate other variables. Then differentiate, find stationary points, and determine whether they give a maximum or minimum, usually by second derivative or by considering endpoints.

优化问题旨在给定约束条件下寻求某个量的最大值或最小值——这是 IB 和 Edexcel 都钟爱的应用题型，常设置在实际情境中。方法是利用给定约束条件消去其他变量，将要优化的量表达为单一变量的函数。然后求导、寻找驻点，并通常通过二阶导数或考虑端点来确定驻点给出最大值还是最小值。

Typical problems include maximising volume of a box made from a cut sheet, minimising surface area for a given volume, or optimising cost functions. For a box with square base x by x and height h, where volume V = x²h is fixed, the surface area A = 2x² + 4xh can be expressed as A(x) = 2x² + 4V/x, and then differentiated. Setting A'(x) = 0 yields the optimal dimensions. Remember to verify the nature of the stationary point (is it a min?) and state the answer in context with units.

典型题型包括最大化从裁剪薄板制成的盒子的体积、在给定体积下最小化表面积，或优化成本函数。对于一个底为 x × x 的正方形、高为 h 的盒子，若体积 V = x²h 固定，表面积 A = 2x² + 4xh 可表示为 A(x) = 2x² + 4V/x，然后求导。令 A'(x) = 0 即可求得最优尺寸。务必验证驻点的性质（是否为最小值？），并在情境中附上单位陈述答案。

Express quantity as f(x), find f'(x) = 0, test with f”(x) or endpoints.

12. Related Rates | 相关变化率

Related rates problems involve finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known. The key is to write an equation linking the variables and then differentiate both sides with respect to time t, using the chain rule implicitly. For a sphere’s volume V = (4/3)πr³, if the radius r changes with time, then dV/dt = 4πr² (dr/dt). Given dr/dt, we find dV/dt.

相关变化率问题涉及通过将某个量与其他变化率已知的量联系起来，求该量的变化率。关键是写出连接各变量的方程，然后对方程两边同时对时间 t 求导，隐式地运用链式法则。对于球体体积 V = (4/3)πr³，若半径 r 随时间变化，则 dV/dt = 4πr² (dr/dt)。给定 dr/dt，即可求得 dV/dt。

Edexcel and IB exams frequently test related rates with geometric contexts: a ladder sliding down a wall, water leaking from a conical tank, or an expanding circle. A ladder of length L leaning against a wall: x² + y² = L². Differentiate: 2x dx/dt + 2y dy/dt = 0. If dx/dt is known, solve for dy/dt. Always identify the given rates and the required rate, write the relationship, differentiate, substitute known values (only after differentiation), and solve. Watch for consistent units and sign conventions.

Edexcel 和 IB 考试经常在几何情境中考查相关变化率：梯子从墙上滑下、水从圆锥形容器泄漏，或一个正在扩张的圆。一根长度为 L 的梯子靠墙放着：x² + y² = L²。求导：2x dx/dt + 2y dy/dt = 0。若已知 dx/dt，即可解出 dy/dt。解题时，始终要识别已知变化率和待求变化率，建立关系式，求导，代入已知数值（仅在求导之后），并求解。注意单位一致和符号约定。

Write equation relating variables → Differentiate w.r.t. t → Substitute and solve.

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