IB Mathematics: Complete Guide to Applications of Derivatives | IB数学：导数的应用全解析

📚 IB Mathematics: Complete Guide to Applications of Derivatives | IB数学：导数的应用全解析

Derivatives are far more than a set of algebraic rules for finding slopes. In IB Mathematics, understanding the applications of derivatives is essential for solving real-world optimisation problems, analysing motion, and interpreting the behaviour of functions. This guide systematically covers every major application you need to master for both Analysis and Approaches (AA) and Applications and Interpretation (AI) courses.

导数远不止是一套求斜率的代数规则。在IB数学中，理解导数的应用是解决现实世界优化问题、分析运动过程以及解读函数行为的关键。本指南系统性地涵盖你在数学分析与方法（AA）以及数学应用与解释（AI）课程中需要掌握的所有主要应用。

1. Tangent and Normal Lines | 切线与法线

The derivative of a function at a point gives the slope of the tangent line at that point. To write the equation of a tangent, simply use the point-slope form: y – f(a) = f'(a)(x – a). The normal line is perpendicular to the tangent, so its slope is the negative reciprocal of the derivative: -1 / f'(a), provided f'(a) ≠ 0.

函数在某一点的导数给出该点切线的斜率。要写出切线方程，只需使用点斜式：y – f(a) = f'(a)(x – a)。法线与切线垂直，因此其斜率为导数的负倒数：-1 / f'(a)，前提是 f'(a) ≠ 0。

For example, if f(x) = x³ and a = 1, then f'(x) = 3x², so f'(1) = 3. The tangent equation is y – 1 = 3(x – 1), which simplifies to y = 3x – 2. The normal has slope -1/3, giving y – 1 = (-1/3)(x – 1).

例如，若 f(x) = x³ 且 a = 1，则 f'(x) = 3x²，所以 f'(1) = 3。切线方程为 y – 1 = 3(x – 1)，化简得 y = 3x – 2。法线斜率为 -1/3，得 y – 1 = (-1/3)(x – 1)。

Always check that the point actually lies on the curve before writing the tangent equation.
Remember: for a horizontal tangent, set f'(x) = 0 and solve for x.

在写出切线方程前，务必确认该点确实在曲线上。
记住：水平切线只需令 f'(x) = 0 并解出 x 即可。

2. Increasing and Decreasing Functions | 函数的增减性

The sign of the first derivative determines whether a function is increasing or decreasing on an interval. If f'(x) > 0 for all x in an interval, then f is strictly increasing there. If f'(x) < 0, then f is strictly decreasing. This test is fundamental for sketching curves and locating extrema.

一阶导数的符号决定了函数在某区间上是递增还是递减。如果区间内所有 x 都满足 f'(x) > 0，则 f 在该区间上严格递增；如果 f'(x) < 0，则 f 严格递减。这一判定方法是绘制函数图像和定位极值的基础。

Consider g(x) = x² – 4x + 3. Its derivative is g'(x) = 2x – 4. Setting 2x – 4 > 0 gives x > 2, so g is increasing on (2, ∞) and decreasing on (-∞, 2). The critical point at x = 2 corresponds to a minimum.

考虑 g(x) = x² – 4x + 3。其导数为 g'(x) = 2x – 4。令 2x – 4 > 0 得 x > 2，因此 g 在 (2, ∞) 上递增，在 (-∞, 2) 上递减。x = 2 处的临界点对应一个极小值。

Sign of f'(x)	Behaviour of f(x)	图形行为
Positive / 正	Increasing / 递增	Curve rises left to right / 曲线从左到右上升
Negative / 负	Decreasing / 递减	Curve falls left to right / 曲线从左到右下降

3. Stationary Points and Local Extrema | 驻点与局部极值

A stationary point occurs where f'(x) = 0. At such a point, the tangent is horizontal. To classify a stationary point as a local maximum, local minimum, or point of inflection, we use either the first derivative test or the second derivative test.

