Exercise 16B.2: Mastering the Chain Rule | 习题16B.2：掌握链式法则

📚 Exercise 16B.2: Mastering the Chain Rule | 习题16B.2：掌握链式法则

The chain rule is an essential tool in calculus for differentiating composite functions. In Exercise 16B.2, students explore a variety of functions requiring the chain rule, often combined with product and quotient rules. This article provides detailed solutions and explanations for key problems from the exercise, helping you build confidence in applying the chain rule correctly.

链式法则是微积分中区分复合函数的基本工具。在习题16B.2中，学生们将处理各种需要链式法则的函数，并经常结合乘积法则和商法则。本文为该练习中的重点题目提供详细解答和解释，帮助你自信地正确应用链式法则。

1. Power of a Linear Function | 线性函数的幂

Problem: y = (5x − 3)⁸

题目：y = (5x − 3)⁸

= dy/du × du/dx. Here dy/du = 8u⁷ and du/dx = 5. Substituting back yields the derivative.

我们将外层函数视为 u⁸，内层函数设为 u = 5x − 3。链式法则表示为 dy/dx = dy/du × du/dx。这里 dy/du = 8u⁷，du/dx = 5。代回原变量即得导数。

dy/dx = 40(5x − 3)⁷

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2. Power of a Quadratic | 二次函数的幂

Problem: y = (x² + 3x)⁴

题目：y = (x² + 3x)⁴

3x)³(2x + 3). This is the final answer; it can be expanded but is typically left in factored form.

设 u = x² + 3x，则 y = u⁴。求导得 dy/du = 4u³，du/dx = 2x + 3。应用链式法则，dy/dx = 4(x² + 3x)³(2x + 3)。这是最终答案；可以展开，但通常保留因式分解形式。

dy/dx = 4(x² + 3x)³(2x + 3)

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3. Trigonometric Function Composition | 三角复合函数

Problem: y = sin(2x + 1)

题目：y = sin(2x + 1)

of sin(u) is cos(u) · du/dx. Here u = 2x + 1, du/dx = 2. Therefore dy/dx = 2 cos(2x + 1).

sin(u) 的导数是 cos(u) · du/dx。这里 u = 2x + 1，du/dx = 2。因此 dy/dx = 2 cos(2x + 1)。

dy/dx = 2 cos(2x + 1)

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4. Exponential Composition | 指数复合函数

Problem: y = e^{cos x}

题目：y = e^{cos x}

For e^u, dy/du = e^u. Let u = cos x, so du/dx = −sin x. The chain rule gives dy/dx = e^{cos x} · (−sin x) = −sin x · e^{cos x}.

对于 e^u，dy/du = e^u。设 u = cos x，则 du/dx = −sin x。链式法则得出 dy/dx = e^{cos x} · (−sin x) = −sin x · e^{cos x}。

dy/dx = −sin x · e^{cos x}

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5. Logarithmic Composition | 对数复合函数

Problem: y = ln(x² + 1)

题目：y = ln(x² + 1)

] = (1/u) · du/dx. With u = x² + 1, du/dx = 2x. Hence dy/dx = (1/(x²+1)) · 2x = 2x/(x²+1).

回想 d/dx [ln(u)] = (1/u) · du/dx。取 u = x² + 1，du/dx = 2x。因此 dy/dx = (1/(x²+1)) · 2x = 2x/(x²+1)。

dy/dx = 2x/(x² + 1)

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6. Square Root Function | 平方根函数

Problem: y = √4x − 1 = (4x − 1)^1/2

题目：y = √4x − 1 = (4x − 1)^1/2

Rewrite the square root as a power. Let u = 4x − 1, y = u^1/2. Then dy/du = (1/2)u^−1/2 and du/dx = 4. Multiply: dy/dx = (1/2)(4x−1)^−1/2 · 4 = 2 / √4x−1.

将平方根改写为幂函数。设 u = 4x − 1，y = u^1/2。则 dy/du = (1/2)u^−1/2，du/dx = 4。相乘得 dy/dx = (1/2)(4x−1)^−1/2 · 4 = 2 / √4x−1。

dy/dx = 2 / √4x − 1

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7. Product of Two Composite Functions | 两个复合函数的乘积（结合乘积法则）

Problem: y = x² · (2x + 1)³

题目：y = x² · (2x + 1)³

This requires both the product rule and the chain rule. Let u = x² and v = (2x+1)³. Then u' = 2x and v' must be found via the chain rule: v' = 3(2x+1)²·2 = 6(2x+1)². Using the product rule (uv)' = u'v + uv' 6(2x+1)². Factorising 2(2x+1)² gives dy/dx = 2(2x+1)² [x(2x+1) + 3x²] = 2(2x+1)²(5x² + x).

这道题需要结合乘积法则和链式法则。设 u = x²，v = (2x+1)³。则 u' = 2x，而 v' 需用链式法则求得：v' = 3(2x+1)²·2 = 6(2x+1)²。应用乘积法则 (uv)' = u'v + uv'，得到 dy/dx = 2x·(2x+1)³ + x²·6(2x+1)²。提取公因式 2(2x+1)²，dy/dx = 2(2x+1)² [x(2x+1) + 3x²] = 2(2x+1)²(5x² + x)。

dy/dx = 2(2x + 1)²(5x² + x)

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8. Quotient Involving Composite | 商式与链式法则结合

Problem: y = (x+1)² / (3x−2)

题目：y = (x+1)² / (3x−2)

Let's use the quotient rule: (u/v)' = (u'v − uv')/v². Set u = (x+1)², v = 3x−2. Compute u' via chain rule: 2(x+1)·1 = 2(x+1). v' ) − 3(x+1)] = (x+1)[6x−4 −3x −3] = (x+1)(3x−7). Final answer: dy/dx = (x+1)(3x−7)/(3x−2)².

我们使用商法则：(u/v)' = (u'v − uv')/v²

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