Unit 2: Differentiation Definition & Fundamental Properties
单元 2：求导 定义与基本性质
Build the derivative from limits, master the core differentiation rules, and connect instantaneous rates of change to the tangent line.
从极限出发构建导数，掌握核心求导法则,并把瞬时变化率与切线联系起来。
10–12% of AP Exam占 AP 考试 10–12%~13–14 Class Periods约 13–14 课时10 Topics10 个小节
Topic 2.1
小节 2.1
Defining Average and Instantaneous Rates of Change at a Point
定义某一点处的平均变化率与瞬时变化率
Big Idea
The derivative starts with a simple question: how fast is something changing right now? We answer it by first computing average rates of change over an interval, then shrinking that interval to zero.
核心思想
导数（derivative）起源于一个简单的问题：某个量此刻变化得有多快？我们先在一个区间上计算平均变化率（average rate of change），再让这个区间收缩为零。
Average Rate of Change (Difference Quotient)
平均变化率（差商）
$$ \text{AROC} = \frac{f(b) - f(a)}{b - a} $$
This is the slope of the secant line connecting $(a, f(a))$ and $(b, f(b))$. It tells you how much $f$ changes, on average, per unit change in $x$ over the interval $[a, b]$.
The instantaneous rate of change at $x = a$ is the limit of the average rate of change as the interval shrinks to zero — that is, as $b \to a$ (or equivalently, as $h \to 0$).
Geometric MeaningThe average rate of change = slope of the secant line through two points.The instantaneous rate of change = slope of the tangent line at a single point.As the two points on the secant line get closer and closer together, the secant line becomes the tangent line.
Defining the Derivative of a Function and Using Derivative Notation
定义函数的导数与使用导数记号
In Topic 2.1 we found the derivative at a single point. Now we generalize: the derivative of $f$ is itself a function that gives the slope of the tangent line at every point where the limit exists.
Notation — All Mean the Same Thing$f'(x)$ — "f prime of x" (Lagrange notation)$\dfrac{dy}{dx}$ — "dy dx" (Leibniz notation — looks like a fraction but is a single symbol)$y'$ — shorthand when $y = f(x)$$\dfrac{d}{dx}[f(x)]$ — operator notation: "the derivative with respect to $x$ of $f(x)$"You must be comfortable reading and writing all of these — the AP exam uses them interchangeably.
记号——含义相同$f'(x)$ ——“f prime of x”（拉格朗日记号）$\dfrac{dy}{dx}$ ——“dy dx”（莱布尼茨记号——形如分数，但实际上是单一符号）$y'$ ——当 $y = f(x)$ 时的简写$\dfrac{d}{dx}[f(x)]$ ——算子记号：“$f(x)$ 对 $x$ 求导”这几种记号都必须熟练读写——AP 考试会混用。
Tangent Line EquationThe derivative at a point gives the slope of the tangent line at that point. The equation of the tangent line to $y = f(x)$ at $x = a$ is:
$$ y - f(a) = f'(a)(x - a) $$
切线方程某点处的导数即该点切线的斜率（slope）。曲线 $y = f(x)$ 在 $x = a$ 处的切线方程为：
$$ y - f(a) = f'(a)(x - a) $$
Worked Example — Finding Derivative from the Definition例题——用定义求导
Find $f'(x)$ for $f(x) = 3x^2$ using the limit definition.
✓ Correct! The limit definition of the derivative at $x = 2$ is $f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0} \frac{(2+h)^3 - 8}{h}$.✓ 正确！$x = 2$ 处导数的极限定义为 $f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = \lim_{h \to 0} \frac{(2+h)^3 - 8}{h}$。
✗ Not quite. Use the limit definition $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$. With $f(x) = x^3$ and $a = 2$, this becomes $\lim_{h \to 0} \frac{(2+h)^3 - 8}{h}$.✗ 还差一点。使用极限定义 $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$。代入 $f(x) = x^3$ 与 $a = 2$，得 $\lim_{h \to 0} \frac{(2+h)^3 - 8}{h}$。
Topic 2.3
小节 2.3
Estimating Derivatives of a Function at a Point
估计函数在某点处的导数
When you don't have a formula for $f$ but have a table of values or a graph, you can still estimate the derivative using a difference quotient with nearby points.
Key TechniqueTo estimate $f'(a)$ from a table, use the two values closest to $a$:
$$ f'(a) \approx \frac{f(b) - f(c)}{b - c} $$
where $b$ and $c$ are the table values flanking $a$ (or closest available).
