Example 4.4.1: Differentiation of a Product | 例题4.4.1：乘积的微分

📚 Example 4.4.1: Differentiation of a Product | 例题4.4.1：乘积的微分

This worked example walks through the differentiation of a product of two functions using the product rule and the chain rule. It is a cornerstone of A-Level Pure Mathematics and frequently appears in AQA exams.

本例题详细讲解如何运用乘积法则和链式法则对两个函数的乘积进行微分。这是A-Level纯数学的核心内容，也是AQA考试中的常考题型。

1. The Question | 题目

Differentiate with respect to \( x \):

对 \( x \) 求导：

y = e^(2x) · cos(3x)

Find \(\frac{dy}{dx}\) and simplify your answer.

求 \(\frac{dy}{dx}\)，并化简结果。

2. Required Rules | 所需法则

To solve this problem, we need two fundamental rules of differentiation:

解决这一问题需要两个基本的微分法则：

Product Rule | 乘积法则: If \(y = uv\), then \(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\).
Chain Rule | 链式法则: If \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\).

In particular, for \(e^{kx}\), the derivative is \(ke^{kx}\); for \(\cos(kx)\), the derivative is \(-k\sin(kx)\).

特别地，对于 \(e^{kx}\)，导数为 \(ke^{kx}\)；对于 \(\cos(kx)\)，导数为 \(-k\sin(kx)\)。

3. Identify \(u\) and \(v\) | 识别 \(u\) 和 \(v\)

We set the function as a product of two simpler functions.

我们将该函数视为两个较简单函数的乘积。

u = e^(2x), v = cos(3x)

Then \(y = u \cdot v\).

于是 \(y = u \cdot v\)。

4. Differentiate \(u\) | 对 \(u\) 求导

For \(u = e^{2x}\), apply the chain rule: the outer derivative of \(e^{2x}\) is \(e^{2x}\), and multiply by the derivative of \(2x\), which is 2.

对于 \(u = e^{2x}\)，使用链式法则：\(e^{2x}\) 的外部导数为 \(e^{2x}\)，再乘以 \(2x\) 的导数 2。

\(\frac{du}{dx} = 2e^{2x}\)

5. Differentiate \(v\) | 对 \(v\) 求导

For \(v = \cos(3x)\), again use the chain rule. The derivative of \(\cos\) is \(-\sin\), and the derivative of \(3x\) is 3.

对于 \(v = \cos(3x)\)，同样使用链式法则。\(\cos\) 的导数是 \(-\sin\)，而 \(3x\) 的导数是 3。

\(\frac{dv}{dx} = -3\sin(3x)\)

6. Apply the Product Rule | 应用乘积法则

Now substitute into the product rule formula:

现在代入乘积法则公式：

\(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\)

因此

\(\frac{dy}{dx} = e^{2x}(-3\sin(3x)) + \cos(3x)(2e^{2x})\)

7. Simplify the Expression | 化简表达式

Factor out the common factor \(e^{2x}\):

提取公因式 \(e^{2x}\)：

\(\frac{dy}{dx} = e^{2x}(2\cos(3x) – 3\sin(3x))\)

This is the simplified derivative.

这就是化简后的导数。

8. Verification | 验证

We can check our answer by differentiating a numerical point or using a graphing calculator. For example, at \(x = 0\),

我们可以通过在某点求导数值或用图形计算器来验证答案。例如，在 \(x = 0\) 处，

\(y'(0) = e^{0}(2\cos(0) – 3\sin(0)) = 2\)

The original function at \(x=0\) has slope 2, consistent with a quick numerical approximation.

原函数在 \(x=0\) 处的斜率为 2，与快速数值近似一致。

9. Common Mistakes | 常见错误

Forgetting the chain rule | 忘记链式法则: Some students write \(\frac{d}{dx}e^{2x} = e^{2x}\) instead of \(2e^{2x}\).
Sign error in differentiating \(\cos\) | 求 \(\cos\) 导数时符号错误: Remember the derivative of \(\cos(ax)\) is \(-a\sin(ax)\).
Not simplifying | 不化简: Always factor out common terms to obtain the neatest form.

中文: 学生常犯错误包括：将 \(e^{2x}\) 的导数误写为 \(e^{2x}\)；忘记 \(\cos\) 导数前的负号；以及不提取公因式导致结果不简洁。

10. Practice Problem | 练习

Try differentiating \(y = e^{3x}\sin(2x)\) using the same method.

尝试用同样的方法对 \(y = e^{3x}\sin(2x)\) 求导。

Answer: \(\frac{dy}{dx} = e^{3x}(3\sin(2x) + 2\cos(2x))\).

答案：\(\frac{dy}{dx} = e^{3x}(3\sin(2x) + 2\cos(2x))\)。

11. Key Takeaways | 要点总结

The product rule is essential for differentiating products of functions.
The chain rule is needed when differentiating composite functions like \(e^{2x}\) and \(\cos(3x)\).
Always simplify your final answer by factoring.

中文: 乘积法则用于处理函数乘积；链式法则用于处理复合函数；最后务必通过提取公因式化简结果。

12. Exam Tip | 考试提示

In AQA exams, this type of question is worth 4–5 marks. Show every step clearly: name \(u\) and \(v\), write down their derivatives, then substitute into the product rule formula. This ensures you receive method marks even if the final simplification is incorrect.

在AQA考试中，此类题型占4–5分。请清晰写出每一步：设出 \(u\) 和 \(v\)，写出它们的导数，然后代入乘积法则公式。这样即使最后化简出错，也能获得方法分。

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Example 4.4.1: Differentiation of a Product | 例题4.4.1：乘积的微分