📚 Example 6.10.3: Chain Rule Differentiation | 链式法则求导示例
x + 2)⁵ with respect to x. Questions of this exact shape appear frequently in AS and A-Level Pure Mathematics papers, particularly in the differentiation sections, where examiners want to see whether you can recognise a 'function of a function' r function wrapped around an inner cubic polynomial. Expanding the bracket by hand would be extremely long and error-prone, so the chain rule provides a short, reliable route to the answer, and it also earns the method marks that examiners award for correct structure.
在本节的 AQA A-Level 数学课程中，我们讲解示例 6.10.3，题目要求对复合函数 y = (4x³ − 7x + 2)⁵ 关于 x 求导。这类题目在 AS 与 A-Level 纯数学试卷中频繁出现，尤其是在微分部分，考官想考察你能否识别“函数的函数”（复合函数）并干净利落地使用链式法则。该表达式的结构是：外层为幂函数，内层为三次多项式。手动展开括号会非常冗长且极易出错，因此链式法则提供了一条清晰高效的解题路径，同时也能拿到考官给出的方法分。
1. Problem Statement | 题目陈述
e expression is a composite function: the inner cubic expression 4x³ − 7x + 2 is raised to the fifth power by the outer function u⁵. Because the whole bracket is raised to a power, the derivative is not simply 5(4x³ − 7x + 2)⁴; there is a second factor coming from the derivative of the inside, and forgetting it is the single most common error in this topic. The bracket cannot be expanded quickly by hand, so the chain rule is the most efficient method available.
已知 y = (4x³ − 7x + 2)⁵，求 dy/dx。该表达式是复合函数：内层的三次表达式 4x³ − 7x + 2 被放进外层幂函数 u⁵ 中。由于整个括号被取幂，导数并非只是 5(4x³ − 7x + 2)⁴，还必须乘上由内部函数求导产生的第二个因子——忘记这个因子是本专题最常见的错误。我们无法快速手动展开括号，因此链式法则是最有效的方法。
2. The Chain Rule Formula | 链式法则公式
ule states that we differentiate the outer function with respect to the inner variable u, then multiply by the derivative of the inner function with respect to x.
若 y = f(u) 且 u = g(x)，链式法则告诉我们：先对外层函数关于内层变量 u 求导，再乘以内部函数关于 x 的导数。
dy/dx = (dy/du) × (du/dx)
In words: differentiate the outside, keep the inside unchanged, then multiply by the derivative of the inside. In Leibniz notation the "du" terms appear to cancel, which is a useful memory aid, although the formal justification is a limit argument. This rule is essential for AQA Pure Mathematics and is regularly combined with the product rule and the quotient rule in later topics such as implicit differentiation, parametric equations and rates of change. It also underpins the differentiation of trigonometric, exponential and logarithmic composites.
用文字表述就是：对外层求导，内层保持不变，再乘以对内层的求导结果。在莱布尼茨记号中，两个 “du” 看起来可以“约掉”，这是一个很好用的记忆技巧，不过严格的证明要依赖极限论证。该法则对 AQA 纯数学至关重要，并且常与乘积法则、商法则结合出现于后续内容，比如隐函数求导、参数方程与变化率问题，也是三角、指数与对数复合函数求导的基础。
3. Identifying the Inner and Outer Functions | 识别内层与外层函数
The first skill the examiner is testing is recognition, not calculation. Look at the structure of the expression: there is a bracket, and the whole bracket carries an exponent. Whatever sits inside the bracket is the inner function, and the power applied to it is the outer function.
考官首先考查的是识别能力，而不是计算能力。观察表达式结构：有一个括号，而且整个括号带有指数。括号里的内容就是内层函数，施加在它上面的幂就是外层函数。
- Outer function: y = u⁵, where u is the whole bracket.
- Inner function: u = 4x³ − 7x + 2.
- Key signal words: "bracket", "power of", "function of a function", "composite".
- 外层函数：y = u⁵，其中 u 代表整个括号。
- 内层函数：u = 4x³ − 7x + 2。
- 关键识别信号：出现括号、整体带幂次、“函数的函数”、复合函数。
If you can write down these two lines correctly at the start of your answer, you have already secured a good proportion of the marks, even before any differentiation has taken place. Examiners see these two statements as evidence of a correct method and will award them accordingly.
