The Chain Rule and Differentiating Composite Functions | 数学考点：复合函数的求导与链式法则

📚 The Chain Rule and Differentiating Composite Functions | 数学考点：复合函数的求导与链式法则

The chain rule is the most heavily examined differentiation technique in A-Level, IB and AP Calculus. Almost every “hard” differentiation question in a past paper hides a composite function, and examiners deliberately design markschemes around whether you remembered to multiply by the derivative of the inside. Once you can reliably identify the inner function and apply the chain rule mechanically, a whole family of questions – trigonometric, exponential, logarithmic, root and connected-rates problems – becomes routine.

链式法则是 A-Level、IB 与 AP 微积分中考查频率最高的求导技巧。历年真题中几乎所有”难题”都隐藏着一个复合函数，而阅卷标准往往就卡在你是否记得乘上内层函数的导数。一旦你能稳定地识别内层函数并机械地套用链式法则，三角函数、指数函数、对数函数、根式以及相关变化率这一整类题目都会变成常规操作。

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1. What Is a Composite Function? | 什么是复合函数

A composite function is created when the output of one function is fed directly into another function. If u = g(x) is the inner function and y = f(u) is the outer function, the composite is written y = f(g(x)), or (f ∘ g)(x). The critical first step in every chain-rule problem is to decompose the expression into an inner part and an outer part.

复合函数是指把一个函数的输出直接作为另一个函数的输入。若 u = g(x) 是内层函数，y = f(u) 是外层函数，则复合函数写作 y = f(g(x))，或 (f ∘ g)(x)。每一道链式法则题目的第一步，都是把表达式拆解成”内层”和”外层”两部分。

Notice that the outer function is always evaluated at the whole inner expression, not at x alone. This is what distinguishes a composite from a product: (3x + 5)⁷ is a composite, while x⁷(3x + 5) is a product.

注意外层函数永远作用于整个内层表达式，而不是单独作用于 x。这正是复合函数与乘积的区别：(3x + 5)⁷ 是复合函数，而 x⁷(3x + 5) 是乘积。

Expression	Inner u = g(x)	Outer y = f(u)
(3x + 5)⁷	u = 3x + 5	y = u⁷
sin(4x)	u = 4x	y = sin u
e^(x²)	u = x²	y = eᵘ
√(x² + 1)	u = x² + 1	y = √u
ln(cos 2x)	u = cos 2x	y = ln u

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2. The Chain Rule: Symbolic Statement | 链式法则的符号表述

If y is a differentiable function of u, and u is a differentiable function of x, then y is a differentiable function of x and the derivative is the product of the two derivatives.

若 y 是 u 的可导函数，u 是 x 的可导函数，则 y 是 x 的可导函数，其导数等于两个导数之积。

dy/dx = (dy/du) × (du/dx)

The same rule written in prime notation, which is often faster to apply, is:

用撇号记法写出的同一法则往往应用更快：

d/dx [ f(g(x)) ] = f′(g(x)) × g′(x)

In words: differentiate the outer function with respect to its own argument, leaving the inner function untouched, then multiply by the derivative of the inner function. Memorise this sentence – it is the single most useful habit for avoiding dropped factors.

用文字表述：先对外层函数求导（内层函数整体保持不变），再乘以内层函数的导数。请把这句话背下来——它是避免漏乘因子最有效的习惯。

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3. Why It Works: The Leibniz Insight | 原理：莱布尼茨形式的直观理解

From the definition of a derivative, a small change δx in x produces a small change δu in u, which in turn produces a small change δy in y. Provided δu ≠ 0, we can write the difference quotient as a product of two quotients, and then let δx → 0.

从导数定义出发，x 的微小变化 δx 引起 u 的微小变化 δu，进而引起 y 的微小变化 δy。只要 δu ≠ 0，就可以把差商写成两个差商之积，然后令 δx → 0。

δy/δx = (δy/δu) × (δu/δx) → dy/dx = (dy/du) × (du/dx)

The δu symbols appear to “cancel” like ordinary fractions, which is exactly why the Leibniz notation is so powerful. In prime notation the same content looks less obvious, which is why students who work purely in f′(x) notation are more likely to forget the inner derivative.

