Exercise 15D: The Chain Rule | 习题15D：链式法则

📚 Exercise 15D: The Chain Rule | 习题15D：链式法则

In IB Mathematics, Exercise 15D typically focuses on the Chain Rule – a fundamental technique in differential calculus that allows us to differentiate composite functions. Mastering this rule is essential for success in both Analysis & Approaches and Applications & Interpretation courses, as it opens the door to handling a wide variety of functions involving powers, trigonometry, exponentials, and logarithms. This article unpacks the chain rule step by step, providing clear explanations, worked examples, and exam-oriented insights that mirror the challenges you will face in Exercise 15D and beyond.

在IB数学中，习题15D通常聚焦于链式法则——微分学中用于对复合函数求导的基本技巧。掌握这一法则对于分析和应用两类课程的成功至关重要，因为它为处理涉及幂、三角、指数和对数的各种函数打开了大门。本文逐步解析链式法则，提供清晰的解释、详细的例题和针对考试的见解，这些都与你在习题15D以及更广范围内将要面对的挑战相呼应。

1. What Is a Composite Function? | 什么是复合函数？

IB exam language, you will often see expressions like (3x − 5)&sup4;, sin(2x), or e^3x². Each of these has an inner function (the part inside the brackets or argument) and an outer function that operates on the inner result. Being able to identify the inner and outer functions quickly is the first skill tested in Exercise 15D.

在深入学习求导之前，识别复合函数至关重要。复合函数可以写成 f(g(x))，意味着一个函数被应用于另一个函数的输出。例如，若 f(u) = √u 且 g(x) = x² + 1，那么 f(g(x)) = √(x² + 1)。在IB考试用语中，你经常会见到诸如 (3x − 5)&sup4;、sin(2x) 或 e^3x² 这样的表达式。每一个都有一个内层函数（括号或自变量内部的部分）和一个作用于内层结果的外层函数。能够迅速识别内层和外层函数是习题15D考查的第一项技能。

position, applying the chain rule correctly is nearly impossible. Practice identifying the “layers” in each function you encounter; it is the foundation for all that follows.

以 y = ln(4x + 7) 为例。这里内层函数是 u = 4x + 7，外层函数是 y = ln(u)。如果不进行这种分解，几乎不可能正确应用链式法则。练习识别你所遇到的每个函数中的“层次”，这是后续所有内容的基础。

---

2. The Chain Rule Formula | 链式法则公式

y = f(u), then dy/dx = dy/du · du/dx. This product of derivatives explains the name “chain”: we multiply the derivatives along a chain of dependencies. In mathematical symbols:

链式法则表述为：若 y = f(g(x))，则导数 dy/dx 等于 f′(g(x)) · g′(x)。等价地，如果设 u = g(x) 且 y = f(u)，那么 dy/dx = dy/du · du/dx。这个导数的乘积解释了“链”的名称：我们沿着依赖链将导数相乘。用数学符号表示：

dy/dx = dy/du × du/dx

In words: differentiate the outer function with respect to the inner variable (treating the inner function as a single variable), then multiply by the derivative of the inner function with respect to x. The expression f′(g(x)) means we first differentiate f as if g(x) were x, and then evaluate at g(x). Many students lose marks because they forget to multiply by g′(x) or mis-apply the outer derivative. Keep the dy/du · du/dx form handy during Exercise 15D to minimise errors.

用语言描述：先对外层函数关于内层变量求导（将内层函数视为单一变量），然后乘以内层函数关于 x 的导数。表达式 f′(g(x)) 意味着我们先对 f 求导，就好像 g(x) 就是 x 一样，然后在 g(x) 处取值。许多学生丢分是因为忘记乘以 g′(x) 或错误应用外层导数。在习题15D中随时使用 dy/du · du/dx 的形式可以有效减少错误。

---

3. Step-by-Step Differentiation Using the Chain Rule | 使用链式法则的逐步求导步骤

equired for Exercise 15D. Suppose we need to differentiate y = (5x³; − 2)&sup4;.