驻点出现在 f'(x) = 0 处。在驻点处，切线是水平的。要对驻点进行分类（局部极大值、局部极小值或拐点），我们可以使用一阶导数判定法或二阶导数判定法。

For a twice-differentiable function f, if f'(c) = 0 and f”(c) > 0, then c is a local minimum. If f”(c) < 0, then c is a local maximum. If f”(c) = 0, the test is inconclusive and you must examine the sign of f’ on both sides of c.

对于二阶可导函数 f，若 f'(c) = 0 且 f”(c) > 0，则 c 是局部极小值点；若 f”(c) < 0，则 c 是局部极大值点。若 f”(c) = 0，则判定法失效，必须考察 c 两侧 f’ 的符号变化。

h(x) = 2x³ – 9x² + 12x + 1, h'(x) = 6x² – 18x + 12 = 6(x – 1)(x – 2)

Setting h'(x) = 0 gives x = 1 and x = 2. The second derivative is h”(x) = 12x – 18. At x = 1, h”(1) = -6 < 0, so (1, 6) is a local maximum. At x = 2, h”(2) = 6 > 0, so (2, 5) is a local minimum.

令 h'(x) = 0 得 x = 1 和 x = 2。二阶导数为 h”(x) = 12x – 18。在 x = 1 处，h”(1) = -6 < 0，所以 (1, 6) 是局部极大值；在 x = 2 处，h”(2) = 6 > 0，所以 (2, 5) 是局部极小值。

4. Interval of Concavity and Points of Inflection | 凹凸区间与拐点

Concavity describes the curvature of a graph. If f”(x) > 0 on an interval, the graph is concave up (shaped like a cup ∪). If f”(x) < 0, the graph is concave down (shaped like a cap ∩). A point of inflection is a point where the concavity changes, which requires f”(x) = 0 and a sign change in f”(x) across that point.

凹凸性描述的是图像的弯曲方向。若某区间上 f”(x) > 0，则图像是凹向上的（形如杯子 ∪）；若 f”(x) < 0，则图像是凹向下的（形如帽子 ∩）。拐点是凹凸性发生改变的点，要求 f”(x) = 0 且 f”(x) 在该点两侧符号发生变化。

For k(x) = x⁴ – 6x², we have k”(x) = 12x² – 12 = 12(x² – 1). Setting k”(x) = 0 gives x = ±1. Since k”(x) changes sign at both x = -1 and x = 1, both points are inflection points.

对于 k(x) = x⁴ – 6x²，有 k”(x) = 12x² – 12 = 12(x² – 1)。令 k”(x) = 0 得 x = ±1。由于 k”(x) 在 x = -1 和 x = 1 两处都改变符号，因此这两个点都是拐点。

An inflection point does not require f'(x) = 0; it only requires a change in concavity.
Always verify a sign change in f”(x), not merely f”(x) = 0.

拐点不要求 f'(x) = 0，只要求凹凸性改变。
务必验证 f”(x) 的符号变化，而不仅仅是 f”(x) = 0。

5. Optimisation Problems | 最优化问题

Optimisation is one of the most important applications of derivatives in IB Mathematics. The goal is to find the maximum or minimum value of a quantity subject to certain constraints. The standard approach is: identify the quantity to optimise, express it as a function of a single variable, find critical points by setting the derivative to zero, and then confirm with the second derivative test or by evaluating endpoints.

最优化是IB数学中导数最重要的应用之一。目标是在给定约束条件下求某个量的最大值或最小值。标准步骤是：确定要优化的量，将其表示为单变量函数，令导数为零求临界点，然后用二阶导数判定法或通过比较端点来确认。

A farmer has 200 metres of fencing and wants to enclose a rectangular field with maximum area. Let the width be x and the length be 100 – x (since perimeter 2x + 2y = 200). The area is A(x) = x(100 – x) = 100x – x². Setting A'(x) = 100 – 2x = 0 gives x = 50. Since A”(x) = -2 < 0, this is a maximum. The maximum area is 50 × 50 = 2500 m².

