📚 A-Level Maths: Core Applications of the Chain Rule in Differentiation | A-Level 数学：链式法则求导核心应用
The chain rule is one of the most powerful and frequently tested differentiation techniques in A-Level Mathematics. It allows us to differentiate composite functions by breaking them into an ‘outer’ and an ‘inner’ function, then multiplying their derivatives together. Mastering this rule is essential for solving a wide range of exam problems, from simple polynomials to complex trigonometric, exponential, and implicit functions.
链式法则是 A-Level 数学中最强大、最常考的求导技巧之一。它通过将复合函数拆分为“外函数”和“内函数”，并将两者的导数相乘，从而完成求导。掌握链式法则对于解决各类考试题目至关重要，无论是简单的多项式，还是复杂的三角函数、指数函数和隐函数。
1. The Chain Rule Formula | 链式法则公式
The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). In Leibniz notation, if y = f(u) where u = g(x), then dy/dx = dy/du × du/dx. This form is particularly useful when dealing with multi-step compositions.
链式法则指出：若 y = f(g(x))，则 dy/dx = f'(g(x)) · g'(x)。用莱布尼茨记号表示，若 y = f(u) 且 u = g(x)，则 dy/dx = dy/du × du/dx。这种形式在处理多层复合时尤其方便。
dy/dx = dy/du × du/dx
Think of it as ‘differentiate the outer function, keep the inner function unchanged, then multiply by the derivative of the inner function’. This ‘outside-inside’ approach is the key to applying the rule correctly.
可以这样理解：“先对外函数求导，内函数保持不变，再乘以内函数的导数”。这种“外-内”思路是正确运用法则的关键。
2. Differentiating Powers of Functions | 函数幂次的求导
The most common application of the chain rule is for functions of the form (ax + b)ⁿ. Here, the outer function is uⁿ and the inner function is u = ax + b. Using the rule, the derivative becomes n(ax + b)ⁿ⁻¹ × a.
链式法则最常见的应用是形如 (ax + b)ⁿ 的函数。此时外函数为 uⁿ，内函数为 u = ax + b。根据法则，导数为 n(ax + b)ⁿ⁻¹ × a。
Example: Differentiate y = (3x + 2)⁴. Let u = 3x + 2, then dy/du = 4u³ and du/dx = 3. Therefore dy/dx = 4(3x + 2)³ × 3 = 12(3x + 2)³.
例：求 y = (3x + 2)⁴ 的导数。令 u = 3x + 2，则 dy/du = 4u³，du/dx = 3。因此 dy/dx = 4(3x + 2)³ × 3 = 12(3x + 2)³。
For more general powers, such as y = (x² + 5x – 1)⁶, the inner derivative is 2x + 5, so dy/dx = 6(x² + 5x – 1)⁵ × (2x + 5). This pattern – bring down the power, reduce it by one, then multiply by the derivative of the base – always works.
对于更一般的幂次，如 y = (x² + 5x – 1)⁶，内函数导数为 2x + 5，所以 dy/dx = 6(x² + 5x – 1)⁵ × (2x + 5)。这个规律——降幂、减一、再乘以底数的导数——始终成立。
3. Chain Rule with Trigonometric Functions | 含三角函数的链式法则
When differentiating sin, cos, and tan of a function of x, the chain rule is indispensable. For y = sin(u), dy/dx = cos(u) · du/dx. Similarly, for y = cos(u), dy/dx = -sin(u) · du/dx, and for y = tan(u), dy/dx = sec²(u) · du/dx.
当对 x 的函数的正弦、余弦、正切求导时，链式法则不可或缺。若 y = sin(u)，则 dy/dx = cos(u) · du/dx。类似地，若 y = cos(u)，则 dy/dx = -sin(u) · du/dx；若 y = tan(u)，则 dy/dx = sec²(u) · du/dx。
Example: Differentiate y = sin(5x – 3). Here u = 5x – 3, so du/dx = 5. Thus dy/dx = 5cos(5x – 3).