让我们将这一过程分解为可复现的步骤，这正是习题15D所需要的。假设我们需要对 y = (5x³; − 2)&sup4; 求导。

• Step 1: Identify the inner function. Here u = 5x³; − 2.
• 步骤1：识别内层函数。这里 u = 5x³; − 2。
• Step 2: Write y in terms of u. y = u&sup4;.
• 步骤2：用 u 表示 y，即 y = u&sup4;。
• Step 3: Differentiate y with respect to u: dy/du = 4u³;.
• 步骤3：对 y 关于 u 求导：dy/du = 4u³;。
• Step 4: Differentiate u with respect to x: du/dx = 15x²;.
• 步骤4：对 u 关于 x 求导：du/dx = 15x²;。
• Step 5: Multiply: dy/dx = 4u³; · 15x² = 60x²; u³;.
• 步骤5：相乘：dy/dx = 4u³; · 15x² = 60x²; u³;。
• Step 6: Substitute back u = 5x³; − 2: dy/dx = 60x²(5x³; − 2)³.
• 步骤6：代回 u = 5x³; − 2：dy/dx = 60x²(5x³; − 2)³。

Following this sequence makes even complicated derivatives manageable. In Exercise 15D you will see functions where the outer function is a power, sine, cosine, tangent, exponential, or logarithm. The same logic applies each time: differentiate the outer, keep the inner unchanged, then multiply by the derivative of the inner.

遵循这一顺序使得即便是复杂的导数也变得易于处理。在习题15D中，你会看到外层函数为幂、正弦、余弦、正切、指数或对数的函数。每次逻辑都相同：先对外层求导，保持内层不变，然后乘以内层的导数。

---

4. Chain Rule with Power Functions | 幂函数的链式法则

One of the most common applications in Exercise 15D involves a power of a linear or polynomial expression. The generalised power rule is: d/dx [ (g(x))ⁿ ] = n (g(x))ⁿ⁻¹ ;, then g′(x) = 2x + 3, and the derivative becomes 5(x² + 3x)&sup4;(2x + 3). Never forget the factor from the inner derivative – it is the most frequent omission in early chain rule practice.

习题15D中最常见的一类应用是线性或多项式表达式的幂。广义幂法则是：d/dx [ (g(x))ⁿ ] = n (g(x))ⁿ⁻¹ · g′(x)。这是链式法则的直接推论。例如，要对 y = (4x + 1)&sup7; 求导，设 u = 4x + 1，那么 y = u&sup7;。dy/du = 7u&sup6;，du/dx = 4，因此 dy/dx = 7(4x + 1)&sup6; · 4 = 28(4x + 1)&sup6;。如果内层函数更复杂，比如 y = (x² + 3x)&sup5;，那么 g′(x) = 2x + 3，导数变为 5(x² + 3x)&sup4;(2x + 3)。永远不要忘记来自内层导数的因子——这是早期链式法则练习中最常见的遗漏。

hen the power is negative or a fraction, as in y = 1/(3x − 2)³ = (3x − 2)⁻³. The chain rule gives dy/dx = −3(3x − 2)⁻⁴ · 3 = −9(3x − 2)⁻⁴. Writing the answer with a positive exponent is often required in IB exams, so rewrite as −9/(3x − 2)&sup4;. This technique appears frequently in Exercise 15D.

当幂为负数或分数时要格外小心，例如 y = 1/(3x − 2)³ = (3x − 2)⁻³。链式法则得出 dy/dx = −3(3x − 2)⁻⁴ · 3 = −9(3x − 2)⁻⁴。IB考试中通常要求用正指数书写答案，所以重写为 −9/(3x − 2)&sup4;。这一技巧在习题15D中频繁出现。

---

5. Chain Rule with Trigonometric Functions | 三角函数的链式法则

)) · g′(x). For example, differentiating y = sin(5x) yields dy/dx = cos(5x) · 5 = 5 cos(5x). If the inner function is a polynomial, say y = cos(x² + 1), the derivative is −sin(x² + 1) · (2x) = −2x sin(x² + 1).

