📚 Example 4.7.1: Differentiating y = sin(3x² + 1) | 例题4.7.1：求导 y = sin(3x² + 1)
in(3x² + 1). This example is carefully chosen to illustrate how the derivative of a sine function combines with the derivative of a quadratic expression inside the argument.
在AQA A-Level数学中，使用链式法则求复合函数的导数是基本技能。例题4.7.1着重于将链式法则应用于三角函数，具体为 y = sin(3x² + 1)。该例题精心设计，用于展示正弦函数的导数与内部二次表达式导数的结合方式。
1. Problem Statement | 问题陈述
nd the derivative dy/dx. This expression is a composite of the sine function and the quadratic polynomial 3x² + 1, and it requires the chain rule to differentiate correctly.
已知函数 y = sin(3x² + 1)，求其导数 dy/dx。该表达式是正弦函数与二次多项式 3x² + 1 的复合形式，需要运用链式法则正确求导。
2. Prerequisite Rules | 预备规则
Before attempting the solution, it is essential to recall two standard derivatives. First, for any constant a and differentiable function u(x), the derivative of sin(u) with respect to x is cos(u) multiplied by du/dx. Second, the derivative of x² is 2x, and the derivative of a constant term is zero. These rules form the backbone of the chain rule application.
在尝试求解之前，必须回顾两个标准导数。首先，对于任意常数 a 和可微函数 u(x)，sin(u) 关于 x 的导数为 cos(u) 乘以 du/dx。其次，x² 的导数是 2x，而常数项的导数为零。这些规则构成链式法则应用的基础。
3. Step-by-Step Solution | 分步解法
Step 1: Identify the outer and inner functions. e . The outer function is sine, and the inner function is u = 3x² + 1. Thus we write y = sin(u).
第一步：识别外函数和内函数。外函数为正弦函数，内函数为 u = 3x² + 1。因此我们写作 y = sin(u)。
Step 2: Differentiate the outer function with respect to u. The derivative of sin(u) with respect to u is cos(u). So dy/du = cos(u).
第二步：对外函数关于 u 求导。sin(u) 关于 u 的导数为 cos(u)。因此 dy/du = cos(u)。
Step 3: Differentiate the inner function with respect to x. Here u = 3x² + 1. Taking the derivative gives du/dx = 6x.
第三步：对内函数关于 x 求导。此处 u = 3x² + 1。求导得到 du/dx = 6x。
Step 4: Apply the chain rule. The chain rule states that dy/dx = (dy/du) × (du/dx). Substituting the results:
第四步：应用链式法则。链式法则表述为 dy/dx = (dy/du) × (du/dx)。代入结果：
dy/dx = cos(u) × 6x = 6x cos(3x² + 1)
Therefore, the derivative of y = sin(3x² + 1) is dy/dx = 6x cos(3x² + 1).
因此，y = sin(3x² + 1) 的导数为 dy/dx = 6x cos(3x² + 1)。
4. Why the Chain Rule Works | 链式法则为什么有效
The chain rule arises from the concept of the derivative as a rate of change. If a variable y changes at a rate dy/du with respect to u, and u changes at a rate du/dx with respect to x, then the total rate of change of y with respect to x is the product of these two rates. This multiplicative relationship is exactly what the chain rule expresses.
链式法则源于导数作为变化率的概念。如果变量 y 关于 u 的变化率为 dy/du，而 u 关于 x 的变化率为 du/dx，那么 y 关于 x 的总变化率就是这两个变化率的乘积。这种乘法关系正是链式法则所表达的。
Graphically, the slope of the composite function at any point is equal to the slope of the outer function evaluated at the inner value, multiplied by the slope of the inner function. This intuitive interpretation helps in understanding why the factor 6x appears outside the cosine term.
从图像上看，复合函数在任意一点的斜率等于外函数在内值处的斜率乘以内函数的斜率。这种直观解释有助于理解为何因子 6x 出现在余弦项外面。
5. Evaluating at a Specific Point | 在特定点求值
function sin(3x² + 1) has a local maximum or minimum near x = 0, where the slope vanishes.
