Example 6.10.3: Combining the Product and Chain Rules | 例 6.10.3：乘积法则与链式法则的综合运用

📚 Example 6.10.3: Combining the Product and Chain Rules | 例 6.10.3：乘积法则与链式法则的综合运用

This worked example develops a complete solution to a typical AQA A-Level Mathematics differentiation problem. It demonstrates how to combine the product rule and the chain rule in a single process, and how to present your working clearly in the Pure Mathematics paper.

本例题完整展示一道典型 AQA A-Level 数学求导题的解法和书写过程。它说明了如何在一个求导问题中综合运用乘积法则与链式法则，以及如何在纯数试卷中清晰地呈现解题步骤。

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1. The Problem Statement | 题目表述

Differentiate with respect to x:

关于 x 求导：

y = (3x² + 1)⁷ sin(2x)

This expression is the product of a composite power function and a trigonometric function. The function is not a single standard derivative, so we must break it into components before applying any rule.

该表达式是一个复合幂函数与一个三角函数的乘积。这个函数不属于任何单一的标准导数形式，因此在应用任何法则之前，我们必须先把它拆分为若干组成部分。

The most efficient approach is to differentiate each factor separately, applying the chain rule where necessary, and then combine the results using the product rule. This mirrors the standard layout expected in AQA mark schemes.

最有效的方法是分别对每个因子求导（在必要时使用链式法则），然后再利用乘积法则将结果组合起来。这种写法也与 AQA 评分标准所期望的规范步骤一致。

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2. Identifying the Required Rules | 确定所需法则

For any two differentiable functions u(x) and v(x), the product rule states:

对于任意两个可导函数 u(x) 与 v(x)，乘积法则表述为：

d/dx [u(x)v(x)] = u(x)·dv/dx + v(x)·du/dx

Here we take u(x) = (3x² + 1)⁷ and v(x) = sin(2x). Both of these components are themselves composite functions, so each must be differentiated using the chain rule before the product rule can be assembled.

在此我们令 u(x) = (3x² + 1)⁷，v(x) = sin(2x)。这两个分量本身都是复合函数，因此在运用乘积法则之前，必须先用链式法则分别对它们求导。

Recognising the correct order of operations is crucial. If you try to apply the product rule first without differentiating the inner parts correctly, you will lose both accuracy and method marks.

正确判断运算顺序至关重要。如果先套用乘积法则却没有正确求出内层部分的导数，你不仅会算错答案，还会丢掉方法分。

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3. Differentiating the First Factor | 对第一个因子求导

Consider u(x) = (3x² + 1)⁷. Write u as an outer power function with inner function g(x) = 3x² + 1. The chain rule gives:

考虑 u(x) = (3x² + 1)⁷。将 u 视为外层幂函数与内层函数 g(x) = 3x² + 1 的复合。链式法则给出：

du/dx = 7(3x² + 1)⁶ × d/dx(3x² + 1) = 7(3x² + 1)⁶ × 6x = 42x(3x² + 1)⁶

Notice that the index is lowered from 7 to 6, and we must multiply by the derivative of the inner function, which is 6x. This inner derivative is the step most frequently omitted by students.

注意指数从 7 降至 6，而且我们必须乘以内层函数的导数 6x。这个内层导数正是学生最常遗漏的一步。

If you prefer, you can write this in the Leibniz notation for the chain rule as du/dx = du/dg × dg/dx, which makes the structure explicit and reduces the chance of error.

如果你习惯使用莱布尼茨记号，也可以将链式法则写成 du/dx = du/dg × dg/dx，这样结构一目了然，能有效降低出错概率。

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4. Differentiating the Second Factor | 对第二个因子求导

Now consider v(x) = sin(2x). The derivative of sine is cosine, but because the argument is 2x rather than x, the chain rule requires us to multiply by the derivative of 2x:

接下来考虑 v(x) = sin(2x)。正弦的导数是余弦，但由于自变量是 2x 而不是 x，链式法则要求我们乘以内层函数 2x 的导数：

dv/dx = 2cos(2x)

The factor 2 is essential. Losing it is a classic error that leaves the final answer incorrect by a factor of 2 in one term. Make it a habit to write the inner derivative down before simplifying.

因子 2 是必不可少的。丢失它是一个经典错误，会使最终答案中某一项相差一个因子 2。请养成先把内层导数写出来再化简的习惯。

In general, for any composite function of the form v = sin(ax + b), the derivative is a·cos(ax + b). Memorising this pattern can save time in the exam, but showing the chain rule step gains more method marks.

一般而言，对于形如 v = sin(ax + b) 的复合函数，其导数为 a·cos(ax + b)。记住这一规律可以在考试中节省时间，但写出链式法则的步骤能获得更多方法分。

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5. Applying the Product Rule | 应用乘积法则

We have now computed the necessary components. The following table summarises them for clarity:

现在我们已经求出了所有必要的分量。下表将其清晰汇总：

u	du/dx	v	dv/dx
(3x² + 1)⁷	42x(3x² + 1)⁶	sin(2x)	2cos(2x)

Substituting these into the product rule formula gives:

将这些结果代入乘积法则公式：

dy/dx = u·dv/dx + v·du/dx

dy/dx = (3x² +

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