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。
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
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。
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 $[a, b]$.
这就是连接 $(a, f(a))$ 与 $(b, f(b))$ 两点的割线（secant line）的斜率。它表示在区间 $[a, b]$ 上，$x$ 每变化一个单位，$f$ 平均变化多少。
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 $h \to 0$).
$x = a$ 处的瞬时变化率（instantaneous rate of change）就是当区间收缩为零时（即 $b \to a$，等价地 $h \to 0$）平均变化率的极限。
tangent line）的斜率。
当割线上的两点不断靠近时，割线就变成了切线。
Worked Example — Average vs. Instantaneous Rate例题——平均变化率 vs. 瞬时变化率
Given $f(x) = x^2 + 1$.
已知 $f(x) = x^2 + 1$。
AROC on $[1, 3]$:
区间 $[1, 3]$ 上的平均变化率：
$$\frac{f(3) - f(1)}{3 - 1} = \frac{(9+1)-(1+1)}{2} = \frac{8}{2} = 4$$Instantaneous rate at $x = 1$:
$x = 1$ 处的瞬时变化率：
$$\begin{aligned} f'(1) &= \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \\[4pt] &= \lim_{h \to 0} \frac{(1+h)^2 + 1 - 2}{h} \\[4pt] &= \lim_{h \to 0} \frac{2h + h^2}{h} \\[4pt] &= \lim_{h \to 0} (2 + h) = 2 \end{aligned}$$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.
在 2.1 中我们求出了某一点处的导数。现在把它推广：$f$ 的导数本身就是一个函数，它在每个极限存在的点处给出切线的斜率。
provided this limit exists
前提是该极限（limit）存在
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.
用极限定义求 $f(x) = 3x^2$ 的导数 $f'(x)$。
$$\begin{aligned} f'(x) &= \lim_{h \to 0} \frac{3(x+h)^2 - 3x^2}{h} \\[4pt] &= \lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 - 3x^2}{h} \\[4pt] &= \lim_{h \to 0} \frac{6xh + 3h^2}{h} \\[4pt] &= \lim_{h \to 0} (6x + 3h) = 6x \end{aligned}$$Worked Example — Equation of a Tangent Line例题——切线方程
Find the equation of the tangent line to $f(x) = x^2$ at $x = 3$.
求 $f(x) = x^2$ 在 $x = 3$ 处的切线方程。
Step 1: $f(3) = 9$
第 1 步：$f(3) = 9$
Step 2: $f'(x) = 2x \;\Longrightarrow\; f'(3) = 6$
第 2 步：$f'(x) = 2x \;\Longrightarrow\; f'(3) = 6$
Step 3: Point-slope form:
第 3 步：点斜式：
$$\begin{aligned} y - 9 &= 6(x - 3) \\ y &= 6x - 9 \end{aligned}$$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.
当没有 $f$ 的解析式，只有一张数值表或一张图像时，仍可以用相邻点的差商（difference quotient）来估计导数。
Worked Example — Estimating from a Table例题——根据表格估计导数
Given table:
已知下表：
| $x$ | 1 | 3 | 4 | 6 |
| $f(x)$ | 2 | 8 | 11 | 20 |
Estimate $f'(3)$ using symmetric values around $x=3$:
用 $x = 3$ 两侧对称的值估计 $f'(3)$：
$$f'(3) \approx \frac{f(4) - f(1)}{4 - 1} = \frac{11 - 2}{3} = 3$$Or, using the closest points on each side:
或者取两侧距离最近的点：
$$f'(3) \approx \frac{f(4) - f(3)}{4 - 3} = \frac{11 - 8}{1} = 3$$Connecting Differentiability and Continuity
可微性与连续性的联系
differentiable），则 $f$ 在 $x = a$ 处连续（continuous）。
等价（逆否命题）：若 $f$ 在 $x = a$ 处不连续，则 $f$ 在 $x = a$ 处不可微。
| Type | What It Looks Like | Example |
|---|---|---|
| Corner / Cusp | 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 |
Worked Example — Checking Differentiability of a Piecewise Function例题——判断分段函数的可微性
Is $f$ differentiable at $x = 2$?
