📚 Worked Example 4.2.2: Chain Rule for a Composite Polynomial | 示例 4.2.2：复合多项式的链式法则
In AQA A-Level Pure Mathematics, the chain rule is one of the most frequently examined differentiation techniques. This worked example develops a complete AQA-style solution for a composite polynomial function, showing how to differentiate from the outside in, how to avoid common errors, and how to present working clearly in an examination.
在 AQA A-Level 纯数学中，链式法则是最常考查的求导技巧之一。本示例围绕一个复合多项式函数展开完整的 AQA 风格解答，展示如何从外到内求导、如何避免常见错误，以及如何在考试中清晰呈现解题过程。
1. AQA Specification Link | AQA 考试大纲链接
The chain rule appears in the AQA Pure Mathematics specification under differentiation. It is introduced in Year 1 and then extended in Year 2 with exponential, logarithmic, trigonometric and parametric functions. Marks are usually awarded for correct structure, substitution and final derivative form.
链式法则出现在 AQA 纯数学考纲的微分部分。它在第一年引入，第二年在指数函数、对数函数、三角函数和参数函数中进一步拓展。考试通常对正确的结构、代换和最终导数形式给分。
2. The Problem Statement | 题目陈述
Example 4.2.2 asks us to differentiate the following function with respect to x:
示例 4.2.2 要求我们对下列函数关于 x 求导：
y = (3x² + 5x − 1)⁴
The expression is a composite function, so we cannot simply bring the power down and subtract 1 from the exponent. The inner quadratic expression has its own derivative that must be included.
这是一个复合函数，因此不能简单地把指数拿下来并减 1。内层的二次表达式有自己的导数，必须包含在结果中。
3. Recognising the Composite Structure | 识别复合结构
the form y = f(g(x)), where f(u) = u⁴ and g(x) = 3x² + 5x − 1. The chain rule then states:
将函数写成 y = f(g(x)) 的形式，其中 f(u) = u⁴，g(x) = 3x² + 5x − 1。链式法则表示：
dy/dx = f'(g(x)) × g'(x)
u/dx. This form is often clearer in exam working because it separates the outer and inner derivatives.
等价地，使用莱布尼茨记号可以写成 dy/dx = dy/du × du/dx。这种形式在考试解题中通常更清晰，因为它把外层导数和内层导数分开。
4. Step 1: Set Up the Substitution | 第一步：设替换变量
Let u equal the inner function. Then rewrite y in terms of u:
设 u 等于内层函数，然后将 y 用 u 表示：
u = 3x² + 5x − 1
y = u⁴
This substitution isolates the outer power from the inner quadratic. AQA examiners expect to see this step clearly, especially when the question carries method marks.
这个代换把外层幂函数和内层二次函数分离开。AQA 考官希望看到这一步清晰写出，尤其是在题目有方法分的情况下。
5. Step 2: Differentiate the Outer Function | 第二步：对外层函数求导
Differentiate y with respect to u. Since y = u⁴, the power rule gives:
对 y 关于 u 求导。因为 y = u⁴，幂法则得到：
dy/du = 4u³
At this intermediate stage, the variable is u, not x. We keep u in the derivative because we have not yet substituted the inner expression back.
在这个中间步骤中，变量是 u 而不是 x。我们暂时保留 u，因为还没有把内层表达式代回。
6. Step 3: Differentiate the Inner Function | 第三步：对内层函数求导
Now differentiate u with respect to x. Applying the power rule term by term:
现在对 u 关于 x 求导。逐项应用幂法则：
du/dx = 6x + 5
This comes from differentiating 3x² to obtain 6x, and 5x to obtain 5. The constant term −1 differentiates to 0, so it does not appear in the result.
这是由 3x² 求导得到 6x，5x 求导得到 5。常数项 −1 求导为 0，所以不会出现在结果中。
7. Step 4: Multiply and Substitute Back | 第四步：相乘并代回
Apply the chain rule by multiplying the two derivatives:
应用链式法则，将两个导数相乘：
dy/dx = dy/du × du/dx = 4u³ × (6x + 5)
titute u = 3x² + 5x − 1 back into the expression. This gives the derivative in terms of x:
现在将 u = 3x² + 5x − 1 代回表达式中，得到以 x 表示的导数：
dy/dx = 4(3x² + 5x − 1)³(6x + 5)
It is important to substitute fully before moving on. Leaving u in the final answer would be an incomplete derivative for an AQA question.