例：求 y = sin(5x – 3) 的导数。这里 u = 5x – 3，du/dx = 5。因此 dy/dx = 5cos(5x – 3)。
Example: Differentiate y = cos²(x). This can be written as (cos x)², so the outer power rule applies first: dy/dx = 2cos x × (-sin x) = -2sin x cos x = -sin(2x). This double-angle simplification is a common exam favourite.
例：求 y = cos²(x) 的导数。可将它写成 (cos x)²，先使用幂法则：dy/dx = 2cos x × (-sin x) = -2sin x cos x = -sin(2x)。这种二倍角化简是考试中常见的考点。
4. Chain Rule with Exponential Functions | 含指数函数的链式法则
For exponential functions of the form y = eᵘ, the derivative is simply dy/dx = eᵘ · du/dx. The exponential function is unique in that its derivative is itself, but the chain rule must still be applied to the exponent.
对于形如 y = eᵘ 的指数函数，其导数为 dy/dx = eᵘ · du/dx。指数函数的导数等于其自身，但链式法则仍须应用于指数部分。
Example: Differentiate y = e^(3x² + 2x). Let u = 3x² + 2x, then du/dx = 6x + 2. Therefore dy/dx = (6x + 2)e^(3x² + 2x).
例：求 y = e^(3x² + 2x) 的导数。令 u = 3x² + 2x，则 du/dx = 6x + 2。因此 dy/dx = (6x + 2)e^(3x² + 2x)。
For other bases, such as aᵘ, the derivative is aᵘ · ln(a) · du/dx. For example, y = 2^(4x) differentiates to 2^(4x) · ln 2 × 4 = 4 · 2^(4x) · ln 2.
对于其他底数，如 aᵘ，其导数为 aᵘ · ln(a) · du/dx。例如，y = 2^(4x) 的导数为 2^(4x) · ln 2 × 4 = 4 · 2^(4x) · ln 2。
5. Chain Rule with Logarithmic Functions | 含对数函数的链式法则
For natural logarithms, y = ln(u) gives dy/dx = (1/u) · du/dx = u’/u. This rule is frequently used in exam questions involving logs of composite expressions.
对于自然对数，y = ln(u) 给出 dy/dx = (1/u) · du/dx = u’/u。这条法则在涉及复合表达式的对数题目中频繁使用。
Example: Differentiate y = ln(x² + 1). Here u = x² + 1, so du/dx = 2x. Thus dy/dx = 2x/(x² + 1).
例：求 y = ln(x² + 1) 的导数。这里 u = x² + 1，du/dx = 2x。因此 dy/dx = 2x/(x² + 1)。
For logarithms with other bases, y = logₐ(u) yields dy/dx = u’/(u · ln a). For example, y = log₁₀(sin x) gives dy/dx = cos x / (sin x · ln 10) = cot x / ln 10.
对于其他底数的对数，y = logₐ(u) 给出 dy/dx = u’/(u · ln a)。例如，y = log₁₀(sin x) 的导数为 cos x / (sin x · ln 10) = cot x / ln 10。
6. Chain Rule with Radicals and Fractional Powers | 含根式与分数幂的链式法则
Radicals can be rewritten as fractional powers, making the chain rule easy to apply. For instance, √u = u^(1/2), so d/dx(√u) = (1/2)u^(-1/2) · du/dx = du/dx / (2√u).
根式可以改写为分数指数幂，从而轻松地应用链式法则。例如，√u = u^(1/2)，所以 d/dx(√u) = (1/2)u^(-1/2) · du/dx = du/dx / (2√u)。
Example: Differentiate y = √(x³ + 4x). Rewrite as y = (x³ + 4x)^(1/2). The derivative is (1/2)(x³ + 4x)^(-1/2) × (3x² + 4), which simplifies to (3x² + 4) / (2√(x³ + 4x)).
例：求 y = √(x³ + 4x) 的导数。改写为 y = (x³ + 4x)^(1/2)。其导数为 (1/2)(x³ + 4x)^(-1/2) × (3x² + 4)，化简为 (3x² + 4) / (2√(x³ + 4x))。
Fractional powers with negative exponents also work the same way. For y = 1/(2x + 1)³, rewrite as y = (2x + 1)⁻³, then dy/dx = -3(2x + 1)⁻⁴ × 2 = -6/(2x + 1)⁴.