$f$ 在 $x = 2$ 处是否可微？
$$f(x) = \begin{cases} x^2, & x \le 2 \\ 4x - 4, & x > 2 \end{cases}$$Step 1 — Continuity:
第 1 步 —— 连续性：
$$\lim_{x \to 2^-} x^2 = 4, \quad \lim_{x \to 2^+}(4x-4) = 4, \quad f(2) = 4 \;\;\checkmark$$Step 2 — Derivatives from each side:
第 2 步 —— 两侧的导数：
$$\begin{aligned} \text{Left: } &\frac{d}{dx}(x^2) = 2x \;\Longrightarrow\; 2(2) = 4 \\[4pt] \text{Right: } &\frac{d}{dx}(4x-4) = 4 \end{aligned}$$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，$f'(2) = 4$。✓
Applying the Power Rule
应用幂法则
power rule）是使用率最高的求导（differentiate）法则。它适用于任意实指数，可以替代对多项式类函数繁琐的极限定义。
for any real number $n$
对任意实数 $n$ 都成立
Worked Example — Power Rule with Rewriting例题——改写后应用幂法则
Find the derivative of $f(x) = 5x^4 - \dfrac{2}{\sqrt{x}}$.
求 $f(x) = 5x^4 - \dfrac{2}{\sqrt{x}}$ 的导数。
Rewrite: $f(x) = 5x^4 - 2x^{-1/2}$
改写：$f(x) = 5x^4 - 2x^{-1/2}$
$$\begin{aligned} f'(x) &= 5 \cdot 4 x^3 - 2\!\left(-\tfrac{1}{2}\right)x^{-3/2} \\[4pt] &= 20x^3 + x^{-3/2} \\[4pt] &= 20x^3 + \frac{1}{x^{3/2}} \end{aligned}$$Derivative Rules: Constant, Sum, Difference, and Constant Multiple
导数法则：常数、求和、求差与常数倍
| 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） | $\dfrac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)$ |
Worked Example — Combining Rules例题——综合运用各法则
Find $\dfrac{dy}{dx}$ if $y = 7x^5 - 3x^3 + 4x - 9$.
求 $y = 7x^5 - 3x^3 + 4x - 9$ 的 $\dfrac{dy}{dx}$。
$$\begin{aligned} \frac{dy}{dx} &= 7 \cdot 5x^4 - 3 \cdot 3x^2 + 4 \cdot 1 - 0 \\[4pt] &= 35x^4 - 9x^2 + 4 \end{aligned}$$Derivatives of cos x, sin x, ex, and ln x
cos x、sin x、ex 与 ln x 的导数
| Function $f(x)$ | Derivative $f'(x)$ |
|---|---|
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $e^x$ | $e^x$ |
| $\ln x$ | $\dfrac{1}{x}$ |
| 函数 $f(x)$ | 导数 $f'(x)$ |
|---|---|
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $e^x$ | $e^x$ |
| $\ln x$ | $\dfrac{1}{x}$ |
Worked Example — Derivatives of Transcendental Functions例题——超越函数的导数
Find $f'(x)$ for $f(x) = 3e^x + 5\sin x - 2\ln x$.
求 $f(x) = 3e^x + 5\sin x - 2\ln x$ 的导数 $f'(x)$。
$$\begin{aligned} f'(x) &= 3e^x + 5\cos x - 2 \cdot \frac{1}{x} \\[4pt] &= 3e^x + 5\cos x - \frac{2}{x} \end{aligned}$$Derivation — Why $\dfrac{d}{dx}[\sin x] = \cos x$ (from the limit definition)推导——为何 $\dfrac{d}{dx}[\sin x] = \cos x$（从极限定义出发）
This proof uses two limits from Unit 1 as building blocks:
本证明用到单元 1 中的两个极限作为基础：
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 1 — Apply the limit definition.
第 1 步 —— 套用极限定义。
$$\frac{d}{dx}[\sin x] \;=\; \lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}$$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.
指数函数（exponential）的法则 $\frac{d}{dx}[e^x] = e^x$ 可推广到任意底 $a > 0$。技巧是把 $a^x$ 改写为含 $e$ 的复合函数，再使用链式法则（chain rule）。
Step 1 — Rewrite using the identity $a = e^{\ln a}$.
第 1 步 —— 利用恒等式 $a = e^{\ln a}$ 进行改写。
For any $a > 0$, raising both sides to the $x$ gives
对任意 $a > 0$，两边同时取 $x$ 次幂得
$$a^x \;=\; \bigl(e^{\ln a}\bigr)^{x} \;=\; e^{x \ln a}.$$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 $\dfrac{du}{dx} = \ln a$:
第 2 步 —— 用链式法则求导。外层函数为 $e^u$，其中 $u = x \ln a$；注意 $\ln a$ 对 $x$ 而言是常数，所以 $\dfrac{du}{dx} = \ln a$：
$$\frac{d}{dx}\bigl[e^{x \ln a}\bigr] \;=\; e^{x \ln a} \cdot \ln a.$$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 $e^{x \ln a} = a^x$：
$$\frac{d}{dx}\bigl[a^x\bigr] \;=\; a^x \ln a. \qquad\Box$$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 $e$.