符号 δu 看起来像普通分数一样”约掉”了，这正是莱布尼茨记法强大的原因。用撇号记法时这一内容不那么直观，所以只习惯 f′(x) 记法的学生更容易漏掉内层导数。

Rigorous proofs handle the case δu = 0 carefully using the continuity of g, but the intuitive form above is sufficient for all examination purposes.

严格证明需要用 g 的连续性单独处理 δu = 0 的情形，但上述直观形式对考试已完全够用。

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4. Linear Inner Functions: (ax + b)ⁿ | 线性内层函数

When the inner function is linear, u = ax + b, its derivative is simply the constant a. This gives the most common and most formulaic case of the chain rule, sometimes called the “general power rule”.

当内层函数是一次函数 u = ax + b 时，其导数为常数 a。这是链式法则中最常见、最模式化的情形，有时称为”广义幂法则”。

d/dx (ax + b)ⁿ = an(ax + b)ⁿ⁻¹

So the whole linear bracket is differentiated exactly as if it were a single variable, and the constant a is brought out in front. This single line covers positive integer powers, negative powers, roots and fractional powers.

也就是说，整个线性括号当作一个整体变量求导，常数 a 提到最前面。这一行公式同时覆盖正整数幂、负整数幂、根式和分数幂。

• d/dx (3x + 5)⁷ = 21(3x + 5)⁶ (English: bring the 7 down, reduce the power, multiply by 3)
• d/dx (2 – 5x)⁴ = 4(2 – 5x)³ × (-5) = -20(2 – 5x)³
• d/dx (4x – 1)⁻² = -2(4x – 1)⁻³ × 4 = -8(4x – 1)⁻³
• d/dx √(6x + 1) = ½(6x + 1)⁻¹ᐟ² × 6 = 3 / √(6x + 1)

The sign of a matters enormously. If the inner function is decreasing, such as 2 – 5x, the derivative of the composite is negative, and examiners routinely deduct marks for losing that minus sign.

a 的符号至关重要。若内层函数递减，例如 2 – 5x，则复合函数的导数为负，阅卷时因漏掉负号而扣分的情况非常普遍。

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5. Trigonometric Composites | 三角函数复合

For trigonometric composites the outer derivative follows the standard results, and the inner derivative is multiplied on. With u = ax + b the patterns are completely regular.

三角函数复合时，外层导数遵循标准结论，再乘上内层导数。当 u = ax + b 时，规律完全规则。

Function	Derivative
sin(ax + b)	a cos(ax + b)
cos(ax + b)	-a sin(ax + b)
tan(ax + b)	a sec²(ax + b)

For a non-linear inner function, apply the same principle but keep the inner derivative symbolic. For example, differentiating y = sin(x³) gives cos(x³) × 3x² = 3x² cos(x³), whereas y = sin³x = (sin x)³ gives 3 sin²x × cos x. The difference between sin³x and sin(x³) trips up a large number of candidates every session.

若内层不是线性函数，原则不变，但内层导数要保持符号形式。例如 y = sin(x³) 求导得 cos(x³) × 3x² = 3x²cos(x³)，而 y = sin³x = (sin x)³ 求导得 3sin²x × cos x。sin³x 与 sin(x³) 的区别每场考试都会让大量考生失分。

• d/dx tan(5x) = 5 sec²(5x)
• d/dx cos(1 – 2x) = 2 sin(1 – 2x)
• d/dx sin(2x + π/3) = 2 cos(2x + π/3)
• d/dx cos²(3x) = 2 cos(3x) × (-3 sin 3x) = -6 sin 3x cos 3x = -3 sin 6x

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6. Exponential and Logarithmic Composites | 指数与对数复合

The exponential function has the elegant property that it is its own derivative, so the chain rule contributes only the inner derivative. The logarithm differentiates to a reciprocal, again multiplied by the inner derivative.

指数函数具有”导数是自身”的优美性质，因此链式法则只需补上内层导数。对数函数求导得到倒数，同样乘以内层导数。

d/dx e^(g(x)) = g′(x) e^(g(x))    d/dx ln(g(x)) = g′(x) / g(x)

Using the change-of-base identity a^x = e^(x ln a), the general exponential rule follows immediately and is worth knowing because it appears in markschemes for both pure and modelling questions.

