📚 Example 4.3.1: Using the Chain Rule | 例题4.3.1：链式法则的应用
This example demonstrates how to apply the chain rule to differentiate a composite function. We will work through a standard AQA A-Level Mathematics question step by step, then discuss common pitfalls and provide practice questions to reinforce your understanding.
本例题演示如何应用链式法则对复合函数求导。我们将逐步解答一道标准的AQA A-Level数学题，然后讨论常见错误并提供练习题以巩固你的理解。
1. The Problem Statement | 问题陈述
a composite function because we have a polynomial (3x² + 1) raised to a power 5. We cannot simply bring down the power as in basic differentiation; we must account for the inner polynomial.
本例要求我们求 y = (3x² + 1)⁵ 关于 x 的导数。这是一个复合函数，因为我们将多项式 (3x² + 1) 作为底数、5 作为指数。我们不能像基本求导那样简单地把指数乘下来，还必须考虑内部的多项式。
2. Why the Chain Rule Works | 为什么链式法则有效
The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) · g' f(g(x)). The overall rate of change is the product of the two individual rates of change.
链式法则指出：若 y = f(g(x))，则 dy/dx = f'(g(x)) · g'(x)。原因是 x 的微小变化会引起 g(x) 的微小变化，进而引起 f(g(x)) 的变化。总的变化率是这两个单独变化率的乘积。
3. Step 1: Identify Inner and Outer Functions | 第一步：识别内函数和外函数
er function is the power function u⁵, and the inner function is the quadratic u = 3x² + 1.
设 u = 3x² + 1（内函数），则 y = u⁵（外函数）。识别出这种拆分是正确应用链式法则的关键。外函数是幂函数 u⁵，内函数是二次函数 u = 3x² + 1。
4. Step 2: Differentiate the Outer Function | 第二步：对外函数求导
ule, dy/du = 5u⁴. We keep the inner function u as a single block; we do not expand it yet.
对 y = u⁵ 关于 u 求导。根据幂法则，dy/du = 5u⁴。我们把内函数 u 视为一个整体，暂时不展开它。
dy/du = 5u⁴
5. Step 3: Differentiate the Inner Function | 第三步：对内函数求导
x. The derivative of 3x² is 6x, and the constant 1 has derivative 0. Therefore du/dx = 6x.
对 u = 3x² + 1 关于 x 求导。3x² 的导数为 6x，常数 1 的导数为 0。因此 du/dx = 6x。
du/dx = 6x
6. Step 4: Multiply the Derivatives | 第四步：将导数相乘
ults, we get dy/dx = 5u⁴ × 6x. Replace u with 3x² + 1 to express the answer in terms of x.
链式法则给出 dy/dx = dy/du × du/dx。代入我们的结果，得到 dy/dx = 5u⁴ × 6x。用 3x² + 1 替换 u，将答案表示成关于 x 的式子。
dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴
7. Common Mistakes | 常见错误
Students often forget to multiply by du/dx and simply write 5(3x² + 1)⁴. Another common mistake is mis-differentiating the inner function, for example writing 3x instead of 6x. Always check both the outer and inner derivatives.
学生常常忘记乘上 du/dx，直接写成 5(3x² + 1)⁴。另一个常见错误是错误地求内函数的导数，例如写成 3x 而不是 6x。务必同时检查外函数和内函数的导数。
Forgetting the inner derivative: the final answer must include the factor 6x.
忘记内函数的导数：最终答案必须包含因子 6x。
Incorrect coefficient or sign: re-check du/dx carefully before multiplying.
系数或符号错误：相乘前仔细检查 du/dx。
8. Alternative Notation | 其他记号
We can also write the chain rule in function notation: if f(x) = (3x² + 1)⁵, then f' s is often quicker for simple problems and matches the notation used in exam mark schemes.
我们也可以用函数记号来写链式法则：若 f(x) = (3x² + 1)⁵，则 f'(x) = 5(3x² + 1)⁴ · 6x = 30x(3x² + 1)⁴。对于简单问题，这通常更快，也符合考试评分标准中的记号。
9. Practice Problems | 练习问题
Try these on your own to build confidence with the chain rule.
请自行尝试以下题目以增强对链式法则的信心。
Differentiate y = (2x³ + 5)⁴.
求 y = (2x³ + 5)⁴ 的导数。
Differentiate y = sin(3x²).
求 y = sin(3x²) 的导数。
Differentiate y = e⁽²ˣ⁺¹⁾.
求 y = e⁽²ˣ⁺¹⁾ 的导数。
10. Solutions | 解答
1) Let u = 2x³ + 5, y = u⁴. Then dy/du = 4u³, du/dx = 6x². So dy/dx = 24x²(2x³ + 5)³.
1) 设 u = 2x³ + 5，y = u⁴。则 dy/du = 4u³，du/dx = 6x²。因此 dy/dx = 24x²(2x³ + 5)³。
2) Let u = 3x², y = sin u. Then dy/du = cos u, du/dx = 6x. So dy/dx = 6x cos(3x²).
2) 设 u = 3x²，y = sin u。则 dy/du = cos u，du/dx = 6x。因此 dy/dx = 6x cos(3x²)。
3) Let u = 2x + 1, y = eᵘ. Then dy/du = eᵘ, du/dx = 2. So dy/dx = 2e⁽²ˣ⁺¹⁾.
3) 设 u = 2x + 1，y = eᵘ。则 dy/du = eᵘ，du/dx = 2。因此 dy/dx = 2e⁽²ˣ⁺¹⁾。
11. Key Takeaways | 要点总结
Remember the core steps: identify the inner and outer functions, differentiate the outer function, differentiate the inner function, and multiply the two results. Always express the final answer in terms of the original variable.
记住核心步骤：识别内函数和外函数，外函数求导，内函数求导，然后将两者相乘。始终用原变量表示最终答案。
Chain rule: dy/dx = dy/du × du/dx.
链式法则：dy/dx = dy/du × du/dx。
Write u before differentiating to avoid confusion.
求导前先写出 u 以避免混淆。
Check answers by re-expanding simpler examples.
对于简单例子，可以通过重新展开来检查答案。
12. Conclusion | 结论
In this article, we have explored Example 4.3.1 from a typical AQA A-Level Mathematics textbook. Mastering the chain rule is essential for differentiating composite functions in exams. Practice with a variety of functions to build fluency and accuracy.
在本文中，我们详细解析了典型AQA A-Level数学教材中的例题4.3.1。掌握链式法则对于在考试中求复合函数的导数至关重要。通过练习不同类型的函数来提升熟练度和准确性。
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