📚 Exercise 6B.3: Advanced Differentiation | 练习6B.3：高级微分
Exercise 6B.3 in the IB Mathematics course is designed to deepen your understanding of differentiation beyond the basic rules. It focuses on the chain rule, product rule, and quotient rule, often combining them to differentiate complex functions. Mastering these techniques is essential for success in both Standard Level and Higher Level exams.
IB数学课程中的练习6B.3旨在加深你对微分基本法则之外的理解。它聚焦于链式法则、乘积法则和商法则，经常将它们结合使用以对复杂函数求导。掌握这些技巧对于标准水平和高级水平的考试都至关重要。
1. Review of Basic Derivatives | 基础导数复习
Before tackling Exercise 6B.3, ensure you are confident with the derivatives of simple functions: the derivative of xⁿ is n xⁿ⁻¹, the derivative of eˣ is eˣ, and the derivative of ln x is 1/x.
在开始练习6B.3之前，请确保你已经熟练掌握简单函数的导数：xⁿ 的导数是 n xⁿ⁻¹，eˣ 的导数是 eˣ，ln x 的导数是 1/x。
You should also know the derivatives of sin x, cos x, and tan x by heart. These foundational results will appear repeatedly inside more complicated expressions.
你还应该熟记 sin x、cos x 和 tan x 的导数。这些基础结果会反复出现在更复杂的表达式中。
Having these at your fingertips allows you to focus on the application of the advanced rules instead of looking up basic formulas each time.
将这些牢记于心，能让你专注于高级法则的应用，而不必每次都查阅基本公式。
2. The Chain Rule | 链式法则
rds, differentiate the outer function, keeping the inner function unchanged, then multiply by the derivative of the inner function.
链式法则用于对复合函数求导。如果 y = f(g(x))，则 dy/dx = f′(g(x)) · g′(x)。简单来说，先对外层函数求导并保持内层函数不变，再乘以内层函数的导数。
) and the inner function as u = 3x + 1. The derivative is cos(3x + 1) · 3 = 3 cos(3x + 1).
例如，要对 y = sin(3x + 1) 求导，可将外层函数视为 sin(u)，内层函数为 u = 3x + 1。其导数为 cos(3x + 1) · 3 = 3 cos(3x + 1)。
)), work from the outside inwards: derivative = (1/cos(2x)) · (-sin(2x)) · 2 = -2 tan(2x).
当函数有多层嵌套时，反复应用链式法则。对于 y = ln(cos(2x))，从外向内求导：导数 = (1/cos(2x)) · (-sin(2x)) · 2 = -2 tan(2x)。
3. The Product Rule | 乘积法则
he product rule states: d/dx [u(x) v(x)] = u′(x) v(x) + u(x) v′(x). A common memory aid is "first times derivative of second, plus second times derivative of first."
当两个函数相乘时，乘积法则表示为：d/dx [u(x) v(x)] = u′(x) v(x) + u(x) v′(x)。一个常用的记忆方法是“第一个函数乘第二个的导数，加上第二个函数乘第一个的导数”。
Consider y = x² eˣ. Let u = x² and v = eˣ. Then u′ = 2x, v′ = eˣ. The derivative is 2x eˣ + x² eˣ, which can be factorised as eˣ (x² + 2x).
考虑 y = x² eˣ。设 u = x²，v = eˣ。则 u′ = 2x，v′ = eˣ。导数为 2x eˣ + x² eˣ，可因式分解为 eˣ (x² + 2x)。
It is good practice to simplify the result by factorising, as this will help with finding stationary points or second derivatives later.
最好通过因式分解来化简结果，因为这将有助于后续寻找驻点或求二阶导数。
4. The Quotient Rule | 商法则
dles divisions of two functions: d/dx (u/v) = (u′ v - u v′) / v². The order matters: it is "tom times derivative of top minus top times derivative of bottom, all over bottom squared."
商法则处理两个函数相除的情形：d/dx (u/v) = (u′ v - u v′) / v²。顺序非常重要：“分母乘分子的导数减去分子乘分母的导数，整体除以分母的平方”。
1·(x²+2) - (x+1)·2x) / (x²+2)² = (x² + 2 - 2x² - 2x) / (x²+2)² = (-x² - 2x + 2) / (x²+2)².