利用换底恒等式 a^x = e^(x ln a)，一般指数函数的求导法则可直接推出；由于它在纯数学与建模题的评分标准中都会出现，值得记住。

d/dx a^x = a^x ln a    d/dx a^(g(x)) = g′(x) a^(g(x)) ln a

• d/dx e^(3x + 1) = 3e^(3x + 1)
• d/dx e^(x²) = 2x e^(x²)
• d/dx ln(5x – 2) = 5 / (5x – 2)
• d/dx ln(x² + 4) = 2x / (x² + 4)
• d/dx 2^(4x) = 4 ln 2 × 2^(4x)

Two special cases deserve attention. First, d/dx e^(kx) = k e^(kx) is the workhorse of growth and decay models. Second, d/dx ln(cos x) = -sin x / cos x = -tan x, a result that frequently appears as part of a larger product or quotient.

两个特例值得关注。第一，d/dx e^(kx) = k e^(kx) 是增长与衰减模型的核心。第二，d/dx ln(cos x) = -sin x / cos x = -tan x，这一结果常作为更复杂乘积或商式的一部分出现。

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7. Roots, Fractional and Negative Powers | 根式、分数幂与负幂

Before differentiating anything containing a root or a denominator, rewrite it as a power with a fractional or negative exponent. This converts an unfamiliar composite into the standard (ax + b)ⁿ form and removes the need for any new rule.

在对含根式或分母的表达式求导前，先把它改写成分数指数或负指数的幂。这样就把陌生的复合形式转化为标准的 (ax + b)ⁿ 形式，无需任何新法则。

Original	Rewrite	Derivative
√(3x + 4)	(3x + 4)¹ᐟ²	3 / (2√(3x + 4))
1 / (2x – 1)³	(2x – 1)⁻³	-6(2x – 1)⁻⁴
√(x² + 1)	(x² + 1)¹ᐟ²	x / √(x² + 1)
1 / √(1 – x)	(1 – x)⁻¹ᐟ²	1 / (2(1 – x)³ᐟ²)

Note that the final answers in the table are written with positive indices, which is the conventional presentation expected in most markschemes. “Correct but unsimplified” answers frequently lose the final accuracy mark.

注意表中最终答案都写成正指数形式，这是多数评分标准期望的规范写法。”答案正确但未化简”经常丢掉最后一个准确性得分点。

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8. Three or More Layers: Nested Chains | 多层嵌套：三层及以上的链

Composite functions can be nested to any depth. If y = f(g(h(x))), the chain rule extends by simply multiplying one factor per layer.

复合函数可以任意深度嵌套。若 y = f(g(h(x)))，链式法则只需按层数逐层相乘即可扩展。

dy/dx = f′(g(h(x))) × g′(h(x)) × h′(x)

The practical technique is to work from the outside inwards: peel off one layer at a time, writing the untouched inner expression in full, and keep going until you reach x.

实用技巧是从外向内逐层剥离：每次只处理最外层一层，被剥掉的外层内部表达式照抄不动，一直到 x 为止。

• y = sin(ln(x² + 1)): outer sin, then ln, then x² + 1. dy/dx = cos(ln(x² + 1)) × 1/(x² + 1) × 2x = 2x cos(ln(x² + 1)) / (x² + 1)
• y = e^(√(3x + 2)): dy/dx = e^(√(3x + 2)) × (1/(2√(3x + 2))) × 3 = 3e^(√(3x + 2)) / (2√(3x + 2))
• y = ln(sin²x): dy/dx = (1/sin²x) × 2 sin x cos x = 2 cot x

When nesting gets deep, apply the product rule or quotient rule only after the chain rule has been fully applied to the outer structure. Mixing the order of operations is a common source of algebraic chaos.

嵌套较深时，应先对外层结构完整应用链式法则，再考虑乘积法则或商法则。颠倒运算顺序常导致代数混乱。

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9. Combining with the Product and Quotient Rules | 与乘积法则、商法则结合

Most examination questions place a composite factor inside a product or a quotient. The discipline required is simple: name your factors first, then differentiate them, then assemble.

