Exercise 9D: Mastering Differentiation Rules | 练习9D：掌握微分法则

📚 Exercise 9D: Mastering Differentiation Rules | 练习9D：掌握微分法则

Differentiation rules form the backbone of calculus in the IB Mathematics syllabus, and Exercise 9D typically challenges you to apply the chain, product and quotient rules with confidence. This article explains each rule step by step, offers worked examples, common pitfalls and exam-style practice designed to sharpen your technique so that you can handle any function thrown at you — from simple polynomials to intricate combinations of trigonometric, exponential and logarithmic expressions.

微分法则是 IB 数学课程中微积分的核心内容，练习 9D 通常要求你熟练运用链式法则、乘积法则和商法则。本文逐步解析每条法则，提供详细范例、常见错误以及模拟考题式的练习，帮助你打磨技巧，从容应对任何函数——从简单的多项式到三角函数、指数函数和对数函数的复杂组合。

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1. Quick Refresher on Basic Derivatives | 基础导数速览

f(x) = xⁿ, the derivative is f’(x) = nxⁿ⁻¹. The derivative of sin x is cos x, cos x gives –sin x, eˣ stays eˣ multiple rule lets you pull a coefficient through: if g(x) = c·f(x), then g’(x) = c·f’(x).

在进入组合法则之前，你必须对手边的标准导数为熟悉。对于幂函数 f(x) = xⁿ，导数是 f’(x) = nxⁿ⁻¹；sin x 的导数为 cos x，cos x 的导数为 –sin x，eˣ 的导数仍是 eˣ，ln x 给出 1/x。常数的导数为零，常数倍法则允许你将系数提到外面：若 g(x) = c·f(x)，则 g’(x) = c·f’(x)。

Function	Derivative
xⁿ	nxⁿ⁻¹
sin x	cos x
cos x	–sin x
eˣ	eˣ
ln x	1/x

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2. Understanding the Chain Rule | 理解链式法则

The chain rule handles composite functions — d inner functions. In function notation, if h(x) = f(g(x)), then h’(x) = f’(g(x)) · g’(x).

链式法则用于处理复合函数——函数中还有函数的情况。若 y = f(u) 且 u = g(x)，则 y 关于 x 的变化率为 dy/dx = (dy/du) × (du/dx)。在莱布尼茨记号中，相当于外层函数导数与内层函数导数相乘。用函数记号表示，若 h(x) = f(g(x))，则 h’(x) = f’(g(x)) · g’(x)。

Think “differentiate the outside, keep the inside, then multiply by the derivative of the inside” is u⁴ and the inner is 3x²+1. The derivative then becomes 4(3x²+1)³ × (6x) = 24x(3x²+1)³.

记住“先对外层求导，保留内层，再乘以内层函数的导数”。以 y = (3x²+1)⁴ 为例，外层是 u⁴，内层是 3x²+1，求导得 4(3x²+1)³ × (6x) = 24x(3x²+1)³。

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3. Step-by-Step Chain Rule Worked Examples | 链式法则逐步范例

Example 1: Differentiate y = sin(2x).
in u → derivative cos u. Inner: u = 2x → derivative 2. So dy/dx = cos(2x) × 2 = 2 cos(2x).

范例 1：对 y = sin(2x) 求导。
外层：sin u → 导数 cos u。内层：u = 2x → 导数 2。故 dy/dx = cos(2x) × 2 = 2 cos(2x)。

Example 2: y = e^x². Outer: e^u → derivative e^u. Inner: u = x² → derivative 2x. Result: 2x e^x².

范例 2：y = e^x²。外层 e^u 的导数仍是 e^u，内层 x² 的导数为 2x，结果为 2x e^x²。

Example 3: = ln(5x³+4). Outer derivative: 1/(5x³+4). Inner derivative: 15x². Answer: (15x²)/(5x³+4).

范例 3：y = ln(5x³+4)。外层导数为 1/(5x³+4)，内层导数为 15x²，答案为 (15x²)/(5x³+4)。

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4. The Product Rule – Two Functions Multiplied | 乘积法则——两个函数相乘

s, differentiate the first function, multiply by the second (left intact), then add the first function times the derivative of the second. The order does not matter as long as both combinations appear.

当你有 y = u(x)·v(x) 时，导数公式为 y’ = u’v + uv’。简言之，先对第一个函数求导并乘以第二个函数（保持原样），再加上第一个函数乘以第二个函数的导数。只要两项同时出现，顺序并不重要。

A handy mnemonic: “first d-second plus second d-first”. It is essential to decide which factor to call u and which v; choose so that u’ becomes simpler, making the algebra manageable.

一个巧记法是“第一乘第二的导数 加 第二乘第一的导数”。选择哪个作 u、哪个作 v 很关键；通常让 u’ 更为简单，从而使后续代数运算可控。

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5. Product Rule Worked Examples | 乘积法则范例解析

Example 1: x² sin x. Let u = x² (u’ = 2x) and v = sin x (v’ = cos x). Then y’ = 2x sin x + x² cos x.

