Example 5.2.1: Differentiation of a Composite Function | 示例5.2.1：复合函数的微分

📚 Example 5.2.1: Differentiation of a Composite Function | 示例5.2.1：复合函数的微分

In this article, we work through Example 5.2.1 from the AQA A-Level Mathematics syllabus. This example focuses on differentiating a composite function using the chain rule, a key technique in calculus that every student must master.

本文将带领大家逐步解析 AQA A-Level 数学大纲中的示例 5.2.1。该示例重点在于使用链式法则对复合函数进行求导，这是微积分中每位学生都必须掌握的核心技巧。

1. The Context of Example 5.2.1 | 示例5.2.1的背景

Example 5.2.1 is part of Chapter 5 which introduces differentiation rules. The example is designed to illustrate how to differentiate a function that is composed of an inner function and an outer function. In particular, it shows a straightforward application of the chain rule.

示例 5.2.1 是第 5 章中介绍微分法则的一部分。该例题旨在说明如何对由一个内层函数和一个外层函数复合而成的函数进行求导。具体来说，它演示了链式法则的直接应用。

The function we will examine is a typical power of a polynomial:

我们将要研究的函数是一个典型的多项式幂函数：

y = (3x² + 2)⁵

This expression is a composite function because we first compute the inner expression u = 3x² + 2 and then raise it to the fifth power.

该表达式是一个复合函数，因为我们首先计算内层表达式 u = 3x² + 2，然后将其提高到五次方。

2. What Is a Composite Function? | 什么是复合函数？

A composite function is formed when one function is applied to the result of another. In algebra, if we have two functions f and g, then the composite function (f ∘ g)(x) means f(g(x)).

复合函数是将一个函数应用于另一个函数的结果而构成的。在代数中，如果有两个函数 f 和 g，则复合函数 (f ∘ g)(x) 表示 f(g(x))。

In our example, the inner function is g(x) = 3x² + 2, and the outer function is f(u) = u⁵. The whole function is therefore f(g(x)) = (3x² + 2)⁵.

在我们的例子中，内层函数是 g(x) = 3x² + 2，外层函数是 f(u) = u⁵。因此整个函数是 f(g(x)) = (3x² + 2)⁵。

Recognising the inner and outer functions is essential before applying the chain rule.

在应用链式法则之前，识别内层和外层函数至关重要。

3. The Chain Rule Formula | 链式法则公式

The chain rule states that if y = f(u) and u = g(x), then the derivative of y with respect to x is given by the product of the derivative of y with respect to u and the derivative of u with respect to x.

链式法则指出，如果 y = f(u) 且 u = g(x)，那么 y 关于 x 的导数等于 y 关于 u 的导数与 u 关于 x 的导数的乘积。

d y/d x = d y/d u × d u/d x

Equivalently, for a function raised to a power, if y = [g(x)]ⁿ, then:

等价地，对于一个幂函数，如果 y = [g(x)]ⁿ，则：

d y/d x = n[g(x)]ⁿ⁻¹ × g′(x)

This is the form we will use for Example 5.2.1.

这就是我们在示例 5.2.1 中将要使用的形式。

4. Step 1: Identify the Outer and Inner Functions | 步骤1：识别外层和内层函数

Before differentiating, break the composite function into two separate parts. The inner function is the expression inside the parentheses, and the outer function is the power operation applied to it.

在求导之前，将复合函数分解为两个独立的部分。内层函数是括号内的表达式，外层函数是应用于它的幂运算。

For y = (3x² + 2)⁵, we identify:

对于 y = (3x² + 2)⁵，我们识别出：

Inner function: u = 3x² + 2

内层函数：u = 3x² + 2
Outer function: y = u⁵

外层函数：y = u⁵

Make sure you clearly label these two parts before moving on.

在继续之前，请务必清楚地标记这两个部分。

5. Step 2: Differentiate the Outer Function | 步骤2：对外层函数求导

First, differentiate the outer function with respect to the inner variable u. Using the power rule, if y = uⁿ, then d y/d u = n·uⁿ⁻¹.

首先，对外层函数关于内层变量 u 求导。根据幂法则，如果 y = uⁿ，则 d y/d u = n·uⁿ⁻¹。

Here n = 5, so:

这里 n = 5，所以：

d y/d u = 5u⁴

Remember to keep the inner function u unchanged at this stage; we will substitute it back later.

注意，在此阶段保持内层函数 u 不变；我们将在稍后将其代回。

6. Step 3: Differentiate the Inner Function | 步骤3：对内层函数求导

Next, differentiate the inner function u = 3x² + 2 with respect to x. The derivative of a sum is the sum of the derivatives.

接下来，对内层函数 u = 3x² + 2 关于 x 求导。和的导数等于各项导数的和。

d u/d x = d(3x²)/d x + d(2)/d x = 6x + 0 = 6x

The constant term 2 has a derivative of zero, so we are left with 6x.

常数项 2 的导数为零，因此我们得到 6x。

7. Step 4: Apply the Chain Rule | 步骤4：应用链式法则

Now multiply the two derivatives together. According to the chain rule, d y/d x = d y/d u × d u/d x.