含负指数的分数幂也同样适用。对于 y = 1/(2x + 1)³，改写为 y = (2x + 1)⁻³，则 dy/dx = -3(2x + 1)⁻⁴ × 2 = -6/(2x + 1)⁴。
7. Nested Chain Rule (Multiple Layers) | 多层嵌套链式法则
Some functions require applying the chain rule multiple times within a single differentiation. For example, y = sin(√(x² + 1)) involves three layers: sine, then square root, then x² + 1.
有些函数需要在一次求导中多次应用链式法则。例如，y = sin(√(x² + 1)) 包含三层：正弦、平方根，然后是 x² + 1。
To differentiate, work from the outside in. First, derivative of sine: cos(√(x² + 1)). Then multiply by the derivative of the inner square root: (1/2)(x² + 1)^(-1/2) × (2x). Combining all parts gives:
求导时需要由外向内逐层进行。首先，正弦的导数为 cos(√(x² + 1))；然后乘以内层平方根的导数：(1/2)(x² + 1)^(-1/2) × (2x)。整合所有部分得到：
dy/dx = cos(√(x² + 1)) × x / √(x² + 1)
For more complex exponentials, such as y = e^(sin(3x)), we differentiate the exponential first, then the sine, then the linear argument:
对于更复杂的指数函数，如 y = e^(sin(3x))，我们先对指数求导，再对正弦求导，最后对线性自变量求导：
dy/dx = e^(sin(3x)) × cos(3x) × 3 = 3cos(3x) e^(sin(3x))
The key is to maintain a systematic step-by-step approach, tracking each layer carefully.
关键在于保持系统化的分步求解，仔细追踪每一层。
8. Combining Chain Rule with Product and Quotient Rules | 链式法则与乘除法则的结合
Exam questions often require combining the chain rule with the product or quotient rule. For example, to differentiate y = x² sin(2x), we use the product rule first, then apply the chain rule to the second term:
考试题目常常需要将链式法则与乘法法则或除法法则结合。例如，求 y = x² sin(2x) 的导数时，先用乘法法则，再对第二项应用链式法则：
dy/dx = 2x · sin(2x) + x² · 2cos(2x) = 2x sin(2x) + 2x² cos(2x).
dy/dx = 2x · sin(2x) + x² · 2cos(2x) = 2x sin(2x) + 2x² cos(2x)。
For quotients, such as y = e^(2x) / (x³ + 1), use the quotient rule: dy/dx = [2e^(2x)(x³ + 1) – e^(2x)(3x²)] / (x³ + 1)². The chain rule appears when differentiating e^(2x).
对于商式，如 y = e^(2x) / (x³ + 1)，使用除法法则：dy/dx = [2e^(2x)(x³ + 1) – e^(2x)(3x²)] / (x³ + 1)²。在求 e^(2x) 的导数时应用了链式法则。
A useful strategy is to identify which rule to apply first. The order of operations in differentiation follows the structure of the expression: products and quotients demand their respective rules first, while nested functions demand the chain rule.
一个有效的策略是先判断应优先应用哪条法则。求导的运算顺序取决于表达式的结构：乘积与商优先使用对应的法则，复合结构则优先使用链式法则。
9. Implicit Differentiation via the Chain Rule | 利用链式法则进行隐函数求导
When differentiating equations where y is defined implicitly, we treat y as a function of x. Thus, any term involving y must be differentiated using the chain rule: d/dx(y²) = 2y · dy/dx, and d/dx(sin y) = cos y · dy/dx.
当对以隐式定义的 y 的方程求导时，我们将 y 视为 x 的函数。因此，任何包含 y 的项都必须使用链式法则求导：d/dx(y²) = 2y · dy/dx，d/dx(sin y) = cos y · dy/dx。
Example: Given x² + y² = 25, differentiating both sides with respect to x gives 2x + 2y · dy/dx = 0, so dy/dx = -x/y. This is a standard result for a circle.