一位农民有200米围栏，想围出一个面积最大的矩形田地。设宽为 x，长为 100 – x（因为周长 2x + 2y = 200）。面积为 A(x) = x(100 – x) = 100x – x²。令 A'(x) = 100 – 2x = 0 得 x = 50。由于 A”(x) = -2 < 0，这是最大值。最大面积为 50 × 50 = 2500 平方米。

Methods: ① find the function ② differentiate ③ set f'(x) = 0 ④ verify ⑤ interpret the result

方法：① 建立函数 ② 求导 ③ 令 f'(x) = 0 ④ 验证 ⑤ 解释结果

6. Related Rates | 相关变化率

Related rates problems involve finding the rate at which one quantity changes based on the rate of change of another related quantity. The chain rule is the core tool: if z depends on y and y depends on t, then dz/dt = (dz/dy) × (dy/dt). These problems are common in geometry, physics, and engineering contexts.

相关变化率问题涉及根据一个相关量的变化率来求另一个量的变化率。链式法则是核心工具：若 z 依赖于 y，而 y 依赖于 t，则 dz/dt = (dz/dy) × (dy/dt)。这类问题常见于几何、物理和工程背景中。

A balloon is being inflated at a rate of 10 cm³/s. Find how fast the radius is increasing when the radius is 5 cm. For a sphere, V = (4/3)πr³, so dV/dr = 4πr². Using the chain rule: dV/dt = (dV/dr) × (dr/dt), hence 10 = 4π(5)² × dr/dt, giving dr/dt = 10/(100π) = 1/(10π) cm/s.

一个气球以 10 cm³/s 的速率被充气。求当半径为 5 cm 时半径增加的速率。对于球体，V = (4/3)πr³，所以 dV/dr = 4πr²。利用链式法则：dV/dt = (dV/dr) × (dr/dt)，因此 10 = 4π(5)² × dr/dt，得 dr/dt = 10/(100π) = 1/(10π) cm/s。

Draw a diagram whenever possible to visualise the relationship between variables.
Identify all given rates and the rate you need to find before differentiating.

尽可能画图以直观理解变量之间的关系。
在求导之前，先明确所有已知变化率和需要求的变化率。

7. Kinematics: Position, Velocity, Acceleration | 运动学：位置、速度与加速度

In kinematics, the derivative of displacement s(t) with respect to time gives velocity v(t) = s'(t), and the derivative of velocity gives acceleration a(t) = v'(t) = s”(t). Speed is the absolute value of velocity. This application connects calculus directly to physics and is a frequent IB exam topic.

在运动学中，位移 s(t) 对时间的导数给出速度 v(t) = s'(t)，速度的导数给出加速度 a(t) = v'(t) = s”(t)。速率是速度的绝对值。这一应用将微积分与物理直接联系，是IB考试中的常考主题。

Suppose s(t) = t³ – 6t² + 9t (metres, t in seconds). Then v(t) = 3t² – 12t + 9 and a(t) = 6t – 12. Setting v(t) = 0 gives t = 1 or t = 3, meaning the particle changes direction at these instants. The acceleration at t = 1 is a(1) = -6 m/s², while at t = 3 it is a(3) = 6 m/s².

设 s(t) = t³ – 6t² + 9t（单位米，t 以秒计）。则 v(t) = 3t² – 12t + 9，a(t) = 6t – 12。令 v(t) = 0 得 t = 1 或 t = 3，意味着粒子在这些时刻改变运动方向。t = 1 时的加速度为 a(1) = -6 m/s²，而 t = 3 时为 a(3) = 6 m/s²。

Quantity / 物理量	Formula / 公式
Displacement / 位移	s(t)
Velocity / 速度	v(t) = s'(t)
Acceleration / 加速度	a(t) = v'(t) = s”(t)

8. Curve Sketching | 函数绘图

Derivatives provide all the information needed to sketch an accurate graph. The first derivative reveals intervals of increase and decrease, while the second derivative reveals concavity. Combined with x-intercepts, y-intercepts, asymptotes, and stationary points, you can produce a complete sketch without a graphing calculator.

