📚 IB Math Exercise 10E: Differentiation Applications Explained | IB数学练习10E：微分应用解析
In IB Mathematics, Exercise 10E typically challenges students with problems involving the application of derivatives to find equations of tangents and normals, analyze increasing and decreasing functions, and solve optimization problems. This article provides a comprehensive review of the key concepts and techniques needed to tackle such exercises confidently, blending conceptual understanding with step-by-step worked methods.
在IB数学中，练习10E通常考察学生运用导数求切线和法线方程、分析函数的增减性以及解决最优化问题的能力。本文全面回顾攻克此类练习所需的核心概念与技巧，将概念理解与分步解题方法相结合，助你自信应对考试。
1. Understanding Derivatives as Slopes | 理解导数即斜率
The derivative f' cally represents the slope of the tangent line to the curve at any point. If we evaluate f'(a), we obtain the gradient of the tangent at x = a.
函数 f(x) 的导数 f'(x) 表示 f 对 x 的瞬时变化率，在几何上它代表曲线在任意点处切线的斜率。计算 f'(a) 即可得到曲线在 x = a 处切线的梯度。
δx. For a straight line tangent to the curve, the slope m = dy/dx evaluated at that point.
用莱布尼茨记号表示，导数写作 dy/dx。对于曲线 y = f(x)，微小变化量 δy 除以微小变化量 δx 的极限就是 dy/dx。与曲线相切的直线，其斜率 m 等于该点处的 dy/dx 值。
m_tangent = f'(x₁) = dy/dx|ₓ₌ₓ₁
This simple idea underpins almost all the topics in Exercise 10E. Master it, and you will find the rest much easier.
这个简单的概念支撑着练习10E的几乎所有内容。掌握它，后续内容便会容易许多。
2. Equation of a Tangent Line | 切线方程
e tangent line follows directly from the point-slope form: y - y₁ = m(x - x₁), where m = f'(x₁). Remember, the point must lie on the original curve, so y₁ = f(x₁).
一旦知道曲线上点 (x₁, y₁) 处的斜率，切线方程即可直接由点斜式得出：y - y₁ = m(x - x₁)，其中 m = f'(x₁)。请记住，该点必须在原曲线上，因此 y₁ = f(x₁)。
to find the tangent to f(x) = x³ - 3x at x = 2, first compute f(2) = 2³ - 3×2 = 2. Then f'(x) = 3x² - 3, so f' = 3×4 - 3 = 9. The tangent equation is y - 2 = 9(x - 2), which simplifies to y = 9x - 16.
例如，求 f(x) = x³ - 3x 在 x = 2 处的切线。首先计算 f(2)=2³-6=2，接着求导 f'(x)=3x²-3，得 f'(2)=9。切线方程为 y - 2 = 9(x - 2)，化简得 y = 9x - 16。
mx + c unless asked otherwise. Avoid leaving it as y - y₁ = m(x - x₁) in final responses.
除非另有要求，最终答案务必写成 y = mx + c 的形式，避免将 y - y₁ = m(x - x₁) 作为最终答案。
3. Equation of a Normal Line | 法线方程
The normal line to a curve at a given point is the line perpendicular to the tangent at that point. Its slope mnormal satisfies mnormal × m_tangent = -1, so mnormal = -1/m_tangent, provided m_tangent ≠ 0.
曲线上给定点处的法线是指在该点与切线垂直的直线。其斜率 mnormal 满足 mnormal × m_tangent = -1，因此 mnormal = -1/m_tangent（假设切线斜率非零）。
Using the same point (x₁, y₁), the normal equation is y - y₁ = (-1/f'(x₁))(x - x₁). If the tangent is horizontal (f'(x₁)=0), the normal is vertical with equation x = x₁.
对于同一点 (x₁, y₁)，法线方程为 y - y₁ = (-1/f'(x₁))(x - x₁)。若切线水平（f'(x₁)=0），则法线为竖直线，方程为 x = x₁。
Continuing the previous example where f'(2)=9, the normal slope is -1/9, giving y - 2 = (-1/9)(x - 2) or y = -⅑x + 2⅑ + ⅑? Properly y = -⅑x + 20/9. Double-check arithmetic: y = -1/9 x + (2 + 2/9) = -1/9 x + 20/9.
