📚 Differentiation Techniques and Mark Scheme Tips | 求导运算的方法与得分要点
Differentiation is one of the most heavily weighted topics in IB Mathematics, appearing in both Analysis and Approaches (AA) and Applications and Interpretation (AI) papers. This article breaks down the essential methods of differentiation and, more importantly, the specific steps examiners look for when awarding marks.
求导运算是IB数学中权重最高的考点之一，在分析与方法（AA）和应用与解释（AI）两套课程中都会大量出现。本文将系统梳理求导的各类方法，并重点剖析阅卷官在给分时关注的关键步骤。
1. The Definition of the Derivative | 导数的定义
The derivative of a function f(x) at a point x is defined as the limit of the difference quotient as h approaches zero. This definition is the foundation of all differentiation techniques and is a common source of exam questions, particularly in Paper 1.
函数 f(x) 在点 x 处的导数定义为差商在 h 趋向于零时的极限。这一定义是所有求导方法的基础，也是考试中常见的出题点，尤其是在卷一试卷中。
f'(x) = lim(h→0) [f(x+h) − f(x)] / h
To earn full marks when using the definition, you must explicitly write the limit notation, substitute correctly, simplify the numerator before dividing by h, and finally evaluate the limit. Many students lose marks by skipping the limit notation or failing to show algebraic simplification.
使用定义求导时，要拿到满分必须明确写出极限符号、正确代入、先化简分子再除以 h，最后求极限。很多学生因省略极限记号或没有展示代数化简过程而失分。
| Mark Point 得分点 | Common Mistake 常见错误 |
| Write lim(h→0) each step 每一步保留极限符号 | Dropping lim notation 省略极限记号 |
| Factor out h before taking limit 取极限前约去 h | Dividing by zero 除以零 |
2. The Power Rule | 幂函数求导法则
The power rule states that for f(x) = xⁿ, the derivative is f'(x) = nxⁿ⁻¹. This is the most frequently used differentiation rule in the IB syllabus and applies to both positive and negative exponents, as well as fractional exponents.
幂函数求导法则指出：若 f(x) = xⁿ，则其导数为 f'(x) = nxⁿ⁻¹。这是IB大纲中最常用的求导法则，适用于正指数、负指数以及分数指数。
d/dx(xⁿ) = nxⁿ⁻¹, where n ∈ ℝ
When differentiating polynomials, apply the rule term by term. For example, f(x) = 3x⁴ − 5x² + 2x − 7 gives f'(x) = 12x³ − 10x + 2. Note that constants differentiate to zero, and the derivative of kx is simply k.
对多项式求导时，需逐项应用该法则。例如，f(x) = 3x⁴ − 5x² + 2x − 7 的导数为 f'(x) = 12x³ − 10x + 2。注意常数项求导为零，kx 的导数就是 k。
To secure method marks, rewrite expressions like 1/x³ as x⁻³ or √x as x^(1/2) before applying the power rule. Failing to rewrite radical or reciprocal forms is one of the most common preventable errors in IB exams.
为稳妥获得方法分，应用幂法则前应将 1/x³ 改写为 x⁻³，将 √x 改写为 x^(1/2)。未改写根式或倒数形式是IB考试中最常见的可避免错误之一。
3. The Product Rule | 乘积法则
The product rule is used when differentiating a product of two functions: if y = uv, then dy/dx = u(dv/dx) + v(du/dx). In IB exams, this rule appears frequently in both Papers 1 and 2, often combined with other rules.
乘积法则用于两个函数相乘的求导：若 y = uv，则 dy/dx = u(dv/dx) + v(du/dx)。在IB考试中，该法则在卷一和卷二中都频繁出现，且常与其他法则结合使用。
d/dx(uv) = u·v’ + v·u’
When applying the product rule, a clear structure helps secure marks. Define u and v explicitly, find u’ and v’ separately, then substitute into the formula. Do not attempt to simplify the product before differentiating, as this introduces algebra errors.
应用乘积法则时，清晰的结构有助于保证得分。先明确定义 u 和 v，分别求出 u’ 和 v’，再代入公式。切勿先展开乘积再求导，这会引入代数错误。
For example, differentiate y = x²·sin x. Let u = x², v = sin x, then u’ = 2x, v’ = cos x. Therefore dy/dx = x²·cos x + 2x·sin x. Examiners award marks for each correctly identified component.