Exam Skill — Show Your Setup!You must write the difference quotient before evaluating. Just writing the numerical answer — even if correct — will not earn full credit on free-response questions.Example of proper presentation:$f'(3) \approx \dfrac{f(4) - f(3)}{4 - 3} = \dfrac{12.8 - 11.2}{1} = 1.6$
The Key TheoremIf $f$ is differentiable at $x = a$, then $f$ is continuous at $x = a$.Equivalently (contrapositive): If $f$ is not continuous at $x = a$, then $f$ is not differentiable at $x = a$.
Caution — The Converse is False!A function can be continuous at a point but not differentiable there. Differentiability is a stronger condition than continuity.
Sharp change in direction; left and right derivatives differ
$f(x) = |x|$ at $x = 0$
Vertical Tangent
Tangent line is vertical (slope → $\pm\infty$)
$f(x) = \sqrt[3]{x}$ at $x = 0$
Discontinuity
Jump, removable, or infinite discontinuity
Piecewise function with a jump
类型
表现
例子
拐角 / 尖点（corner / cusp）
方向急剧改变；左、右导数不相等
$f(x) = |x|$ 在 $x = 0$
垂直切线（vertical tangent）
切线为竖直方向（斜率 → $\pm\infty$）
$f(x) = \sqrt[3]{x}$ 在 $x = 0$
不连续点
跳跃、可去或无穷不连续点
含跳跃的分段函数
How to Check Differentiability1. Is $f$ continuous at the point? If not → not differentiable.2. Do the left-hand and right-hand derivatives exist and are they equal?
$$ \lim_{h \to 0^-} \frac{f(a+h) - f(a)}{h} = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h} $$
If both conditions hold, the function is differentiable at $x = a$.
Left $=$ Right $= 4$, so $f$ is differentiable at $x=2$ and $f'(2) = 4$. ✓
左 $=$ 右 $= 4$，因此 $f$ 在 $x=2$ 处可微，$f'(2) = 4$。✓
If $f$ is differentiable at $x = 5$, which of the following must be true?
若 $f$ 在 $x = 5$ 处可微，下列哪一项必然成立？
$f'(5) = 0$
$f$ is continuous at $x = 5$$f$ 在 $x = 5$ 处连续
$f$ has no sharp corners anywhere$f$ 处处没有尖角
$f$ is defined for all real numbers$f$ 对所有实数都有定义
✓ Correct! Differentiability implies continuity. If a function is differentiable at a point, it must be continuous there.✓ 正确！可微蕴含连续。若函数在某点可微，则它在该点必然连续。
✗ Not quite. Differentiability at a point implies continuity at that point. It does not mean the derivative is zero, nor that the function is smooth everywhere or defined everywhere.✗ 还差一点。某点可微仅蕴含该点连续，并不意味着导数为零，也不意味着函数处处光滑或处处有定义。
Topic 2.5
小节 2.5
Applying the Power Rule
应用幂法则
Why This Matters
The power rule is the single most-used differentiation rule. It works for any real exponent and replaces the tedious limit definition for polynomial-type functions.
$$ \frac{d}{dx}\left[x^n\right] = n \cdot x^{n-1} $$
for any real number $n$
对任意实数 $n$ 都成立
Rewriting Before DifferentiatingMany functions must be rewritten in the form $x^n$ before applying the power rule:$\sqrt{x} = x^{1/2}$ → $\frac{d}{dx}[\sqrt{x}] = \frac{1}{2}x^{-1/2}$$\dfrac{1}{x^3} = x^{-3}$ → $\frac{d}{dx}\left[\frac{1}{x^3}\right] = -3x^{-4}$$\sqrt[3]{x^2} = x^{2/3}$ → $\frac{d}{dx}\left[\sqrt[3]{x^2}\right] = \frac{2}{3}x^{-1/3}$
✗ Not quite. First rewrite as $3x^{-1/2}$, then apply the power rule: $3 \cdot (-\frac{1}{2}) x^{-3/2} = -\frac{3}{2x^{3/2}}$. Watch the sign from the negative exponent!✗ 还差一点。先改写为 $3x^{-1/2}$，再应用幂法则：$3 \cdot (-\frac{1}{2}) x^{-3/2} = -\frac{3}{2x^{3/2}}$。注意负指数带来的符号！
Topic 2.6
小节 2.6
Derivative Rules: Constant, Sum, Difference, and Constant Multiple
导数法则：常数、求和、求差与常数倍
Fundamental Differentiation Rules
基本求导法则
Rule
Formula
Constant
$\dfrac{d}{dx}[c] = 0$
Constant Multiple
$\dfrac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$
Sum
$\dfrac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$
Difference
$\dfrac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)$
法则
公式
常数（constant）
$\dfrac{d}{dx}[c] = 0$
常数倍（constant multiple）
$\dfrac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)$
求和（sum）
$\dfrac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$
求差（difference）
$\dfrac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)$
Key TakeawayYou can differentiate polynomial functions term by term. Constants "come along for the ride" (constant multiple rule), and the derivative of any standalone constant is zero.
