📚 Exercise 13B: Differentiation Techniques | 习题13B：微分技巧
In IB Mathematics, Exercise 13B typically challenges students with a variety of differentiation problems that go beyond simple power rule applications. This set of exercises often covers the product rule, quotient rule, chain rule, and differentiation of trigonometric, exponential, and logarithmic functions. Mastering these techniques is essential for tackling more advanced topics such as kinematics, optimisation, and differential equations. This article provides a comprehensive revision of the key methods and typical question types you will encounter in Exercise 13B.
在IB数学中，习题13B通常通过大量超出简单幂函数求导的题目来考察学生。这组练习通常涵盖乘积法则、商法则、链式法则，以及三角函数、指数函数和对数函数的求导。掌握这些技巧对于处理运动学、最优化和微分方程等更高级的主题至关重要。本文将全面复习你在习题13B中会遇到的核心方法和典型题型。
1. Revision of Basic Derivatives | 基本导数回顾
r rule states d/dx (xⁿ) = n xⁿ⁻¹. You also need to know the derivatives of sin x, cos x, eˣ n x by heart. These form the building blocks of every chain, product, or quotient problem.
在深入复合法则之前，请确保你能够熟练回忆起基本导数公式。对于任意常数c，d/dx (c) = 0。幂函数法则指出 d/dx (xⁿ) = n xⁿ⁻¹。你还需要熟记 sin x、cos x、eˣ 和 ln x 的导数。它们是每个链式、乘积或商法则问题的基石。
- d/dx (x⁵) = 5x⁴
- d/dx (sin x) = cos x
- d/dx (cos x) = -sin x
- d/dx (eˣ) = eˣ
- d/dx (ln x) = 1/x
确保你能将这些基本公式脱口而出，因为后续所有法则都建立在此之上。
2. The Product Rule | 乘积法则
. It is helpful to identify u and v clearly before starting. Remember that the derivative involves two separate terms, and both must be simplified fully.
当你对形如 y = u(x)·v(x) 的函数求导时，请使用乘积法则：dy/dx = u·(dv/dx) + v·(du/dx)。在开始之前明确确定 u 和 v 会很有帮助。请记住导数包含两个独立的项，并且两者都必须完全化简。
in x. Then du/dx = 2x, dv/dx = cos x. Applying the rule gives dy/dx = x² cos x + 2x sin x.
示例：y = x² sin x。设 u = x²，v = sin x。则 du/dx = 2x，dv/dx = cos x。应用法则得到 dy/dx = x² cos x + 2x sin x。
Always check if the final expression can be factored. For instance, extracting a common factor of x yields x(x cos x + 2 sin x), which is often required in IB mark schemes.
请经常检查最终表达式是否可以因式分解。例如，提取公因子 x 得到 x(x cos x + 2 sin x)，这在IB评分方案中经常要求。
3. The Quotient Rule | 商法则
u(x)/v(x), the quotient rule is dy/dx = [v·(du/dx) − u·(dv/dx)] / v². A common mnemonic is 'low d high minus high d low over low squared' Pay special attention to the subtraction order; getting it reversed is a frequent mistake.
对于 y = u(x)/v(x)，商法则为 dy/dx = [v·(du/dx) − u·(dv/dx)] / v²。一个常见的记忆口诀是“分母乘分子的导数减去分子乘分母的导数，再除以分母的平方”。特别注意减法的顺序；顺序颠倒是一个常见错误。
dy/dx = [cos x·2x − (x² + 1)·(-sin x)] / (cos x)² = [2x cos x + (x² + 1) sin x] / cos² x.
示例：y = (x² + 1)/cos x。这里 u = x² + 1，v = cos x。则 du/dx = 2x，dv/dx = -sin x。代入公式得到 dy/dx = [cos x·2x − (x² + 1)·(-sin x)] / (cos x)² = [2x cos x + (x² + 1) sin x] / cos² x。
When simplifying, do not expand the denominator unnecessarily. Leaving it as cos² x is perfectly acceptable, and you can further split terms using trigonometric identities if required.
