Mastering the Chain Rule: IB Maths Exercise 6C.1 | 掌握链式法则：IB数学练习6C.1

📚 Mastering the Chain Rule: IB Maths Exercise 6C.1 | 掌握链式法则：IB数学练习6C.1

The chain rule is one of the most powerful differentiation techniques in calculus, and it is absolutely essential for IB Mathematics, whether you're taking Analysis and Approaches (AA) or Applications and Interpretation (AI). In many textbooks, Exercise 6C.1 focuses intently on building fluency with the chain rule through a carefully structured set of problems. This article unpacks the chain rule from first principles, walks through worked examples that mirror the style of Exercise 6C.1, and equips you with strategies to handle any chain rule question in your IB exams with confidence.

链式法则是微积分中最强大的微分技巧之一，对于IB数学（无论是分析和方法AA还是应用和解释AI）都绝对必要。在很多教材中，练习6C.1通过精心设计的一组题目，集中训练学生熟练掌握链式法则。本文将从基本原理出发解析链式法则，通过类似练习6C.1的例题进行讲解，并帮助你掌握策略，从而在IB考试中充满信心地应对任何链式法则相关题目。

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1. What Is Differentiation? A Quick Refresher | 什么是微分？快速回顾

Differentiation is the process of finding the derivative of a function. The derivative, written as f'(x) or dy/dx, measures the instantaneous rate of change of the function and gives the slope of the tangent line at any point. In IB Mathematics, you learn to differentiate basic functions such as power functions, exponentials, logarithms, and trigonometric functions using a set of standard rules.

微分是求函数导数的过程。导数写作 f'(x) 或 dy/dx，它衡量函数的瞬时变化率，并给出任意切线的斜率。在IB数学中，你将学会使用一系列标准法则对幂函数、指数函数、对数函数和三角函数等基本函数求导。

The most basic rule is the power rule: for f(x) = xⁿ, the derivative is f' term-by-term. However, many functions cannot be differentiated just by these simple rules – they involve compositions of functions, and that is exactly where the chain rule comes in.

最基本的法则是幂法则：对于 f(x) = xⁿ，其导数为 f'(x) = n xⁿ⁻¹。此外，和、差以及常数倍可以逐项求导。然而，许多函数不能仅依靠这些简单法则求导——它们涉及函数的复合，而这正是链式法则登场的地方。

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2. The Need for the Chain Rule | 链式法则的必要性

er function f(u) = u⁵. The chain rule allows us to handle such nested functions correctly.

考虑函数 h(x) = (2x + 1)⁵。如果我们试图直接使用幂法则，只会得到 5(2x+1)⁴作为导数，这是不完整的。该函数并非 x 的简单多项式；它是由两个函数复合而成的：内层函数 u(x) = 2x + 1 和外层函数 f(u) = u⁵。链式法则使我们能够正确应对这类嵌套函数。

In Exercise 6C.1, you are likely to encounter many composite functions like (ax + b)ⁿ, sin(kx), e^cx, and ln(g(x)). Without the chain rule, differentiation of these expressions would be impossible using only basic rules. Mastering it opens the door to a wide range of calculus problems in the IB syllabus.

在练习6C.1中，你很可能会遇到大量类似 (ax + b)ⁿ、sin(kx)、e^cx 和 ln(g(x)) 的复合函数。没有链式法则，仅靠基本规则根本无法对这些表达式求导。掌握它为你打开了IB课程中广泛微积分问题的大门。

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3. Statement of the Chain Rule | 链式法则的表述

If y is a function of u, and u is a function of x, then y can be considered as a composite function of x. The chain rule in Leibniz notation states:

如果 y 是 u 的函数，而 u 是 x 的函数，那么 y 可以看作 x 的复合函数。链式法则的莱布尼茨记法表述为：

dy/dx = (dy/du) × (du/dx)

In function notation, if h(x) = f(g(x)), then the derivative is:

用函数记法表示，如果 h(x) = f(g(x))，那么导数为：

h'(x) = f'(g(x)) × g'(x)

This means we first differentiate the outer function with respect to the inner function, leaving the inner function unchanged, and then multiply by the derivative of the inner function. This elegant rule will be applied repeatedly throughout your IB course.

这意味着我们先对外层函数关于内层函数求导（保持内层函数不变），然后再乘以内层函数的导数。这条简洁的法则将在你的IB课程中被反复应用。

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4. How to Apply the Chain Rule Step by Step | 如何逐步应用链式法则

Step 1: Identify the inner function u = g(x) and the outer function f(u).

步骤1：识别内层函数 u = g(x) 和外层函数 f(u)。

Step 2: Differentiate the outer function with respect to u, giving f'(u).

步骤2：对外层函数关于 u 求导，得到 f'(u)。

Step 3: Substitute the original inner function back into f'(u) to get f'(g(x)).

步骤3：将原来的内层函数代回 f'(u) 中，得到 f'(g(x))。

Step 4: Differentiate the inner function u with respect to x to get g'(x).

步骤4：将内层函数 u 对 x 求导得到 g'(x)。

Step 5: Multiply the results: h'(x) = f'(g(x)) × g'(x).

步骤5：将结果相乘：h'(x) = f'(g(x)) × g'(x)。

Let's apply this to y = (3x – 5)⁴. Here u = 3x – 5, outer function = u⁴. Then dy/du = 4u³, du/dx = 3, so dy/dx = 4(3x – 5)³ × 3 = 12(3x – 5)³. Practicing this procedure will make solving Exercise 6C.1 problems much easier.