例如，求 y = x²·sin x 的导数。令 u = x²，v = sin x，则 u’ = 2x，v’ = cos x。因此 dy/dx = x²·cos x + 2x·sin x。阅卷官会对每一个正确识别出的分量给分。
4. The Quotient Rule | 商法则
The quotient rule handles differentiation of a ratio of two functions: if y = u/v, then dy/dx = (v·u’ − u·v’) / v². This rule is essential for functions that cannot be rewritten as products, such as rational expressions with denominators raised to powers.
商法则处理两个函数之比的求导：若 y = u/v，则 dy/dx = (v·u’ − u·v’) / v²。对于无法改写为乘积形式的函数（如分母含幂次的有理式），该法则必不可少。
d/dx(u/v) = (v·u’ − u·v’) / v²
The most common error is misordering the numerator: the minus sign must appear between v·u’ and u·v’, not the reverse. A useful memory aid is “low d-high minus high d-low over low squared.”
最常见的错误是分子顺序颠倒：减号必须在 v·u’ 和 u·v’ 之间，顺序不能反。一个实用的记忆口诀是“分母平方，上导下不导减下导上不导”。
Alternatively, rewrite u/v as u·v⁻¹ and apply the product rule. In many IB problems, this is acceptable and sometimes shorter. However, be cautious: the quotient rule is safer when v appears as a complex expression, as the product rule approach requires differentiating v⁻¹ via the chain rule.
另一种方法是将 u/v 改写为 u·v⁻¹ 并应用乘积法则。在许多IB题目中这种做法可行且更简洁。但需注意：当 v 为复杂表达式时，商法则更稳妥，因为乘积法则需要对 v⁻¹ 使用链式法则求导。
5. The Chain Rule | 链式法则
The chain rule is arguably the most important differentiation rule in the IB syllabus. It handles composite functions: if y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). This rule underpins many longer questions involving exponential, logarithmic, and trigonometric functions.
链式法则可以说是IB大纲中最重要的求导法则。它用于处理复合函数：若 y = f(g(x))，则 dy/dx = f'(g(x)) · g'(x)。该法则支撑着许多涉及指数函数、对数函数和三角函数的综合题。
dy/dx = dy/du × du/dx
A reliable technique is the substitution method: let u = g(x), rewrite y in terms of u, differentiate y with respect to u, multiply by du/dx, then substitute back. For example, y = (3x² + 1)⁵: let u = 3x² + 1, so y = u⁵, dy/du = 5u⁴, du/dx = 6x, thus dy/dx = 30x(3x² + 1)⁴.
可靠的技巧是代换法：令 u = g(x)，将 y 用 u 表示，对 u 求导，乘以 du/dx，再代回。例如 y = (3x² + 1)⁵：令 u = 3x² + 1，则 y = u⁵，dy/du = 5u⁴，du/dx = 6x，因此 dy/dx = 30x(3x² + 1)⁴。
In marking schemes, the chain rule typically awards one mark for identifying the inner function and one mark for multiplying by its derivative. Never skip the multiplication step, as this is where the method mark is earned.
在评分标准中，链式法则通常对识别内层函数给1分，对乘以该内层函数的导数给1分。切勿跳过乘法这一步，这正是方法分的所在。
6. Implicit Differentiation | 隐函数求导
Implicit differentiation is required when y cannot be expressed explicitly as a function of x. Differentiate both sides of the equation with respect to x, treating y as a function of x and applying the chain rule to every term containing y.
当 y 无法表示为 x 的显函数时，需要使用隐函数求导。对等式两边关于 x 求导，将 y 视为 x 的函数，对每个含 y 的项应用链式法则。
d/dx(yⁿ) = n·yⁿ⁻¹·(dy/dx)
The key mark-scoring steps are: writing dy/dx after differentiating y-terms, collecting all dy/dx terms on one side, and factoring out dy/dx before solving. For example, for x² + y² = 25: 2x + 2y·(dy/dx) = 0, hence dy/dx = −x/y.
拿分关键步骤包括：对 y 项求导后写出 dy/dx，将所有含 dy/dx 的项移到同一边，先提取公因式再求解。例如，对 x² + y² = 25：2x + 2y·(dy/dx) = 0，故得 dy/dx = −x/y。
Implicit differentiation often appears in questions about tangent lines to curves, where you will also need to substitute coordinates to find the gradient at a specific point. Practise problems involving x³ + y³ = 6xy and eʸ = x² + y to master this skill.