在继续之前必须完全代回。如果在最终答案中保留 u，AQA 题目中会被视为导数不完整。
8. Final Answer in Full | 最终答案的完整形式
The derivative can also be written with the linear factor first:
这个导数也可以把线性因式写在前面：
dy/dx = 4(6x + 5)(3x² + 5x − 1)³
Both forms are acceptable to AQA. Unless the question explicitly asks for an expanded answer, leaving the derivative in factorised form is usually safer and clearer because it avoids unnecessary algebra errors.
这两种形式 AQA 都接受。除非题目明确要求展开，否则保持因式分解形式通常更安全、更清晰，因为可以避免不必要的代数错误。
9. Evaluating at a Point | 在某点处求值
dient at x = 1, substitute x = 1 into the derivative. First evaluate the inner expression:
如果后续问题要求求 x = 1 处的梯度，将 x = 1 代入导数。先计算内层表达式：
u = 3(1)² + 5(1) − 1 = 7
Then substitute into dy/dx:
然后代入 dy/dx：
dy/dx = 4 × 7³ × (6(1) + 5) = 4 × 343 × 11 = 15092
1. In a geometrical context, it is the gradient of the tangent to the curve at that point.
这个值给出了 y 关于 x 在 x = 1 处的瞬时变化率。在几何背景下，它表示该点处曲线的切线斜率。
10. Common Errors and Mark Scheme Advice | 常见错误与评分建议
One of the most common AQA errors is forgetting to multiply by du/dx. A student might write only 4(3x² + 5x − 1)³, but this omits the inner derivative and loses the final accuracy mark.
AQA 考试中最常见的错误之一是忘记乘 du/dx。有学生可能只写 4(3x² + 5x − 1)³，但这遗漏了内层导数，会失去最后的准确分。
Another common error is applying the power rule incorrectly by differentiating the inner function first or mixing the order of multiplication. The correct structure is always: derivative of the outer function with respect to u, then derivative of the inner function with respect to x.
另一个常见错误是错误地应用幂法则，先对内层函数求导或混淆乘法顺序。正确结构始终是：先对 y 关于 u 求外层导数，再对 u 关于 x 求内层导数。
Some students expand (3x² + 5x − 1)⁴ before differentiating. This is possible but very tedious and increases the chance of algebraic mistakes. The chain rule is the intended method and is much more efficient.
有些学生先展开 (3x² + 5x − 1)⁴ 再求导。这是可行的，但非常繁琐，并且增加代数错误的机会。链式法则才是预期方法，而且效率高得多。
11. Practice Question | 练习题
Try this AQA-style question to test your understanding:
尝试下面的 AQA 风格题目，检验你的理解：
Differentiate y = (2x³ − 4x + 1)⁵ with respect to x.
he value of dy/dx at x = 0. Use the same four-step structure shown in this worked example.
然后求 x = 0 处 dy/dx 的值。使用与本示例相同的四步结构。
− 4. The final derivative is 5(2x³ − 4x + 1)⁴(6x² − 4), and at x = 0 the gradient is −20.
供检查使用：设 u = 2x³ − 4x + 1，则 dy/du = 5u⁴，du/dx = 6x² − 4。最终导数为 5(2x³ − 4x + 1)⁴(6x² − 4)，在 x = 0 处梯度为 −20。
12. Summary | 总结
The chain rule requires differentiating from the outside in, then multiplying by the derivative of the inside. Keep the substitution clear and substitute back fully before stating the final answer.
链式法则要求从外到内求导，然后乘以内层函数的导数。保持代换步骤清晰，并在给出最终答案之前完全代回。
In AQA examinations, method marks are awarded for correct structure, so showing dy/du and du/dx separately is valuable. Factorised derivative form is usually sufficient unless expansion is specifically requested.
在 AQA 考试中，方法分是根据正确结构给出的，因此分别写出 dy/du 和 du/dx 非常有价值。除非特别要求展开，否则因式分解形式的导数通常就足够了。
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