如果 y = (x + 1) / (x² + 2)，设 u = x + 1，v = x² + 2。则 u′ = 1，v′ = 2x。导数为 (1·(x²+2) - (x+1)·2x) / (x²+2)² = (x² + 2 - 2x² - 2x) / (x²+2)² = (-x² - 2x + 2) / (x²+2)²。
Be careful with algebraic signs when expanding and simplifying the numerator. Many errors occur because of missing a negative sign.
在展开和化简分子时要格外注意代数符号。许多错误都是由于遗漏负号造成的。
5. Combining Rules | 综合运用法则
In Exercise 6B.3, you will often need to use more than one rule in a single differentiation problem. For example, a product may contain a composite function, requiring the product rule together with the chain rule.
在练习6B.3中，你经常需要在一个求导问题中使用多种法则。例如，一个乘积中可能包含复合函数，这就需要同时使用乘积法则和链式法则。
cos(2x). The full derivative is 3x² sin(2x) + x³ · 2 cos(2x) = 3x² sin(2x) + 2x³ cos(2x).
以 y = x³ sin(2x) 为例。应用乘积法则，设 u = x³，v = sin(2x)。此处 v′ 需要链式法则，得到 v′ = cos(2x) · 2 = 2 cos(2x)。完全导数为 3x² sin(2x) + x³ · 2 cos(2x) = 3x² sin(2x) + 2x³ cos(2x)。
Similarly, a quotient might have a composite denominator. In such cases, identify the overall structure first, then apply the appropriate subordinate rule inside the main differentiation.
类似地，一个商的分母可能是复合函数。在这种情况下，先识别整体结构，然后在主求导过程中应用恰当的次级法则。
6. Derivatives of Trigonometric Functions | 三角函数的导数
ves you must know are: d/dx (sin x) = cos x, d/dx (cos x) = -sin x, d/dx (tan x) = sec² x.
三角函数在IB考试题中频繁出现。你必须掌握的导数有：d/dx (sin x) = cos x，d/dx (cos x) = -sin x，d/dx (tan x) = sec² x。
For example, differentiate y = cos(5x - π/3): y′ = -sin(5x - π/3) · 5 = -5 sin(5x - π/3).
当角度是 x 的线性函数时，例如 sin(ax+b)，使用链式法则。其导数为 a cos(ax+b)。例如，对 y = cos(5x - π/3) 求导：y′ = -sin(5x - π/3) · 5 = -5 sin(5x - π/3)。
be treated as (sin x)². The derivative is 2 sin x · cos x = sin(2x), using the chain rule.
三角函数的幂如 sin² x 应视为 (sin x)²。利用链式法则，其导数为 2 sin x · cos x = sin(2x)。
7. Exponential and Logarithmic Differentiation | 指数与对数微分
Exponential functions of the form eᶠ⁽ˣ keeping the exponential intact and multiplying by the derivative of the exponent: d/dx (eᶠ⁽ˣ⁾) = eᶠ⁽ˣ⁾ · f′(x).
形如 eᶠ⁽ˣ⁾ 的指数函数求导时，保持指数函数不变，再乘上指数的导数：d/dx (eᶠ⁽ˣ⁾) = eᶠ⁽ˣ⁾ · f′(x)。
For y = e3x², we have f(x) = 3x², so f′(x) = 6x. Thus dy/dx = 6x e3x².
对于 y = e3x²，有 f(x) = 3x²，f′(x) = 6x。因此 dy/dx = 6x e3x²。
unctions, d/dx (ln(f(x))) = f′(x) / f(x). If y = ln(x² + 4), then dy/dx = (2x) / (x² + 4).
对于对数函数，d/dx (ln(f(x))) = f′(x) / f(x)。如果 y = ln(x² + 4)，那么 dy/dx = (2x) / (x² + 4)。
These rules are direct consequences of the chain rule, and they are among the most commonly tested applications.