化简时不要不必要地展开分母。将其保留为 cos² x 是完全可行的，如有需要你还可以利用三角恒等式进一步拆分各项。
4. The Chain Rule | 链式法则
The chain rule handles composite functions y = f(g(x)). It states dy/dx = f'(g(x)) · g'(x). Think of it as ' of it as 'derivative of the outside function with the inside unchanged, multiplied by the derivative of the inside'. Setting out the substitution clearly reduces errors.
链式法则处理复合函数 y = f(g(x))。它指出 dy/dx = f'(g(x)) · g'(x)。可以将其理解为“外部函数的导数（内部不变）乘以内部函数的导数”。清晰地写出代换过程可以减少错误。
Then dy/du = 4u³, du/dx = 6x. By the chain rule, dy/dx = 4(3x² + 5)³ · 6x = 24x(3x² + 5)³.
示例：y = (3x² + 5)⁴。令 u = 3x² + 5，则 y = u⁴。那么 dy/du = 4u³，du/dx = 6x。根据链式法则，dy/dx = 4(3x² + 5)³ · 6x = 24x(3x² + 5)³。
For nested functions like sin(e²ˣ), apply the chain rule twice: derivative of sin is cos, multiplied by derivative of e²ˣ which itself requires a chain rule, giving 2e²ˣ. The result is cos(e²ˣ) · 2e²ˣ.
对于像 sin(e²ˣ) 这样的嵌套函数，需要两次应用链式法则：sin 的导数是 cos，乘以 e²ˣ 的导数，而 e²ˣ 本身也需要链式法则，得到 2e²ˣ。最终结果是 cos(e²ˣ) · 2e²ˣ。
5. Derivatives of Trigonometric Functions | 三角函数的导数
s x differentiates to sec² x. Likewise, sec x = 1/cos x gives sec x tan x, and cot x gives –csc² x.
除 sin 和 cos 之外，IB 期望你能够求 tan x、sec x、csc x 和 cot x 的导数，可以通过记忆或利用商法则自行推导。tan x = sin x / cos x 求导后得到 sec² x。类似地，sec x = 1/cos x 的导数为 sec x tan x，而 cot x 的导数为 –csc² x。
x)·5 = 5 sec²(5x). It is crucial to keep the angle unchanged in the outer derivative part.
这些导数经常出现在链式法则问题中，例如 y = tan(5x)。其导数为 sec²(5x)·5 = 5 sec²(5x)。在外部求导部分必须保持角度不变，这一点至关重要。
6. Derivatives of Exponential and Logarithmic Functions | 指数函数与对数函数的导数
The derivative of eˣ is unique because it remains eˣ. For eᵏˣ, the chain rule gives k eᵏˣ. When the base is not e, use logarithmic differentiation or rewrite aˣ as eˣˡⁿᵃ, leading to d/dx(aˣ) = aˣ ln a.
eˣ 的导数具有独特性，因为它仍然是 eˣ。对于 eᵏˣ，链式法则给出 k eᵏˣ。当底数不是 e 时，可以使用对数求导法或将 aˣ 改写为 eˣˡⁿᵃ，得到 d/dx(aˣ) = aˣ ln a。
g base a, convert to natural log: logₐ x = ln x / ln a, thus its derivative is 1/(x ln a).
自然对数的导数为 1/x，但对于 ln(kx)，链式法则得到 1/(kx) · k = 1/x。注意常数 k 会被约掉，因此 d/dx ln(5x) = 1/x。对于以 a 为底的对数，转换为自然对数：logₐ x = ln x / ln a，因此其导数为 1/(x ln a)。
Combining these with other functions, such as y = x²eˣ, requires the product rule. Proceed systematically: u = x², v = eˣ; then dy/dx = x²eˣ + 2x eˣ = x eˣ (x + 2).