沿用上例 f'(2)=9，法线斜率为 -1/9，得 y - 2 = (-1/9)(x - 2)，化为斜截式为 y = -1/9 x + 20/9。请仔细核对运算。
IB exam questions often ask for both the tangent and the normal in the same problem, so practice switching between them quickly.
IB 考题常要求在同一问题中同时求出切线和法线，因此需熟练快速切换。
4. Increasing and Decreasing Functions | 函数的递增与递减
f for any two points x₁ < x₂, f(x₁) ≤ f(x₂). But with derivatives we use a simpler test: f'(x) > 0 on an interval implies f is strictly increasing there; f'(x) < 0 implies strictly decreasing.
若在区间内对于任意 x₁ < x₂ 都有 f(x₁) ≤ f(x₂)，则称函数 f 递增。但借助导数我们使用更简洁的判别法：区间内 f'(x) > 0 意味着函数严格递增；f'(x) < 0 意味着严格递减。
To find intervals of increase/decrease, first locate the derivative f'(x), then solve f'(x) = 0 to find critical numbers. Use a sign chart to determine where f'(x) is positive or negative.
要寻找增减区间，首先求出导数 f'(x)，解方程 f'(x)=0 找到临界数。然后用符号表确定 f'(x) 在哪些区间为正或为负。
Example: f(x) = x³ - 3x + 1 has f'(x) = 3x² - 3 = 3(x - 1)(x + 1). Critical points at x = -1 and x = 1. For x < -1, f'(x) > 0 (increasing); between -1 and 1, f'(x) < 0 (decreasing); for x > 1, f'(x) > 0 (increasing again).
例如：f(x)=x³-3x+1，f'(x)=3x²-3=3(x-1)(x+1)，临界点为 x=-1 和 x=1。当 x < -1 时 f'(x)>0（递增）；-1 < x < 1 时 f'(x)<0（递减）；x > 1 时 f'(x)>0（再次递增）。
Always express intervals using correct IB notation: open intervals (a, b) or inequalities, and remember to state “strictly increasing/decreasing” when f' ≠ 0.
请始终使用正确的 IB 表示法写出区间：开区间 (a, b) 或不等式形式，并注意当 f' ≠ 0 时应表述为“严格递增/递减”。
5. Critical Points and Stationary Points | 临界点与驻点
A critical point (or critical number) is an x-value in the domain of f where f'(x) = 0 or f'(x) does not exist. A stationary point is a specific type of critical point where f'(x) = 0, meaning the tangent is horizontal.
临界点（或临界数）是函数定义域内满足 f'(x)=0 或 f'(x) 不存在的 x 值。驻点是临界点的一种特殊情形，满足 f'(x)=0，即切线水平。
At a stationary point, the function could have a local maximum, a local minimum, or a horizontal point of inflection. Don't assume it's a turning point until you classify it.
在驻点处，函数可能存在局部最大值、局部最小值或水平拐点。在未分类之前，不要贸然断定它是一个极值点。
To find stationary points, solve f' substitute back into f(x) to get the y-coordinate. For example, f(x) = 2x³ - 9x² + 12x: f'(x) = 6x² - 18x + 12 = 6(x - 1)(x - 2). Stationary points at (1, f(1)) and (2, f(2)).
要找出驻点，解 f'(x)=0 得 x，代入 f(x) 得到 y 坐标。例如 f(x)=2x³-9x²+12x，f'(x)=6x²-18x+12=6(x-1)(x-2)，驻点为 (1, f(1)) 和 (2, f(2))。
6. Classifying Stationary Points: First Derivative Test | 判断驻点：一阶导数检验
The first derivative test examines the sign of f'(x) on either side of a critical number c. If f' changes from positive to negative at c, then f has a local maximum at c. If f' changes from negative to positive, it's a local minimum. If there is no sign change, then c is a point of inflection with horizontal tangent.
