📚 Exercise 10D.1: Chain Rule for Composite Functions | 练习 10D.1：复合函数的链式法则
Exercise 10D.1 in the IB Mathematics Analysis and Approaches syllabus is a foundational set of problems designed to build fluency with the chain rule. Mastering this exercise enables you to differentiate a vast range of functions by breaking them down into simpler pieces and applying the correct derivative patterns.
IB 数学分析与方法课程中的练习 10D.1 是一组基础题，旨在培养对链式法则的熟练应用。掌握这部分内容后，你将能够通过将复杂函数分解为更简单的部分并应用正确的求导模式，来对大量函数进行微分。
1. What Is a Composite Function? | 什么是复合函数？
ated when one function is applied to the output of another. It is written as f(g(x)) or (f ∘ tion is (3x + 1)⁴. In this structure, g is the inner function and f is the outer function.
当一个函数被应用于另一个函数的输出时，就产生了复合函数，记作 f(g(x)) 或 (f ∘ g)(x)。例如，设 u = g(x) = 3x + 1，f(u) = u⁴，则复合函数为 (3x + 1)⁴。在这个结构中，g 是内层函数，f 是外层函数。
ing. Common inner functions include linear expressions, trigonometric arguments and powers.
识别函数嵌套的层次是应用链式法则的第一个关键步骤。练习 10D.1 经常要求你在求导之前先确定 u = g(x)。常见的内层函数包括线性表达式、三角函数的参数以及幂指数。
2. The Chain Rule Formula | 链式法则公式
ivative is given by dy/dx = dy/du × du/dx. In Leibniz notation this highlights the idea of 'derivative of outer function times derivative of inner function'. In function notation, if h(x) = f(g(x)), then h'(x) = f'(g(x)) · g'(x).
如果 y = f(u) 且 u = g(x)，那么导数为 dy/dx = dy/du × du/dx。莱布尼茨记法清晰地表达了“外层函数的导数乘以内层函数的导数”的思想。用函数记法表示，若 h(x) = f(g(x))，则 h'(x) = f'(g(x)) · g'(x)。
ative of the inside). Exercise 10D.1 will have you repeat this pattern many times until it becomes automatic.
一个快速的心智模型是：导数 =（将外层函数求导后在内层函数处求值）×（内层函数的导数）。练习 10D.1 会让你反复演练这个模式，直到熟能生巧。
3. Step-by-Step Differentiation Using the Chain Rule | 使用链式法则逐步求导
ately. (4) Multiply them and substitute back u = g(x) to express the answer in terms of x.
对于练习 10D.1 中的每个问题，遵循以下四个步骤：（1）识别内层函数 u = g(x)。（2）用 u 重写外层函数，即 y = f(u)。（3）分别求出 dy/du 和 du/dx。（4）将它们相乘，并将 u = g(x) 代回，用 x 表达最终结果。
ng gives dy/dx = 3u² × 5 = 15(5x - 2)². This method prevents missing the inner derivative.
对于 y = (5x - 2)³，设 u = 5x - 2，则 y = u³。于是 dy/du = 3u²，du/dx = 5。两者相乘得 dy/dx = 3u² × 5 = 15(5x - 2)²。这种方法可以避免漏乘内层导数。
4. Example 1: Polynomial Compositions | 例 1：多项式复合
x + 4)⁴(4x - 1). Notice that the derivative of the inner quadratic is placed as a factor.
考虑 y = (2x² - x + 4)⁵。设 u = 2x² - x + 4，则 y = u⁵。dy/du = 5u⁴，du/dx = 4x - 1。结果为 dy/dx = 5(2x² - x + 4)⁴(4x - 1)。注意内层二次函数的导数作为一个因子保留。
In Exercise 10D.1 you will encounter many polynomial powers. Always check that you have differentiated the inside completely, especially when it contains more than one term. Dropping a constant or misapplying the power rule inside is a common slip.
