📚 Exercise 12A.1: Differentiation from First Principles | 练习12A.1：从第一原理求导
Exercise 12A.1 introduces the formal definition of the derivative using the limit of a difference quotient. This foundational skill underpins the entire calculus syllabus in the IB Mathematics: Analysis and Approaches course. In this set of exercises, you will compute derivatives directly from the formula f'(x) = lim_{h → 0} [f(x + h) − f(x)] / h, simplify algebraic expressions carefully, and interpret the result as the gradient of a tangent or an instantaneous rate of change.
练习12A.1介绍了使用差商的极限来严格定义导数的概念。这项基础技能支撑着IB数学分析与方法课程中整个微积分体系。在本组练习中，你将直接利用公式 f'(x) = lim_{h → 0} [f(x + h) − f(x)] / h 计算导数，仔细化简代数表达式，并将结果解释为切线斜率或瞬时变化率。
1. Understanding the First Principle Formula | 理解第一原理公式
The notation f'(x) is read as “f prime of x” and represents the derivative of the function f with respect to x. The definition f'(x) = lim_{h → 0} [f(x + h) − f(x)] / h tells us that we should evaluate the slope of a secant line over an interval of width h, then take the limit as h approaches zero to obtain the slope of the tangent at a single point.
记号 f'(x) 读作 “f 撇 x”，表示函数 f 关于 x 的导数。定义 f'(x) = lim_{h → 0} [f(x + h) − f(x)] / h 告诉我们先计算宽度为 h 的区间上割线的斜率，然后让 h 趋近于零，取极限得到单点处切线的斜率。
- The numerator f(x + h) − f(x) is the change in the function value. 分子 f(x + h) − f(x) 是函数值的变化量。
- h is the change in x. h 是 x 的变化量。
- The limit process removes the division-by-zero issue when h = 0. 极限过程消除了 h = 0 时的除零问题。
In Exercise 12A.1 you will see simple polynomial functions such as f(x) = x², f(x) = x³, and linear combinations like f(x) = 2x² + 3x. The key challenge is to expand (x + h)ⁿ correctly and then cancel h from the numerator and denominator before letting h tend to 0.
在练习12A.1中你会遇到简单的多项式函数，如 f(x) = x²、f(x) = x³ 以及像 f(x) = 2x² + 3x 这样的线性组合。关键挑战在于先正确展开 (x + h)ⁿ，然后在让 h 趋于 0 之前将分子和分母中的 h 约掉。
2. Step-by-Step Process for f(x) = x² | 函数 f(x) = x² 的分步过程
Let us work through the very first type of question you will encounter. Start with f(x) = x². Compute f(x + h) = (x + h)² = x² + 2xh + h². The difference quotient becomes [x² + 2xh + h² − x²] / h = (2xh + h²) / h.
让我们逐步解决你遇到的第一个类型的题目。从 f(x) = x² 开始。计算 f(x + h) = (x + h)² = x² + 2xh + h²。差商变为 [x² + 2xh + h² − x²] / h = (2xh + h²) / h。
Factor h out of the numerator: h(2x + h) / h. Cancel h (provided h ≠ 0) to obtain 2x + h. Now apply the limit as h → 0: lim_{h → 0} (2x + h) = 2x. Thus f'(x) = 2x.
从分子中提取公因式 h：h(2x + h) / h。约去 h（只要 h ≠ 0）得到 2x + h。现在取 h → 0 时的极限：lim_{h → 0} (2x + h) = 2x。因此 f'(x) = 2x。
This purely algebraic procedure is repeated throughout Exercise 12A.1. The result 2x matches the well-known power rule, but the exercise requires you to show the derivation from first principles instead of quoting the rule.
这个纯代数过程在练习12A.1中反复出现。结果 2x 与熟知的幂法则吻合，但练习要求你展示从第一原理的推导过程，而不是直接引用法则。
3. Applying Limits to Simplify Expressions | 应用极限化简表达式
After cancelling the common factor h, you are left with an expression that is continuous at h = 0. This is deliberate: the original difference quotient is undefined at h = 0 because of division by zero, but algebraic simplification creates a new expression that agrees with the original for all h ≠ 0 and is defined at h = 0. The limit can then be evaluated by direct substitution.
约去公因式 h 后，你会得到一个在 h = 0 处连续的表达式。这是有意为之：原始的差商在 h = 0 处因为除零而没有定义，但代数化简生成了一个对所有 h ≠ 0 与原始表达式一致、且在 h = 0 处有定义的新表达式。这样就可以通过直接代入来求极限。
In symbols, if we have (f(x + h) − f(x)) / h = g(x, h) for h ≠ 0, and after simplification g(x, h) can be written as a polynomial or rational function that is defined at h = 0, then lim_{h → 0} g(x, h) is simply the value of the simplified expression with h = 0.
