📚 Exercise 6B.2: The Chain Rule and Differentiation of Composite Functions | 练习6B.2：链式法则与复合函数求导
In IB Mathematics, Exercise 6B.2 typically focuses on one of the most powerful differentiation tools: the chain rule. Mastering this rule enables you to differentiate composite functions efficiently, which is essential for tackling more complex calculus problems in both Analysis and Applications courses. This article breaks down the chain rule step by step, provides worked examples across various function types, and highlights common pitfalls to avoid, ensuring you gain confidence for your exams.
在IB数学中，练习6B.2通常聚焦于最强有力的微分工具之一：链式法则。掌握这一法则能让你高效地对复合函数求导，这对于应对分析与方法、应用与解释课程中更复杂的微积分问题至关重要。本文将逐步分解链式法则，提供各类函数的详细示例，并指出需要避免的常见错误，帮助你在考试中建立信心。
1. Review of Basic Derivatives | 基本导数回顾
d/dx (cos x) = -sin x, and d/dx (tan x) = sec² x. For exponentials and logarithms, d/dx (eˣ) = eˣ d d/dx (ln x) = 1/x. These results form the foundation upon which the chain rule operates.
在应用链式法则之前，你必须熟练掌握基本函数的导数。回顾幂函数法则：d/dx (xⁿ) = n xⁿ⁻¹ 对任意实数 n 成立。三角函数的导数包括 d/dx (sin x) = cos x、d/dx (cos x) = -sin x 以及 d/dx (tan x) = sec² x。对于指数和对数函数，d/dx (eˣ) = eˣ 且 d/dx (ln x) = 1/x。这些结果构成了链式法则运行的基础。
2. Understanding Composite Functions | 理解复合函数
tation, if f(x) = sin x and g(x) = 2x+1, then the composite f(g(x)) = sin(2x+1) is read as 'f of g of x' itical first step in using the chain rule correctly. Look for expressions like (3x²-5)⁴, etan x, or ln(cos x), where a 'core' function is wrapped by another operation.
当一个函数被放入另一个函数内部时，就形成了复合函数。用符号表示，若 f(x) = sin x 且 g(x) = 2x+1，则复合函数 f(g(x)) = sin(2x+1) 读作“f 在 g(x) 的值”。其中内层函数为 g(x) = 2x+1，外层函数为 f(u) = sin u。识别这种结构是正确使用链式法则的关键第一步。要留意诸如 (3x²-5)⁴、etan x 或 ln(cos x) 这样的表达式，其中一个“核心”函数被另一个运算包裹。
3. The Chain Rule Formula | 链式法则公式
If y = f(u) and u = g(x), then dy/dx = dy/du × du/dx.
This classical Leibniz notation highlights the idea of 'connecting' rates of change. In function notation, the rule becomes:
d/dx [f(g(x))] = f'(g(x)) × g'(x).
In words: differentiate the outer function, leaving the inner function untouched, then multiply by the derivative of the inner function. This sequence—outer derivative evaluated at inner, times inner derivative—is the rhythm you should internalize for every chain rule problem.
这种经典的莱布尼茨记号强调了“连接”变化率的思想。用函数记号表示，法则变为：外函数求导（保持内函数不变）再乘以内函数的导数。这一顺序——先求外函数的导数并在内函数处取值，再乘以内函数的导数——是你在每个链式法则问题中应当内化的节奏。
4. Applying the Chain Rule to Power Functions | 用链式法则求幂函数的导数
s the generalized power rule. For any differentiable function u(x), d/dx [uⁿ] = n uⁿ⁻¹ × u' d the inner is u = 3x+1. Then dy/dx = 4(3x+1)³ × 3 = 12(3x+1)³. Another common case is y = √x²+5, which can be written as (x²+5)1/2. Applying the chain rule gives dy/dx = (1/2)(x²+5)-1/2 × 2x = x / √x²+5. Notice how the derivative of the inner function must always be multiplied at the end.