范例 1：y = x² sin x。令 u = x²（u’ = 2x），v = sin x（v’ = cos x）。则 y’ = 2x sin x + x² cos x。

Example 2: y = eˣ ln x. u = eˣ (u’ = eˣ), v = ln x (v’ = 1/x). So y’ = eˣ ln x + eˣ/x = eˣ (ln x + 1/x).

范例 2：y = eˣ ln x。u = eˣ（u’ = eˣ），v = ln x（v’ = 1/x），于是 y’ = eˣ ln x + eˣ/x = eˣ (ln x + 1/x)。

Example 3: y = (3x²+1)(x³–5x). u = 3x²+1 (u’ = 6x), v = x³–5x (v’ = 3x²–5). Result: y’ = (6x)(x³–5x) + (3x²+1)(3x²–5). Expand and simplify as required.

范例 3：y = (3x²+1)(x³–5x)。u = 3x²+1 (u’ = 6x)，v = x³–5x (v’ = 3x²–5)。结果：y’ = (6x)(x³–5x) + (3x²+1)(3x²–5)。可根据需要展开并化简。

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6. The Quotient Rule – When Division Comes In | 商法则——处理除法求导

For y = u(x) / v(x), provided v(x) ≠ 0, the derivative is y’ = (u’v – product rule but with a subtraction; the denominator is the square of the bottom function.

对于 y = u(x) / v(x)，且 v(x) ≠ 0，导数为 y’ = (u’v – uv’) / v²。注意分子中的减号和分母的平方。分子形式与乘积法则相似，但改为相减；分母则是下方函数的平方。

Many students find it easier to remember as “low d-high minus high d-low over low squared”, where “low” is the denominator and “high” the numerator. Always double-check that you haven’t swapped the subtraction order.

许多学生用“分母乘分子导数 减 分子乘分母导数 除以 分母平方”来记忆。记得始终确认减法顺序没有被颠倒。

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7. Quotient Rule Worked Examples | 商法则范例解析

Example 1: y = x / sin x. u = x, u’ = 1; v = sin x, v’ = cos x. Then y’ = (1·sin x – x·cos x) / sin²x.

范例 1：y = x / sin x。u = x, u’ = 1；v = sin x, v’ = cos x。于是 y’ = (1·sin x – x·cos x) / sin²x。

Example 2: y = (x²+1) / eˣ. u = x²+1 (u’ = 2x), v = eˣ (v’ = eˣ). Result: y’ = (2x·eˣ – (x²+1)eˣ) / e²ˣ = (2x – x² – 1) / eˣ.

范例 2：y = (x²+1) / eˣ。u = x²+1 (u’ = 2x)，v = eˣ (v’ = eˣ)。结果：y’ = (2x·eˣ – (x²+1)eˣ) / e²ˣ = (2x – x² – 1) / eˣ。

Example 3: y = ln x / x. u = ln x (u’ = 1/x), v = x (v’ = 1). Derivative: y’ = ((1/x)·x – ln x·1) / x² = (1 – ln x) / x².

范例 3：y = ln x / x。u = ln x (u’ = 1/x)，v = x (v’ = 1)。导数：y’ = ((1/x)·x – ln x·1) / x² = (1 – ln x) / x²。

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8. Combining Rules – Multilayer Functions | 法则综合——多层嵌套函数

in Exercise 9D often require you to use two or three rules in one derivative. For y = x³ e^{sin x}, you need the product rule and the chain rule for the exponential part. Plan the order: identify the overall structure first — product, quotient, or chain — then apply the inner rules as sub-derivatives.

练习 9D 中真正的 IB 考题常会要求你在同一个导数中运用两条甚至三条法则。对于 y = x³ e^{sin x}，你既需要乘积法则，也需要对指数部分使用链式法则。规划求导顺序：先确定整体结构——乘积、商还是复合——再将内层法则作为子导数逐步应用。

As an illustration, for y = x³ e^{sin x}, let u = x³ and v = e^{sin x}. Then u’ = 3x². For v’, use chain rule: derivative of e^{sin x} is e^{sin x} × cos x. Finally, y’ = 3x² e^{sin x} + x³ e^{sin x} cos x = x² e^{sin x} (3 + x cos x).

以此为例，对于 y = x³ e^{sin x}，设 u = x³，v = e^{sin x}，则 u’ = 3x²。v’ 需使用链式法则：e^{sin x} 的导数为 e^{sin x} × cos x。最终 y’ = 3x² e^{sin x} + x³ e^{sin x} cos x = x² e^{sin x} (3 + x cos x)。

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9. Trig, Expo and Log Combinations | 三角、指数与对数混合求导

re you require the product rule for the two main factors, then the chain rule inside both cos(2x) and ln(x²+1). Derivative: –sin(2x)·2 · ln(x²+1) + cos(2x) · (1/(x²+1))·2x. Simplify to –2 sin(2x) ln(x²+1) + (2x cos(2x))/(x²+1).

