📚 Example 5.5.3: The Chain Rule from AQA A-Level Mathematics | 示例 5.5.3：AQA A-Level 数学中的链式法则
on such as y = (3x² + 2)⁵ or y = sin(2x + 1), where one function is nested inside another.
AQA A-Level 数学教材中的示例 5.5.3 介绍了一个核心微分技巧：对复合函数应用链式法则。该示例通常要求你微分类似 y = (3x² + 2)⁵ 或 y = sin(2x + 1) 的函数，其中一种函数嵌套在另一种函数内部。
This example is fundamental because the chain rule reappears in almost every later topic: connected rates of change, implicit differentiation, parametric equations, and even integration by substitution. Mastering it here saves you from serious difficulties later in the course.
该示例至关重要，因为链式法则几乎会在后续所有主题中重现：相关变化率、隐函数微分、参数方程，甚至换元积分法。在这里掌握它，能让你避免在课程后期遇到严重困难。
1. Understanding Composite Functions | 理解复合函数
y = (3x² + 2)⁵, the inner function is g(x) = 3x² + 2 and the outer function is f(u) = u⁵.
在尝试示例 5.5.3 之前，理解什么是复合函数至关重要。复合函数写为 y = f(g(x))，其中 g(x) 是内层函数，f(u) 是外层函数。例如，在 y = (3x² + 2)⁵ 中，内层函数是 g(x) = 3x² + 2，外层函数是 f(u) = u⁵。
The chain rule, also called the 'function of a function' as dy/dx = dy/du × du/dx. In the notation used by AQA, this is often written as dy/dx = f'(g(x)) × g'(x).
链式法则也称为复合函数法则，是用于微分此类表达式的工具。数学上，它表示为 dy/dx = dy/du × du/dx。在 AQA 使用的记法中，这通常写作 dy/dx = f'(g(x)) × g'(x)。
2. The Chain Rule Formula | 链式法则公式
mula booklet and is essential for Paper 1 and Paper 2. For y = f(g(x)), the derivative is:
链式法则的公式在 AQA 公式手册中给出，并且对 Paper 1 和 Paper 2 都至关重要。对于 y = f(g(x))，其导数为：
dy/dx = f'(g(x)) × g'(x) = dy/du × du/dx
aling with a power function, the rule becomes: if y = [g(x)]ⁿ, then dy/dx = n[g(x)]ⁿ⁻¹ × g'(x). This is the format most commonly seen in Example 5.5.3.
当处理幂函数时，该法则变为：若 y = [g(x)]ⁿ，则 dy/dx = n[g(x)]ⁿ⁻¹ × g'(x)。这是示例 5.5.3 中最常见的格式。
posite functions, the rule extends naturally. If y = sin(g(x)), then dy/dx = cos(g(x)) × g'(x). Similarly, if y = cos(g(x)), then dy/dx = −sin(g(x)) × g'(x), and if y = e^(g(x)), then dy/dx = g'(x) × e^(g(x)).
对于三角函数复合函数，该法则自然延伸。若 y = sin(g(x))，则 dy/dx = cos(g(x)) × g'(x)。类似地，若 y = cos(g(x))，则 dy/dx = −sin(g(x)) × g'(x)，若 y = e^(g(x))，则 dy/dx = g'(x) × e^(g(x))。
3. Identifying the Outer and Inner Functions | 识别外层函数与内层函数
The first step in solving Example 5.5.3 is to identify which part of the function is 'outer' and which is 'inner'. A
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图