在IB考试中，三角函数几乎总是带有一个线性内层函数或更复杂的自变量。标准的导数公式为：d/dx [sin(g(x))] = cos(g(x)) · g′(x)；d/dx [cos(g(x))] = −sin(g(x)) · g′(x)；d/dx [tan(g(x))] = sec²(g(x)) · g′(x)。例如，对 y = sin(5x) 求导得到 dy/dx = cos(5x) · 5 = 5 cos(5x)。如果内层函数是多项式，比如 y = cos(x² + 1)，导数为 −sin(x² + 1) · (2x) = −2x sin(x² + 1)。

n²θ, the result follows. Such connections highlight why the chain rule is indispensable for linking pure calculus with trigonometric identities.

习题15D经常包含证明给定的函数满足涉及三角导数的微分方程。例如，证明 y = tan(3x) 得出 dy/dx = 3(1 + tan²(3x))。使用链式法则，dy/dx = sec²(3x) · 3，并且由于 sec²θ = 1 + tan²θ，结果便可得证。这种联系凸显了链式法则在连接纯微积分与三角恒等式方面不可或缺的作用。

---

广告 / Sponsored

6. Chain Rule with Exponential and Logarithmic Functions | 指数和对数函数的链式法则

re particularly chain-rule friendly, but they still require the inner derivative. If y = e^g(x), then dy/dx = e^g(x) · g′(x). The function e^u is its own derivative, but the multiplier g′(x) must not be dropped. For y = e^2x, dy/dx = 2e^2x; for y = e^−x², dy/dx = −2x e^−x² x + 1). If the logarithm has a different base, use the change-of-base formula or recall that d/dx [log_a(x)] = 1/(x ln a) and then apply the chain rule.

以 e 为底的指数函数尤其适合链式法则，但仍需要内层导数。若 y = e^g(x)，那么 dy/dx = e^g(x) · g′(x)。e^u 的导数就是它本身，但乘数 g′(x) 绝不能丢掉。对于 y = e^2x，dy/dx = 2e^2x；对于 y = e^−x²，dy/dx = −2x e^−x²。对数函数遵循模式 d/dx [ln(g(x))] = g′(x) / g(x)。对于 y = ln(3x + 1)，dy/dx = 3/(3x + 1)。如果对数的底数不同，可使用换底公式，或记住 d/dx [log_a(x)] = 1/(x ln a)，然后应用链式法则。

x). Here g(x) = sin x, so dy/dx = cos x / sin x = cot x. Another typical example is y = e^{cos x}, giving dy/dx = −sin x e^{cos x}. These problems test your ability to combine the chain rule with other differentiation rules and to simplify the result, which is a core IB skill.

在习题15D中，你可能会遇到像 y = ln(sin x) 这样的混合函数。这里 g(x) = sin x，所以 dy/dx = cos x / sin x = cot x。另一个典型例子是 y = e^{cos x}，得到 dy/dx = −sin x e^{cos x}。这些问题考察你将链式法则与其他求导法则相结合并化简结果的能力，这是IB的核心技能。

---

7. Repeated Chain Rule (Multiple Layers) | 多次链式法则（多层复合）

dx = dy/du · du/dv · dv/dx = 2u · cos(v) · 4 = 8 sin(v) cos(v) = 4 sin(2v) = 4 sin(8x) after further simplification. Recognising such hidden compositions is vital, as IB often writes powers of trigonometric functions in shorthand like sinⁿ(x) meaning (sin x)ⁿ.

习题15D中的某些函数包含三层甚至更多层的嵌套。例如，y = sin²(4x) = (sin(4x))²;。这里最外层是平方，中间层是正弦，最内层是 4x。求导时需从外向内操作。设 v = 4x，u = sin(v)，y = u²;。那么 v · dv/dx = 2u · cos(v) · 4 = 8 sin(v) cos(v) = 4 sin(2v) = 4 sin(8x)（进一步化简后）。识别这种隐藏的复合至关重要，因为IB经常将三角函数的幂写为简写形式如 sinⁿ(x)，表示 (sin x)ⁿ。

Another multi-layer example: y = e^√(x²+1). The layers are: outermost e^u, middle u = √v, innermost v = x²+1. Then dy/dx = e^√(x²+1) · (1/(2√(x²+1))) · 2x = x e^√(x²+1) / √(x²+1). Practice breaking down complex expressions methodically; this prevents panic during exam-style questions in Exercise 15D.

