IB Math SL Differentiation: Complete Guide to Derivatives — IB数学SL微分：导数完全指南

一、导数的直观理解：切线斜率与瞬时变化率 | The Intuitive Meaning of Derivatives: Tangent Slope and Instantaneous Rate of Change

导数（derivative）是微积分的核心概念之一。在 IB Math SL 课程中，导数被定义为函数在某一点的瞬时变化率。从几何角度看，导数就是函数图像在该点处切线的斜率。想象一条弯曲的公路：如果你只关心起点和终点的平均速度，那是平均变化率；但如果你想精确知道某一瞬间的速度表读数，那就是瞬时变化率 – 导数。

The derivative is one of the core concepts of calculus. In the IB Math SL syllabus, the derivative is defined as the instantaneous rate of change of a function at a given point. Geometrically, the derivative is the slope of the tangent line to the function’s graph at that point. Imagine a winding road: if you only care about the average speed from start to finish, that is the average rate of change; but if you want to know exactly what the speedometer reads at a single instant, that is the instantaneous rate of change – the derivative.

用数学语言表达：对于函数 f(x)，在 x = a 处的导数定义为极限

In mathematical language: for a function f(x), the derivative at x = a is defined as the limit

f'(a) = lim_h→0 [f(a + h) – f(a)] / h

这个极限的含义是：我们取一个非常小的增量 h，计算函数值的变化量 f(a+h) – f(a)，除以 h 得到变化率，然后让 h 趋近于零。当 h 无限接近零时，这个比值就趋近于切线的真实斜率。

The meaning of this limit is: we take a very small increment h, compute the change in the function value f(a+h) – f(a), divide by h to get the rate of change, and then let h approach zero. As h gets arbitrarily close to zero, this ratio approaches the true slope of the tangent line.

IB 考试中经常要求学生从定义出发计算简单函数的导数，例如 f(x) = x² 在 x = 3 处的导数。理解极限定义不仅是为了应付”from first principles”题型，更是为了建立对导数本质的深刻认识。

IB exams frequently require students to compute the derivative of simple functions from the definition, for example f(x) = x² at x = 3. Understanding the limit definition is not just for tackling “from first principles” questions – it builds a deep understanding of what a derivative truly represents.

二、基本求导法则：幂函数、指数函数与三角函数的导数公式 | Basic Differentiation Rules: Power Rule, Exponential and Trigonometric Derivatives

IB Math SL 要求学生熟练掌握以下基本函数的导数公式：

IB Math SL requires students to master the following basic derivative formulas:

幂函数法则（Power Rule）：若 f(x) = xⁿ，则 f'(x) = n·x^n-1。这是最基础也是最常用的求导法则。例如 (x⁵)’ = 5x⁴，(x^1/2)’ = (1/2)x^-1/2 = 1/(2√x)。特别注意：常数的导数为零，因为常函数的变化率为零。

Power Rule: If f(x) = xⁿ, then f'(x) = n·x^n-1. This is the most fundamental and widely used differentiation rule. For example, (x⁵)’ = 5x⁴, (x^1/2)’ = (1/2)x^-1/2 = 1/(2√x). Special note: the derivative of a constant is zero, because a constant function has zero rate of change.

指数函数（Exponential Functions）：若 f(x) = e^x，则 f'(x) = e^x。e^x 是唯一一个导数等于自身的函数，这一独特性质使其在微积分中占据核心地位。对于一般指数函数 f(x) = a^x，f'(x) = a^x·ln a。

Exponential Functions: If f(x) = e^x, then f'(x) = e^x. The function e^x is the only function whose derivative equals itself – this unique property places it at the heart of calculus. For a general exponential function f(x) = a^x, f'(x) = a^x·ln a.

三角函数（Trigonometric Functions）：IB 公式表中直接给出四个标准导数：(sin x)’ = cos x，(cos x)’ = -sin x，(tan x)’ = sec² x。熟练掌握这些是解相关题目的前提。

Trigonometric Functions: The IB formula booklet provides four standard derivatives directly: (sin x)’ = cos x, (cos x)’ = -sin x, (tan x)’ = sec² x. Mastering these is a prerequisite for solving related problems.

