Differentiation Example 5.2.3 | 微分示例 5.2.3

📚 Differentiation Example 5.2.3 | 微分示例 5.2.3

This article walks through Example 5.2.3 from the AQA A-Level Mathematics syllabus, focusing on applying differentiation techniques to a compound function. We will break down the problem step by step, explain the rules used, and highlight common pitfalls to avoid in the exam.

本文详细解析 AQA A-Level 数学大纲中的示例 5.2.3，重点在于对复合函数应用微分技巧。我们将分步骤拆解问题，解释所使用的方法，并强调考试中常见的易错点。

1. Understanding the Problem | 理解题目

Before attempting any differentiation problem, it is essential to identify the type of function involved. A compound function is one in which a function is applied to the result of another function. In Example 5.2.3, we typically encounter an expression of the form \(f(x) = (ax + b)^n\), where \(a\), \(b\), and \(n\) are constants.

在尝试任何微分问题之前，必须首先识别所涉及函数的类型。复合函数是指将一个函数应用于另一个函数结果的形式。在示例 5.2.3 中，我们通常会遇到形如 \(f(x) = (ax + b)^n\) 的表达式，其中 \(a\)、\(b\) 和 \(n\) 为常数。

2. The Chain Rule | 链式法则

The chain rule is the fundamental tool for differentiating composite functions. It states that if \(y = (u(x))^n\), then \(\frac{dy}{dx} = n \cdot (u(x))^{n-1} \cdot \frac{du}{dx}\). In Leibniz notation, this is written as \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\).

链式法则是求解复合函数微分的基本工具。它指出，若 \(y = (u(x))^n\)，则 \(\frac{dy}{dx} = n \cdot (u(x))^{n-1} \cdot \frac{du}{dx}\)。用莱布尼茨记号表达为 \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)。

3. A Concrete Example | 具体示例

Let us consider Example 5.2.3: Differentiate \(y = (3x + 2)^5\) with respect to \(x\). Here, we identify the inner function \(u = 3x + 2\) and the outer power function \(n = 5\).

我们来看示例 5.2.3：对 \(y = (3x + 2)^5\) 关于 \(x\) 求导。这里，我们设内层函数 \(u = 3x + 2\)，外层幂指数 \(n = 5\)。

4. Step-by-Step Differentiation | 逐步求导

First, differentiate the outer function with respect to \(u\): \(\frac{dy}{du} = 5(3x + 2)^4\). Next, differentiate the inner function with respect to \(x\): \(\frac{du}{dx} = 3\). Finally, multiply the two derivatives together to obtain the result.

首先，对 \(u\) 求外函数的导数：\(\frac{dy}{du} = 5(3x + 2)^4\)。接着，对 \(x\) 求内函数的导数：\(\frac{du}{dx} = 3\)。最后，将两个导数相乘得到最终结果。

\(\frac{dy}{dx} = 5(3x + 2)^4 \times 3 = 15(3x + 2)^4\)

5. Alternative Notation | 替代记号

It is also acceptable to use function notation. We can write \(f(x) = (3x + 2)^5\) and state \(f'(x) = 15(3x + 2)^4\). Both forms are widely accepted in AQA examinations, provided that the method is clearly communicated.

使用函数记号同样可行。我们可以设 \(f(x) = (3x + 2)^5\)，并写出 \(f'(x) = 15(3x + 2)^4\)。只要方法表达清晰，这两种形式在 AQA 考试中均被广泛接受。

6. Application of the Product Rule | 乘积法则的应用

Some variations of Example 5.2.3 involve a product of two functions, such as \(y = x^2(2x + 1)^3\). In such cases, we must combine the product rule with the chain rule. The product rule states \(\frac{d}{dx}[u \cdot v] = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}\).

示例 5.2.3 的某些变体涉及两个函数的乘积，例如 \(y = x^2(2x + 1)^3\)。此时，必须将乘积法则与链式法则结合使用。乘积法则指出 \(\frac{d}{dx}[u \cdot v] = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}\)。

Let \(u = x^2\) and \(v = (2x + 1)^3\). Then \(\frac{du}{dx} = 2x\) and \(\frac{dv}{dx} = 3(2x + 1)^2 \cdot 2 = 6(2x + 1)^2\). Applying the product rule gives \(\frac{dy}{dx} = x^2 \cdot 6(2x + 1)^2 + (2x + 1)^3 \cdot 2x\).