在练习 10D.1 中，你会遇到许多多项式的幂。一定要确保对内层函数完全求导，尤其是当它包含多个项时。漏掉常数项或对内层错误应用幂法则是一个常见的失误。
5. Example 2: Trigonometric Compositions | 例 2：三角函数复合
= 4x. dy/du = cos u, du/dx = 4. So dy/dx = 4 cos(4x). More generally, d/dx[sin(g(x))] = g'(x) cos(g(x)). The same pattern holds for cosine: d/dx[cos(g(x))] = -g'(x) sin(g(x)).
对于 y = sin(4x)，u = 4x，dy/du = cos u，du/dx = 4，因此 dy/dx = 4 cos(4x)。更一般地，d/dx[sin(g(x))] = g'(x) cos(g(x))。对于余弦函数，模式类似：d/dx[cos(g(x))] = -g'(x) sin(g(x))。
² u. Hence dy/dx = 2x sec²(x²). Exercise 10D.1 includes tan, cot, sec, and csc composites.
当内层函数更复杂时，如 y = tan(x²)，则 u = x²，du/dx = 2x。tan u 的导数是 sec² u。因此 dy/dx = 2x sec²(x²)。练习 10D.1 包含正切、余切、正割和余割的复合函数。
6. Example 3: Exponential and Logarithmic Functions | 例 3：指数与对数函数
For y = e3x+2, the inner function is u = 3x + 2. dy/du = eu, du/dx = 3, giving dy/dx = 3e3x+2. In general, d/dx[e^(g(x))] = g'(x) e^(g(x)). The chain rule for natural exponential is remarkably tidy.
对于 y = e3x+2，内层函数为 u = 3x + 2。dy/du = eu，du/dx = 3，得到 dy/dx = 3e3x+2。一般地，d/dx[e^(g(x))] = g'(x) e^(g(x))。自然指数函数的链式法则形式非常简洁。
(5x² - 1). Remember the domain restriction: the argument of the logarithm must remain positive.
对于 y = ln(5x² - 1)，设 u = 5x² - 1。dy/du = 1/u，du/dx = 10x。因此 dy/dx = (10x) / (5x² - 1)。请记住定义域的限制：对数的真数必须为正。
7. Common Mistakes in Exercise 10D.1 | 练习 10D.1 中的常见错误
The most frequent error is forgetting to multiply by the derivative of the inner function. Students often differentiate sin(2x) as cos(2x) instead of 2 cos(2x). Always ask: "Where is the du/dx factor?"
最常见的错误是忘记乘以内层函数的导数。学生常常将 sin(2x) 的导数误写为 cos(2x)，而正确结果是 2 cos(2x)。永远要问自己：“du/dx 因子在哪里？”
irectly. The chain rule still works with u = x⁻¹, but careful selection avoids extra work.
另一个陷阱是内、外层函数的错误识别。对于 y = (1/x)²，容易把 1/x 当作外层，但把它重写为 x⁻² 就能直接应用幂法则。若用链式法则，设 u = x⁻¹ 同样可行，但合理选择可以避免额外的计算。
ot, and csc. For y = cos(5x), the derivative is -5 sin(5x). The negative sign is integral.
此外，在处理三角函数复合时，不要忘记 cosine、cotangent 和 cosecant 导数中的负号。对于 y = cos(5x)，导数是 -5 sin(5x)。这个负号不可或缺。
8. Combining Chain Rule with Product and Quotient Rules | 链式法则与乘积、商法则的结合
with the chain rule embedded for the second factor. Let u = x, v = (2x+1)³. Then dy/dx = u'v + uv'. Here v' = 3(2x+1)²·2 = 6(2x+1)².