如果你能在解题开头正确写下这两行式子，即使还没开始求导，也已经拿到了相当一部分分数。考官会把这两句视为方法正确的证据，并据此给分。
4. Step 1: Substitute u for the Inner Function | 第一步：令 u 等于内层函数
s y = u⁵, which is a basic power function you have differentiated many times before. Writing the substitution explicitly is strongly recommended in exams because it makes your reasoning visible and makes it far easier to catch your own mistakes.
令 u = 4x³ − 7x + 2。这个代换把一个复杂的复合表达式拆成两个简单而熟悉的部分。原函数随即变成 y = u⁵，也就是你已经反复求导过很多次的基本幂函数。考试中强烈建议把代换步骤写清楚，因为这能让你的推理过程可见，也更容易发现自己的错误。
u = 4x³ − 7x + 2 → y = u⁵
Note that u⁵ is only a legitimate temporary form: the final answer must always be written back in terms of x, because the question asked for dy/dx as a function of x, not as a function of u. Keeping u in the final line is a classic way to lose the final accuracy mark.
注意 u⁵ 只是临时的形式：最终答案必须换回用 x 表示，因为题目要求的是关于 x 的 dy/dx，而不是关于 u 的函数。最后一行还留着 u，是丢掉最后一个准确分点的经典方式。
5. Step 2: Differentiate the Outer Function | 第二步：对外层函数求导
u = 5u⁴. At this stage the inner function has not yet been touched; it is simply carried along inside the u term.
对 y = u⁵ 关于 u 求导。使用标准幂法则：把指数作为乘数放到前面，再把指数减一。得到 dy/du = 5u⁴。在这一步，内层函数还没有被处理，它只是被装在 u 里一起带着走。
y = u⁵ → dy/du = 5u⁴
Students sometimes try to differentiate directly with respect to x here and write 5(4x³ − 7x + 2)⁴, then stop. That expression is only the first factor of the answer. The chain rule is not finished until it has been multiplied by du/dx, which is why the next step is compulsory rather than optional.
有些同学会试图直接对 x 求导，写出 5(4x³ − 7x + 2)⁴ 就停下。这个表达式只是答案的第一个因子。链式法则在乘上 du/dx 之前并没有完成，所以下一步是必须的，而不是可选的。
6. Step 3: Differentiate the Inner Function | 第三步：对内层函数求导
espect to x, term by term. The derivative of 4x³ is 12x², the derivative of −7x is −7, and the constant +2 differentiates to zero.
现在对 u = 4x³ − 7x + 2 关于 x 逐项求导。4x³ 的导数是 12x²，−7x 的导数是 −7，常数 +2 求导为 0。
u = 4x³ − 7x + 2 → du/dx = 12x² − 7
Two small slips are worth guarding against here. First, multiplying the coefficient by the index incorrectly, for example writing 12x³ instead of 12x². Second, dropping the sign on the linear term and writing +7 rather than −7. A ten-second re-check of this single line before moving on will protect the accuracy mark at the end of the question.
这里有两点小失误要特别防范。第一，系数与指数相乘时出错，比如把 12x² 写成 12x³。第二，丢掉一次项的负号，把 −7 写成 +7。在继续下一步之前，用十秒钟回头检查这一行，就能守住最后的准确分。
7. Step 4: Multiply the Two Derivatives | 第四步：把两个导数相乘
Apply the chain rule by multiplying the two results together, exactly as the formula requires.
按照公式要求，把刚才的两个结果相乘，即可应用链式法则。
dy/dx = (dy/du) × (du/dx) = 5u⁴ × (12x² − 7)
te u = 4x³ − 7x + 2 back into the expression so that the answer is entirely in terms of x.
最后把 u = 4x³ − 7x + 2 代回表达式，使答案完全用 x 表示。
dy/dx = 5(4x³ − 7x + 2)⁴ (12x² − 7)
The order of the two factors does not matter mathematically, but writing the bracket first and the polynomial second mirrors the structure of the original question and is the format most mark schemes show. An equivalent answer is (60x² − 35)(4x³ − 7x + 2)⁴, obtained by multiplying 5 into the second bracket, and both forms are normally accepted.