自然对数函数（Natural Logarithm）：若 f(x) = ln x，则 f'(x) = 1/x（x > 0）。这是从 e^x 的导数通过反函数求导法则推导出来的，但在 SL 考试中可以直接使用。

Natural Logarithm: If f(x) = ln x, then f'(x) = 1/x (x > 0). This is derived from the derivative of e^x via the inverse function rule, but it can be used directly in SL exams.

三、和差法则与常数倍法则：多项式函数的快速求导 | Sum, Difference and Constant Multiple Rules: Fast Differentiation of Polynomials

导数的线性性质使得我们可以拆解复杂函数为简单部分分别求导。三个核心法则：

The linearity property of derivatives allows us to break down complex functions into simpler parts and differentiate each separately. Three core rules:

和法则（Sum Rule）：[f(x) + g(x)]’ = f'(x) + g'(x)。和的导数等于导数之和。

Sum Rule: [f(x) + g(x)]’ = f'(x) + g'(x). The derivative of a sum equals the sum of the derivatives.

差法则（Difference Rule）：[f(x) – g(x)]’ = f'(x) – g'(x)。差的导数等于导数之差。

Difference Rule: [f(x) – g(x)]’ = f'(x) – g'(x). The derivative of a difference equals the difference of the derivatives.

常数倍法则（Constant Multiple Rule）：[c·f(x)]’ = c·f'(x)。常数可以提到导数符号外面。

Constant Multiple Rule: [c·f(x)]’ = c·f'(x). Constants can be factored out of the derivative.

综合运用这三条法则加上幂函数法则，任何多项式函数都可以快速求导。例如 f(x) = 3x⁴ – 5x³ + 2x² – 7x + 9，逐项求导：

Combining these three rules with the power rule, any polynomial function can be differentiated rapidly. For example, f(x) = 3x⁴ – 5x³ + 2x² – 7x + 9, differentiating term by term:

f'(x) = 3·4x³ – 5·3x² + 2·2x – 7·1 + 0 = 12x³ – 15x² + 4x – 7

IB 考试中的多项式求导通常不会超过四次，但要注意负指数和分数指数的处理 – 它们在幂函数法则下完全适用。

Polynomial differentiation in IB exams typically does not go beyond the fourth degree, but watch out for negative and fractional exponents – they work perfectly under the power rule.

四、乘积法则：两个函数相乘的求导策略 | The Product Rule: Differentiating the Product of Two Functions

当我们需要对两个函数相乘的形式求导时，不能简单地分别求导再相乘。正确的法则是：

When we need to differentiate the product of two functions, we cannot simply differentiate each and multiply. The correct rule is:

[f(x)·g(x)]’ = f'(x)·g(x) + f(x)·g'(x)

即：第一项的导数乘第二项，加上第一项乘第二项的导数。记忆口诀：”前导后不导加前不导后导”。

That is: derivative of the first times the second, plus the first times derivative of the second. Memory aid: “derivative of first times second unchanged, plus first unchanged times derivative of second.”

经典例题：对 h(x) = x²·sin x 求导。令 f(x) = x²，f'(x) = 2x；g(x) = sin x，g'(x) = cos x。代入乘积法则：h'(x) = 2x·sin x + x²·cos x。

Classic example: Differentiate h(x) = x²·sin x. Let f(x) = x², f'(x) = 2x; g(x) = sin x, g'(x) = cos x. Apply the product rule: h'(x) = 2x·sin x + x²·cos x.

当遇到三个函数连乘时，可以将前两个视为一个整体再应用乘积法则： [f(x)·g(x)·h(x)]’ = f'(x)·g(x)·h(x) + f(x)·g'(x)·h(x) + f(x)·g(x)·h'(x)。注意其中每一项恰好都有且仅有一个函数被求导。

When dealing with the product of three functions, treat the first two as one unit and apply the product rule: [f(x)·g(x)·h(x)]’ = f'(x)·g(x)·h(x) + f(x)·g'(x)·h(x) + f(x)·g(x)·h'(x). Notice that exactly one function is differentiated in each term.