我们把这个流程应用到 y = (3x – 5)⁴。此处 u = 3x – 5，外层函数为 u⁴。那么 dy/du = 4u³，du/dx = 3，因此 dy/dx = 4(3x – 5)³ × 3 = 12(3x – 5)³。练习这个程序将使解决练习6C.1的题目容易得多。

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5. Common Derivatives You Must Know | 你必须掌握的基本导数

Before tackling Exercise 6C.1, ensure you are confident with the following basic derivatives, as they often appear as either the inner or outer function:

在攻克练习6C.1之前，要确保你对以下基本导数了如指掌，因为它们常作为内层或外层函数出现：

Function (English/中文)	Derivative (English/中文)
xⁿ | xⁿ	n xⁿ⁻¹
eˣ | eˣ	eˣ
aˣ | aˣ (a > 0)	aˣ ln a
ln x | ln x	1/x
sin x | sin x	cos x
cos x | cos x	– sin x
tan x | tan x	sec² x

These results form the building blocks of chain rule problems. Always keep them accessible in your mind or on your formula booklet for quick reference during exercises and exams.

这些结果是链式法则问题的基石。时刻将它们记在脑海中或放在公式手册里，以便在练习和考试中快速查阅。

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6. Worked Examples Similar to Exercise 6C.1 | 类似练习6C.1的例题解析

Below are several detailed examples that closely mirror the type of questions you'll find in a typical Exercise 6C.1. Work through each one carefully.

以下是几个详细的例题，非常接近你在典型练习6C.1中会遇到的题型。请仔细研读每一题。

Example 1: Differentiate y = (5x + 2)⁷.

例1：求 y = (5x + 2)⁷ 的导数。

x + 2, so y = u⁷. Then dy/du = 7u⁶, du/dx = 5. Hence dy/dx = 7(5x + 2)⁶ × 5 = 35(5x + 2)⁶.

解：设 u = 5x + 2，则 y = u⁷。那么 dy/du = 7u⁶，du/dx = 5。因此 dy/dx = 7(5x + 2)⁶ × 5 = 35(5x + 2)⁶。

Example 2: Find the derivative of f(x) = √4x – 1.

例2：求 f(x) = √4x – 1 的导数。

Solution: Rewrite as (4x – 1)^½. Let u = 4x – 1. Then f'(x) = (½)(4x – 1)^–½ × 4 = 2 / √4x – 1.

解：改写为 (4x – 1)^½。设 u = 4x – 1。则 f'(x) = (½)(4x – 1)^–½ × 4 = 2 / √4x – 1。

Example 3: Differentiate y = 1/(3x² + 1).

例3：求 y = 1/(3x² + 1) 的导数。

Solution: Write as (3x² + 1)⁻¹. Let u = 3x² + 1, then dy/du = –1 · u⁻², du/dx = 6x. So dy/dx = –(6x) / (3x² + 1)².

解：写成 (3x² + 1)⁻¹。设 u = 3x² + 1，则 dy/du = –1 · u⁻²，du/dx = 6x。所以 dy/dx = –(6x) / (3x² + 1)²。

These examples illustrate how the chain rule generalizes the power rule to any linear or non-linear inner function, a core skill tested in Exercise 6C.1.

这些例子说明了链式法则如何将幂法则推广到任意线性或非线性的内层函数，这是练习6C.1中测试的核心技能。

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7. The Chain Rule with Trigonometric Functions | 链式法则与三角函数

vative is cos(u) × k = k cos(kx). This pattern extends to cos, tan, and their reciprocals.

三角函数经常出现在复合函数内部。对于 y = sin(kx)，外层函数为 sin(u)，内层函数为 u = kx。导数为 cos(u) × k = k cos(kx)。这一模式可推广到 cos、tan 及其倒数函数。

For a more complex example like y = cos(x²), let u = x², then dy/dx = –sin(u) × 2x = – as a factor while the trigonometric derivative remains unchanged apart from its argument.

对于更复杂的例子，如 y = cos(x²)，设 u = x²，则 dy/dx = –sin(u) × 2x = –2x sin(x²)。注意到链式法则只是将内层函数的导数作为一个因子乘上，而三角函数的导数除了自变量外保持不变。

When dealing with products like x sin(3x), you will need to combine the chain rule with the product rule. This is exactly the kind of challenge that appears towards the end of Exercise 6C.1.

当处理像 x sin(3x) 这样的乘积时，你需要将链式法则与乘法法则结合使用。这正是练习6C.1末尾会出现的一类挑战。

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8. The Chain Rule with Exponential and Logarithmic Functions | 链式法则与指数和对数函数

The natural exponential function eˣ rivative is itself. Therefore, for y = e^(g(x)), the chain rule gives dy/dx = e^(g(x)) × g'(x). For example, if y = e^2x, then dy/dx = e^2x × 2 = 2e^2x.

自然指数函数 eˣ 的特殊之处在于它的导数就是其本身。因此，对于 y = e^(g(x))，链式法则给出 dy/dx = e^(g(x)) × g'(x)。例如，若 y = e^2x，则 dy/dx = e^2x × 2 = 2e^2x。

For general exponential functions y = a^(g(x)), the derivative is a^(g(x)) ln a × g'(x). Similarly, for logarithms: if

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