例：已知 x² + y² = 25，两边对 x 求导得到 2x + 2y · dy/dx = 0，因此 dy/dx = -x/y。这是圆的标准结论。
Implicit differentiation is heavily tested in A-Level papers, especially with products and quotients of x and y terms. Remember to multiply each y-derived term by dy/dx, and treat y itself as ‘inside’ the differentiation.
隐函数求导在 A-Level 考试中出现频率很高，尤其涉及 x 与 y 项的乘积和商式。切记每个对含有 y 的项求导后都要乘以 dy/dx，并将 y 视为“内层”函数。
10. Related Rates and Connected Rates of Change | 相关变化率与联变率
The chain rule is the foundation of related rates problems. If a quantity y depends on u, which in turn depends on time t, then dy/dt = dy/du × du/dt. This is a direct application of the chain rule in real-world contexts.
链式法则是相关变化率问题的基础。若一个量 y 依赖于 u，而 u 又依赖于时间 t，则 dy/dt = dy/du × du/dt。这是链式法则在实际情境中的直接应用。
Example: The radius r of a circle is increasing at a rate of 0.5 cm/s. Find the rate at which the area A is increasing when r = 4 cm. Since A = πr², we have dA/dt = dA/dr × dr/dt = 2πr × 0.5 = πr. At r = 4, dA/dt = 4π cm²/s.
例：圆的半径 r 以 0.5 cm/s 的速率增加。求当 r = 4 cm 时面积 A 的增加速率。由于 A = πr²，我们有 dA/dt = dA/dr × dr/dt = 2πr × 0.5 = πr。当 r = 4 时，dA/dt = 4π cm²/s。
In such problems, always identify the intermediate variable connecting the two quantities of interest, then apply the chain rule to relate their rates.
在这类问题中，务必找出联系两个相关量的中间变量，然后运用链式法则将变化率联系起来。
11. Common Mistakes and Exam Tips | 常见错误与考试提示
One frequent error is forgetting to multiply by the derivative of the inner function. For example, students may write d/dx of (2x + 1)³ as 3(2x + 1)² instead of 6(2x + 1)². Always check whether you have included the inner derivative.
一个常见错误是忘记乘以内函数的导数。例如，学生可能会将 (2x + 1)³ 的导数写成 3(2x + 1)²，而不是 6(2x + 1)²。始终检查是否包含了内函数的导数。
Another tip: when in doubt, use substitution. Let u equal the inner function, differentiate separately, then multiply the results. This reduces errors, especially for complex expressions.
另一个技巧：拿不准时使用代换。令 u 等于内函数，分别求导后再相乘。这能减少错误，尤其对于复杂的表达式。
Always simplify final answers where possible. Combining like terms and using trigonometric identities (such as sin 2x = 2sin x cos x) can help present answers in the form examiners expect.
尽可能化简最终答案。合并同类项并使用三角恒等式（如 sin 2x = 2sin x cos x）有助于将答案化为考官期望的形式。
12. Summary | 总结
The chain rule is a versatile and essential tool for differentiating composite functions. It applies to polynomials, trigonometric, exponential, logarithmic, radical, and implicit functions, and forms the basis for related rates problems. The general procedure is simple: differentiate the outside, keep the inside, multiply by the derivative of the inside.
链式法则是对复合函数求导的一件灵活而必备的工具。它适用于多项式、三角函数、指数函数、对数函数、根式函数和隐函数，并且是相关变化率问题的基础。一般步骤很简单：外函数求导，内函数保持不变，再乘以内函数的导数。
To succeed in exams, practise identifying the inner and outer functions quickly, combine the chain rule with product and quotient rules confidently, and always simplify your final answer. With consistent practice, the chain rule will become second nature.
要在考试中取得成功，请练习快速识别内函数和外函数，自信地将链式法则与乘法、除法法则结合使用，并始终化简最终答案。通过持之以恒的练习，链式法则将变得驾轻就熟。
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