IB Math: Single-Parameter Chain Rule | IB数学：单参数链式法则

📚 IB Math: Single-Parameter Chain Rule | IB数学：单参数链式法则

The single-parameter chain rule is one of the most frequently tested differentiation techniques in IB Mathematics, appearing in both Analysis and Approaches (AA) and Applications and Interpretation (AI), at Standard Level (SL) and Higher Level (HL).

单参数链式法则是IB数学中最常考到的微分技巧之一，在分析与方法（AA）以及应用与解释（AI）的标准级（SL）和高等级（HL）中都会出现。

This guide breaks down the rule from first principles, gives you a clear step-by-step method, and provides worked examples modelled on real IB exam questions.

本指南从基本原理出发拆解这一法则，为你提供清晰的逐步解题方法，并给出仿照IB真题风格编写的例题解答。

1. What is the Single-Parameter Chain Rule? | 什么是单参数链式法则？

The chain rule is used to differentiate a composite function — a function inside another function. In the single-parameter case, there is only one independent variable, usually x, and one intermediate variable, often u.

链式法则用于对复合函数求导——即一个函数嵌套在另一个函数之中。在单参数情形下，只有一个自变量（通常为 x）和一个中间变量（通常为 u）。

if y = f(u) and u = g(x), then dy/dx = (dy/du) × (du/dx)

In words: differentiate the outer function, leaving the inner function unchanged, then multiply by the derivative of the inner function.

用语言表述：先对外层函数求导，保持内层函数不变，再乘以内层函数的导数。

2. The Rule in Leibniz Notation | 莱布尼茨记号下的法则

Leibniz notation makes the chain rule look almost like fraction cancellation, though this is only a mnemonic — the derivatives are not fractions.

莱布尼茨记号使链式法则看起来几乎像分数的约分，但这只是一种记忆辅助——导数并不是分数。

dy/dx = dy/du × du/dx

Example 1

Let y = (3x² + 2)⁵. Here u = 3x² + 2, so y = u⁵.

设 y = (3x² + 2)⁵。此处 u = 3x² + 2，因此 y = u⁵。

dy/du = 5u⁴, du/dx = 6x.

因此 dy/du = 5u⁴，du/dx = 6x。

dy/dx = 5u⁴ × 6x = 30x(3x² + 2)⁴.

所以 dy/dx = 5u⁴ × 6x = 30x(3x² + 2)⁴。

3. Function Notation: f(g(x)) | 函数记号：f(g(x))

In function notation, the chain rule is written as:

在函数记号下，链式法则可以写为：

d/dx [ f(g(x)) ] = f′(g(x)) × g′(x)

Example 2

Differentiate h(x) = sin(2x³).

求 h(x) = sin(2x³) 的导数。

Here f(u) = sin(u), g(x) = 2x³.

这里 f(u) = sin(u)，g(x) = 2x³。

f′(u) = cos(u), g′(x) = 6x².

所以 f′(u) = cos(u)，g′(x) = 6x²。

h′(x) = cos(2x³) × 6x² = 6x² cos(2x³).

因此 h′(x) = cos(2x³) × 6x² = 6x² cos(2x³)。

4. Step-by-Step Method | 逐步解题法

To apply the single-parameter chain rule reliably, follow these five steps.

要可靠地运用单参数链式法则，请遵循以下五个步骤。

Step 1: Identify the outer function f and the inner function g. | 第一步：识别外层函数 f 和内层函数 g。
Step 2: Differentiate the outer function, keeping g(x) intact. | 第二步：对 g(x) 保持不动，对外层函数求导。
Step 3: Differentiate the inner function g(x) with respect to x. | 第三步：对内层函数 g(x) 关于 x 求导。
Step 4: Multiply the two results together. | 第四步：将两个结果相乘。
Step 5: Simplify, if possible, using trigonometric identities or algebraic manipulation. | 第五步：若可能，利用三角恒等式或代数化简简化结果。

5. Common Functions Requiring the Chain Rule | 常见需要链式法则的函数

Outer Function f(u)	Derivative f′(u)	Example with Inner g(x)
uⁿ	n uⁿ⁻¹	(2x+1)⁷
eᵘ	eᵘ	e^(3x²)
ln(u)	1/u	ln(x²+1)
sin(u)	cos(u)	sin(√x)
cos(u)	−sin(u)	cos(ln x)
tan(u)	sec²(u)	tan(eˣ)

6. Worked Example: Algebraic Function | 例题：代数函数

Differentiate y = √(x² + 3x).

求 y = √(x² + 3x) 的导数。

Rewrite the square root as a power: y = (x² + 3x)^{1/2}.

