Exercise 6F: Chain Rule and Implicit Differentiation | 练习6F：链式法则与隐函数求导

📚 Exercise 6F: Chain Rule and Implicit Differentiation | 练习6F：链式法则与隐函数求导

Exercise 6F in the IB Mathematics: Analysis and Approaches (AA) textbook is dedicated to mastering differentiation techniques that go beyond basic power rule applications. This exercise set builds fluency in the chain rule, implicit differentiation, and their use with trigonometric, exponential, and logarithmic functions. A solid command of these skills is vital for tackling IB exams and internal assessments. In this article, we will break down the key ideas, solve representative problems from Exercise 6F, and flag common mistakes to help you excel.

IB数学分析与方法(AA)教材中的练习6F聚焦于掌握超越基本幂函数求导的微分技巧。这套练习帮助学生熟练掌握链式法则、隐函数求导及其在三角函数、指数函数和对数函数中的应用。扎实掌握这些技能对于应对IB考试和内部评估至关重要。本文将详细梳理核心概念，解答练习6F中的典型题目，并指出常见错误，助你取得优异表现。

1. The Chain Rule: Core Concept | 链式法则：核心概念

ect to x is given by dy/dx = (dy/du) × (du/dx). In function notation, this reads [f(g(x))]' = f'(g(x)) × g' iply by the derivative of the inner function. Exercise 6F tests this concept relentlessly.

链式法则是求复合函数导数的首要工具。用莱布尼茨符号表示，若 y = f(u) 且 u = g(x)，则 y 对 x 的导数为 dy/dx = (dy/du) × (du/dx)。用函数记号可写为 [f(g(x))]' = f'(g(x)) × g'(x)。这个法则告诉我们：先对外层函数求导，内层保持不变，再乘以内层函数的导数。练习6F会反复检验这一概念。

If y = f(u) and u = g(x), then dy/dx = f'(u) × g'(x)

omposition makes differentiation smooth and systematic, which is exactly what the problems in Exercise 6F demand.

掌握该法则不仅仅是记住公式，更要能迅速识别内层与外层结构。当你看到类似 (2x+1)⁵ 的表达式时，立即拆解为 u = 2x+1 与 y = u⁵。这种心理分解会使求导过程流畅而系统化，这正是练习6F中题目所要求的能力。

2. Applying the Chain Rule to Polynomials | 对多项式应用链式法则

4(3x+2)³ × 3 = 12(3x+2)³. Exercise 6F is packed with such examples to build automaticity.

许多学生最初通过多项式复合接触链式法则。例如求 y = (3x+2)⁴ 的导数。令 u = 3x+2，则 y = u⁴，那么 dy/du = 4u³，du/dx = 3。相乘得 dy/dx = 12(3x+2)³。熟练后可直接写出：导数 = 4(3x+2)³ × 3 = 12(3x+2)³。练习6F充满了这类例子，旨在建立自动化反应。

Another typical problem is y = √5x²−1, which can be rewritten as y = (5x²−1)^1/2. The outer function is u^1/2 and the inner is 5x²−1. The derivative is dy/dx = (1/2)(5x²−1)^−1/2 × 10x = 5x / √5x²−1. Pay close attention to the power reduction and the factor from the inner derivative. These steps must become second nature for IB exam success.

另一类典型题目是 y = √5x²−1，可改写为 y = (5x²−1)^1/2。外层为 u^1/2，内层为 5x²−1。导数为 dy/dx = (1/2)(5x²−1)^−1/2 × 10x = 5x / √5x²−1。要特别注意幂次的减半以及内层导数带来的因子。为在IB考试中取得成功，这些步骤必须成为条件反射。

3. Differentiating Trigonometric Functions | 三角函数的求导

1). When differentiating y = tan(2x), remember d(tan u)/du = sec²u, so dy/dx = 2 sec²(2x).

练习6F中大量出现涉及三角函数的链式法则题目。对于 y = sin(kx)，导数为 k cos(kx)；对于 y = cos(kx)，导数为 −k sin(kx)。因此，y = sin(3x+1) 的导数 dy/dx = 3 cos(3x+1)。求导 y = tan(2x) 时，记住 d(tan u)/du = sec²u，于是 dy/dx = 2 sec²(2x)。

(4x) cos(4x), this simplifies to −4 sin(8x). Such simplifications are often rewarded in IB mark schemes.

对三角函数平方则增加了一层嵌套：y = cos²(4x) 即 y = (cos(4x))²。两次应用链式法则，dy/dx = 2 cos(4x) × (−sin(4x)) × 4 = −8 cos(4x) sin(4x)。利用倍角公式 sin(8x) = 2 sin(4x) cos(4x)，可化简为 −4 sin(8x)。这类化简在IB评分标准中常会得分。

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4. Chain Rule with Natural Exponential | 自然指数函数与链式法则

The natural exponential function e^u differentiates to e^u × du/dx. For instance, if y = e^5x², then dy/dx = e^5x² × 10x = 10x e^5x². Notice that the exponential factor itself remains unchanged; only the exponent's derivative multiplies it. Exercise 6F ensures you can handle expressions like e^{sin x} effortlessly.