隐函数求导常出现在求曲线切线斜率的问题中，此时还需要代入具体坐标来计算某点处的斜率。建议通过练习 x³ + y³ = 6xy 和 eʸ = x² + y 等题型来掌握这一技能。
7. Parametric Differentiation | 参数方程求导
When x and y are both expressed in terms of a parameter t, the derivative dy/dx is found using dy/dx = (dy/dt)/(dx/dt). This method combines differentiation with algebraic manipulation and frequently appears in IB Paper 2 questions.
当 x 和 y 都表示为参数 t 的函数时，导数 dy/dx 通过 dy/dx = (dy/dt)/(dx/dt) 求得。该方法将求导与代数运算相结合，在IB卷二题目中经常出现。
dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0
To earn full marks, first find dy/dt and dx/dt separately, then form the quotient. If asked for the second derivative d²y/dx², you must differentiate dy/dx with respect to t and then divide by dx/dt again.
要拿满分，需先分别求出 dy/dt 和 dx/dt，再构造商式。若要求二阶导数 d²y/dx²，需先对 dy/dx 关于 t 求导，再除以 dx/dt。
For example, given x = t² + 1 and y = t³ − t: dx/dt = 2t, dy/dt = 3t² − 1, so dy/dx = (3t² − 1)/(2t). Exam questions often ask to find the gradient at a specific parameter value — simply substitute that t value.
例如，已知 x = t² + 1，y = t³ − t：dx/dt = 2t，dy/dt = 3t² − 1，所以 dy/dx = (3t² − 1)/(2t)。考试题常要求在特定参数值处求斜率——直接代入该 t 值即可。
8. Higher-Order Derivatives | 高阶导数
The second derivative f”(x) is obtained by differentiating f'(x) again. In IB mathematics, the second derivative is used to determine concavity and to classify stationary points. It is a required skill in both AA and AI courses.
二阶导数 f”(x) 是对 f'(x) 再次求导得到的。在IB数学中，二阶导数用于判定函数的凹凸性以及分类驻点。AA和AI课程都要求掌握这一技能。
f”(x) = d/dx[f'(x)], representing d²y/dx²
When differentiating a polynomial multiple times, each application of the power rule reduces the degree by one. For instance, f(x) = x⁴ gives f'(x) = 4x³, f”(x) = 12x², f”'(x) = 24x, and f⁗(x) = 24.
对多项式多次求导时，每应用一次幂法则，次数降低一次。例如 f(x) = x⁴：f'(x) = 4x³，f”(x) = 12x²，f”'(x) = 24x，f⁗(x) = 24。
In marking schemes, each correct differentiation earns a separate mark. Write each step on a new line rather than combining multiple derivatives into one line, as this makes it easier for examiners to award partial credit.
在评分标准中，每次正确的求导各得一分。每一步换行书写，不要将多次求导合并到一行，这样便于阅卷官分配部分分数。
9. Derivatives of Exponential and Logarithmic Functions | 指数函数与对数函数的导数
The derivatives of eˣ and ln x are fundamental results that must be memorised: d/dx(eˣ) = eˣ and d/dx(ln x) = 1/x. For general bases, use d/dx(aˣ) = aˣ·ln a and d/dx(logₐx) = 1/(x·ln a).
eˣ 和 ln x 的导数是必须记住的基本结论：d/dx(eˣ) = eˣ，d/dx(ln x) = 1/x。对于一般底数，有 d/dx(aˣ) = aˣ·ln a，d/dx(logₐx) = 1/(x·ln a)。
d/dx(eˣ) = eˣ, d/dx(ln x) = 1/x
When the exponent or argument is not simply x, apply the chain rule. For example, d/dx(e^(2x)) = 2e^(2x), and d/dx(ln(3x² + 1)) = 6x/(3x² + 1). In exams, this combination of exponential or logarithmic rules with chain rule is extremely common.
当指数或对数真数不是单纯的 x 时，需要应用链式法则。例如 d/dx(e^(2x)) = 2e^(2x)，d/dx(ln(3x² + 1)) = 6x/(3x² + 1)。在考试中，指对数法则与链式法则的结合非常常见。
A common trick is to simplify using logarithm laws before differentiating: rewrite ln(x²·eˣ) as 2ln x + x before applying derivative rules. This reduces the complexity of the differentiation significantly.
一个常用技巧是先用对数运算法则化简再求导：将 ln(x²·eˣ) 改写为 2ln x + x 再求导。这样能显著降低求导的复杂度。
10. Derivatives of Trigonometric Functions | 三角函数的导数
The derivatives of the six trigonometric functions are standard results in the IB formula booklet. The most frequently examined are d/dx(sin x) = cos x, d/dx(cos x) = −sin x, and d/dx(tan x) = sec²x.