现在将两个导数相乘。根据链式法则，d y/d x = d y/d u × d u/d x。

Substituting the expressions we found:

代入我们找到的表达式：

d y/d x = 5u⁴ × 6x = 30x·u⁴

Finally, replace u with the original inner expression 3x² + 2 to obtain the derivative in terms of x.

最后，将 u 替换为原始的内层表达式 3x² + 2，从而得到关于 x 的导数。

d y/d x = 30x(3x² + 2)⁴

This is the final answer for Example 5.2.1.

这就是示例 5.2.1 的最终答案。

8. Simplifying the Result | 化简结果

The derivative obtained is already quite simple, but sometimes it can be expanded or factored further depending on the context. In this case, the factor 30x is outside the parentheses, and the power 4 remains inside.

得到的导数已经相当简洁，但有时根据具体情况可以进一步展开或因式分解。在本例中，因子 30x 在括号外面，而幂次 4 保留在括号内。

If required, one could expand (3x² + 2)⁴ using the binomial theorem, but for most calculus problems the factored form is preferred because it is easier to evaluate or use for further analysis.

如果需要，可以使用二项式定理展开 (3x² + 2)⁴，但在大多数微积分问题中，因式分解的形式更受欢迎，因为它更容易求值或用于进一步分析。

Always check whether your answer can be simplified by looking for common factors or algebraic identities.

始终通过寻找公因子或代数恒等式来检查你的答案是否可以化简。

9. Common Mistakes to Avoid | 应避免的常见错误

Students often make mistakes when applying the chain rule. Here are some common pitfalls specifically related to Example 5.2.1:

学生在应用链式法则时常犯错误。以下是针对示例 5.2.1 的一些常见陷阱：

Forgetting the inner derivative: Some students simply write d y/d x = 5(3x² + 2)⁴ and forget the factor 6x. Always multiply by d u/d x.

忘记内层函数的导数：有些学生直接写 d y/d x = 5(3x² + 2)⁴，忘记了因子 6x。始终要乘以 d u/d x。
Reducing the exponent incorrectly: When differentiating u⁵, the exponent becomes 4, not 5. Ensure you subtract 1 correctly.

指数减错：对 u⁵ 求导时，指数变为 4，而不是 5。确保正确减去 1。
Confusing inner and outer functions: In this example, some students mistakenly differentiate 3x² + 2 as the outer function and then apply the power rule to x². Clearly identify which part is the outer function before differentiating.

混淆内层和外层函数：在本例中，有些学生错误地对 3x² + 2 作为外层函数求导，然后对 x² 应用幂法则。在求导前明确哪部分是外层函数。

10. Further Applications | 进一步应用

The chain rule is not limited to polynomials. It can be applied to trigonometric, exponential, and logarithmic functions as well. For example, if y = sin(2x³), then the inner function is 2x³ and the outer function is sin(u).

链式法则不仅限于多项式。它也可以应用于三角函数、指数函数和对数函数。例如，如果 y = sin(2x³)，那么内层函数是 2x³，外层函数是 sin(u)。

In general, any composite function can be differentiated using the same four-step process: identify, differentiate outer, differentiate inner, and multiply.

一般来说，任何复合函数都可以使用相同的四步过程求导：识别、外层求导、内层求导、相乘。

This skill is essential for solving more advanced problems in AQA A-Level mathematics, such as implicit differentiation and parametric equations.

这一技能对于解决 AQA A-Level 数学中更高级的问题（如隐函数求导和参数方程）至关重要。

11. Practice Questions | 练习问题

To reinforce the skill demonstrated in Example 5.2.1, try the following similar problems:

为了巩固在示例 5.2.1 中展示的技能，请尝试以下类似的题目：

Differentiate y = (5x − 1)³

求导 y = (5x − 1)³
Differentiate y = (2x² − 3)⁶

求导 y = (2x² − 3)⁶
Differentiate y = (x³ + 4x)⁻²

求导 y = (x³ + 4x)⁻²

For each problem, identify the inner and outer functions first, then apply the chain rule step by step.

对于每个问题，首先识别内层和外层函数，然后逐步应用链式法则。

12. Conclusion | 总结

Example 5.2.1 provides a clear illustration of how to differentiate a composite function using the chain rule. By mastering this method, you can handle a wide range of differentiation problems in AQA A-Level mathematics.

示例 5.2.1 清晰地展示了如何使用链式法则对复合函数求导。通过掌握这一方法，你可以处理 AQA A-Level 数学中各种求导问题。

Remember the key formula: d y/d x = d y/d u × d u/d x. Always follow the steps: identify, differentiate outer, differentiate inner, and multiply.

记住关键公式：d y/d x = d y/d u × d u/d x。始终遵循步骤：识别、外层求导、内层求导、相乘。

Keep practising with the provided exercises, and you will soon find the chain rule intuitive and powerful.

继续使用所提供的练习进行练习，你很快就会发现链式法则既直观又强大。

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Example 5.2.1: Differentiation of a Composite Function | 示例5.2.1：复合函数的微分