将根号改写为幂：y = (x² + 3x)^{1/2}。

Let u = x² + 3x. Then y = u^{1/2}.

令 u = x² + 3x，则 y = u^{1/2}。

dy/du = (1/2) u^{-1/2}, du/dx = 2x + 3.

因此 dy/du = (1/2) u^{-1/2}，du/dx = 2x + 3。

dy/dx = (1/2)(x² + 3x)^{-1/2}(2x + 3) = (2x + 3) / (2√(x² + 3x)).

所以 dy/dx = (1/2)(x² + 3x)^{-1/2}(2x + 3) = (2x + 3) / (2√(x² + 3x))。

7. Worked Example: Exponential with a Twist | 例题：带变化的指数函数

Differentiate f(x) = e^(sin x).

求 f(x) = e^(sin x) 的导数。

Outer function: eᵘ; inner function: u = sin x.

外层函数为 eᵘ，内层函数为 u = sin x。

f′(u) = eᵘ, u′ = cos x.

f′(u) = eᵘ，u′ = cos x。

f′(x) = e^(sin x) × cos x = cos x · e^(sin x).

因此 f′(x) = e^(sin x) × cos x = cos x · e^(sin x)。

8. Worked Example: Logarithmic Function | 例题：对数函数

Differentiate y = ln(1 + x²).

求 y = ln(1 + x²) 的导数。

Here u = 1 + x², so dy/du = 1/u and du/dx = 2x.

此处 u = 1 + x²，所以 dy/du = 1/u，du/dx = 2x。

dy/dx = (1 / (1 + x²)) × 2x = 2x / (1 + x²).

因此 dy/dx = (1 / (1 + x²)) × 2x = 2x / (1 + x²)。

9. Chain Rule with Trigonometric Powers | 链式法则与三角幂函数

Differentiate y = sin⁴(x). Note that sin⁴(x) means (sin x)⁴.

求 y = sin⁴(x) 的导数。注意 sin⁴(x) 表示 (sin x)⁴。

Outer: u⁴, inner: u = sin x.

外层为 u⁴，内层为 u = sin x。

dy/du = 4u³, du/dx = cos x.

所以 dy/du = 4u³，du/dx = cos x。

dy/dx = 4(sin x)³ × cos x = 4 sin³ x cos x.

因此 dy/dx = 4(sin x)³ × cos x = 4 sin³ x cos x。

10. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Mistake 1: Forgetting the Inner Derivative

Students often write the derivative of (2x+1)⁵ as 5(2x+1)⁴, omitting the factor 2.

学生经常将 (2x+1)⁵ 的导数写成 5(2x+1)⁴，漏掉了因子 2。

Always multiply by du/dx.

永远不要忘记乘上 du/dx。

Mistake 2: Confusing sinⁿx with sin(xⁿ)

sin²x = (sin x)², while sin(x²) = sin(x × x). Their derivatives are different.

sin²x = (sin x)²，而 sin(x²) = sin(x × x)。两者的导数不同。

d/dx [sin²x] = 2 sin x cos x, but d/dx [sin(x²)] = 2x cos(x²)

Mistake 3: Rewriting √x incorrectly

Note that √x = x^{1/2}, not x² or x^{-2}. The power 1/2 is essential.

注意 √x = x^{1/2}，而不是 x² 或 x^{-2}。指数 1/2 至关重要。

11. Exam-Style Question | 真题风格练习

Find the derivative of each function with respect to x:

求下列各函数关于 x 的导数：

a) y = (5x − 2)³
b) y = e^(cos x)
c) y = ln(3x + 7)
d) y = tan(2x)

Solutions

a) Let u = 5x − 2, dy/du = 3u², du/dx = 5, so dy/dx = 3(5x − 2)² × 5 = 15(5x − 2)².

a) 令 u = 5x − 2，dy/du = 3u²，du/dx = 5，所以 dy/dx = 3(5x − 2)² × 5 = 15(5x − 2)²。

b) dy/dx = −sin x · e^(cos x).

b) dy/dx = −sin x · e^(cos x)。

c) dy/dx = 3 / (3x + 7).

c) dy/dx = 3 / (3x + 7)。

d) dy/dx = 2 sec²(2x).

d) dy/dx = 2 sec²(2x)。

12. Quick Revision Checklist | 快速复习清单

Identify the inner function first. | 先识别内层函数。
Use Leibniz notation to track your steps. | 使用莱布尼茨记号追踪计算步骤。
Remember to multiply by the inner derivative. | 记得乘以内层导数。
Practise with past IB questions to build speed. | 用IB真题练习以提升速度。
Check your answer by expanding or using technology. | 通过展开或使用技术工具检查答案。

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IB Math: Single-Parameter Chain Rule | IB数学：单参数链式法则