六个三角函数的导数是IB公式手册中的标准结论。最常考的是 d/dx(sin x) = cos x，d/dx(cos x) = −sin x，以及 d/dx(tan x) = sec²x。
d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(tan x) = sec²x
When the angle is not x but a function of x, such as sin(2x) or cos(πx), the chain rule must be applied. For sin(2x), the derivative is 2cos(2x). Students often forget to multiply by the derivative of the inner angle function — this is the exact point where marks are lost.
当角度不是 x 而是 x 的函数时（如 sin(2x) 或 cos(πx)），必须使用链式法则。对 sin(2x) 求导得到 2cos(2x)。学生常常忘记乘以内层角函数的导数——这正是失分的环节。
The reciprocal trig functions also appear: d/dx(csc x) = −csc x·cot x, d/dx(sec x) = sec x·tan x, and d/dx(cot x) = −csc²x. Practise recognising all six forms quickly, as speed matters in timed exam conditions.
倒数三角函数同样会出现：d/dx(csc x) = −csc x·cot x，d/dx(sec x) = sec x·tan x，d/dx(cot x) = −csc²x。在限时考试条件下，快速识别全部六种形式至关重要。
11. Differentiating Products of Trigonometric and Algebraic Functions | 三角与代数函数的乘积求导
Combination questions require applying the product rule together with trigonometric derivatives. For example, differentiating y = x²·sin x uses the product rule while differentiating y = sin x·cos x may use either the product rule or the double-angle identity first.
综合题要求将乘积法则与三角导数结合使用。例如，对 y = x²·sin x 求导需用乘积法则，而对 y = sin x·cos x 求导可以先使用二倍角公式或直接使用乘积法则。
d/dx(x·sin x) = sin x + x·cos x
A strategic question to ask yourself: can an identity simplify the product? For sin x·cos x, rewriting as ½·sin(2x) turns a product rule question into a simple chain rule question. This is a legitimate score-saving technique that examiners reward.
遇到乘积时先自问：有没有恒等式可以简化？对 sin x·cos x，改写为 ½·sin(2x) 可以将乘积法则的问题转化为简单的链式法则问题。这是阅卷官认可的有效节省分数的技巧。
| Expression 表达式 | Method 方法 | Derivative 导数 |
| x²·eˣ | Product rule 乘积法则 | x²eˣ + 2xeˣ |
| x³·ln x | Product rule 乘积法则 | 3x²ln x + x² |
| eˣ·sin x | Product rule 乘积法则 | eˣ(sin x + cos x) |
12. Typical Exam Questions and Common Pitfalls | 典型考题与常见陷阱
In Paper 1, differentiation questions are often algebraic and require exact answers. Common question types include finding the gradient of a tangent, determining whether a function is increasing or decreasing, and locating stationary points using f'(x) = 0.
在卷一（Paper 1）中，求导题通常偏代数化并要求精确答案。常见题型包括求切线斜率、判断函数的增减性，以及通过 f'(x) = 0 定位驻点。
In Paper 2, differentiation is frequently embedded in longer questions involving optimization problems, motion along a line, or curve sketching. These multi-part questions reward showing all intermediate steps clearly, as part marks are allocated per stage.
在卷二（Paper 2）中，求导常嵌入更长的题目中，涉及优化问题、直线运动或曲线作图。这类多部分题目要求清晰展示所有中间步骤，因为各步骤分别设有得分点。
The most common pitfalls are: forgetting the chain rule when differentiating composite functions, misapplying the quotient rule sign order, omitting dy/dx in implicit differentiation, and failing to simplify answers where expected. Memorise the standard derivatives and always check whether a rule combination is needed.
最常见的陷阱包括：复合函数求导时忘记链式法则、商法则中符号顺序用错、隐函数求导时遗漏 dy/dx，以及在需要化简答案的地方没有化简。牢记标准导数表，并始终检查是否需要组合运用多个法则。
Finally, in exam conditions, time allocation matters. Spend no more than two minutes identifying which rules are involved before writing. Once you have the plan, execute each step on a separate line. This structured approach ensures you collect marks even if a small arithmetic error occurs at the end.
最后，在考试条件下，时间分配至关重要。动笔前花不超过两分钟确定涉及哪些法则。一旦有了思路，每个步骤单独写一行。这种结构化方法能确保即使最后出现细微算术错误，你仍然能获得各步骤分数。
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