假设我们需要求 y = sin(3x² + 1) 在 x = 0 处的切线斜率。将 x = 0 代入导数得到 dy/dx = 6(0) cos(3(0)² + 1) = 0 × cos(1) = 0。因此，在 x = 0 处梯度为 0，意味着切线是水平的。这可以通过查看图像验证：函数 sin(3x² + 1) 在 x = 0 附近有局部最大值或最小值，此时斜率为零。
If we choose x = √π/6 + 1 = π/2 + 1, and cos(π/2 + 1) = -sin(1) (using trigonometric identities). Then dy/dx = 6√π/6 × (-sin(1)) = -6√π/6 sin(1). This shows how the derivative value depends on evaluating the cosine at the transformed argument.
如果我们选择 x = √π/6，则 3x² + 1 = π/2 + 1，且 cos(π/2 + 1) = -sin(1)（利用三角恒等式）。则 dy/dx = 6√π/6 × (-sin(1)) = -6√π/6 sin(1)。这表明导数值如何取决于在变换后的自变量处计算余弦值。
6. Common Mistakes and How to Avoid Them | 常见错误及避免方法
Forgetting the inner derivative: s(3x² + 1) without the factor 6x. Always remember that the chain rule requires multiplying by the derivative of the inner function.
忘记内层导数：常见错误是写成 dy/dx = cos(3x² + 1) 而没有因子 6x。务必记住链式法则需要乘以内层函数的导数。
Incorrect sign when differentiating sin: The derivative of sin(u) is cos(u), not -cos(u). Students often confuse it with the derivative of cos(u), which is -sin(u). Keep the sign rules clear.
求导 sin 时符号错误：sin(u) 的导数是 cos(u)，而不是 -cos(u)。学生常将其与 cos(u) 的导数（即 -sin(u)）混淆。请牢记符号规则。
Miscopying the inner function: If the inner function were sin(3x³ + 1), the derivative would be 9x² cos(3x³ + 1). Ensure you read the exponent correctly before differentiating.
抄错内层函数：如果内层函数是 sin(3x³ + 1)，导数应为 9x² cos(3x³ + 1)。求导前务必正确识别指数。
7. Generalisation to Other Composite Functions | 推广到其他复合函数
o any differentiable outer function f and inner function g. If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). In our example, f(u) = sin(u) and g(x) = 3x² + 1.
链式法则推广到任意可微的外函数 f 和内函数 g。若 y = f(g(x))，则 dy/dx = f'(g(x)) · g'(x)。在我们的例子中，f(u) = sin(u)，g(x) = 3x² + 1。
ance, if y = cos(5x³), then dy/dx = -sin(5x³) × 15x² = -15x² sin(5x³). Similarly, if y = e2x², then dy/dx = 4x e2x². This pattern is consistent across all composite functions.
例如，若 y = cos(5x³)，则 dy/dx = -sin(5x³) × 15x² = -15x² sin(5x³)。类似地，若 y = e2x²，则 dy/dx = 4x e2x²。此模式在所有复合函数中一致成立。
8. Connection with Integration | 与积分之间的联系
indefinite integral of 6x cos(3x² + 1) with respect to x is sin(3x² + 1) + C, where C is the constant of integration. This relationship is often tested in A-Level exams, where candidates may be asked to recognise the derivative pattern in integration questions.
微分与积分是互逆运算。已知 dy/dx = 6x cos(3x² + 1) 意味着 6x cos(3x² + 1) 对 x 的不定积分为 sin(3x² + 1) + C，其中 C 是积分常数。这种关系在A-Level考试中经常出现，考生可能需要识别积分题中的导数模式。
sin(u) + C. This example beautifully ties together the chain rule and substitution method.