一阶导数检验观察临界数 c 左右两侧 f'(x) 的符号。若 f' 在 c 处由正变负，则 f 在 c 处有局部最大值；若由负变正，则为局部最小值；若符号不变，则 c 为具有水平切线的拐点。
Using the previous example f'(x) = 6(x - 1)(x - 2). At x = 1, for x slightly less than 1, f' (since (x-1) negative, (x-2) negative, product positive); for x slightly greater than 1, f' negative gives a local maximum at x=1. At x=2, negative → positive yields a local minimum.
沿用上例 f'(x)=6(x-1)(x-2)。在 x=1 处，当 x 略小于1时，f'(x) 为正（(x-1)负，(x-2)负，乘积正）；略大于1时 f'(x) 为负。因此由正变负，x=1 处为局部最大值。在 x=2 处，由负变正，为局部最小值。
This method is reliable even when the second derivative is zero or difficult to compute. Always show your sign sketch or table.
即使二阶导数为零或难以计算，此方法仍然可靠。请务必画出符号草图或表格。
7. Classifying Stationary Points: Second Derivative Test | 判断驻点：二阶导数检验
If f'(c) = 0 and f''(c) exists, then the second derivative test gives a quick classification: if f''(c) > 0, the function is concave up and has a local minimum at c. If f''(c) < 0, it is concave down and has a local maximum. If f''(c) = 0, the test is inconclusive; revert to the first derivative test.
若 f'(c)=0 且 f''(c) 存在，则二阶导数检验可快速分类：若 f''(c) > 0，函数上凹，在 c 处取得局部最小值；若 f''(c) < 0，函数下凹，取得局部最大值；若 f''(c)=0，则此法无法判断，应回到一阶导数检验。
For f(x) = 2x³ - 9x² + 12x, f''(x) = 12x - 18. At x=1, f''(1) = -6 < 0 → local maximum. At x=2, f''(2) = 6 > 0 → local minimum. This matches the first derivative test.
对于 f(x)=2x³-9x²+12x，f''(x)=12x-18。在 x=1 处，f''(1)=-6<0 → 局部最大值；在 x=2 处，f''(2)=6>0 → 局部最小值。这与一阶导数检验结果一致。
Use the second derivative test as a time-saver, but be cautious when f''(c)=0. IB marks often require a justification, so write “f''(c) < 0, hence local maximum” explicitly.
利用二阶导数检验可节省时间，但当 f''(c)=0 时需谨慎。IB 评分通常要求给出理由，所以应明确写出“f''(c) < 0，故为局部最大值”。
8. Concavity and Points of Inflection | 凹性与拐点
A curve is concave up (convex) where f''(x) > 0, and concave down (concave) where f'' ction occurs where the concavity changes sign, and at a non-stationary inflection point, f''(x)=0 but f'(x) may not be zero.
曲线在 f''(x) > 0 处上凹（凸），在 f''(x) < 0 处下凹（凹）。拐点出现在凹性改变符号的位置，在非驻点拐点处，f''(x)=0 但 f'(x) 可以不为零。
To find inflection points, set f''(x) = 0 and test for sign change. For example, f(x) = x³ - 3x²: f'(x)=3x²-6x, f''(x)=6x-6. Setting f''(x)=0 gives x=1. Test x<1 (f'' negative) and x>1 (f'' positive), so concavity changes and x=1 is an inflection point.
寻找拐点时，令 f''(x)=0 并检验符号变化。例如 f(x)=x³-3x²，f'(x)=3x²-6x，f''(x)=6x-6。令 f''(x)=0 得 x=1。检验 x<1（f'' 为负）和 x>1（f'' 为正），凹性改变，故 x=1 为拐点。
IB questions often ask you to “determine the nature of the stationary points and find any points of inflection.” Keep these procedures distinct in your mind.
IB 考题常要求“判断驻点的性质并求出所有拐点”。请将这些步骤在心中清晰区分。
9. Optimization Problems: Setting Up | 最优化问题：建立模型
Optimization problems ask for the maximum or minimum value of a quantity under given constraints. The first step is to identify the variable to optimize (e.g., area, volume, cost) and express it as a function of a single variable using the constraint.
最优化问题要求在给定约束下求某个量的最大值或最小值。第一步是确定待优化变量（如面积、体积、成本），并利用约束条件将其表示为单一变量的函数。
Typical steps: draw a diagram if applicable, introduce variables, write the quantity Q to be optimized as a function of one variable x, determine the domain of x (often physical bounds), then differentiate.