将这些函数与其他函数组合，例如 y = x²eˣ，需要使用乘积法则。系统地处理：u = x²，v = eˣ；则 dy/dx = x²eˣ + 2x eˣ = x eˣ (x + 2)。
7. Higher-Order Derivatives | 高阶导数
Exercise 13B frequently includes finding second derivatives, denoted d²y/dx² or f''(x). Once you obtain the first derivative, simply differentiate again. The second derivative indicates the concavity of a function and is crucial for identifying maxima and minima in the second derivative test.
习题13B经常包含求二阶导数，记作 d²y/dx² 或 f''(x)。一旦求得一阶导数，只需再次求导即可。二阶导数表示函数的凹凸性，对于在二阶导数检验法中确定极大值和极小值至关重要。
ve to zero finds stationary points, while the sign of the second derivative classifies them.
示例：给定 y = x³ − 6x² + 9x，一阶导数为 dy/dx = 3x² − 12x + 9。再次求导得到 d²y/dx² = 6x − 12。令一阶导数为零可以求出驻点，而二阶导数的符号则对它们进行分类。
x/dt), and the second derivative requires d²y/dx² = d/dx(dy/dx) = [d/dt(dy/dx)] / (dx/dt).
对于三角函数，请注意符号：sin x 的二阶导数是 -sin x。对于参数方程，如果 x = f(t) 且 y = g(t)，那么 dy/dx = (dy/dt) / (dx/dt)，而二阶导数则需要 d²y/dx² = d/dx(dy/dx) = [d/dt(dy/dx)] / (dx/dt)。
8. Combining Rules: A Systematic Approach | 法则组合：系统方法
² + 1)¹⁰ cos(3x) requires the product rule and chain rule simultaneously. A structured approach helps: first identify the overall structure (product, quotient, or composite) and then apply the rules from the outside in.
许多习题13B的题目会结合两到三个法则。例如，y = (x² + 1)¹⁰ cos(3x) 需要同时使用乘积法则和链式法则。结构化的方法会有所帮助：首先确定整体结构（乘积、商或复合），然后从外向内应用法则。
3 sin(3x). Then substitute into the product rule. Working step by step prevents confusion.
将函数拆分为 u = (x² + 1)¹⁰ 和 v = cos(3x)。对 u 使用链式法则：du/dx = 10(x² + 1)⁹·2x = 20x(x² + 1)⁹。对 v，dv/dx = -3 sin(3x)。然后代入乘积法则。逐步进行可以避免混乱。
Another common combination is a quotient with an exponential numerator, such as y = e²ˣ / (x + 1). Use quotient rule with u = e²ˣ, v = x + 1. Then du/dx = 2e²ˣ, dv/dx = 1. The result is [(x+1)·2e²ˣ − e²ˣ·1] / (x+1)² = e²ˣ(2x + 1) / (x+1)².
另一种常见的组合是指数函数分子的商，例如 y = e²ˣ / (x + 1)。使用商法则，u = e²ˣ，v = x + 1。那么 du/dx = 2e²ˣ，dv/dx = 1。结果为 [(x+1)·2e²ˣ − e²ˣ·1] / (x+1)² = e²ˣ(2x + 1) / (x+1)²。
9. Applications in Tangents and Normals | 切线与法线中的应用
t the gradient m. The tangent equation is y − y₁ = m(x − x₁). The normal gradient is -1/m.
在给定点求切线或法线方程是习题13B中嵌入的经典IB题型。要求切线，先计算 dy/dx，并在给定的 x 坐标处求值以获得斜率 m。切线方程为 y − y₁ = m(x − x₁)。法线斜率则为 -1/m。
− 1 = 3(x − 1), or y = 3x − 2. The normal has gradient -1/3, giving y − 1 = -1/3 (x − 1).
示例：对于 y = x³ 在 x = 1 处，y = 1。导数为 3x²，因此 m = 3。切线为 y − 1 = 3(x − 1)，即 y = 3x − 2。法线斜率为 -1/3，得到 y − 1 = -1/3 (x − 1)。
itution. Always write the final line in the required form (ax + by + d = 0 or y = mx + c).