两个因子的先后顺序在数学上无所谓，但把括号写在前面、多项式写在后面，与原题结构一致，也是大多数评分标准展示的写法。等价的答案还有 (60x² − 35)(4x³ − 7x + 2)⁴，它是把 5 乘进后面括号得到的，两种形式通常都被接受。
8. The Complete Worked Solution | 完整解题过程
Below is the full solution laid out exactly as it should appear in an examination script, with every line earning its method mark. The table afterwards summarises the four quantities involved so that you can revise the pattern quickly.
下面按照考试答题应有的样子写出完整解答，每一行都在争取自己的方法分。后面的表格总结了涉及的四个量，方便你快速复盘这个模式。
y = (4x³ − 7x + 2)⁵
Let u = 4x³ − 7x + 2, so y = u⁵
dy/du = 5u⁴ and du/dx = 12x² − 7
dy/dx = 5u⁴ (12x² − 7) = 5(4x³ − 7x + 2)⁴ (12x² − 7)
| Quantity | Expression | Chinese |
|---|---|---|
| Inner function u | 4x³ − 7x + 2 | 内层函数 |
| Outer function y | u⁵ | 外层函数 |
| dy/du | 5u⁴ | 外层导数 |
| du/dx | 12x² − 7 | 内层导数 |
| dy/dx | 5(4x³ − 7x + 2)⁴ (12x² − 7) | 最终答案 |
te. For example, at x = 1 the bracket equals 4 − 7 + 2 = −1, so the fourth power is (−1)⁴ = 1 and 12x² − 7 = 5, giving dy/dx = 5 × 1 × 5 = 25.
如果题目改问曲线在某一点处的梯度，就在代数全部完成之后最后代 x 的值。例如在 x = 1 处，括号内为 4 − 7 + 2 = −1，四次方为 (−1)⁴ = 1，且 12x² − 7 = 5，于是 dy/dx = 5 × 1 × 5 = 25。
9. Why Expanding the Bracket is a Bad Idea | 为什么不应展开括号
A common temptation is to expand (4x³ − 7x + 2)⁵ into a long polynomial and differentiate it term by term. This is technically possible using the binomial theorem, but the expansion of a trinomial raised to the fifth power produces a large number of terms, many of them with awkward coefficients, and every extra line is another chance to make an arithmetic slip.
一个常见的诱惑是把 (4x³ − 7x + 2)⁵ 展开成一个长多项式再逐项求导。用二项式定理在技术上确实可行，但三项式取五次方会展开出很多项，其中不少项的系数相当难算，而且每多写一行，就多一次算错的机会。
dx = 0 or to find stationary points. A fully expanded polynomial would hide the structure and make those follow-up parts much harder. The chain rule therefore wins on speed, on accuracy and on usefulness for later parts of the same question.
还要注意，链式法则给出的答案本身已经是完全因式分解的形式，这正是解决后续问题所需要的——比如令 dy/dx = 0 求驻点。若写成完全展开的多项式，结构就被掩盖了，后续小问会难得多。因此无论是速度、准确度，还是对同题后续小问的实用性，链式法则都更胜一筹。
10. Common Variations of This Question | 本题的常见变式
Examiners rarely test exactly the same expression twice, but the underlying pattern is highly stable. Recognising the variations is the fastest way to prepare for this topic.
考官很少重复考查完全相同的表达式，但其内在模式非常稳定。识别这些变式是准备该专题最快捷的方法。
- Higher or negative powers: y = (4x³ − 7x + 2)⁻³ or y = (4x³ − 7x + 2)1/2 The method is unchanged, but the index arithmetic becomes the most likely source of error.
- 7x + 2)⁵ requires the product rule, with the chain rule applied inside the second factor.
- Gradient at a point: after differentiating, substitute a given x-value to obtain a numerical gradient or the equation of a tangent.
- s: set dy/dx = 0 and solve, remembering that a product is zero when either factor is zero.
- rigonometric composites: y = sin(3x² + 1) or y = cos⁴ x follow exactly the same structure.