导数为绘制精确函数图像提供了全部所需信息。一阶导数显示增减区间，二阶导数显示凹凸性。结合 x 截距、y 截距、渐近线和驻点，你可以在没有图形计算器的情况下绘制出完整的图像。

To sketch f(x) = x³ – 3x² + 2:

要绘制 f(x) = x³ – 3x² + 2 的图像：

Find where the curve crosses the axes: solve f(x) = 0 for x-intercepts.
Find critical points: f'(x) = 3x² – 6x = 3x(x – 2), so x = 0 and x = 2.
Classify them: f”(x) = 6x – 6. At x = 0, f”(0) = -6, so a maximum at (0, 2). At x = 2, f”(2) = 6, so a minimum at (2, -2).
Check the sign of f'(x) to confirm the shape between critical points.

求曲线与坐标轴的交点：解 f(x) = 0 得到 x 截距。
求临界点：f'(x) = 3x² – 6x = 3x(x – 2)，所以 x = 0 和 x = 2。
分类：f”(x) = 6x – 6。在 x = 0 处，f”(0) = -6，因此 (0, 2) 为极大值点；在 x = 2 处，f”(2) = 6，因此 (2, -2) 为极小值点。
检查 f'(x) 的符号以确认临界点之间的图像形状。

9. Marginal Analysis in Economics | 经济学中的边际分析

In economics, derivatives represent marginal quantities. Marginal cost is the derivative of the total cost function C'(x), marginal revenue is R'(x), and marginal profit is P'(x). Profit is maximised when marginal revenue equals marginal cost, i.e., R'(x) = C'(x), provided the second derivative confirms a maximum.

在经济学中，导数代表边际量。边际成本是总成本函数的导数 C'(x)，边际收益是 R'(x)，边际利润是 P'(x)。当边际收益等于边际成本时，即 R'(x) = C'(x)，利润达到最大化，前提是二阶导数确认该点为最大值。

If a company has cost C(x) = 500 + 20x + 0.5x² and revenue R(x) = 100x – x², then marginal cost is C'(x) = 20 + x and marginal revenue is R'(x) = 100 – 2x. Setting 20 + x = 100 – 2x gives x = 80/3 ≈ 26.67 units. The profit function P(x) = R(x) – C(x) = -1.5x² + 80x – 500, and P”(x) = -3 < 0, confirming a maximum.

若一家公司的成本为 C(x) = 500 + 20x + 0.5x²，收益为 R(x) = 100x – x²，则边际成本为 C'(x) = 20 + x，边际收益为 R'(x) = 100 – 2x。令 20 + x = 100 – 2x 得 x = 80/3 ≈ 26.67 个单位。利润函数 P(x) = R(x) – C(x) = -1.5x² + 80x – 500，且 P”(x) = -3 < 0，确认该点为最大值。

10. L’Hôpital’s Rule and Limits | 洛必达法则与极限

Although strictly speaking L’Hôpital’s Rule belongs to limits, it is included in the applications of derivatives because it uses differentiation to evaluate indeterminate forms such as 0/0 or ∞/∞. The rule states that if lim f(x)/g(x) is indeterminate and both derivatives exist, then lim f(x)/g(x) = lim f'(x)/g'(x).

严格来说洛必达法则属于极限范畴，但它也被纳入导数的应用，因为它利用微分来求 0/0 或 ∞/∞ 等不定式的极限。该法则指出：若 lim f(x)/g(x) 是不定式且两个导数都存在，则 lim f(x)/g(x) = lim f'(x)/g'(x)。

Consider lim (x → 0) sin(x)/x. Direct substitution gives 0/0, so we apply L’Hôpital’s Rule: differentiating numerator and denominator gives lim (x → 0) cos(x)/1 = 1. This elegant result is a cornerstone of calculus.

