📚 Example 2.5.1: Differentiating a Composite Function | 复合函数求导示例 2.5.1
h respect to x. This type of question tests your understanding of the chain rule, a core technique for differentiating composite functions.
在这份 AQA A-Level 数学复习指南中，我们将逐步讲解示例 2.5.1。题目要求对 y = (3x² - 5x + 2)⁴ 关于 x 求导。这类问题考查你对链式法则的理解，这是求复合函数导数的核心技术。
1. Understanding the Problem | 理解题目
mposite function: an outer power function is applied to the inner quadratic 3x² - 5x + 2. We cannot expand the fourth power easily, so the chain rule is the efficient method.
表达式 y = (3x² - 5x + 2)⁴ 是一个复合函数：外层的幂函数作用于内层的二次式 3x² - 5x + 2。我们很难展开四次幂，因此链式法则是高效的方法。
In AQA exams, questions like this often appear in Pure Mathematics papers under differentiation. You need to show clear working to gain method marks even if a small slip occurs later.
在 AQA 考试中，这类题目常出现在纯数学试卷的微分部分。你需要展示清晰的过程，即使后面有小错误，也能获得方法分。
2. The Chain Rule in Words | 链式法则的文字表述
The chain rule states that the derivative of a composite function is the derivative of the outer function with respect to the inner function, multiplied by the derivative of the inner function with respect to x.
链式法则指出，复合函数的导数等于外层函数对内层函数求导，再乘以内层函数对 x 求导。
= (dy/du) × (du/dx). This is the standard Leibniz form you should quote in your solution.
如果记 y = f(u)，u = g(x)，那么 dy/dx = (dy/du) × (du/dx)。这是你应该在解答中引用的标准莱布尼茨形式。
The rule makes intuitive sense: the differential dy depends on a small change in u, and du depends on a small change in x. Multiplying these two rates gives the overall rate dy/dx.
这个法则直观上很容易理解：dy 取决于 u 的微小变化，而 du 取决于 x 的微小变化。将这两个变化率相乘，就得到整体变化率 dy/dx。
3. Identifying Inner and Outer Functions | 识别内层函数与外层函数
2. Choosing the right u is the most important step in applying the chain rule accurately.
这里外层函数是 f(u) = u⁴，内层函数是 u = 3x² - 5x + 2。选择正确的 u 是准确应用链式法则的最重要一步。
If you choose the wrong inner function, the derivative will be incorrect. Look for the part of the expression that is 'inside' the power, bracket, root, or trigonometric function.
如果内层函数选错，导数就会错误。要找出表达式中位于幂、括号、根号或三角函数'内部'的部分。
function. If you let u = 3x² - 5x + 2, then y = u⁴ gives the original expression exactly.
一个有效的检查方法是把 u 代回外层函数。如果设 u = 3x² - 5x + 2，那么 y = u⁴ 能完全还原为原始表达式。
4. Step 1 - Assign u | 第一步：设 u
ution makes the composite structure explicit and prepares the function for differentiation.
设 u = 3x² - 5x + 2，那么原表达式变为 y = u⁴。这个代换使复合结构清晰，并为求导做好准备。
Always write this substitution clearly in your working. In AQA exams, an explicit statement of u can earn method marks.
始终在你的步骤中清楚写下这个代换。在 AQA 考试中，明确写出 u 可以获得方法分。
Do not rush this step. If u is written down incorrectly, every later step will be affected even if the chain rule method itself is correct.
这一步不要匆忙。如果 u 写错了，即使后面的链式法则方法正确，所有后续步骤也会受到影响。
5. Step 2 - Differentiate u | 第二步：求 u 的导数
² is 6x, the derivative of -5x is -5, and the derivative of 2 is 0, we get du/dx = 6x - 5.
对 u = 3x² - 5x + 2 关于 x 求导。因为 3x² 的导数是 6x，-5x 的导数是 -5，2 的导数是 0，所以得到 du/dx = 6x - 5。
- d/dx (3x²) = 6x
- d/dx (-5x) = -5
- d/dx (2) = 0
n xⁿ⁻¹. For x², n = 2, so the derivative is 2x, multiplied by the coefficient 3 gives 6x.
这些基础结果来自幂法则：d/dx (xⁿ) = n xⁿ⁻¹。对于 x²，n = 2，所以导数是 2x，乘以系数 3 得到 6x。
Remember that constants differentiate to zero. The term +2 disappears, which is a common source of error if you try to differentiate too quickly.
记住常数项的导数为零。+2 项会消失，如果你求导太快，这是常见的错误来源。
6. Step 3 - Differentiate the Outer Function | 第三步：求外层函数的导数
rentiate y = u⁴ with respect to u. Using the power rule with n = 4, we obtain dy/du = 4u³.
现在对 y = u⁴ 关于 u 求导。使用幂法则，n = 4，得到 dy/du = 4u³。
At this stage, do not substitute u back yet. Keeping dy/du in terms of u makes the chain rule multiplication clearer.
在这个阶段，先不要代回 u。把 dy/du 保留为 u 的表达式能让链式法则的乘法更清晰。
This step is straightforward for power functions, but the same principle applies when the outer function is sine, cosine, exponential, or logarithmic. You differentiate the outer layer as if the inner expression were just a single variable.
对于幂函数来说这一步很简单，但当外层函数是正弦、余弦、指数或对数时，原理是一样的。你只需要把内层表达式当作一个单一变量来对外层求导。
7. Step 4 - Combine the Derivatives | 第四步：合并导数
have dy/du = 4u³ and du/dx = 6x - 5, so dy/dx = 4u³ × (6x - 5) = 4(3x² - 5x + 2)³(6x - 5).
应用 dy/dx = (dy/du) × (du/dx)。我们有 dy/du = 4u³，du/dx = 6x - 5，所以 dy/dx = 4u³ × (6x - 5) = 4(3x²
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