五、商法则：两个函数相除的求导技巧 | The Quotient Rule: Differentiating the Division of Two Functions

商法则用于求两个函数相除的导数，是 IB SL 考试中的必考内容：

The quotient rule is used to differentiate the division of two functions and is a guaranteed topic in IB SL exams:

[f(x)/g(x)]’ = [f'(x)·g(x) – f(x)·g'(x)] / [g(x)]²

记忆公式：”分子导数乘分母，减去分子乘分母导数，再除以分母的平方”。注意分子中的减号 – 这是学生最容易出错的地方，容易写成加号。

Memory formula: “derivative of numerator times denominator, minus numerator times derivative of denominator, all over denominator squared.” Pay attention to the minus sign in the numerator – this is where students most commonly make mistakes, often writing a plus sign instead.

典型例题：对 y = x²/(x+1) 求导。令 f(x) = x²，f'(x) = 2x；g(x) = x+1，g'(x) = 1。代入商法则：

Typical example: Differentiate y = x²/(x+1). Let f(x) = x², f'(x) = 2x; g(x) = x+1, g'(x) = 1. Apply the quotient rule:

y’ = [2x·(x+1) – x²·1] / (x+1)² = [2x² + 2x – x²] / (x+1)² = [x² + 2x] / (x+1)²

IB SL 公式表直接提供了商法则，考试时无需背诵 – 但不建议完全依赖公式表，因为考场上翻阅公式表会浪费宝贵时间。建议将乘积法则和商法则练到肌肉记忆的程度。

The IB SL formula booklet directly provides the quotient rule – no need to memorize for exams. However, it is not advisable to rely entirely on the booklet, as flipping through it during the exam wastes precious time. It is recommended to practice the product and quotient rules to the point of muscle memory.

六、链式法则：复合函数求导的核心工具 | The Chain Rule: The Core Tool for Differentiating Composite Functions

链式法则（Chain Rule）是 IB Math SL 微积分中最强大的求导工具。它处理的是复合函数 – 一个函数嵌套在另一个函数内部的情况：

The chain rule is the most powerful differentiation tool in IB Math SL calculus. It handles composite functions – situations where one function is nested inside another:

若 y = f(g(x))，则 dy/dx = f'(g(x))·g'(x)

If y = f(g(x)), then dy/dx = f'(g(x))·g'(x)

通俗解释：”外层函数的导数（保持内层不变）乘以内层函数的导数”。许多课本称之为”外导乘内导”。

In plain terms: “the derivative of the outer function (keeping the inner function unchanged) multiplied by the derivative of the inner function.” Many textbooks call this “derivative of outside times derivative of inside.”

经典示例一：y = (3x² + 2x)⁵。令 u = 3x² + 2x（内层函数），则 y = u⁵（外层函数）。dy/du = 5u⁴，du/dx = 6x + 2。因此 dy/dx = 5(3x² + 2x)⁴·(6x + 2)。

Classic example 1: y = (3x² + 2x)⁵. Let u = 3x² + 2x (inner function), then y = u⁵ (outer function). dy/du = 5u⁴, du/dx = 6x + 2. Therefore dy/dx = 5(3x² + 2x)⁴·(6x + 2).

经典示例二：y = sin(2x + 1)。外层是 sin u，导数为 cos u；内层 u = 2x + 1，导数为 2。应用链式法则：y’ = cos(2x + 1)·2 = 2cos(2x + 1)。

Classic example 2: y = sin(2x + 1). Outer is sin u, derivative cos u; inner u = 2x + 1, derivative 2. Apply the chain rule: y’ = cos(2x + 1)·2 = 2cos(2x + 1).

经典示例三：y = e^3x²。外层 e^u，导数为 e^u；内层 u = 3x²，导数为 6x。因此 y’ = e^3x²·6x。

Classic example 3: y = e^3x². Outer e^u, derivative e^u; inner u = 3x², derivative 6x. Therefore y’ = e^3x²·6x.