例如，要积分 ∫ 6x cos(3x² + 1) dx，可以使用换元法 u = 3x² + 1，du = 6x dx，将积分简化为 ∫ cos(u) du = sin(u) + C。这个例子完美地将链式法则与换元法联系起来。
9. Graphical Interpretation | 图像解释
, the sign of the slope depends on the sign of cos(3x² + 1). The curve oscillates more rapidly than sin(x) due to the quadratic argument, and the amplitude of the derivative grows linearly with |x|.
导数 dy/dx 给出曲线上每一点切线的斜率。对于 y = sin(3x² + 1)，导数是 6x 与 cos(3x² + 1) 的乘积。当 x = 0，斜率为 0；对于正 x，斜率的符号取决于 cos(3x² + 1) 的符号。由于二次自变量，曲线比 sin(x) 振荡得更快，而导数的振幅随 |x| 线性增大。
ly a scaled sine curve. The envelope of the derivative is a pair of lines y = ±6x, because |cos(3x² + 1)| ≤ 1, so |dy/dx| ≤ 6|x|.
该图像帮助学生理解为什么导数不是简单缩放的正弦曲线。导数的包络线是一对直线 y = ±6x，因为 |cos(3x² + 1)| ≤ 1，所以 |dy/dx| ≤ 6|x|。
10. Exam-Style Question | 考试风格问题
). This value represents the instantaneous rate of change of y with respect to x at x = 2.
为巩固理解，考虑以下变体：求 y = sin(3x² + 1) 在 x = 2 处的导数。使用我们的导数，dy/dx = 6(2) cos(3(2)² + 1) = 12 cos(13)。因为 13 弧度即 13，dy/dx ≈ 12 × cos(13) ≈ 12 × 0.907 = 10.884（使用计算器）。该值表示 y 关于 x 在 x = 2 处的瞬时变化率。
tal line. This type of question asks for both the derivative and the use of the point-slope form.
另一个变体：如果 y = sin(3x² + 1)，且要求 x = 0 处的切线方程，我们已经知道斜率为 0。x = 0 处的 y 值为 sin(1)。因此切线为 y = sin(1)，一条水平线。这类问题既要求导数，也要求使用点斜式。
11. The Role of Notation | 记号的作用
rule, especially in multi-step calculations. Some students prefer the function notation f'(g(x))·g' Both are acceptable, but consistency is key. When substituting u = 3x² + 1, write clearly:
在AQA考试中，在处理链式法则时使用莱布尼茨记号 dy/dx 非常重要，尤其是在多步计算中。有些学生喜欢函数记号 f'(g(x))·g'(x)。两种都可以，但一致性是关键。在代换 u = 3x² + 1 时，请清晰书写：
u = 3x² + 1, dy/dx = dy/du × du/dx
This notation clearly separates the intermediate steps and reduces the chance of algebra mistakes.
这种记号清晰地区分了中间步骤，减少了代数错误的可能性。
12. Summary and Revision Tips | 总结与复习提示
: identify the outer and inner functions, differentiate each separately, and multiply the results. Remember to keep the nnder argument unchanged when differentiating the sine function, and always multiply by the derivative of the inner function.
例题4.7.1展示了 y = sin(3x² + 1) 的导数为 dy/dx = 6x cos(3x² + 1)。关键步骤为：识别外函数和内函数，分别求导，然后相乘。记住在求导正弦函数时保持内部参数不变，并始终乘以内层函数的导数。
For revision, practise with different inner functions: sin(2x), sin(5x³ + x), sin(1/x), etc. Master the chain rule early because it forms the basis for implicit differentiation, parametric differentiation, and further applications in mechanics and statistics. Good luck with your A-Level Mathematics studies!
复习时，请练习不同内函数：sin(2x)、sin(5x³ + x)、sin(1/x) 等。尽早掌握链式法则，因为它是隐函数求导、参数方程求导以及力学和统计学中进一步应用的基础。祝你在A-Level数学学习中取得好成绩！
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