典型步骤：如适用，先画示意图；引入变量；将待优化量 Q 表示为单一变量 x 的函数；确定 x 的取值范围（通常受实际意义限制）；然后求导。
x. Area A = x(10 - x) = 10x - x² with domain 0 < x < 10. Now we can find the maximum area.
例如，周长为 20 米的矩形：设长为 x，宽为 10-x。面积 A = x(10-x) = 10x - x²，定义域 0 < x < 10。现在可求最大面积。
10. Optimization Problems: Solving and Verifying | 最优化问题：求解与验证
After obtaining A(x) = 10x - x², differentiate: A'(x) = 10 - 2x. Set A'(x) = 0 → x = 5. Check second derivative A'' m area = 5(5) = 25 m². Always verify that the solution lies in the domain and makes sense.
得到 A(x)=10x-x² 后，求导：A'(x)=10-2x。令其为零得 x=5。检查二阶导数 A''(x)=-2<0，确认为最大值。最大面积 = 5×5 = 25 平方米。请务必验证解落在定义域内且符合实际。
r use chain rule. Always simplify the function before differentiating if possible. And don't forget to answer the original question: state the maximum area, not just x=5.
在更复杂的问题中，你可能需要求导乘积、商或使用链式法则。如有可能，求导前先简化函数。切勿忘记回答原始问题：陈述最大面积为多少，而不仅仅是 x=5。
Common pitfalls: confusing the constraint with the objective, forgetting to check endpoints (especially if the stationary point gives a minimum but maximum occurs at endpoint), and poor algebraic simplification.
常见误区：混淆约束条件与目标函数；忘记检查端点（特别是当驻点给出最小值而最大值出现在端点时）；代数化简不当。
11. Common Exam Mistakes and How to Avoid Them | 常见考试错误与避免方法
- Mistake: Forgetting the derivative rule for negative powers. Fix: Rewrite 1/x³ as x⁻³ and use power rule.
- 错误：忽略负指数求导法则。对策：将 1/x³ 改写为 x⁻³，再使用幂函数求导法则。
- Mistake: Using second derivative test when f''(c)=0 and concluding it's an inflection point. Fix: Use first derivative test or sign of f'' around c.
- 错误：当 f''(c)=0 时使用二阶导数检验并直接断定为拐点。对策：使用一阶导数检验或分析 c 附近 f'' 的符号。
- Mistake: Setting derivative equal to zero for increasing/decreasing intervals without sign test. Fix: Always test intervals between critical points.
- 错误：仅仅令导数为零来判断增减区间，未进行符号检验。对策：务必检验临界点之间的区间符号。
- Mistake: Not simplifying optimization functions before differentiating, leading to algebraic errors.
- 错误：求导前未简化优化目标函数，导致代数错误。
Carefully read Exercise 10E questions; they often embed these traps intentionally. Slow down and check your work.
仔细阅读练习10E的题目，它们常有意设置这些陷阱。放慢速度并检查解答。
12. Connecting Concepts to IB Exam Success | 将概念与IB考试成功串联起来
Exercise 10E is not just a set of isolated drills; it builds the foundation for Paper 1 and Paper 2 long questions that integrate graphing, calculus, and real-world contexts. Mastery here helps you tackle problems like “Find the value of x that minimizes the total surface area for a fixed volume.”
练习10E并非一组孤立的训练，它为试卷一和试卷二中结合图像、微积分与实际情境的大题奠定基础。掌握这些内容能帮助你解决诸如“求使固定体积下总表面积最小的 x 值”等问题。
When revising, attempt problems without looking at solutions first. Then check your working step by step. Write out the derivative, critical points, sign diagram, and final answer neatly. IB examiners reward clear, logical structure.
复习时，先尝试不看答案独立解题，然后逐步核对过程。清晰写出导数、临界点、符号图和最终答案。IB 考官青睐条理分明、逻辑严密的解答。
Finally, practice with a timer. Many Exercise 10E variations appear under time pressure, so build speed in finding tangents, normals, and optimizing efficiently.
最后，限时练习。许多练习10E的变体在时间压力下出现，因此要加快求切线、法线和求解最优化的速度。
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