链式法则最简单且最频繁的应用涉及广义幂函数法则。对于任意可微函数 u(x)，d/dx [uⁿ] = n uⁿ⁻¹ × u'。例如，求 y = (3x+1)⁴ 的导数。这里外函数是 u⁴，内函数是 u = 3x+1。于是 dy/dx = 4(3x+1)³ × 3 = 12(3x+1)³。另一个常见例子是 y = √x²+5，可写成 (x²+5)1/2。应用链式法则得 dy/dx = (1/2)(x²+5)-1/2 × 2x = x / √x²+5。请注意，内函数的导数必须在最终步骤中相乘。
5. Trigonometric Functions and the Chain Rule | 三角函数与链式法则
For y = sin(ax+b), the derivative is a cos(ax+b). More generally, d/dx [sin u] = cos u × u', d/dx [cos u] = -sin u × u', and d/dx [tan u] = sec² u × u' ensure the argument is measured in radians for these derivative formulas to hold without extra constants.
当三角函数的变量不只是 x 时，链式法则就不可或缺。一般地，d/dx [sin u] = cos u × u'，d/dx [cos u] = -sin u × u'，d/dx [tan u] = sec² u × u'。考虑 y = cos(3x²)。令 u = 3x²，则 dy/dx = -sin(3x²) × 6x = -6x sin(3x²)。对于 y = tan(x/2)，我们得到 dy/dx = sec²(x/2) × (1/2) = (1/2) sec²(x/2)。务必确保变量以弧度为单位，上述导数公式才无需附加常数。
6. Exponential and Logarithmic Functions | 指数与对数函数
The derivative of eᵘ is remarkably straightforward: d/dx [eᵘ] = eᵘ × u'. For instance, if y = e5x, then dy/dx = 5e5x. If y = ex², then dy/dx = 2x ex². For the natural logarithm, d/dx [ln u] = u' ln(2x+1), the derivative is 2/(2x+1). When dealing with bases other than e, such as y = 23x, rewrite using eln2 × 3x and differentiate: dy/dx = 3 ln2 × 23x. Knowing these patterns helps you handle any exponential or logarithmic composite rapidly.
eᵘ 的导数非常直接：d/dx [eᵘ] = eᵘ × u'。例如，若 y = e5x，则 dy/dx = 5e5x。若 y = ex²，则 dy/dx = 2x ex²。对于自然对数，d/dx [ln u] = u' / u。因此，对于 y = ln(2x+1)，导数为 2/(2x+1)。当底数不是 e 时，例如 y = 23x，可改写为 eln2 × 3x 再求导：dy/dx = 3 ln2 × 23x。掌握这些模式能助你快速处理任何指数或对数复合函数。
7. Combining Chain Rule with Product Rule | 结合链式法则与乘积法则
n rule inside another differentiation rule. For y = x² sin(3x), the product rule says (uv)' = u'v + uv'. Let u = x² and v = sin(3x). Then u' = 2x, while v' os(3x). Always apply the chain rule first on each factor that is a composite function before substituting into the product formula.
典型的IB题目常要求在其他求导法则中嵌入链式法则。对于 y = x² sin(3x)，乘积法则指出 (uv)' = u'v + uv'。令 u = x² 且 v = sin(3x)。则 u' = 2x，而 v' 需要链式法则：sin(3x) 的导数为 3 cos(3x)。将其组合得 dy/dx = (2x)(sin(3x)) + (x²)(3 cos(3x)) = 2x sin(3x) + 3x² cos(3x)。在代入乘积公式之前，务必先对每个复合函数因子应用链式法则。
8. Combining Chain Rule with Quotient Rule | 结合链式法则与商法则
Similarly, the quotient rule often works hand in hand with the chain rule. Consider y = e2x / (x+1). By the quotient rule, dy/dx = [ (e2x)' (x+1) - e2x(x+1)' ] / (x+1)². The derivative of e2x uses the chain rule: 2e2x, while (x+1)' = 1. Hence dy/dx = [ 2e2x(x+1) - e2x ] / (x+1)² = e2x(2x+1) / (x+1)². Pay close attention to parentheses when simplifying; a small slip can change the entire numerator.