用符号表示，若对 h ≠ 0 有 (f(x + h) − f(x)) / h = g(x, h)，而化简后 g(x, h) 可写为在 h = 0 处有定义的多项式或有理函数，那么 lim_{h → 0} g(x, h) 就是化简表达式中令 h = 0 所得的值。
This is why expanding brackets and factorising are such critical skills for Exercise 12A.1. Without careful algebra, the limit cannot be evaluated safely.
这就是为什么展开括号和因式分解对练习12A.1来说如此重要。没有细致的代数运算，就无法安全地求出极限。
4. From Secant to Tangent: Geometric Interpretation | 从割线到切线：几何解释
The point P (x, f(x)) and Q (x + h, f(x + h)) lie on the curve y = f(x). The gradient of the chord PQ is exactly the difference quotient. As h becomes smaller, Q slides along the curve towards P, and the chord approaches the tangent line at P.
点 P (x, f(x)) 和 Q (x + h, f(x + h)) 位于曲线 y = f(x) 上。弦 PQ 的斜率正是差商。当 h 越来越小时，Q 沿着曲线向 P 滑动，弦趋近于 P 处的切线。
The derivative f'(x) is the limiting slope of these secants. This geometric idea is vital for understanding why the derivative gives the “instantaneous” rate of change rather than an average over an interval. Exercise 12A.1 may include sketches of a curve with a secant turning into a tangent to reinforce this visual understanding.
导数 f'(x) 就是这些割线斜率的极限。这一几何思想对于理解为何导数给出的是”瞬时”变化率而非区间上的平均值至关重要。练习12A.1可能会包含曲线草图，展示割线变为切线，以强化这一直观理解。
| Quantity 量 | Formula 公式 | Meaning 含义 |
| Average rate of change 平均变化率 | [f(x + h) − f(x)] / h | Gradient of secant PQ 割线 PQ 的斜率 |
| Instantaneous rate 瞬时变化率 | lim_{h → 0} [f(x + h) − f(x)] / h | Gradient of tangent at P 点 P 处切线的斜率 |
5. Differentiating f(x) = xⁿ for Positive Integers | 对正整数幂函数 f(x) = xⁿ 求导
Once you are comfortable with f(x) = x², Exercise 12A.1 extends the idea to f(x) = x³, f(x) = x⁴, and sometimes f(x) = 1/x (which is x⁻¹). For f(x) = x³, use the binomial expansion: (x + h)³ = x³ + 3x²h + 3xh² + h³. Then f(x + h) − f(x) = 3x²h + 3xh² + h³, and division by h gives 3x² + 3xh + h². The limit as h → 0 yields 3x².
熟练掌握 f(x) = x² 之后，练习12A.1将这一思想扩展到 f(x) = x³、f(x) = x⁴，有时还有 f(x) = 1/x（即 x⁻¹）。对于 f(x) = x³，使用二项式展开：(x + h)³ = x³ + 3x²h + 3xh² + h³。于是 f(x + h) − f(x) = 3x²h + 3xh² + h³，除以 h 得到 3x² + 3xh + h²。h → 0 时的极限给出 3x²。
This pattern reveals the power rule: if f(x) = xⁿ, then f'(x) = n xⁿ⁻¹. While the IB eventually expects you to use the rule directly, Exercise 12A.1 ensures you can prove it for small integers from the limit definition, which deepens your understanding of why the rule works.
这一规律揭示了幂法则：若 f(x) = xⁿ，则 f'(x) = n xⁿ⁻¹。尽管 IB 最终期望你直接使用该法则，但练习12A.1确保你能对较小的整数 n 从极限定义出发证明它，从而加深你对法则背后原理的理解。
6. Handling More Complex Examples: f(x) = 2x³ + 5x | 处理更复杂的例子：f(x) = 2x³ + 5x
Linearity of the limit operator allows us to differentiate term-by-term. For f(x) = 2x³ + 5x, compute the difference quotient for the whole function. f(x + h) = 2(x + h)³ + 5(x + h). Expand carefully: 2(x³ + 3x²h + 3xh² + h³) + 5x + 5h = 2x³ + 6x²h + 6xh² + 2h³ + 5x + 5h.