当函数包含乘积或复合时，你依然遵循相同的步骤，但导数求值可能需要仔细代换。请始终将最终结果写成要求的形式（ax + by + d = 0 或 y = mx + c）。
10. Common Mistakes and How to Avoid Them | 常见错误与避免方法
One of the most frequent errors is forgetting to multiply by the derivative of the inner function when using the chain rule. Always ask: 'What is inside the main function?' and check whether you have handled it. Another pitfall is misplacing the minus sign in the quotient rule, writing u'v − uv' instead of v u' − u v'.
最常见的错误之一是在使用链式法则时忘记乘以内层函数的导数。请始终自问：“主体函数内部是什么？”并检查你是否处理了它。另一个陷阱是商法则中放置减号顺序错误，写成了 u'v − uv' 而非 v u' − u v'。
Simplification errors are also common. In product rule answers, students often leave expressions like 2x sin x + x² cos x, missing the opportunity to factor out x. IB examiners expect fully simplified answers where possible. Always look for common factors or trigonometric identities.
化简错误也很常见。在乘积法则的答案中，学生常常保留像 2x sin x + x² cos x 这样的表达式，错过了提取 x 的机会。IB阅卷人期望尽可能完全化简的答案。请始终留意公因子或三角恒等式。
Finally, many students confuse eˣ with ln x differentiation. Remember: d/dx eˣ = eˣ whole solution. Writing a quick reference card for tricky derivatives can boost accuracy.
最后，许多学生会混淆 eˣ 和 ln x 的求导。请记住：d/dx eˣ = eˣ，而 d/dx ln x = 1/x。将它们弄混会毁掉整个解答。为容易出错的导数制作一张速查卡可以提高准确性。
11. Practice Tips for Exercise 13B | 习题13B练习技巧
Start by categorising the problems: do all product rule questions first, then move to quotient rule, and finally tackle mixed combinations. This builds confidence with each rule individually before facing the more complex scenarios IB often sets. Time yourself to mimic exam pressure.
首先对题目进行分类：先做完所有乘积法则的题目，然后转向商法则，最后处理混合组合。这样可以在面对IB常设置的更复杂情景之前，分别建立对每种法则的信心。给自己计时以模拟考试压力。
both your derivative formula and a numerical approximation on the original function. This 'sanity check' catches algebraic slips. Also, use a graphing calculator to compare slopes.
解答之后，通过测试一个特定值（例如 x = 0 或 x = 1）来验证你的导数，同时使用你的导数公式和对原函数进行数值近似。这种“合理性检查”能发现代数差错。此外，使用图形计算器比较斜率也是个好方法。
Do not skip word problems that ask for rates of change. These contextualise differentiation, linking back to real-world interpretations such as velocity or population growth. They are a staple of the IB exam and often appear in the back portion of Exercise 13B.
不要跳过要求变化率的应用题。这些问题将微分置于语境中，与速度或人口增长等现实世界诠释相联系。它们是IB考试的重点，经常出现在习题13B的后半部分。
12. Summary | 总结
Exercise 13B is designed to solidify your ability to differentiate products, quotients, and composites involving polynomials, trigonometric functions, exponentials, and logarithms. Through systematic practice, you will internalise the product, quotient, and chain rules, and learn to combine them efficiently. Remember to check for simplification, correct application of signs, and completeness of the chain rule. Mastery of this exercise builds a strong foundation for integration, optimisation, and beyond.
习题13B旨在巩固你对包含多项式、三角函数、指数函数和对数函数的乘积、商及复合函数求导的能力。通过系统练习，你将掌握乘积法则、商法则和链式法则，并学会高效地将它们组合使用。请记住检查化简、正确应用符号以及链式法则的完整性。掌握这一练习将为积分、最优化以及更深入的内容打下坚实基础。
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