Exercise 7B.2: Mastering the Chain Rule | 练习 7B.2：掌握链式法则

📚 Exercise 7B.2: Mastering the Chain Rule | 练习 7B.2：掌握链式法则

In IB Mathematics, Exercise 7B.2 focuses on one of the most fundamental differentiation techniques—the Chain Rule. This exercise provides a variety of functions requiring the Chain Rule, from basic polynomials to trigonometric, exponential, and logarithmic compositions. Mastery of this rule is essential for solving complex derivatives and tackling real-world applications such as related rates and optimization. This article revisits the Chain Rule concept, offers step-by-step examples, and highlights common pitfalls, aligning with the practice problems in Exercise 7B.2 to help you achieve full marks.

在 IB 数学中，练习 7B.2 聚焦于最基本的求导技巧之一——链式法则（Chain Rule）。该练习涵盖从简单多项式到三角、指数和对数复合函数的多种类型，要求运用链式法则求导。掌握这一法则对于求解复杂导数和处理相关变化率、最优化等实际应用至关重要。本文回顾链式法则的核心概念，提供分步示例，并指出常见错误，紧扣练习 7B.2 中的习题，助你冲刺满分。

1. Recall: The Chain Rule Formula | 回顾：链式法则公式

= dy/du · du/dx. Alternatively, for a composite function y = f(g(x)), the derivative is f'(g(x)) · g'(x). This rule allows us to differentiate 'functions inside functions' by taking the derivative of the outer function evaluated at the inner function, then multiplying by the derivative of the inner function.

链式法则指出，若 y = f(u) 且 u = g(x)，则 dy/dx = dy/du · du/dx。或者，对于复合函数 y = f(g(x))，其导数为 f'(g(x)) · g'(x)。这一法则通过先对外层函数求导（保留内层函数），再乘以内层函数的导数，解决了“函数套函数”的求导问题。

dy/dx = f'(g(x)) · g'(x)

(dy/du) · (du/dx), which is especially practical when dealing with multiple substitutions.

也可用莱布尼茨记号表示为 dy/dx = (dy/du) · (du/dx)，这在涉及多重代换时特别实用。

2. Identifying the Inside and Outside Functions | 分清“内层”与“外层”函数

(3x + 5) = 4(3x + 5)³ · 3 = 12(3x + 5)³. This pattern appears repeatedly in Exercise 7B.2.

应用链式法则时，需先分清内层函数 u = g(x) 和外层函数 y = f(u)。例如，对于 y = (3x + 5)⁴，内层是 u = 3x + 5，外层是 u⁴。求导得 dy/dx = 4(u)³ · d/dx(3x + 5) = 4(3x + 5)³ · 3 = 12(3x + 5)³。这种模式在练习 7B.2 中反复出现。

Another typical example: y = √5x² - 2. Rewrite as (5x² - 2)^1/2, so the outside is u^1/2 and the inside is 5x² - 2. The derivative is (1/2)(5x² - 2)^-1/2 · 10x = 5x / √5x² - 2.

另一个典型例题：y = √5x² - 2。将其改写为 (5x² - 2)^1/2，外层为 u^1/2，内层为 5x² - 2。求导得 (1/2)(5x² - 2)^-1/2 · 10x = 5x / √5x² - 2。

3. Basic Chain Rule Examples from Exercise 7B.2 | 练习 7B.2 中的基础链式例题

Let' 30x²(2x³ - 7)⁴. In the exercise, you will encounter similar polynomials raised to a power.

我们来求导 y = (2x³ - 7)⁵。设 u = 2x³ - 7，则 y = u⁵。dy/du = 5u⁴，du/dx = 6x²。因此 dy/dx = 5(2x³ - 7)⁴ · 6x² = 30x²(2x³ - 7)⁴。在练习中，你会遇到类似的幂次多项式函数。

For y = 1/(x² + 4)³, rewrite as (x² + 4)^-3. Then derivative = -3(x² + 4)^-4 · 2x = -6x / (x² + 4)⁴. Pay close attention to the negative exponent handling.

对于 y = 1/(x² + 4)³，改写为 (x² + 4)^-3。则导数为 -3(x² + 4)^-4 · 2x = -6x / (x² + 4)⁴。务必留意负指数的处理。

4. Chain Rule with Trigonometric Functions | 三角函数的链式法则

r tan(x²), the derivative is sec²(x²) · 2x. Exercise 7B.2 frequently tests these patterns.

求导复合三角函数时，如 sin(kx) 或 cos(f(x))，必须使用链式法则。例如，d/dx[sin(5x)] = cos(5x) · 5 = 5cos(5x)。若遇到 tan(x²)，其导数为 sec²(x²) · 2x。练习 7B.2 经常考查这类模式。

A slightly more involved example: differentiate y = cos(√x). Outside: cos(u), inside: u = √x = x^1/2. Derivative: -sin(√x) · (1/2)x^-1/2 = -sin(√x) / (2√x). The chain rule effortlessly handles the square root inside the cosine.