Watch the Sign!The derivative of $\cos x$ is $\mathbf{-}\sin x$ — note the negative sign. Many students forget this.Memory aid: the "co-" functions (cosine, cotangent, cosecant) all pick up a negative sign when differentiated.
Recognizing Limit DefinitionsSometimes a limit on the exam is secretly the definition of a derivative in disguise:
$$ \lim_{h \to 0} \frac{\sin\left(\frac{\pi}{3} + h\right) - \sin\left(\frac{\pi}{3}\right)}{h} $$
This is $f'(\pi/3)$ where $f(x) = \sin x$, so the answer is $\cos(\pi/3) = 1/2$.
✓ Correct! This limit is the definition of $f'(2)$ where $f(x) = e^x$. Since $\frac{d}{dx}[e^x] = e^x$, we get $f'(2) = e^2$.✓ 正确！该极限就是 $f(x) = e^x$ 时 $f'(2)$ 的定义。由于 $\frac{d}{dx}[e^x] = e^x$，所以 $f'(2) = e^2$。
✗ Not quite. Recognize this as $f'(2)$ where $f(x) = e^x$. The derivative of $e^x$ is $e^x$, so the answer is $e^2$.✗ 还差一点。把它认作 $f(x) = e^x$ 时的 $f'(2)$。$e^x$ 的导数仍是 $e^x$，因此答案是 $e^2$。
The first comes from the Squeeze Theorem on $\cos h \le \frac{\sin h}{h} \le 1$. The second follows from the first via the identity $\cos h - 1 = -\dfrac{\sin^2 h}{\cos h + 1}$.
第一个来自夹逼定理 $\cos h \le \frac{\sin h}{h} \le 1$。第二个由第一个借助恒等式 $\cos h - 1 = -\dfrac{\sin^2 h}{\cos h + 1}$ 推出。
Step 2 — Expand using the sum-of-angles identity $\sin(x+h) = \sin x \cos h + \cos x \sin h$:
第 2 步 —— 用和角公式展开$\sin(x+h) = \sin x \cos h + \cos x \sin h$：
$$= \lim_{h\to 0}\frac{\sin x \cos h + \cos x \sin h - \sin x}{h}$$
Step 3 — Group the $\sin x$ terms.
第 3 步 —— 合并含 $\sin x$ 的项。
$$= \lim_{h\to 0}\frac{\sin x\,(\cos h - 1) + \cos x \,\sin h}{h}
\;=\; \sin x \cdot \lim_{h\to 0}\frac{\cos h - 1}{h} \;+\; \cos x \cdot \lim_{h\to 0}\frac{\sin h}{h}$$
($\sin x$ and $\cos x$ are constants with respect to $h$, so they pull out of the $h$-limit.)
（$\sin x$ 与 $\cos x$ 对 $h$ 而言都是常数，所以可以提到极限外面。）
Step 4 — Substitute the two known limits.
第 4 步 —— 代入两个已知极限。
$$= \sin x \cdot 0 \;+\; \cos x \cdot 1 \;=\; \cos x$$
Therefore $\;\dfrac{d}{dx}[\sin x] = \cos x$. The derivation for $\dfrac{d}{dx}[\cos x] = -\sin x$ is structurally identical, using $\cos(x+h) = \cos x\cos h - \sin x \sin h$.
因此$\;\dfrac{d}{dx}[\sin x] = \cos x$。$\dfrac{d}{dx}[\cos x] = -\sin x$ 的推导结构完全相同，只需使用 $\cos(x+h) = \cos x\cos h - \sin x \sin h$。
Generalization — $\dfrac{d}{dx}[a^x] = a^x \ln a$ for any base $a > 0$推广——对任意底 $a > 0$ 有 $\dfrac{d}{dx}[a^x] = a^x \ln a$
The exponential rule $\frac{d}{dx}[e^x] = e^x$ generalizes to any base $a > 0$. The trick is to rewrite $a^x$ as a composition involving $e$, then apply the chain rule.