链式法则与乘积法则可以联合使用。例如 y = x²·e^3x，需要先用乘积法则，对 e^3x 这部分再用链式法则：y’ = 2x·e^3x + x²·3e^3x = e^3x(2x + 3x²)。这种多法则联合使用是 IB 考试中 Paper 2 长题目的常见考点。

The chain rule can be combined with the product rule. For example, y = x²·e^3x: first apply the product rule, then use the chain rule on the e^3x part: y’ = 2x·e^3x + x²·3e^3x = e^3x(2x + 3x²). This multi-rule combination is a common feature in IB Paper 2 long questions.

七、高阶导数：二阶导数与运动学应用 | Higher-Order Derivatives: Second Derivatives and Kinematics Applications

导数本身也是函数，可以继续求导，得到二阶导数、三阶导数等。在 IB SL 中，重点掌握二阶导数：

Derivatives are themselves functions and can be differentiated further, yielding second derivatives, third derivatives, and so on. In IB SL, the focus is on the second derivative:

f”(x) = d²y/dx² = d/dx [f'(x)]

在物理相关的应用题中，如果位移函数为 s(t)，则：一阶导数 s'(t) = v(t) 表示速度（velocity）；二阶导数 s”(t) = a(t) 表示加速度（acceleration）。

In physics-based application problems, if the displacement function is s(t), then: the first derivative s'(t) = v(t) represents velocity; the second derivative s”(t) = a(t) represents acceleration.

典型 IB 题目：一个质点的位移函数为 s(t) = t³ – 6t² + 9t（单位：米，时间 t 单位：秒，0 ≤ t ≤ 5）。求：(a) 速度函数 v(t)；(b) t = 2 时的加速度；(c) 质点静止的时刻。

Typical IB question: A particle’s displacement function is s(t) = t³ – 6t² + 9t (units: meters, time t in seconds, 0 ≤ t ≤ 5). Find: (a) the velocity function v(t); (b) the acceleration at t = 2; (c) the times when the particle is at rest.

解答：(a) v(t) = s'(t) = 3t² – 12t + 9；(b) a(t) = v'(t) = 6t – 12，a(2) = 0 m/s²；(c) 静止意味着 v(t) = 0，即 3t² – 12t + 9 = 0，解得 t = 1 或 t = 3 秒。

Solution: (a) v(t) = s'(t) = 3t² – 12t + 9; (b) a(t) = v'(t) = 6t – 12, a(2) = 0 m/s²; (c) “at rest” means v(t) = 0, so 3t² – 12t + 9 = 0, giving t = 1 or t = 3 seconds.

八、切线方程与法线方程：导数的基本几何应用 | Tangent and Normal Equations: Basic Geometric Applications of Derivatives

给定函数 y = f(x) 和曲线上一点 (a, f(a))，该点处的切线方程可以用点斜式直接写出：

Given a function y = f(x) and a point (a, f(a)) on the curve, the tangent line equation at that point can be written directly using the point-slope form:

y – f(a) = f'(a)·(x – a)

其中 f'(a) 是切线斜率。法线（normal line）是垂直于切线的直线，其斜率为 -1/f'(a)（前提 f'(a) ≠ 0）。因此法线方程为：

where f'(a) is the slope of the tangent. The normal line is perpendicular to the tangent, with slope -1/f'(a) (provided f'(a) ≠ 0). Therefore the normal line equation is:

y – f(a) = -1/f'(a)·(x – a)

典型例题：求曲线 y = x³ – 2x 在点 (2, 4) 处的切线和法线方程。先求导数 f'(x) = 3x² – 2，f'(2) = 10。切线：y – 4 = 10(x – 2)，即 y = 10x – 16。法线：y – 4 = -1/10(x – 2)，即 y = -0.1x + 4.2。

Typical example: Find the tangent and normal equations to the curve y = x³ – 2x at the point (2, 4). First find the derivative f'(x) = 3x² – 2, f'(2) = 10. Tangent: y – 4 = 10(x – 2), i.e. y = 10x – 16. Normal: y – 4 = -1/10(x – 2), i.e. y = -0.1x + 4.2.

