← 返回资讯屏轩教育 · 行业资讯

Worked Example 6.3.1: Differentiating y = sin(3x) using the Chain Rule | 例题 6.3.1：用链式法则求 y = sin(3x) 的导数

2026-08-21 20:57 · 本文部分内容由AI辅助创作

📚 Worked Example 6.3.1: Differentiating y = sin(3x) using the Chain Rule | 例题 6.3.1：用链式法则求 y = sin(3x) 的导数

sin(3x) by applying the chain rule. Understanding this technique is essential for success in Paper 1 (Pure Mathematics) and for later topics such as integration by substitution and differential equations.

在这篇 AQA A-Level 数学复习文章中，我们将详细讲解例题 6.3.1，重点是用链式法则求三角函数 y = sin(3x) 的导数。掌握这一技巧对在试卷一（纯数学）中取得好成绩以及后续的换元积分法和微分方程等内容都至关重要。

1. The Chain Rule: Core Idea | 链式法则：核心思想

unction of u, and u is a function of x. The derivative of y with respect to x is given by:

链式法则用于求复合函数的导数。如果 y = f(u) 且 u = g(x)，那么 y 是 u 的函数，u 是 x 的函数。y 对 x 的导数由下式给出：

dy/dx = (dy/du) × (du/dx)

This formula tells us to differentiate the outer function with respect to the inner variable u, then multiply by the derivative of the inner function with respect to x.

这个公式告诉我们，先对外层函数关于内层变量 u 求导，再乘以内层函数对 x 的导数。

2. Recognising a Composite Function in y = sin(3x) | 识别 y = sin(3x) 中的复合函数

e sine function, so we cannot simply differentiate term by term without accounting for this inner linear term.

在例题 6.3.1 中，给定的函数是 y = sin(3x)。这里外层函数是 sin(u)，内层函数是 u = 3x。表达式 3x 位于正弦函数内部，因此我们不能忽略这个内层线性项而直接逐项求导。

bstitution makes the composite structure explicit and prepares the function for the chain rule.

如果我们令 u = 3x，那么 y = sin(u)。这个代换使复合结构更加明确，并为使用链式法则做好准备。

3. Step-by-Step Differentiation of y = sin(3x) | y = sin(3x) 的分步求导

e the inner function: if u = 3x, then du/dx = 3. Third, multiply the two results together.

现在我们分三个清晰的步骤应用链式法则。第一步，对外层函数求导：如果 y = sin(u)，那么 dy/du = cos(u)。第二步，对内层函数求导：如果 u = 3x，那么 du/dx = 3。第三步，将两个结果相乘。

dy/dx = cos(u) × 3 = 3cos(3x)

nner function 3x gives the final derivative. Therefore, the derivative of y = sin(3x) is y' = 3cos(3x).

将 u 替换回原来的内层函数 3x，就得到最终导数。因此，y = sin(3x) 的导数是 y' = 3cos(3x)。

4. Key Result for sin(ax + b) | sin(ax + b) 的关键结论

Example 6.3.1 illustrates a broader rule that AQA examiners expect you to quote fluently. For any constants a and b, the derivative of sin(ax + b) with respect to x is a cos(ax + b). The coefficient a comes from the derivative of the inner linear function ax + b.

例题 6.3.1 展示了一个更广泛的规则，AQA 考官希望你能熟练运用。对于任意常数 a 和 b，sin(ax + b) 对 x 的导数是 a cos(ax + b)。系数 a 来自内层线性函数 ax + b 的导数。

Similarly, the derivative of cos(ax + b) is -a sin(ax + b). Notice the negative sign, which is a common source of lost marks in A-Level exams.

类似地，cos(ax + b) 的导数是 -a sin(ax + b)。注意这里的负号，这是 A-Level 考试中常见的失分点。

广告 / Sponsored

🛒
TI-Nspire CX II CAS 图形计算器 A-Level数学
¥1298
📱 去京东购买
京东精选

5. Extending the Chain Rule to tan(ax + b) | 将链式法则推广到 tan(ax + b)

ivative of tan(u) gives another standard result. Since d/du [tan(u)] = sec²(u), we obtain:

对于正切函数，链式法则结合已知的 tan(u) 导数可以得到另一个标准结果。因为 d/du [tan(u)] = sec²(u)，所以我们有：

d/dx [tan(ax + b)] = a sec²(ax + b)

In AQA mark schemes, you may write this as a sec²(ax + b) or equivalently a/cos²(ax + b). Both forms are acceptable, but you must include the inner derivative a.

在 AQA 的评分标准中，你可以写成 a sec²(ax + b)，或者等价地写成 a/cos²(ax + b)。两种形式都可以接受，但必须包含内层导数 a。

6. Applying the Chain Rule to Exponential and Logarithmic Functions | 链式法则在指数函数和对数函数中的应用

ons, which often appear in AQA exam questions alongside trigonometric functions. For y = ekx, the derivative is dy/dx = k ekx ve is dy/dx = 1/x, but only if kx is positive; in general, d/dx [ln(ax + b)] = a/(ax + b).

同样的链式法则结构也适用于复合指数函数和对数函数，这些函数经常与三角函数一起出现在 AQA 考试题中。对于 y = ekx，其导数是 dy/dx = k ekx。对于 y = ln(kx)，导数在 kx 为正时是 dy/dx = 1/x；一般来说，d/dx [ln(ax + b)] = a/(ax + b)。

Recognising these patterns saves time and reduces errors, because you do not need to perform a formal substitution each time.

识别这些模式可以节省时间并减少错误，因为你不需要每次都进行正式的代换。

7. Common Mistakes in AQA Exams | AQA 考试中的常见错误

Many students lose marks on chain rule questions for predictable reasons. The most common error is forgetting to multiply by the derivative of the inner function, especially when the inner function is 3x, 2x + 1, or 5 - x.

许多学生在链式法则题目上因一些可预见的原因而失分。最常见的错误是忘记乘以内层函数的导数，尤其是当内层函数为 3x、2x + 1 或 5 - x 时。