极限算子的线性性质允许我们逐项求导。对于 f(x) = 2x³ + 5x，计算整个函数的差商。f(x + h) = 2(x + h)³ + 5(x + h)。仔细展开：2(x³ + 3x²h + 3xh² + h³) + 5x + 5h = 2x³ + 6x²h + 6xh² + 2h³ + 5x + 5h。
Subtract f(x) = 2x³ + 5x to obtain 6x²h + 6xh² + 2h³ + 5h. Divide by h: 6x² + 6xh + 2h² + 5. Taking the limit as h → 0 gives f'(x) = 6x² + 5. Notice that the derivative of 2x³ is 6x² and the derivative of 5x is 5, matching the term-by-term differentiation rule.
减去 f(x) = 2x³ + 5x 得到 6x²h + 6xh² + 2h³ + 5h。除以 h：6x² + 6xh + 2h² + 5。取 h → 0 的极限得到 f'(x) = 6x² + 5。注意到 2x³ 的导数是 6x²，5x 的导数是 5，这符合逐项求导法则。
7. Common Pitfalls and Algebraic Mistakes | 常见陷阱与代数错误
One frequent error is forgetting to distribute the negative sign when subtracting f(x). For f(x) = x² + 1, students sometimes write f(x + h) − f(x) = (x + h)² + 1 − x² + 1, which is incorrect because the +1 must also be subtracted. The correct expression is (x + h)² + 1 − (x² + 1) = (x + h)² + 1 − x² − 1 = (x + h)² − x².
一个常见错误是在减去 f(x) 时忘记分配负号。对于 f(x) = x² + 1，学生有时会写 f(x + h) − f(x) = (x + h)² + 1 − x² + 1，这是错误的，因为 +1 也必须被减去。正确的表达式是 (x + h)² + 1 − (x² + 1) = (x + h)² + 1 − x² − 1 = (x + h)² − x²。
Another pitfall is cancelling h before fully factorising the numerator. If the numerator is 3x²h + h², writing (3x²h + h²)/h = 3x² + h² is wrong because h²/h = h, not h². Always factor h from every term: h(3x² + h)/h = 3x² + h.
另一个陷阱是在完全分解分子之前就约去 h。如果分子是 3x²h + h²，写成 (3x²h + h²)/h = 3x² + h² 是错误的，因为 h²/h = h，而不是 h²。务必从每一项中提取公因式 h：h(3x² + h)/h = 3x² + h。
Sloppy notation, such as writing “lim” before the fraction has been simplified, can lead to incorrect limits. Always simplify the difference quotient to a point where direct substitution h = 0 is valid, then write the limit symbol and substitute.
潦草的书写习惯，比如在分式尚未化简前就写上 “lim”，可能导致错误的极限。始终先将差商化简到可直接代入 h = 0 的形式，然后再写下极限符号并代入。
8. Connecting First Principles to Differentiation Rules | 将第一原理与求导法则联系起来
After completing Exercise 12A.1, you will notice patterns that justify the standard rules. The sum rule [f(x) + g(x)]’ = f'(x) + g'(x) is a direct consequence of the limit of a sum being the sum of limits. The constant multiple rule [c·f(x)]’ = c·f'(x) emerges because the constant c can be factored out of the difference quotient: [c·f(x+h) − c·f(x)]/h = c·[f(x+h) − f(x)]/h.
完成练习12A.1后，你会发现一些规律，它们为标准法则提供了依据。和法则 [f(x) + g(x)]’ = f'(x) + g'(x) 直接源于和的极限等于极限的和这一性质。常数倍法则 [c·f(x)]’ = c·f'(x) 的出现是因为常数 c 可以从差商中提出：[c·f(x+h) − c·f(x)]/h = c·[f(x+h) − f(x)]/h。
The exercise thus bridges the gap between the rigorous definition and the efficient rules you will use later for more complicated functions, such as exponentials, logarithms, and trigonometric functions. Building this algebraic confidence is essential before moving to derivatives of sin x, cos x, or eˣ, where the limit manipulations rely on specifically known limits.
因此，该练习在严格定义和随后你将用于更复杂函数（如指数函数、对数函数和三角函数）的高效法则之间架起了一座桥梁。在进入 sin x、cos x 或 eˣ 的导数之前，建立这种代数自信至关重要，因为那些极限操作依赖于已知的特定极限。
9. Applications: Instantaneous Rate of Change | 应用：瞬时变化率
IB questions often ask you to find the gradient of a curve at a specific point or the speed of a particle at a given time. For example, if the displacement of a particle is given by s(t) = t² + 2t metres after t seconds, then the instantaneous velocity at t = 3 is s'(3). First find s'(t) from first principles (you will get 2t + 2), then substitute t = 3 to obtain 8 m/s.