考试中可能出现”求切线在 x 轴的截距”或”求法线与坐标轴围成的三角形面积”等变体，本质都是先求 f'(a) 再代入几何公式。

Exam questions may present variants such as “find the x-intercept of the tangent” or “find the area of the triangle formed by the normal and the coordinate axes” – the essence is always to first compute f'(a) and then apply geometric formulas.

九、函数的递增与递减区间：用一阶导数判断单调性 | Increasing and Decreasing Intervals: Using the First Derivative to Determine Monotonicity

一阶导数的正负号直接反映了原函数的增减趋势：

The sign of the first derivative directly reflects the increasing or decreasing trend of the original function:

当 f'(x) > 0 时，f(x) 在该区间上单调递增；当 f'(x) < 0 时，f(x) 在该区间上单调递减；当 f'(x) = 0 时，f(x) 在该点处可能取得极值（驻点 stationary point）。

When f'(x) > 0, f(x) is increasing on that interval; when f'(x) < 0, f(x) is decreasing on that interval; when f'(x) = 0, f(x) may have an extremum at that point (a stationary point).

解题步骤：(1) 求 f'(x)；(2) 解方程 f'(x) = 0 找驻点；(3) 用驻点和定义域边界将数轴分段；(4) 在每个区间内取一个测试点代入 f'(x)，判断正负号；(5) 整理成增减性表格（sign diagram）。

Solution steps: (1) find f'(x); (2) solve f'(x) = 0 to find stationary points; (3) divide the number line into intervals using stationary points and domain boundaries; (4) pick a test point in each interval and evaluate f'(x) to determine sign; (5) organize into a sign diagram.

示例：分析 f(x) = x³ – 3x 的单调性。f'(x) = 3x² – 3 = 3(x² – 1) = 3(x+1)(x-1)。驻点：x = -1 和 x = 1。测试：x = -2 时 f'(-2) = 9 > 0（递增）；x = 0 时 f'(0) = -3 < 0（递减）；x = 2 时 f'(2) = 9 > 0（递增）。因此函数在 (-∞, -1) 递增，在 (-1, 1) 递减，在 (1, ∞) 递增。

Example: Analyze the monotonicity of f(x) = x³ – 3x. f'(x) = 3x² – 3 = 3(x² – 1) = 3(x+1)(x-1). Stationary points: x = -1 and x = 1. Test: at x = -2, f'(-2) = 9 > 0 (increasing); at x = 0, f'(0) = -3 < 0 (decreasing); at x = 2, f'(2) = 9 > 0 (increasing). Therefore the function is increasing on (-∞, -1), decreasing on (-1, 1), and increasing on (1, ∞).

十、极值与最优化问题：一阶与二阶导数联合判定 | Extrema and Optimization Problems: Joint Application of First and Second Derivatives

IB SL 考试中的最优化问题通常分为两步：(1) 用一阶导数找候选极值点；(2) 用二阶导数判定极值类型。

Optimization problems in IB SL exams typically involve two steps: (1) use the first derivative to find candidate extremum points; (2) use the second derivative to classify the type of extremum.

二阶导数判定法（Second Derivative Test）：当 f'(a) = 0 时，若 f”(a) > 0，则 (a, f(a)) 是局部极小值点（图像呈 U 形）；若 f”(a) < 0，则 (a, f(a)) 是局部极大值点（图像呈倒 U 形）；若 f''(a) = 0，则该判定法失效，需用一阶导数变号法进一步判断（可能是拐点 inflection point）。

Second Derivative Test: When f'(a) = 0, if f”(a) > 0, then (a, f(a)) is a local minimum (U-shaped graph); if f”(a) < 0, then (a, f(a)) is a local maximum (inverted U-shaped graph); if f''(a) = 0, the test is inconclusive and the first derivative sign-change method must be used (it may be an inflection point).

最优化应用题（Optimization）是 Paper 2 的重点：(1) 根据题意建立目标函数（如面积、体积、利润）；(2) 用约束条件将多变量函数转化为单变量函数；(3) 求导、找驻点；(4) 验证该驻点确实是最优解（通常还需检查区间端点）。

Applied optimization problems are a key focus of Paper 2: (1) formulate the objective function from the problem statement (e.g., area, volume, profit); (2) use constraints to reduce a multi-variable function to a single variable; (3) differentiate and find stationary points; (4) verify that the stationary point is indeed the optimal solution (usually also check endpoints of the interval).