设 \(u = x^2\)，\(v = (2x + 1)^3\)。则 \(\frac{du}{dx} = 2x\)，\(\frac{dv}{dx} = 3(2x + 1)^2 \cdot 2 = 6(2x + 1)^2\)。应用乘积法则得到 \(\frac{dy}{dx} = x^2 \cdot 6(2x + 1)^2 + (2x + 1)^3 \cdot 2x\)。

7. Simplifying the Result | 化简结果

After applying the product rule, it is often necessary to simplify the expression. Factoring out the common factor \((2x + 1)^2 \cdot x\) from both terms yields a more compact form.

应用乘积法则后，通常需要化简表达式。从两项中提取公因式 \((2x + 1)^2 \cdot x\) 可以得到更简洁的形式。

\(\frac{dy}{dx} = x(2x + 1)^2 \cdot [6x + 2(2x + 1)] = x(2x + 1)^2(10x + 2)\)

This simplification is useful for finding stationary points or solving equations involving the derivative.

这种化简对于求驻点或解包含导数的方程非常有用。

8. The Quotient Rule | 商法则

If the function in Example 5.2.3 is presented as a quotient, such as \(y = \frac{(x + 1)^4}{x^2}\), the quotient rule becomes necessary. The rule states \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} – u \cdot \frac{dv}{dx}}{v^2}\).

如果示例 5.2.3 中的函数以商的形式呈现，例如 \(y = \frac{(x + 1)^4}{x^2}\)，则需要使用商法则。该法则指出 \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} – u \cdot \frac{dv}{dx}}{v^2}\)。

9. Worked Quotient Example | 商的示例演练

Set \(u = (x + 1)^4\) and \(v = x^2\). Then \(\frac{du}{dx} = 4(x + 1)^3\) and \(\frac{dv}{dx} = 2x\). Substituting into the quotient rule formula gives the derivative.

设 \(u = (x + 1)^4\)，\(v = x^2\)。则 \(\frac{du}{dx} = 4(x + 1)^3\)，\(\frac{dv}{dx} = 2x\)。将其代入商法则公式即可得到导数。

\(\frac{dy}{dx} = \frac{x^2 \cdot 4(x + 1)^3 – (x + 1)^4 \cdot 2x}{x^4}\)

10. Common Mistakes | 常见错误

One common mistake is forgetting to multiply by the derivative of the inner function when using the chain rule. Another frequent error is incorrectly applying the power rule to the entire bracket before differentiating the inside. Both errors lead to incorrect answers in the exam.

一个常见错误是使用链式法则时忘记乘以内函数的导数。另一个常见错误是在对整个括号应用幂规则之后再对内层求导。这两种错误都会导致考试中答案错误。

Always identify the inner function first. | 始终先识别内层函数。
Double-check the coefficient of the inner function. | 仔细检查内函数的系数。
Simplify the final answer where possible. | 尽可能化简最终答案。

11. Exam-Style Question | 考试风格题目

A typical AQA question might ask: “Find the derivative of \(y = \sqrt{(4x – 1)^3}\).” Rewriting the square root as a fractional index is the first step. Then apply the chain rule carefully.

一个典型的 AQA 题目可能为：”求 \(y = \sqrt{(4x – 1)^3}\) 的导数。” 首先将平方根改写为分数指数。然后小心应用链式法则。

\(y = (4x – 1)^{3/2}\)

\(\frac{dy}{dx} = \frac{3}{2}(4x – 1)^{1/2} \cdot 4 = 6(4x – 1)^{1/2}\)

12. Summary and Revision Tips | 总结与复习建议

Example 5.2.3 illustrates the essence of differentiation for compound functions. Mastery of the chain rule, along with the product and quotient rules, is essential for success in AQA A-Level Mathematics. Practice with a variety of functions, including those with negative and fractional powers, to build confidence.

示例 5.2.3 阐述了复合函数微分的核心。熟练掌握链式法则以及乘积法则和商法则，对于在 AQA A-Level 数学中取得成功至关重要。练习各种函数（包括负指数和分数指数）有助于建立信心。

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Differentiation Example 5.2.3 | 微分示例 5.2.3