这些法则都是链式法则的直接推论，也是考试中最常考查的应用之一。
8. Second-Order Derivatives | 二阶导数
Exercise 6B.3 may also ask for second derivatives, denoted f′′(x) or d²y/dx². The second derivative is simply the derivative of the first derivative.
练习6B.3还可能要求求二阶导数，记为 f′′(x) 或 d²y/dx²。二阶导数就是一阶导数的导数。
educes the degree by one. For y = 5x⁴ - 2x³ + x, y′ = 20x³ - 6x² + 1 and y′′ = 60x² - 12x.
对于多项式函数，每次求导都会使次数减一。对于 y = 5x⁴ - 2x³ + x，y′ = 20x³ - 6x² + 1，y′′ = 60x² - 12x。
When the first derivative involves products or quotients, the second derivative may require the product or quotient rule again. Always simplify the first derivative before differentiating a second time where possible.
当一阶导数包含乘积或商时，二阶导数可能需要再次使用乘积法则或商法则。尽可能在第二次求导之前先化简一阶导数。
9. Worked Examples from Exercise 6B.3 | 练习6B.3例题解析
= (2x + 1)³ using the chain rule. Let u = 2x + 1, then y = u³. dy/dx = 3u² · 2 = 6(2x+1)².
例题1：利用链式法则对 y = (2x + 1)³ 求导。设 u = 2x + 1，则 y = u³。dy/dx = 3u² · 2 = 6(2x+1)²。
Example 2: Find dy/dx for y = eˣ cos x. Use the product rule: u = eˣ, v = cos x. dy/dx = eˣ cos x + eˣ (-sin x) = eˣ (cos x - sin x).
例题2：求 y = eˣ cos x 的导数。使用乘积法则：u = eˣ，v = cos x。dy/dx = eˣ cos x + eˣ (-sin x) = eˣ (cos x - sin x)。
s is a chain rule case: outer = ln(u), inner u = sin x. dy/dx = (1/sin x) · cos x = cot x.
例题3：对 y = ln(sin x) 求导。这是一个链式法则的情形：外层 = ln(u)，内层 u = sin x。dy/dx = (1/sin x) · cos x = cot x。
Example 4: Find the derivative of y = (x² + 3x) / (eˣ). Using the quotient rule: u = x²+3x, v = eˣ. u′ = 2x+3, v′ = eˣ. dy/dx = [(2x+3)eˣ - (x²+3x)eˣ] / e²ˣ = eˣ(2x+3 - x² - 3x)/e²ˣ = ( -x² - x + 3) / eˣ.
例题4：求 y = (x² + 3x) / (eˣ) 的导数。使用商法则：u = x²+3x，v = eˣ。u′ = 2x+3，v′ = eˣ。dy/dx = [(2x+3)eˣ - (x²+3x)eˣ] / e²ˣ = eˣ(2x+3 - x² - 3x)/e²ˣ = ( -x² - x + 3) / eˣ。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
A frequent error is forgetting to multiply by the derivative of the inner function when applying the chain rule. Always ask: "Is there a function inside another function?"
一个常见错误是在应用链式法则时忘记乘以内层函数的导数。永远要问自己：“是否存在一个函数套在另一个函数里面？”
In the product rule, students sometimes differentiate both functions and multiply them. Remember it is u′v + uv′, not u′v′.
在乘积法则中，学生有时会对两个函数分别求导然后相乘。请记住公式是 u′v + uv′，而不是 u′v′。
For the quotient rule, the subtraction order is critical. Writing the formula as (u′v - uv′)/v² on your formula booklet will prevent sign mistakes.
对于商法则，减法的顺序至关重要。将公式写成 (u′v - uv′)/v² 记录在公式手册上可以防止符号错误。
When an expression can be simplified before differentiation – for example, splitting a fraction into separate terms – do it. It often makes the derivative far easier.
当一个表达式可以在求导前化简时——例如，将分式拆分成独立的项——一定要这样做。这通常会让求导变得容易得多。
Finally, practise writing clear, step-by-step solutions. In IB exams, method marks are awarded for correct application of the rules, so show your working.
最后，要练习写出清晰、分步的求解过程。在IB考试中，正确应用法则可以获得方法分，因此要展示你的演算步骤。
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