经典题目：用 100 米长的篱笆围一个长方形场地，一边靠墙不需要篱笆。求最大面积。设垂直于墙的边长为 x，平行于墙的边长为 y，则约束为 2x + y = 100。面积为 A = xy = x(100 – 2x) = 100x – 2x²。A'(x) = 100 – 4x = 0，得 x = 25。A”(25) = -4 < 0，极大值。最大面积为 25 × 50 = 1250 m²。

Classic problem: Use 100 meters of fencing to enclose a rectangular field with one side against a wall (needing no fence). Find the maximum area. Let the side perpendicular to the wall be x, and the side parallel to the wall be y. The constraint is 2x + y = 100. The area is A = xy = x(100 – 2x) = 100x – 2x². A'(x) = 100 – 4x = 0, giving x = 25. A”(25) = -4 < 0, indicating a maximum. Maximum area = 25 × 50 = 1250 m².

十一、IB Math SL 考试中的导数常见题型与评分要点 | Common Derivative Question Types in IB Math SL and Marking Scheme Insights

IB Math SL 考试中导数相关题目占总分的约 15-20%，分布在 Paper 1（无计算器）和 Paper 2（有计算器）。常见题型包括：

Derivative-related questions account for approximately 15-20% of the total marks in IB Math SL exams, distributed across Paper 1 (no calculator) and Paper 2 (with calculator). Common question types include:

题型一：From First Principles（Paper 1 经典题） – 用极限定义求导数，通常给 5-6 分。关键步骤：写出差分商的极限表达式 → 代数化简 → 取极限 → 得到结果。阅卷人会检查你每一步是否清晰展示。

Type 1: From First Principles (classic Paper 1 question) – use the limit definition to find a derivative, typically worth 5-6 marks. Key steps: write the limit expression for the difference quotient → algebraic simplification → take the limit → obtain the result. Examiners check whether each step is clearly shown.

题型二：复合函数求导 + 切线方程（Paper 1 & 2） – 结合链式法则和点斜式，通常 6-8 分。常见错误：忘记链式法则的”内导”部分，或写错法线斜率（应为 -1/m 而不是 1/m）。

Type 2: Composite function differentiation + tangent equation (Paper 1 & 2) – combines the chain rule and point-slope form, typically 6-8 marks. Common errors: forgetting the “inner derivative” in the chain rule, or writing the normal slope incorrectly (should be -1/m, not 1/m).

题型三：增减性与极值分析（Paper 2 大题） – 给出函数，要求完整的单调性和极值分析，并画出示意草图。评分项包括：正确求导（2-3 分）、解方程找驻点（1 分）、符号表（2 分）、极值判定（1-2 分）、草图标注关键点（1-2 分）。

Type 3: Monotonicity and extremum analysis (Paper 2 long question) – given a function, a complete monotonicity and extremum analysis is required, along with a sketch. Mark allocation: correct differentiation (2-3 marks), solving for stationary points (1 mark), sign diagram (2 marks), extremum classification (1-2 marks), sketch with key points labeled (1-2 marks).

题型四：应用题 / 最优化（Paper 2） – 几何或物理背景下的极值问题，15-18 分的综合大题。评分重点：正确建立函数关系式（3-4 分）、求导（2 分）、解方程（1-2 分）、验证最优解（2 分）、回答原始问题的解释性语句（1-2 分）。

Type 4: Applications / Optimization (Paper 2) – extremum problems in geometric or physical contexts, comprehensive 15-18 mark questions. Marking focus: correct formulation of the functional relationship (3-4 marks), differentiation (2 marks), solving the equation (1-2 marks), verifying the optimal solution (2 marks), interpretative statement answering the original question (1-2 marks).