- 高次或负次幂：y = (4x³ − 7x + 2)⁻³ 或 y = (4x³ − 7x + 2)1/2。方法完全不变，但指数运算成了最容易出错的地方。
- 法则结合：y = x²(4x³ − 7x + 2)⁵ 需要乘积法则，其中第二个因子内部再用链式法则。
- 某点处的梯度：求导后代入给定的 x 值，得到数值梯度或切线方程。
- 驻点问题：令 dy/dx = 0 并求解，注意乘积为零意味着任一因子为零。
- 三角复合函数：y = sin(3x² + 1) 或 y = cos⁴ x 的结构完全相同。
In every case the routine is identical: identify inner and outer, differentiate each, multiply, and rewrite in terms of the original variable. Mastering this one example therefore unlocks a very large family of exam questions.
无论哪种变式，套路都一样：识别内层与外层，分别求导，相乘，最后换回原变量。掌握这一道例题，就等于打开了一大类考题的大门。
11. Common Mistakes and Examiner Tips | 常见错误与考官提示
Most lost marks in chain rule questions come from a small number of repeated errors rather than from genuine conceptual confusion. Being aware of them in advance is far more effective than discovering them in the exam hall.
链式法则题目中丢分，多半来自少数几个反复出现的错误，而不是真正的概念困惑。提前知道这些坑，比在考场上才发现要有效得多。
- Forgetting the second factor du/dx entirely, giving 5(4x³ − 7x + 2)⁴ as the final answer.
- Differentiating the inner function incorrectly, for example writing 4x² − 7 or 12x² + 7.
- Leaving the answer in terms of u instead of substituting back to x.
- Expanding the bracket first and making arithmetic errors under time pressure.
- Applying the power rule to the whole expression without treating the bracket as a single inner unit.
- 完全忘记第二个因子 du/dx，直接把 5(4x³ − 7x + 2)⁴ 当作最终答案。
- 内层函数求导出错，例如写成 4x² − 7 或 12x² + 7。
- 答案仍以 u 表示，没有代回 x。
- 先展开括号，在时间压力下出现算术错误。
- 把幂法则直接套在整个表达式上，没有把括号看作一个内层整体。
Examiner tip: the mark scheme for this style of question typically awards one mark for the correct substitution, one for the outer derivative, one for the inner derivative, and one for the correctly combined final answer in terms of x. Writing each of those lines separately, rather than compressing the work into a single line, maximises your chance of being credited for what you actually did correctly.
考官提示：这类题目的评分标准通常给这样的分点——正确代换 1 分，外层导数 1 分，内层导数 1 分，正确组合并换回 x 的最终答案 1 分。把每一步分行写出，而不是把整个推理压缩成一行，能最大程度地确保你真正做对的部分被认可给分。
12. Exam-Style Extension and Summary | 考试风格延伸与总结
nt to C at the point where x = 1. From the derivative above, the gradient at x = 1 is 25, and the y-coordinate is (−1)⁵ = −1. Using y − y₁ = m(x − x₁) gives y + 1 = 25(x − 1), which simplifies to y = 25x − 26.
本题的一个典型延伸版本可能是：曲线 C 的方程为 y = (4x³ − 7x + 2)⁵。求 dy/dx，并由此求 C 在 x = 1 处的切线方程。由上面的导数可知，x = 1 处的梯度为 25，而 y 坐标为 (−1)⁵ = −1。用 y − y₁ = m(x − x₁) 得到 y + 1 = 25(x − 1)，化简为 y = 25x − 26。
er function with respect to u, differentiate the inner function with respect to x, multiply the two results, and finally rewrite everything in terms of x. In this example the answer is dy/dx = 5(4x³ − 7x + 2)⁴ (12x² − 7). Practise the same five-step routine on powers, roots, trigonometric and exponential composites, and this will become one of the most dependable sources of marks in your Pure Mathematics papers.
总结一下：只要一个函数被嵌套在另一个函数内部，就要用链式法则。写出内层函数 u，先对外层关于 u 求导，再对内层关于 x 求导，把两个结果相乘，最后把所有内容换回 x 表示。本题的答案是 dy/dx = 5(4x³ − 7x + 2)⁴ (12x² − 7)。把同样的五步套路练在幂函数、根式、三角与指数复合函数上，它将成为你纯数试卷中最稳定的得分来源之一。
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