📚 Exercise 6B: Differentiation Rules | 练习6B：微分法则
In IB Mathematics, Exercise 6B typically focuses on applying the product, quotient and chain rules to differentiate more complex functions. This article revisits those core rules step by step, providing worked examples and common error analysis. Studying these techniques thoroughly will sharpen your calculus skills and prepare you for Paper 1 and Paper 2 questions where analytic differentiation is required.
在IB数学中，练习6B通常集中于应用乘积法则、商法则和链式法则对较复杂的函数求导。本文逐步重温这些核心法则，并提供详细的例题与常见错误分析。透彻掌握这些技巧将提升你的微积分能力，为需要解析求导的试卷一和试卷二问题做好准备。
1. Quick Recap of Basic Derivatives | 基础导数速览
d trigonometric functions. For any constant c and real n, d/dx (xⁿ) = n xⁿ⁻¹. Also d/dx (eˣ) = eˣ, d/dx (ln x) = 1/x, d/dx (sin x) = cos x, and d/dx (cos x) = –sin x. These building blocks will be combined in Exercise 6B.
在深入法则之前，请确保你能即时回忆幂函数、指数函数、对数函数和三角函数的导数。对于任意常数c和实数n，d/dx (xⁿ) = n xⁿ⁻¹。还有 d/dx (eˣ) = eˣ，d/dx (ln x) = 1/x，d/dx (sin x) = cos x，以及 d/dx (cos x) = –sin x。这些基本模块将在练习6B中组合使用。
2. The Product Rule Statement | 乘积法则的表述
ferentiable functions, the derivative of their product is given by: d/dx [u(x) · v(x)] = u'(x) v(x) + u(x) v'(x). It is often memorised as “first times derivative of second plus second times derivative of first”. Always identify u and v clearly before differentiating.
如果 u(x) 和 v(x) 是可导函数，它们乘积的导数由下式给出：d/dx [u(x) · v(x)] = u'(x) v(x) + u(x) v'(x)。通常记作“前导后不导加后导前不导”。在求导前务必清晰确定 u 和 v。
3. Product Rule Worked Example | 乘积法则示例
Differentiate f(x) = x² sin x. Let u = x², so u' = 2x. Let v = sin x, so v' = cos x. Applying the product rule: f ' s x) = 2x sin x + x² cos x. No further simplification is needed, but you could factor x: f '(x) = x(2 sin x + x cos x).
求 f(x) = x² sin x 的导数。令 u = x²，则 u' = 2x。令 v = sin x，则 v' = cos x。应用乘积法则：f '(x) = (2x)(sin x) + (x²)(cos x) = 2x sin x + x² cos x。无需进一步化简，但也可提取公因子 x：f '(x) = x(2 sin x + x cos x)。
4. The Quotient Rule Statement | 商法则的表述
ns u(x) and v(x) with v(x) ≠ 0, the derivative of their quotient is: d/dx [u(x)/v(x)] = [u'(x) v(x) – u(x) v' minus sign, and the denominator is the square of the original denominator. Order matters: “low d-high minus high d-low over low squared”.
对于可导函数 u(x) 和 v(x)（v(x) ≠ 0），它们商的导数为：d/dx [u(x)/v(x)] = [u'(x) v(x) – u(x) v'(x)] / [v(x)]²。分子类似于乘积法则但带减号，分母为原分母的平方。顺序很重要：“下导上减上导下，除以下的平方”。
5. Quotient Rule Worked Example | 商法则示例
Find the derivative of g(x) = eˣ / (x + 1). Set u = eˣ, u' = eˣ; v = x + 1, v' = 1. Then g'(x) = [eˣ (x + 1) – eˣ (1)] / (x + 1)² = eˣ (x + 1 – 1) / (x + 1)² = x eˣ / (x + 1)². Always check if the numerator simplifies; here the 1 cancels neatly.
求 g(x) = eˣ / (x + 1) 的导数。设 u = eˣ，u' = eˣ；v = x + 1，v' = 1。则 g'(x) = [eˣ (x + 1) – eˣ (1)] / (x + 1)² = eˣ (x + 1 – 1) / (x + 1)² = x eˣ / (x + 1)²。始终检查分子是否可以化简；这里1被优雅地抵消了。
6. The Chain Rule Concept | 链式法则的概念
In Leibniz notation it looks like fractions cancelling. Alternatively, d/dx [f(g(x))] = f '(g(x)) · g'(x). Think “derivative of outer function evaluated at inner, times derivative of inner”.
