📚 Example 3.6.1: Using the Chain Rule | 例题 3.6.1：链式法则的应用
ically asks you to find dy/dx for a function of this type. This article works through the method step by step, reviews the underlying rule, and highlights exam technique.
在 AQA A-Level 数学中，微分是纯数学的核心工具。链式法则使你能够对复合函数求导，例如 y = (3x² + 5)⁴。例题 3.6.1 通常要求你求出这类函数的 dy/dx。本文将逐步讲解求解方法，复习基本法则，并强调考试技巧。
1. What the Chain Rule Is | 链式法则是什么
ule states that dy/dx is the product of dy/du and du/dx. In words, differentiate the outer function with respect to the inner variable, then multiply by the derivative of the inner function. This rule breaks a complicated derivative into two simpler parts.
如果 y = f(u) 且 u = g(x)，那么链式法则表明 dy/dx 等于 dy/du 与 du/dx 的乘积。也就是说，先对外层函数关于内层变量求导，再乘以内层函数的导数。这个法则将一个复杂的导数拆分成两个较简单的部分。
dy/dx = f'(g(x)) × g'(x)
2. Why the Chain Rule Matters | 链式法则为什么重要
Many AQA exam questions involve powers of linear or quadratic expressions, trigonometric functions such as sin(2x), exponentials like e5x, and logarithms. Without the chain rule, these cannot be differentiated correctly. It is one of the most frequently tested differentiation skills in AS and A-Level papers.
许多 AQA 考试题目涉及线性或二次表达式的幂、诸如 sin(2x) 的三角函数、像 e5x 的指数函数以及对数函数。如果没有链式法则，这些函数无法被正确求导。它是 AS 和 A-Level 试卷中最常考查的微分技能之一。
3. Worked Example 3.6.1 | 例题 3.6.1 解析
ur goal is to find dy/dx. The expression inside the brackets, 3x² + 5, is not simply x, so we cannot just apply the basic power rule directly. The chain rule is required.
考虑函数 y = (3x² + 5)⁴。我们的目标是求出 dy/dx。括号内的表达式 3x² + 5 不只是 x，因此我们不能直接应用基本幂法则。这里需要链式法则。
y = (3x² + 5)⁴
4. Step 1: Identify the Outer and Inner Functions | 第 1 步：识别外层与内层函数
unction is u⁴, and the inner function is 3x² + 5. This substitution is the key to applying the chain rule correctly.
设 u = 3x² + 5。那么 y = u⁴。外层函数是 u⁴，内层函数是 3x² + 5。这个代换是正确应用链式法则的关键。
u = 3x² + 5, y = u⁴
5. Step 2: Differentiate the Outer Function | 第 2 步：对外层函数求导
³. This step uses the standard power rule: bring down the power and reduce the exponent by 1.
对 y 关于 u 求导。因为 y = u⁴，导数为 dy/du = 4u³。这一步使用标准幂法则：将指数移下来，并将指数减 1。
dy/du = 4u³
6. Step 3: Differentiate the Inner Function | 第 3 步：对内层函数求导
erivative is du/dx = 6x. The constant 5 differentiates to 0, and 3x² differentiates to 6x.
对 u 关于 x 求导。因为 u = 3x² + 5，导数为 du/dx = 6x。常数 5 求导为 0，3x² 求导为 6x。
du/dx = 6x
7. Step 4: Multiply the Results | 第 4 步：相乘得到结果
Substitute u = 3x² + 5 back into the expression to obtain the final answer in terms of x.
应用链式法则，将 dy/du 与 du/dx 相乘。得到 dy/dx = 4u³ × 6x。将 u = 3x² + 5 代回表达式，得到以 x 表示的最终答案。
dy/dx = 24x(3x² + 5)³
8. Common Mistakes | 常见错误
³ instead of 24x(3x² + 5)³. Another mistake is expanding the bracket before differentiating, which is slow and error-prone.
忘记乘以内层导数是最常见的错误。学生经常写成 dy/dx = 4(3x² + 5)³，而不是 24x(3x² + 5)³。另一个错误是在求导前先展开括号，这样既慢又容易出错。
- Missing the inner derivative du/dx | 遗漏内层导数 du/dx
- Using the wrong power after differentiation | 求导后使用错误的幂
- Expanding brackets first instead of using the chain rule | 先展开括号，而不是使用链式法则
- Sign errors when the inner function has negative coefficients | 内层函数有负系数时出现符号错误
9. AQA Exam Tips | AQA 考试技巧
du and du/dx separately before multiplying. This earns method marks even if the final answer is incorrect.
在 AQA 试卷中，链式法则题目通常占 2 到 4 分。请清楚地写出代换 u = ...。在相乘之前分别写出 dy/du 和 du/dx。即使最终答案不正确，这样做也能获得方法分。
10. Practice Question | 练习题
e the same four-step method: identify u, differentiate y with respect to u, differentiate u with respect to x, then multiply and substitute back.
尝试对 y = (2x³ − 7)⁵ 求导。使用同样的四步方法：识别 u，对 y 关于 u 求导，对 u 关于 x 求导，然后相乘并代回。
Answer: dy/dx = 30x²(2x³ − 7)⁴
11. Extended Chain Rule with Trigonometry | 链式法则与三角函数的扩展
s(3x + 1). This pattern appears frequently in AQA trigonometric differentiation questions.
同样的方法适用于 y = sin(3x + 1)。设 u = 3x + 1，则 y = sin u。那么 dy/du = cos u，du/dx = 3，因此 dy/dx = 3 cos(3x + 1)。这种模式经常出现在 AQA 的三角函数求导题中。
dy/dx = 3 cos(3x + 1)
12. Summary | 总结
The chain rule is essential for differentiating composite functions. Identify the outer and inner functions, differentiate each, multiply the results, and substitute back. Use clear working to secure AQA method marks, and always check for the inner derivative.
链式法则对于求复合函数的导数至关重要。识别外层和内层函数，分别求导，将结果相乘，然后代回。使用清晰的解题步骤以获取 AQA 方法分，并始终检查是否包含内层导数。
Published by TutorHao | 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