IB 题目常要求你求曲线在某特定点的斜率或粒子在给定时刻的速度。例如，若粒子的位移由 s(t) = t² + 2t 给出（t 秒后的米数），则 t = 3 时的瞬时速度为 s'(3)。首先从第一原理求出 s'(t)（你会得到 2t + 2），然后代入 t = 3 得到 8 米/秒。
The phrase “from first principles” is a common command term in IB exams. It explicitly requires the limit approach, not just quoting the power rule. Exercise 12A.1 prepares you to respond correctly to such instructions, showing all steps of expansion, simplification, and limit evaluation for full marks.
“从第一原理”是 IB 考试中常见的指令词。它明确要求使用极限方法，而不仅仅是引用幂法则。练习12A.1让你准备好对这种指令做出正确回应，展示展开、化简和求极限的所有步骤以获得满分。
10. Practice Exercise Walkthrough | 练习走查
Consider a typical question: Use first principles to differentiate f(x) = x² − 3x. Write the difference quotient: [(x + h)² − 3(x + h) − (x² − 3x)] / h. Expand to get [x² + 2xh + h² − 3x − 3h − x² + 3x] / h = [2xh + h² − 3h] / h = h(2x + h − 3) / h = 2x + h − 3. As h → 0, f'(x) → 2x − 3.
考虑一道典型题目：用第一原理求 f(x) = x² − 3x 的导数。写出差商：[(x + h)² − 3(x + h) − (x² − 3x)] / h。展开得到 [x² + 2xh + h² − 3x − 3h − x² + 3x] / h = [2xh + h² − 3h] / h = h(2x + h − 3) / h = 2x + h − 3。当 h → 0 时，f'(x) → 2x − 3。
Another example: f(x) = 1/x. Compute f(x + h) − f(x) = 1/(x + h) − 1/x = (x − (x + h)) / (x(x + h)) = −h / (x(x + h)). Divide by h to obtain −1 / (x(x + h)). Take the limit as h → 0 to get f'(x) = −1/x². This rational example demonstrates the importance of common denominators.
- Always state the definition first: f'(x) = lim_{h → 0} [f(x+h) − f(x)]/h. 一定要先写出定义：f'(x) = lim_{h → 0} [f(x+h) − f(x)]/h。
- Show expansion and cancellation clearly. 清晰地展示展开和约分过程。
- Write a concluding statement: “Hence, by first principles, f'(x) = …” 写出总结陈述：”因此，由第一原理，f'(x) = …”
11. Notation and Final Limit Evaluation | 记号与最后的极限计算
IB examiners look for precise notation. Always use the “lim” symbol with “h → 0” underneath. Once the expression is simplified to a form like 4x + 2h, write: f'(x) = lim_{h → 0} (4x + 2h) = 4x. Avoid plugging h = 0 before the limit sign unless you have shown that the simplified expression is continuous at 0.
IB 考官看重精确的符号使用。务必使用带 “h → 0” 下标的 “lim” 符号。一旦表达式化简为如 4x + 2h 的形式，就写：f'(x) = lim_{h → 0} (4x + 2h) = 4x。除非你已经说明化简后的表达式在 0 处连续，否则不要在极限符号之前就代入 h = 0。
Also, distinguish between the derivative function f'(x) and the derivative evaluated at a point f'(a). Exercise 12A.1 mainly asks for the derivative function, but some parts may ask for the gradient at a specific x-value. In those cases, first find f'(x) and then substitute the given x-coordinate.
此外，要区分导函数 f'(x) 和在某一点的导数值 f'(a)。练习12A.1主要要求求出导函数，但有些部分可能会要求求特定 x 处的梯度。在这些情况下，先求出 f'(x)，再代入给定的 x 坐标。
12. Why First Principles Matter in the IB | 为何第一原理对 IB 至关重要
You may wonder why you must learn the limit process when the power rule is so much faster. The IB includes first principles to ensure you understand the conceptual foundation of calculus. It also tests algebraic manipulation, limit reasoning, and the ability to follow a formal mathematical argument. These skills transfer to proof-based topics and to understanding definitions of integrals and series later in the course.
你可能会问，既然幂法则快捷得多，为什么还要学习极限过程。IB 包含第一原理是为了确保你理解微积分概念的基础。它也考查代数处理、极限推理以及遵循形式数学论证的能力。这些技能会迁移到基于证明的课题，以及后续课程中对积分和级数定义的理解。
Furthermore, some functions encountered later, such as piecewise functions or functions defined via limits, can only be differentiated at certain points by returning to the first principles definition. Mastery of Exercise 12A.1 builds a solid platform for these advanced topics.
此外，之后遇到的一些函数，比如分段函数或通过极限定义的函数，其某些点处的导数只能通过回到第一原理定义来求得。掌握练习12A.1可为这些进阶主题搭建坚实的平台。
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