📚 Exercise 6F.1: Advanced Differentiation Techniques | 练习6F.1：高级微分技巧
This article provides a comprehensive review of the differentiation skills required to successfully complete Exercise 6F.1. The focus is on mastering the chain rule, product rule, quotient rule, and derivatives of transcendental functions such as exponential, logarithmic, and trigonometric functions. By understanding these core techniques and practicing methodically, you will be well‑prepared for both the exercise and the broader IB Mathematics syllabus.
本文全面回顾完成练习6F.1所需的微分技巧，重点掌握链式法则、乘积法则、商法则以及超越函数（指数函数、对数函数和三角函数）的导数。通过理解这些核心技术并有条理地练习，你将为该练习和更广泛的IB数学大纲做好充分准备。
1. Revision of Basic Derivatives | 基础导数复习
ny constant c, d/dx (c) = 0. The power rule states that for f(x) = xⁿ, the derivative is f '(x) = n xⁿ⁻¹. Basic derivatives such as d/dx (sin x) = cos x and d/dx (eˣ) = eˣ should be at your fingertips.
在攻克练习6F.1之前，牢固掌握基本导数公式至关重要。对于任意常数c，d/dx (c) = 0。幂法则指出，若f(x) = xⁿ，则导数f '(x) = n xⁿ⁻¹。像d/dx (sin x) = cos x以及d/dx (eˣ) = eˣ这样的基础导数应能信手拈来。
The power rule can be extended to negative and fractional powers: d/dx (x⁻²) = –2x⁻³, and d/dx (√x) = 1/(2√x). These forms appear frequently in the exercise.
幂法则可推广到负指数和分数指数：d/dx (x⁻²) = –2x⁻³，以及d/dx (√x) = 1/(2√x)。这些形式在练习中频繁出现。
2. The Chain Rule | 链式法则
ate composite functions, i.e., functions of the form y = f(g(x)). It states that dy/dx = f '(g(x)) × g' ational rule is indispensable for most advanced differentiation problems in Exercise 6F.1.
链式法则用于对复合函数（即形如y = f(g(x))的函数）进行求导。其表述为dy/dx = f '(g(x)) × g'(x)，或用莱布尼茨记法表示为dy/dx = dy/du × du/dx，其中u = g(x)。这一基础法则在练习6F.1的大多数高级微分问题中都是不可或缺的。
u⁴ and du/dx = 6x. Applying the chain rule gives dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴.
例如，要对y = (3x² + 1)⁵求导，令u = 3x² + 1，则y = u⁵。于是dy/du = 5u⁴，du/dx = 6x。应用链式法则得dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴。
dy/dx = dy/du × du/dx
3. The Product Rule | 乘积法则
roduct of two differentiable functions, y = u(x) v(x), the product rule applies: dy/dx = u'(x) v(x) + u(x) v' ucial to identify u and v clearly and differentiate each separately before combining them.
当函数为两个可微函数的乘积y = u(x) v(x)时，使用乘积法则：dy/dx = u'(x) v(x) + u(x) v'(x)。关键是要清晰地识别出u和v，分别对它们求导后再组合起来。
For instance, if y = x² sin x, let u = x² and v = sin x. Then u' = 2x, v' = cos x. The derivative is y' = (2x)(sin x) + (x²)(cos x) = 2x sin x + x² cos x.
例如，若y = x² sin x，令u = x²，v = sin x。则u' = 2x，v' = cos x。导数为y' = (2x)(sin x) + (x²)(cos x) = 2x sin x + x² cos x。
If there are three factors, extend the rule: (uvw)' = u'vw + uv'w + uvw'. However, the exercise mostly focuses on two‑factor products.
如果有三个因子，可将法则扩展：(uvw)' = u'vw + uv'w + uvw'。然而练习中主要关注两个因子的乘积。
4. The Quotient Rule | 商法则
For a quotient y = u(x) / v(x), the derivative is given by dy/dx = [u'(x) v(x) – u(x) v' Remember to subtract and square the denominator correctly; order matters in the numerator.
对于商函数y = u(x) / v(x)，导数由dy/dx = [u'(x) v(x) – u(x) v'(x)] / [v(x)]²给出。务必正确地进行减法并对分母平方；分子中的顺序至关重要。
Consider y = eˣ / x. Set u = eˣ, so u' = eˣ; v = x, giving v' = 1. Then dy/dx = (eˣ · x – eˣ · 1) / x² = eˣ (x – 1) / x².
考虑y = eˣ / x。设u = eˣ，则u' = eˣ；v = x，得v' = 1。那么dy/dx = (eˣ · x – eˣ · 1) / x² = eˣ (x – 1) / x²。
An alternative memory aid is 'low d high minus high d low, over the square of what's below', which can prevent sign errors.
一个替代记忆法是“分母乘分子导减去分子乘分母导，除以分母的平方”，这可以避免符号错误。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
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