一个稍复杂的例子：求 y = cos(√x) 的导数。外层：cos(u)，内层：u = √x = x^1/2。导数为 -sin(√x) · (1/2)x^-1/2 = -sin(√x) / (2√x)。链式法则轻松处理余弦内部的平方根。

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5. Chain Rule with Exponentials and Logarithms | 指数与对数函数的链式法则

The exponential function e^(g(x)) differentiates to e^(g(x)) · g'(x). For example, d/dx[e^3x²] = e^3x² · 6x. In Exercise 7B.2 you might see y = e^{sin x}, whose derivative is e^{sin x} · cos x. The pattern 'e to the something' always keeps the same exponential term multiplied by the derivative of the exponent.

指数函数 e^(g(x)) 的导数为 e^(g(x)) · g'(x)。例如，d/dx[e^3x²] = e^3x² · 6x。在练习 7B.2 中你可能遇到 y = e^{sin x}，其导数为 e^{sin x} · cos x。这种“e 的某次方”的模式总是保留原指数项并乘以 exponent 的导数。

For natural logs, d/dx[ln(f(x))] = f' 1). Another common one is y = ln(cos x), which gives derivative = -sin x / cos x = -tan x.

对于自然对数，d/dx[ln(f(x))] = f'(x)/f(x)，假设 f(x) > 0。练习中的一个例子：ln(x³ + 1) 的导数为 (3x²)/(x³ + 1)。另一个常见的是 y = ln(cos x)，其导数为 -sin x / cos x = -tan x。

6. Combining Chain Rule with Product and Quotient Rules | 链式法则与乘积法则、商法则的结合

). So final derivative is 2x sin(4x) + 4x² cos(4x). Always respect the correct order of operations.

练习 7B.2 中含有需要同时运用链式法则和乘积法则的函数。例如 y = x² · sin(4x)，先用乘积法则：dy/dx = (2x) · sin(4x) + x² · d/dx[sin(4x)]，再对 sin(4x) 用链式法则得 4cos(4x)。最终导数为 2x sin(4x) + 4x² cos(4x)。务必遵循正确的运算顺序。

Similarly, for quotients like y = e^2x / (x + 1), use the quotient rule: derivative = [ (2e^2x)(x+1) - e^2x(1) ] / (x+1)². Notice that the derivative of e^2x requires the chain rule, producing 2e^2x. Practice these mixed problems to build fluency.

类似地，对于商如 y = e^2x / (x + 1)，使用商法则：导数 = [ (2e^2x)(x+1) - e^2x(1) ] / (x+1)²。注意 e^2x 的导数需要链式法则，得到 2e^2x。多练习这类混合题型可提升熟练度。

7. Repeated Chain Rule for Nested Functions | 多层嵌套函数的链式法则

Some problems in 7B.2 feature triple compositions like y = sin(e^x²). Here you apply the chain rule multiple times: differentiate outer sin(…) to cos(e^x²), multiply by derivative of e^x² which is e^x² · 2x. So dy/dx = cos(e^x²) · e^x² · 2x. Always work from outside in, layer by layer.

练习 7B.2 中的某些题目涉及三层嵌套，如 y = sin(e^x²)。此时需多次使用链式法则：先对外层 sin(…) 求导得 cos(e^x²)，乘以内层 e^x² 的导数 e^x² · 2x。最终 dy/dx = cos(e^x²) · e^x² · 2x。牢记由外到内逐层求导。

+ 1)]. The derivative is (1/tan(3x+1)) · sec²(3x+1) · 3. You must recognise the chain rule applied twice: once for ln, once for tan, and the linear inner function 3x+1 contributes a factor of 3. Keep track of each multiplication.

一个更复杂的嵌套：y = ln[tan(3x + 1)]。其导数为 (1/tan(3x+1)) · sec²(3x+1) · 3。这里需要两次链式法则：一次给 ln，一次给 tan，而内层线性函数 3x+1 贡献了一个因子 3。每次乘法都不要遗漏。

8. Using Leibniz Notation for Clarity | 使用莱布尼茨记号保持清晰

ec(x)) · sec(x)tan(x) = tan(x). This approach reduces errors by breaking down the process.

莱布尼茨形式 dy/dx = dy/du · du/dx 在复杂求导中非常有用。如 y = ln(sec(x))，设 u = sec(x)，则 y = ln(u)。dy/du = 1/u，du/dx = sec(x)tan(x)。相乘得 dy/dx = (1/sec

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