考虑 lim (x → 0) sin(x)/x。直接代入得到 0/0，因此应用洛必达法则：对分子分母分别求导得 lim (x → 0) cos(x)/1 = 1。这一优雅结果是微积分的基石。

Only apply L’Hôpital’s Rule when an indeterminate form truly exists.
If the resulting limit is still indeterminate, the rule may be applied repeatedly.

只有在真正存在不定式时才可应用洛必达法则。
如果一次应用后仍为不定式，可重复使用该法则。

11. Finding Maximum and Minimum Values on a Closed Interval | 闭区间上的最大值与最小值

The Extreme Value Theorem guarantees that a continuous function attains both a maximum and a minimum on a closed interval [a, b]. To find these absolute extrema, evaluate the function at all critical points within the interval and at the endpoints a and b, then compare the values.

极值定理保证连续函数在闭区间 [a, b] 上必能同时取得最大值和最小值。要找到这些绝对极值，需计算函数在区间内所有临界点以及端点 a 和 b 处的函数值，然后比较大小。

For f(x) = x³ – 3x on [-2, 3], the derivative f'(x) = 3x² – 3 = 3(x – 1)(x + 1), giving critical points x = ±1, both within the interval. Evaluating: f(-2) = -2, f(-1) = 2, f(1) = -2, f(3) = 18. Therefore, the absolute maximum is 18 at x = 3, and the absolute minimum is -2, attained at x = -2 and x = 1.

对于 f(x) = x³ – 3x 在 [-2, 3] 上，导数 f'(x) = 3x² – 3 = 3(x – 1)(x + 1)，临界点为 x = ±1，均落在区间内。计算函数值：f(-2) = -2，f(-1) = 2，f(1) = -2，f(3) = 18。因此绝对最大值为 x = 3 处的 18，绝对最小值为 -2，在 x = -2 和 x = 1 处取得。

Steps: ① find f'(x) ② solve f'(x) = 0 ③ compute f at critical points and endpoints ④ compare

步骤：① 求 f'(x) ② 解 f'(x) = 0 ③ 计算临界点和端点的 f 值 ④ 比较

12. Differentiability and Continuity | 可导性与连续性

A solid understanding of the relationship between differentiability and continuity is essential. If a function is differentiable at a point, it must be continuous there. However, a continuous function is not necessarily differentiable; a classic example is f(x) = |x| at x = 0, which is continuous but has a sharp corner, so the derivative does not exist.

深入理解可导性与连续性的关系至关重要。若函数在某点可导，则它在该点必然连续。但连续函数不一定可导；一个典型例子是 f(x) = |x| 在 x = 0 处，该函数连续但有一个尖角，因此导数不存在。

In IB questions, you may be asked to identify points where a piecewise function is not differentiable, such as where two pieces meet with different slopes. Computing the left-hand and right-hand derivatives separately and checking whether they are equal is the standard method.

在IB题目中，你可能会被要求找出分段函数不可导的点，例如两段函数以不同斜率相接的位置。分别计算左导数和右导数并检查它们是否相等是标准方法。

Continuity is a necessary but not sufficient condition for differentiability.
A vertical tangent or a cusp also implies that the derivative does not exist at that point.

连续性是可导性的必要条件，但不是充分条件。
垂直切线或尖点也意味着该点导数不存在。

Mastering the applications of derivatives transforms calculus from abstract rules into a powerful toolkit for solving problems across mathematics, physics, economics, and engineering. Regular practice with optimisation, related rates, and curve sketching will build the confidence needed for IB exam success. Remember to always interpret your mathematical results in the context of the original problem.

掌握导数的应用将微积分从抽象规则转变为解决数学、物理、经济和工程问题的强大工具。通过定期练习最优化、相关变化率和函数绘图，你将为IB考试成功建立信心。切记始终在原始问题的背景下解释你的数学结果。

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

IB Mathematics: Complete Guide to Applications of Derivatives | IB数学：导数的应用全解析