同样地，商法则常与链式法则携手工作。考虑 y = e2x / (x+1)。根据商法则，dy/dx = [ (e2x)' (x+1) - e2x(x+1)' ] / (x+1)²。e2x 的导数需用链式法则：2e2x，而 (x+1)' = 1。因此 dy/dx = [ 2e2x(x+1) - e2x ] / (x+1)² = e2x(2x+1) / (x+1)²。化简时要特别注意括号，一个小失误就可能改变整个分子。
9. The Chain Rule in Implicit Differentiation | 隐函数求导中的链式法则
n of x. This illustrates the chain rule: outer derivative 2y times the derivative of the inner function y with respect to x, namely dy/dx. Solving for dy/dx yields -x/y. Recognizing this hidden use of the chain rule is essential for relating rates of change in related-rates problems.
隐函数求导（常见于HL）在根本上依赖链式法则。对于方程 x² + y² = 25，两边同时对 x 求导。y² 的导数为 2y × (dy/dx)，因为 y 是 x 的函数。这展示了链式法则：外函数导数 2y 乘以内函数 y 对 x 的导数，即 dy/dx。解出 dy/dx 得 -x/y。认识到这种链式法则的隐藏用法对于解决相关变化率问题至关重要。
10. Higher-Order Derivatives with the Chain Rule | 高阶导数中的链式法则
ion. Suppose s(t) = sin(2t) gives the displacement of a particle. The velocity is v(t) = s'(t) = 2 cos(2t) using the chain rule. To find acceleration, differentiate again: a(t) = v' s pattern is common: repeated chain rule applications accumulate powers of the inner derivative.
在计算复合函数的高阶导数时，链式法则也会登场。假设 s(t) = sin(2t) 给出粒子的位移。速度 v(t) = s'(t) = 2 cos(2t) 使用了链式法则。为求加速度，再次求导：a(t) = v'(t) = -4 sin(2t)，对 cos(2t) 再次应用链式法则。注意内函数导数中的因子 2 每次都会乘入，导致加速度中出现 2² = 4。这种模式很常见：反复应用链式法则会累积内层导数的幂次。
11. Common Mistakes and How to Avoid Them | 常见错误及如何避免
d of 2 cos(2x). Another pitfall occurs when the inner function is misidentified; for y = ex²+1 when differentiating cos u or when the inner derivative is negative. A systematic approach—write down u, find u', then write f'(u) × u'—will prevent most blunders.
最常见的一个错误是完全忘记乘以内层函数的导数。学生常把 sin(2x) 的导数写成 cos(2x)，而不是 2 cos(2x)。另一个陷阱是错误识别内层函数；对于 y = ex²+1³，内层函数是 (x²+1)³ 而非 x²+1，这需要两次应用链式法则。此外，在求 cos u 的导数或内层导数为负时，要注意符号错误。系统性的方法——写下 u，求出 u'，然后写出 f'(u) × u'——能避免大多数疏漏。
12. Practice Tips for Exercise 6B.2 | 练习6B.2的解题技巧
Begin by differentiating simple power composites like (ax+b)ⁿ to build confidence, then progress to trigonometric, exponential, and mixed forms. Always explicitly state u and u' before writing the final derivative. After finding an answer, mentally check that the outer derivative's structure is correct and that the inner derivative is present as a factor. With exercises such as differentiating ln(sin x) or ecos x, practice peeling back the layers one at a time. Consistent drill using Exercise 6B.2 style problems will make the chain rule second nature well before your IB exams.
从对简单的幂函数复合如 (ax+b)ⁿ 求导开始建立信心，然后进阶到三角函数、指数函数及混合形式
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