另一个多层例子：y = e^√(x²+1)。各层分别为：最外层 e^u，中层 u = √v，最内层 v = x²+1。那么 dy/dx = e^√(x²+1) · (1/(2√(x²+1))) · 2x = x e^√(x²+1) / √(x²+1)。练习有条理地分解复杂表达式，这样做可以避免在习题15D的考试类题目中手忙脚乱。

---

8. The Chain Rule and Other Differentiation Rules | 链式法则与其他求导法则

side the product you need the chain rule for sin(3x). Let u = x², v = sin(3x). Then u' = 2x, v' = 3 cos(3x). By the product rule, dy/dx = u'v + uv' ajor rule governs the overall structure, then apply the chain rule to any composite parts.

链式法则经常需要与乘法法则和商法则结合使用。在习题15D中，你可能会遇到诸如对 y = x² sin(3x) 求导的问题。这里首先用到乘法法则，但在乘积内部需要对 sin(3x) 使用链式法则。设 u = x²;，v = sin(3x)。那么 u' = 2x，v' = 3 cos(3x)。根据乘法法则，dy/dx = u'v + uv' = 2x sin(3x) + x²; · 3 cos(3x) = 2x sin(3x) + 3x² cos(3x)。顺序很关键：先判断由哪一主要法则控制整体结构，然后对任何复合部分应用链式法则。

Similarly, with the quotient rule, for y = e^2x ive of the numerator, while the denominator is simple. The quotient rule gives dy/dx = [2e^2x(x+1) − e^2x(1)] / (x+1)². Such layered problems are excellent preparation for IB Paper 2 questions where multiple skills are assessed simultaneously.

类似地，对于商法则，y = e^2x / (x+1)，链式法则提供了分子的导数，而分母则较简单。商法则给出 dy/dx = [2e^2x(x+1) − e^2x(1)] / (x+1)²。这种层次化的问题是应对IB试卷二的绝佳准备，因为那时需要同时运用多种技能。

---

9. Common Mistakes and How to Avoid Them | 常见错误及如何避免

unction. A student might write d/dx [sin(4x)] = cos(4x) and lose the factor 4. Always ask: “Have I accounted for the rate of change of the inside?” ture. Third, failing to apply the chain rule multiple times when three or more functions are composed. Fourth, incorrectly differentiating the outer function: for instance, treating d/dx [ (2x+1)³ ] as 3(2)² instead of 3(2x+1)² · 2. This error comes from not keeping the inner function intact before multiplying. Finally, algebraic simplification errors after differentiating – always take an extra moment to combine like terms and factor gracefully.

习题15D暴露出几个可预见的陷阱。第一，忘记乘以内层函数的导数。学生可能会写出 d/dx [sin(4x)] = cos(4x) 而丢失因子 4。始终要问自己：“我是否考虑了内部的变化率？”第二，错误识别内层函数，尤其是在处理像 sin²x 这样的幂时。将其视为 (sin x)² 可以明晰结构。第三，当三个或更多函数复合时未能多次应用链式法则。第四，对外层函数求导不正确：例如，将 d/dx [ (2x+1)³ ] 当作 3(2)² 而不是 3(2x+1)² · 2。这一错误的根源是在相乘之前没有保持内层函数不变。最后，求导后的代数化简错误——总是多花一点时间恰当地合并同类项并提取公因式。

To avoid these, adopt a systematic approach: underline the inner function, write out u and du/dx explicitly, and double-check your final answer by substituting a small value of x into the original function and its derivative to see if the gradient makes sense – although this is not a proof, it can catch gross errors.

为避免这些错误，采用系统方法：在内层函数下面划线，明确写出 u 和 du/dx，并通过将 x 的一个较小值代入原函数及其导数来检验最终答案，看看斜率是否合理——虽然这不是证明，但可以捕捉到重大错误。

---

10. Practice Problems from Exercise 15D | 习题15D中的典型练习

Let us work through several problems that closely resemble those found in Exercise 15D of a standard IB Mathematics textbook.

让我们来完成几道与标准IB数学教材习题15D中题目非常相似的练习。

Problem 1: Differentiate y = (2x²; − 5)&sup6;.
题目1：对 y = (2x²; − 5)&sup6; 求导。

Solution: Let u = 2x²; −

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）