Step 2 — Differentiate using the chain rule. The outer function is $e^u$ with $u = x \ln a$; note $\ln a$ is a constant with respect to $x$, so $\dfrac{du}{dx} = \ln a$:
Step 3 — Convert back to $a^x$. Using $e^{x \ln a} = a^x$ from Step 1:
第 3 步 —— 换回 $a^x$。利用第 1 步的 $e^{x \ln a} = a^x$：
$$\frac{d}{dx}\bigl[a^x\bigr] \;=\; a^x \ln a. \qquad\Box$$
Sanity check at $a = e$.
Plugging $a = e$ into $a^x \ln a$ gives $e^x \ln e = e^x \cdot 1 = e^x$, matching the rule for $e^x$. The factor $\ln a$ is the price you pay for picking a base other than $e$.
Mnemonic: "derivative of the first times the second, plus the first times the derivative of the second"
口诀：“前导后不导加前不导后导”——即第一项的导数乘第二项，加上第一项乘第二项的导数。
Common MistakeThe derivative of a product is NOT the product of the derivatives:
$$ \frac{d}{dx}[f \cdot g] \neq f' \cdot g' $$
This is one of the most frequent errors on the AP exam.
常见错误乘积的导数不是导数的乘积：
$$ \frac{d}{dx}[f \cdot g] \neq f' \cdot g' $$
这是 AP 考试中最常见的错误之一。
Worked Example — Product Rule例题——乘积法则
Find $\dfrac{d}{dx}\bigl[x^2 \cdot \sin x\bigr]$.
求 $\dfrac{d}{dx}\bigl[x^2 \cdot \sin x\bigr]$。
Let $f = x^2,\; g = \sin x$, so $f' = 2x,\; g' = \cos x$:
Mnemonic: "Low d-High minus High d-Low, over Low squared" (Lo·dHi − Hi·dLo) / Lo²
口诀：“下导上减上导下，除以下的平方”（Lo·dHi − Hi·dLo）/ Lo²
When to Use Which Rule?Product rule when you see $f(x) \cdot g(x)$.Quotient rule when you see $\dfrac{f(x)}{g(x)}$.Pro tip: Sometimes you can avoid the quotient rule by rewriting a fraction as a product with a negative exponent, then using the product rule or power rule instead.
Finding the Derivatives of Tangent, Cotangent, Secant, and Cosecant
求正切、余切、正割与余割的导数
These derivatives are derived from the quotient rule applied to the basic trig functions. You should memorize the results.
这些导数都是把商的法则用于基本三角函数（trig functions）推出的，结果必须背下来。
Derivatives of All Six Trig Functions
六个三角函数的导数
Function
Derivative
$\sin x$
$\cos x$
$\cos x$
$-\sin x$
$\tan x$
$\sec^2 x$
$\cot x$
$-\csc^2 x$
$\sec x$
$\sec x \tan x$
$\csc x$
$-\csc x \cot x$
函数
导数
$\sin x$
$\cos x$
$\cos x$
$-\sin x$
$\tan x$
$\sec^2 x$
$\cot x$
$-\csc^2 x$
$\sec x$
$\sec x \tan x$
$\csc x$
$-\csc x \cot x$
Pattern to RememberNotice the "co-" pattern: every cofunction's derivative has a negative sign.$\cos x → -\sin x$ | $\cot x → -\csc^2 x$ | $\csc x → -\csc x \cot x$Also note the "sec/csc families" pair nicely: $\sec$ goes with $\sec \cdot \tan$, while $\csc$ goes with $\csc \cdot \cot$.
规律注意“co-”规律：每个余函数（cofunction）的导数都带负号。$\cos x → -\sin x$ | $\cot x → -\csc^2 x$ | $\csc x → -\csc x \cot x$另外，“正割族”与“余割族”各自配对得很好：$\sec$ 配 $\sec \cdot \tan$，$\csc$ 配 $\csc \cdot \cot$。
Derive $\dfrac{d}{dx}[\tan x]$ using the quotient rule on $\tan x = \dfrac{\sin x}{\cos x}$:
把 $\tan x = \dfrac{\sin x}{\cos x}$ 用商的法则求导：
$$\begin{aligned}
\frac{d}{dx}[\tan x] &= \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} \\[4pt]
&= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \\[4pt]
&= \frac{1}{\cos^2 x} \qquad\text{(Pythagorean identity)} \\[4pt]
&= \sec^2 x
\end{aligned}$$
Worked Example — Using Trig Derivatives例题——使用三角函数的导数
Find $f'(x)$ for $f(x) = 3\sec x + 2\tan x$.