链式法则用于求复合函数 f(g(x)) 的导数。如果 y = f(u) 且 u = g(x)，那么 dy/dx = dy/du · du/dx。在莱布尼茨记号中这看起来像是分数约分。或者写作 d/dx [f(g(x))] = f '(g(x)) · g'(x)。可记为“外函数在内函数处的导数，乘以内函数的导数”。
7. Chain Rule Worked Examples | 链式法则示例
fferentiate h(x) = (3x² + 1)⁵. Outer: u⁵ gives 5u⁴; inner: 3x²+1 gives derivative 6x. So h' x). Outer: ln u gives 1/u; inner: sin x gives cos x. So dy/dx = (1/sin x) · cos x = cot x.
(a) 求 h(x) = (3x² + 1)⁵ 的导数。外层：u⁵ 得 5u⁴；内层：3x²+1 得导数 6x。所以 h'(x) = 5(3x²+1)⁴ · 6x = 30x (3x²+1)⁴。(b) y = ln(sin x)。外层：ln u 得 1/u；内层：sin x 得 cos x。因此 dy/dx = (1/sin x) · cos x = cot x。
8. Combining the Rules | 法则的综合运用
6B often requires using two or more rules in the same problem. For instance, k(x) = x² e³ˣ involves product rule with a chain rule on e³ˣ. Here u=x², v=e³ˣ. Then u'=2x, v' requires chain rule: derivative of e³ˣ is 3e³ˣ. So k'(x) = 2x·e³ˣ + x²·3e³ˣ = e³ˣ (2x + 3x²). Recognising when to apply each rule is essential.
练习6B常要求在同一个问题中使用两种或更多法则。例如，k(x) = x² e³ˣ 既用到乘积法则，又在 e³ˣ 上用到链式法则。这里 u=x², v=e³ˣ。则 u'=2x，v' 需用链式法则：e³ˣ 的导数为 3e³ˣ。因此 k'(x) = 2x·e³ˣ + x²·3e³ˣ = e³ˣ (2x + 3x²)。识别何时应用每种法则至关重要。
9. Exponential and Logarithmic Combinations | 指数与对数组合
properties: y = ln x² + ln(cos x) = 2 ln x + ln(cos x). Then differentiate term by term: y' = 2/x + (–sin x)/cos x = 2/x – n x. This avoids a messy chain/quotient tangle and is a favourite shortcut in IB problems.
当求导形如 y = ln(x² cos x) 的表达式时，先利用对数性质重写：y = ln x² + ln(cos x) = 2 ln x + ln(cos x)。然后逐项求导：y' = 2/x + (–sin x)/cos x = 2/x – tan x。这避免了复杂的链式/商法则纠缠，是IB问题中常用的快捷方法。
10. Watching the Order in the Quotient Rule | 商法则中的顺序注意
n incorrectly. If your function is 1/cos x, writing u=1, v=cos x gives derivative [0·cos x – 1·(– rsing roles would produce a negative of that. Always set u as numerator, v as denominator.
一个常见错误是交换 u 和 v，从而错误地翻转符号。如果函数是 1/cos x，令 u=1, v=cos x，导数为 [0·cos x – 1·(–sin x)] / cos²x = sin x / cos²x = sec x tan x。若颠倒角色将产生其相反数。始终设 u 为分子，v 为分母。
11. Practising with Negative and Fractional Powers | 含负指数和分数指数的练习
nd reciprocals as power functions to use the chain rule more easily. For instance, y = 1 / √4x+1 can be written as (4x+1)⁻¹’². Then y' = –½ (4x+1)⁻³’² · 4 = – x+1)³’². Writing in exponent form reduces errors and makes the chain rule straightforward.
将根式和倒数重写为幂函数形式，以更方便地使用链式法则。例如，y = 1 / √4x+1 可写为 (4x+1)⁻¹’²。则 y' = –½ (4x+1)⁻³’² · 4 = –2 / (4x+1)³’²。写成指数形式可减少错误，并使链式法则的应用直截了当。
12. Summary and Exercise 6B Mindset | 总结与练习6B心态
Mastering Exercise 6B means you can confidently differentiate any product, quotient, or composite of standard functions. Always scan the expression: can logs simplify it? Should you rewrite roots? Identify the outermost operation first, then work inwards. Check your answers by factoring or by testing with a graphing calculator. Consistent practice builds the fluency that IB examiners expect.
掌握练习6B意味着你能够有信心地对任何标准函数的乘积、商或复合形式求导。始终审视表达式：对数可否简化它？是否该重写根式？先识别最外层运算，然后向内推进。通过提取公因子或使用图形计算器检验答案。持续练习能培养IB考官所期望的熟练度。
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