十二、常见错误与避坑指南：IB 考生高频失分点汇总 | Common Mistakes and Pitfall Guide: High-Frequency Mark-Losing Points for IB Students

基于历年 IB 考官的评卷报告，以下是最常见的导数失分点：

Based on IB examiner reports from past years, the following are the most common mark-losing points on derivative questions:

错误一：幂函数法则中的指数处理错误。常见于 f(x) = 1/x 和 f(x) = √x 的求导。正确写法：1/x = x^-1，导数为 -1·x^-2 = -1/x²；√x = x^1/2，导数为 (1/2)·x^-1/2 = 1/(2√x)。

Mistake 1: Exponent handling errors in the power rule. Common when differentiating f(x) = 1/x and f(x) = √x. Correct approach: 1/x = x^-1, derivative = -1·x^-2 = -1/x²; √x = x^1/2, derivative = (1/2)·x^-1/2 = 1/(2√x).

错误二：链式法则漏掉内层导数。例如 (sin 2x)’ 写成 cos 2x 而漏掉乘以 2。正确：cos 2x × 2 = 2cos 2x。

Mistake 2: Forgetting the inner derivative in the chain rule. For example, writing (sin 2x)’ as cos 2x and forgetting to multiply by 2. Correct: cos 2x × 2 = 2cos 2x.

错误三：乘积法则中的符号顺序混乱。记住公式是 f’g + fg’，不是 f’g’。区分乘积法则和商法则 – 商法则分子中的减号尤其容易被记错。

Mistake 3: Confusing the order of terms in the product rule. Remember the formula is f’g + fg’, not f’g’. Distinguish the product rule from the quotient rule – the minus sign in the quotient rule numerator is especially prone to errors.

错误四：忘记标注驻点的类型。求出 f'(x) = 0 的解后必须判定是极大值还是极小值。IB 阅卷标准要求明确标注（写明”local maximum”或”local minimum”），仅写”turning point”不算完整。

Mistake 4: Forgetting to classify the type of stationary point. After solving f'(x) = 0, you must determine whether each point is a maximum or minimum. IB marking standards require explicit labeling (writing “local maximum” or “local minimum”); merely writing “turning point” is incomplete.

错误五：最优化问题缺少验证步骤。找到驻点后必须用二阶导数或一阶导数变号法确认这是极大值还是极小值。同时检查区间端点 – 有时最优解在端点而非驻点。

Mistake 5: Missing the verification step in optimization problems. After finding a stationary point, you must confirm whether it is a maximum or minimum using the second derivative test or the first derivative sign-change method. Also check interval endpoints – sometimes the optimal solution lies at an endpoint, not a stationary point.

Summary | 总结

IB Math SL 的导数（微分）部分涵盖从基础极限定义到复杂最优化应用题的完整知识链。核心能力要求包括：熟练掌握幂函数、指数函数、三角函数和对数函数的基本导数公式；灵活运用乘积法则、商法则和链式法则处理复合函数和复杂表达式；能够将一阶导数用于切线方程、单调性分析和极值判定；熟练使用二阶导数进行极值类型判定和运动学建模。Paper 1 侧重”from first principles”和无计算器的代数运算能力，Paper 2 侧重应用题建模和最优化综合问题。建议考生通过大量分类练习建立对各类题型的条件反射，同时特别注意阅卷评分标准中的步骤分要求 – 即使最终答案错误，清晰展示的中间步骤仍然可以获得大部分分数。

The IB Math SL differentiation (calculus) section covers a complete knowledge chain from the basic limit definition to complex optimization application problems. Core competency requirements include: mastering the basic derivative formulas for power, exponential, trigonometric, and logarithmic functions; flexibly applying the product rule, quotient rule, and chain rule to handle composite functions and complex expressions; using the first derivative for tangent equations, monotonicity analysis, and extremum identification; and proficiently using the second derivative for extremum classification and kinematic modeling. Paper 1 emphasizes “from first principles” and algebraic manipulation without a calculator, while Paper 2 emphasizes application problem modeling and comprehensive optimization problems. It is recommended that candidates build conditioned responses to each question type through extensive categorized practice, while paying special attention to the step-mark requirements in the marking scheme – even if the final answer is wrong, clearly demonstrated intermediate steps can still earn most of the available marks.

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