大多数考题会把复合因子放在乘积或商式中。所需的自律很简单：先给各因子命名，再分别求导，最后组装。

For y = u(x) v(x), the product rule states:

对于 y = u(x)v(x)，乘积法则为：

dy/dx = u′v + uv′

For y = u(x) / v(x), the quotient rule states:

对于 y = u(x) / v(x)，商法则为：

dy/dx = (u′v – uv′) / v²

• y = x² e^(3x): u = x², v = e^(3x), u′ = 2x, v′ = 3e^(3x), so dy/dx = 2x e^(3x) + 3x² e^(3x) = x e^(3x)(2 + 3x)
• y = e^(2x) sin 3x: dy/dx = 2e^(2x) sin 3x + 3e^(2x) cos 3x = e^(2x)(2 sin 3x + 3 cos 3x)
• y = (x² + 1)⁵ / x: quotient rule with a chained numerator gives dy/dx = [10x(x² + 1)⁴ × x – (x² + 1)⁵] / x² = (x² + 1)⁴(9x² – 1) / x²

Factorising the final answer is not optional decoration; it is almost always required to reach the final mark, and it makes subsequent parts of a structured question far easier.

最后一步因式分解并非可有可无的装饰，而是拿到最后一分的关键，同时也让后续小题变得容易得多。

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10. Connected Rates of Change | 相关变化率

The chain rule with respect to time is the mathematical engine behind connected-rates problems. Because every quantity may depend on t, rates multiply along the chain linking the variables.

关于时间的链式法则是相关变化率问题的数学引擎。由于每个量都可能依赖于 t，各变化率沿着连接变量的链相乘。

dV/dt = (dV/dr) × (dr/dt)

A standard sphere problem: the volume is V = (4/3)πr³, so dV/dr = 4πr². If the radius grows at 0.5 cm s⁻¹ when r = 6 cm, then dV/dt = 4π(36) × 0.5 = 72π cm³ s⁻¹.

一个标准球体问题：体积 V = (4/3)πr³，故 dV/dr = 4πr²。若 r = 6 cm 时半径以 0.5 cm s⁻¹ 增长，则 dV/dt = 4π(36) × 0.5 = 72π cm³ s⁻¹。

The method generalises: write down the geometric relation connecting the variables, differentiate both sides with respect to t (using the chain rule on every non-t variable), then substitute the known values. If only one variable is changing, dV/dx etc. are treated as ordinary derivatives.

方法可以推广：先写出连接各变量的几何关系式，对 t 两边求导（对所有非 t 变量使用链式法则），再代入已知数值。若只有一个变量在变化，dV/dx 等即按普通导数处理。

Typical relations worth memorising include the sphere V = (4/3)πr³, the cone V = (1/3)πr²h, the cylinder V = πr²h, and the circle A = πr². For similar-triangle cone problems, eliminate one variable before differentiating.

值得记住的常用关系包括球体 V = (4/3)πr³、圆锥 V = (1/3)πr²h、圆柱 V = πr²h 以及圆面积 A = πr²。遇到相似三角形圆锥问题，应先消去一个变量再求导。

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11. Worked Exam-Style Examples | 真题风格例题精讲

Example A. Differentiate y = (5x³ – 2x)⁴. The inner function is u = 5x³ – 2x with u′ = 15x² – 2, and the outer is u⁴. Therefore dy/dx = 4(5x³ – 2x)³(15x² – 2).

例 A：求 y = (5x³ – 2x)⁴ 的导数。内层 u = 5x³ – 2x，u′ = 15x² – 2；外层为 u⁴。故 dy/dx = 4(5x³ – 2x)³(15x² – 2)。

Example B. Differentiate y = ln(sin 3x). Working outwards: ln gives a reciprocal, sin gives cosine, and the linear argument gives a factor of 3. Hence dy/dx = (1/sin 3x) × cos 3x × 3 = 3 cot 3x.

例 B：求 y = ln(sin 3x) 的导数。由外向内：ln 给出

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