练习 10D.1 偶尔会延伸到混合题型。对于 y = x · (2x+1)³，你必须同时使用乘积法则，并在第二部分嵌入链式法则。令 u = x，v = (2x+1)³。则 dy/dx = u'v + uv'。其中 v' = 3(2x+1)²·2 = 6(2x+1)²。
Similarly, for y = sin(2x)/ex, use the quotient rule: dy/dx = [cos(2x)·2 · ex - sin(2x)·ex] / e2x. Each term inside the numerator may involve its own chain rule, so write down each component clearly.
类似地，对于 y = sin(2x)/ex，使用商法则：dy/dx = [cos(2x)·2 · ex - sin(2x)·ex] / e2x。分子中的每一项都可能涉及各自的链式法则，因此需要清晰地写出每个组成部分。
9. Higher-Order Derivatives and the Chain Rule | 高阶导数与链式法则
as a product: constant 3 times cos(3x). The derivative is 3 · (−sin(3x) · 3) = −9 sin(3x).
求复合函数的二阶导数需要再次应用链式法则，并且往往还要用到乘积法则。对于 y = sin(3x)，一阶导数为 3 cos(3x)。对此再求导时，将 3 cos(3x) 视为常数 3 乘以 cos(3x) 的乘积，其导数为 3 · (−sin(3x) · 3) = −9 sin(3x)。
Exercise 10D.1 may not always include second derivatives, but building this skill early helps with concavity and kinematics problems later. Keep track of every factor produced by each differentiation step.
练习 10D.1 不一定总包含二阶导数，但尽早建立这一技能有助于后续学习凹凸性和运动学问题。追踪每一步求导产生的每一个因子是非常重要的。
10. Efficiency Tips for Exercise 10D.1 | 练习 10D.1 的高效解题技巧
= derivative of the outside (keeping the inside unchanged) times derivative of the inside.
在做练习时，一开始可以明确写出 u = g(x)，但要逐渐学会心算应用法则。一个实用的速记法是：导数 = 外层导数（内层保持不变）乘以 内层导数。
Check your answer by selecting a test value of x and using numerical approximation if possible. For example, estimate f'(1) using a small difference quotient and compare with your expression. This confirms the algebraic form.
可以通过选取一个 x 的测试值并尽可能使用数值逼近来检查答案。例如，用微小的差商估算 f'(1)，并与你求得的表达式进行比较。这能验证代数形式的正确性。
11. Real-World Connections | 实际关联
s(ωt), directly from the chain rule. In economics, marginal cost may involve composite functions of production levels.
链式法则能够对依赖于其他变化率的变化率进行建模。在物理学中，如果单摆的位置为 s(t) = sin(ωt)，速度 v(t) = ω cos(ωt) 即为链式法则的直接结果。在经济学中，边际成本可能涉及生产水平的复合函数。
oblems, which appear later in the IB course, rely entirely on the chain rule. If a balloon' f volume V = (4/3)π r³ is dV/dt = 4π r² dr/dt. The factor dr/dt comes from the chain rule.
IB 课程后续的相关变化率问题完全依赖于链式法则。如果气球的半径 r 以恒定速率增加，体积 V = (4/3)π r³ 的变化率为 dV/dt = 4π r² dr/dt。因子 dr/dt 正是来自链式法则。
12. Summary and Key Takeaways | 总结与关键要点
Exercise 10D.1 firmly establishes the chain rule as an essential differentiation tool. Always decompose the function, differentiate the outside with respect to the inside, and multiply by the inside derivative. This rule unlocks derivatives for exponential, logarithmic, trigonometric, and radical composite functions.
练习 10D.1 将链式法则牢固地确立为一项基本的求导工具。始终要分解函数，对外层关于内层求导，再乘以内层导数。链式法则为指数、对数、三角和根式等复合函数的求导打开了大门。
Consistent practice with the problems in this exercise will make the process second nature. Once mastered, you will be ready to handle implicit differentiation, parametric equations, and integration by substitution with confidence.
通过持续练习本练习中的问题，这一过程将变得自然而然。一旦掌握，你将能够自信地应对隐函数求导、参数方程以及换元积分法。
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