练习 9D 的一道典型挑战：对 y = cos(2x) · ln(x²+1) 求导。这里你需要对两个主要因子使用乘积法则，再在 cos(2x) 和 ln(x²+1) 内部分别使用链式法则。导数为：–sin(2x)·2 · ln(x²+1) + cos(2x) · (1/(x²+1))·2x。化简得 –2 sin(2x) ln(x²+1) + (2x cos(2x))/(x²+1)。

Remember to factor and simplify sensibly. IB examiners often expect the final answer to be expressed as a single fraction or in a compact factorised form. Consistent practice with these mixed forms guarantees you won’t be caught off guard.

请留意合理地提取公因式与化简。IB 阅卷人通常希望最终答案表示为单一分式或简洁的因式分解形式。持续练习这类混合题型能确保你在考试时从容不迫。

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10. Avoid These Common Pitfalls | 避免这些常见错误

Forgetting to multiply by the inner derivative in chain rule problems is the number one mistake. Write down u, du, dv explicitly as a habit. Swapping the subtraction in the quotient rule — always do “u’v – uv’”, never the reverse. Misapplying the product rule by writing y’ = u’v’ instead of u’v+uv’ loses marks instantly.

在链式法则中忘记乘以内层导数是最常见的错误。养成明确写下 u、du、dv 的习惯。在商法则中颠倒减法——始终是“u’v – uv’”，绝不要反过来。误用乘积法则写成 y’ = u’v’ 而不是 u’v+uv’，会立即失分。

Also, be careful with derivative of ln(x²) — it is (1/x²)·2x = 2/x, not simply 1/x². For e^kx, it’s k e^kx. Memorising these special cases empowers you to move quickly through Exercise 9D.

同时要小心 ln(x²) 的求导——应该是 (1/x²)·2x = 2/x，而非仅仅是 1/x²。e^kx 的导数是 k e^kx。熟记这些特殊情况能让你在练习 9D 中快速推进。

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11. IB-Style Practice with Hints | 带提示的 IB 风格练习

Try these quick problems before the exam: (a) Differentiate y = (2x³–7)¹². (Hint: pure chain rule.) Answer: 12(2x³–7)¹¹ · 6x² = 72x²(2x³–7)¹¹. (b) y = x e^{cos x}. (Hint: product + chain.) Answer: e^{cos x} – x sin x e^{cos x} = e^{cos x}(1 – x sin x).

考试前快速练习以下几题：(a) 对 y = (2x³–7)¹² 求导。（提示：纯链式法则。）答案：12(2x³–7)¹¹ · 6x² = 72x²(2x³–7)¹¹。(b) y = x e^{cos x}。（提示：乘积+链式。）答案：e^{cos x} – x sin x e^{cos x} = e^{cos x}(1 – x sin x)。

(c) y = (tan x)/x. (Hint: quotient rule, derivative of tan x is sec²x.) Answer: (x sec²x – (d) y = (sin x)³ ln(5x). (Product + chain + chain.) Take your time; write u and v clearly.

(c) y = (tan x)/x。（提示：商法则，tan x 的导数为 sec²x。）答案：(x sec²x – tan x) / x²。(d) y = (sin x)³ ln(5x)。（乘积+链式+链式。）慢慢来，清晰地设出 u 和 v。

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12. Key Formulae Summary and Final Tips | 核心公式总结与临门一脚

ule: dy/dx = f’(g(x))·g’(x); Product Rule: (uv)’ = u’v + uv’; Quotient Rule: (u/v)’ = (u’v – uv’)/v². Pair them with the standard derivatives of xⁿ, sin x, cos x, eˣ, ln x, tan x, sec x etc.

铭记以下公式：链式法则 dy/dx = f’(g(x))·g’(x)；乘积法则 (uv)’ = u’v + uv’；商法则 (u/v)’ = (u’v – uv’)/v²。将它们与 xⁿ、sin x、cos x、eˣ、ln x、tan x、sec x 等基本导数配合使用。

When you face Exercise 9D, scan the function: is it a composition, a product, or a quotient? Tackle the outermost rule first, then dive into the inner derivatives. Always simplify your final answer. With enough practice, you will differentiate any function swiftly and accurately, turning a potentially tricky exercise into a confident scoring opportunity.

当你面对练习 9D 时，先扫描函数结构：是复合、乘积还是商？优先处理最外层法则，再深入内层求导。始终化简最终答案。经过充分练习，你将能快速、准确地求出任何函数的导数，把原本棘手的习题转变为稳得分的良机。

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