求 $f(x) = 3\sec x + 2\tan x$ 的导数 $f'(x)$。
$$\begin{aligned}
f'(x) &= 3\sec x\tan x + 2\sec^2 x
\end{aligned}$$
$\dfrac{d}{dx}[\csc x]$ equals:
$\dfrac{d}{dx}[\csc x]$ 等于：
$\csc x \cot x$
$\sec x \cot x$
$-\sec x \tan x$
$-\csc x \cot x$
✓ Correct! The derivative of $\csc x$ is $-\csc x \cot x$. Remember, the "co-" functions always pick up a negative sign.✓ 正确！$\csc x$ 的导数是 $-\csc x \cot x$。记住，所有“co-”函数求导后都带负号。
✗ Not quite. The derivative of $\csc x$ is $-\csc x \cot x$. The negative sign is key — all "co-" function derivatives are negative.✗ 还差一点。$\csc x$ 的导数是 $-\csc x \cot x$。负号是关键——所有“co-”函数求导后都是负的。
Exam Strategy
考试策略
How Unit 2 Appears on the AP Exam
单元 2 在 AP 考试中的考法
MC
MCQ — Common Question Styles
Compute derivatives using power, product, quotient rules, or trig/exponential/log formulas.
Evaluate limits that are secretly derivative definitions in disguise.
Estimate a derivative from a table of values or from a graph.
Determine where a function is or is not differentiable, and explain why.
Find the equation of a tangent line at a given point.
MC
选择题——常见题型
计算导数，使用幂、乘积、商的法则，或三角/指数/对数函数公式。
求极限，识别出它其实是导数定义的“伪装”。
估计导数：根据数值表或图像。
判断函数在何处可微、何处不可微，并说明原因。
求某点处的切线方程。
FR
FRQ — Common Question Styles
Table problems: Estimate a derivative using a difference quotient from given data — you must show the setup.
Product/Quotient rule with given values: Given $f(a)$, $f'(a)$, $g(a)$, $g'(a)$, compute $(fg)'(a)$ or $(f/g)'(a)$.
Tangent line: Find the equation and use it to approximate a function value (local linearization).
Justify: Explain whether a function is differentiable at a point, citing continuity and equal one-sided derivatives.
Top 5 Mistakes That Lose Points1. Not showing the difference quotient setup when estimating a derivative from a table.2. Taking the derivative of a product by differentiating each factor separately (must use product rule).3. Sign errors with $\cos x$ derivatives and the quotient rule's subtraction.4. Forgetting to square the denominator in the quotient rule.5. Using incorrect notation — e.g., writing $\frac{dy}{dx}$ when you mean $\frac{d}{dx}[f(x)]$, or dropping primes.
4. At which of the following points is $f(x) = |x - 3|$ NOT differentiable?
4. $f(x) = |x - 3|$ 在以下哪个点不可微？
$x = 0$
$x = 1$
$x = 3$
$x = -3$
✓ Correct! The absolute value function has a sharp corner at $x = 3$ where the left derivative is $-1$ and the right derivative is $+1$. Since these are not equal, $f$ is not differentiable there.✓ 正确！绝对值函数在 $x = 3$ 处出现尖角，左导数为 $-1$，右导数为 $+1$。两者不相等，所以 $f$ 在该点不可微。
✗ Not quite. $|x - 3|$ has a corner at $x = 3$. The left-hand derivative is $-1$ and the right-hand derivative is $+1$ — they don't match, so $f$ is not differentiable at $x = 3$.✗ 还差一点。$|x - 3|$ 在 $x = 3$ 处有拐角。左导数为 $-1$，右导数为 $+1$——两者不相等，所以 $f$ 在 $x = 3$ 处不可微。
✓Write an equation of the tangent line at a given point写出给定点处的切线方程
✓Identify points where a function fails to be differentiable识别函数不可微的点
✓Select the appropriate differentiation rule for a given expression为给定表达式选择恰当的求导法则
✓Interpret the derivative as slope of the tangent line把导数解读为切线的斜率
Next Step下一步
AP-Style Practice ProblemsAP 风格练习题
Exam-level practice for this unit — multiple-choice plus extended-response items modeled on the AP rubric. Built for top-score prep; go here after you've worked through the notes and the in-page quizzes above.
本单元的考试级练习——按 AP 评分标准编写的选择题与拓展作答题。为冲刺高分而设计，建议在完成上面的笔记与页面内小测后再做。