自然指数函数 e^u 的导数为 e^u × du/dx。例如，若 y = e^5x²，则 dy/dx = e^5x² × 10x = 10x e^5x²。注意指数因子本身不变，只是其指数的导数相乘。练习6F确保你能轻松处理 e^{sin x} 这类表达式。

When the base is not e, proceed with caution. For y = 2^x³, rewrite as y = e^{x³ ln 2}. Then dy/dx = e^{x³ ln 2} × (3x² ln 2) = 2^x³ × 3x² ln 2. Alternatively, using the standard derivative formula for a^x yields the same result. Whichever method you choose, the chain rule is always at work.

当底数不是 e 时，需谨慎处理。对于 y = 2^x³，可改写为 y = e^{x³ ln 2}。则 dy/dx = e^{x³ ln 2} × (3x² ln 2) = 2^x³ × 3x² ln 2。或者使用 a^x 的标准导数公式，结果一致。无论选用何种方法，链式法则始终在发挥作用。

5. Repeated Chain Rule | 重复使用链式法则

2x, giving du/dx = −sin(v) × 2 = −2 sin(2x). Putting everything together, dy/dx = cos(cos(2x)) × (−2 sin(2x)).

有些函数因包含三层或更多嵌套而需要多次应用链式法则。考虑 y = sin(cos(2x))。最外层函数为 sin(u)，其中 u = cos(2x)。sin(u) 的导数为 cos(u) × du/dx。而 du/dx 本身也需链式法则：u = cos(v)，v = 2x，得 du/dx = −sin(v) × 2 = −2 sin(2x)。综合起来，dy/dx = cos(cos(2x)) × (−2 sin(2x))。

Exercise 6F includes progressively layered problems like y = ln(sin(e^x)) to build this multi-step reasoning. In such cases, start from the outside and work systematically inward, recording the derivative of each layer. With practice, this process becomes intuitive and quick.

练习6F包含逐步嵌套的题目，如 y = ln(sin(e^x))，以培养多步推理能力。面对此类题目，从最外层开始有计划地向内推进，记录每一层的导数。经过练习，这个过程会变得直观而迅速。

6. Introduction to Implicit Differentiation | 隐函数求导简介

e equation x² + y² = 25, differentiating gives 2x + 2y (dy/dx) = 0, leading to dy/dx = −x/y.

当 x 与 y 的关系以隐式给出时，如 x² + y² = 25，不能简单解出 y 再求导。此时我们对方程两边关于 x 求导，并把 y 视为 x 的函数。每当对含 y 的项求导时，便通过链式法则乘以 dy/dx。对于方程 x² + y² = 25，求导得 2x + 2y (dy/dx) = 0，从而 dy/dx = −x/y。

This technique is heavily exploited in Exercise 6F. It is essential to remember that derivatives of y terms always carry dy/dx, and constants differentiate to zero. The result is often an expression involving both x and y, which is perfectly acceptable.

练习6F大量运用这一技巧。务必记住，y 的项求导后总带有 dy/dx，而常数求导为零。结果通常是同时包含 x 和 y 的表达式，这完全正常。

7. Step-by-Step Implicit Differentiation Examples | 隐函数求导实例分步解析

ich simplifies to (2y − x²) / (y² − 2x). Such multi-step algebra is common in Exercise 6F.

一道经典题目是 x³ + y³ = 6xy。逐项求导：3x² + 3y² (dy/dx) = 6y + 6x (dy/dx)。将所有含 dy/dx 的项移到一边：3y² (dy/dx) − 6x (dy/dx) = 6y − 3x²。提取 dy/dx：dy/dx (3y² − 6x) = 6y − 3x²，最后 dy/dx = (6y − 3x²) / (3y² − 6x)，化简为 (2y − x²) / (y² − 2x)。这种多步代数运算在练习6F中经常出现。

Another common example is e^y + x y = 1. The derivative of e^y with respect to x is e^y (dy/dx). The term x y requires the product rule: derivative = y + x (dy/dx). Thus, e^y (dy/dx) + y + x (dy/dx) = 0. Isolate dy/dx: (e^y + x)(dy/dx) = −y, giving dy/dx = −y / (e^y + x). Pay careful attention to the product rule when x and y are multiplied together.

另一个常见例子是 e^y + x y = 1。e^y 对 x 求导为 e^y (dy/dx)。项 x y 需用乘积法则：导数为 y + x (dy/dx)。因此 e^y (dy/dx) + y + x (dy/dx) = 0。分离 dy/dx 得 (e^y + x)(dy/dx) = −y，故 dy/dx = −y / (e^y + x)。当 x 与 y 相乘时，一定要留意乘积法则。

8. Finding Tangents and Normals Implicitly | 隐式求切线与法线

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