把 $a = e$ 代入 $a^x \ln a$，得 $e^x \ln e = e^x \cdot 1 = e^x$，正与 $e^x$ 的法则一致。$\ln a$ 这个因子就是“底不取 $e$”所要付出的代价。
$\displaystyle\frac{d}{dx}\bigl[3^x\bigr] = 3^x \ln 3$. At $x = 2$: $3^2 \ln 3 = 9 \ln 3 \approx 9.89$.
$\displaystyle\frac{d}{dx}\bigl[3^x\bigr] = 3^x \ln 3$。 在 $x = 2$ 处： $3^2 \ln 3 = 9 \ln 3 \approx 9.89$。
The Product Rule
乘积法则
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。
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 $f' = 2x,\; g' = \cos x$:
令 $f = x^2,\; g = \sin x$，则 $f' = 2x,\; g' = \cos x$：
$$\begin{aligned} \frac{d}{dx}[x^2 \sin x] &= f'g + fg' \\[4pt] &= 2x\sin x + x^2 \cos x \end{aligned}$$Worked Example — Product Rule with Table Values例题——结合表格数值使用乘积法则
Given: $u(3)=5,\; u'(3)=-2,\; v(3)=4,\; v'(3)=7$. Find $\dfrac{d}{dx}[u(x)\cdot v(x)]$ at $x=3$.
已知：$u(3)=5,\; u'(3)=-2,\; v(3)=4,\; v'(3)=7$。求 $\dfrac{d}{dx}[u(x)\cdot v(x)]$ 在 $x=3$ 处的值。
$$\begin{aligned} &= u'(3)\cdot v(3) + u(3)\cdot v'(3) \\[4pt] &= (-2)(4) + (5)(7) \\[4pt] &= -8 + 35 = 27 \end{aligned}$$The Quotient Rule
商的法则
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
Worked Example — Quotient Rule例题——商的法则
Find $\dfrac{d}{dx}\!\left[\dfrac{\sin x}{x^2}\right]$.
求 $\dfrac{d}{dx}\!\left[\dfrac{\sin x}{x^2}\right]$。
Let $f = \sin x,\; g = x^2$, so $f' = \cos x,\; g' = 2x$:
令 $f = \sin x,\; g = x^2$，则 $f' = \cos x,\; g' = 2x$：
$$\begin{aligned} \frac{d}{dx}\!\left[\frac{\sin x}{x^2}\right] &= \frac{\cos x \cdot x^2 - \sin x \cdot 2x}{(x^2)^2} \\[4pt] &= \frac{x^2\cos x - 2x\sin x}{x^4} \\[4pt] &= \frac{x\cos x - 2\sin x}{x^3} \end{aligned}$$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。
| 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 |
|---|---|
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
| $\cot x$ | $-\csc^2 x$ |
| $\sec x$ | $\sec x \tan x$ |
| $\csc x$ | $-\csc x \cot x$ |
Derivation — Why d/dx[tan x] = sec²x推导——为何 d/dx[tan x] = sec²x
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 $f'(x)$。
$$\begin{aligned} f'(x) &= 3\sec x\tan x + 2\sec^2 x \end{aligned}$$How Unit 2 Appears on the AP Exam
单元 2 在 AP 考试中的考法
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.
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。
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.
表格题：用已知数据通过差商估计导数——必须写出过程。
带数值的乘积/商的法则：已知 $f(a)$、$f'(a)$、$g(a)$、$g'(a)$，求 $(fg)'(a)$ 或 $(f/g)'(a)$。
切线：求切线方程并用它估算函数值（局部线性化）。
解释：说明函数在某点是否可微，依据连续性与左右导数相等。
Flashcards — Click to Flip
闪卡——点击翻面
of the derivative?导数的极限定义？
$$ f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} $$
$\frac{d}{dx}[x^n] = \;?$幂法则：
$\frac{d}{dx}[x^n] = \;?$
$$ n x^{n-1} $$
Works for any real $n$.对任意实数 $n$ 都成立。
$$ \cos x $$
$$ -\sin x $$
Don't forget the negative!别漏负号！
$$ e^x $$
The only function equal to its own derivative.唯一与自己的导数相等的函数。
$$ \frac{1}{x} $$
$\frac{d}{dx}[f \cdot g] = \;?$乘积法则：
$\frac{d}{dx}[f \cdot g] = \;?$
$$ f' g + f g' $$
"d-first × second + first × d-second"“前导×后 + 前×后导”
$\frac{d}{dx}\!\left[\frac{f}{g}\right] = \;?$商的法则：
$\frac{d}{dx}\!\left[\frac{f}{g}\right] = \;?$
$$ \frac{f'g - fg'}{g^2} $$
"Lo dHi − Hi dLo over Lo²"“下导上 − 上导下，除以下²”
$$ \sec^2 x $$
$$\text{Differentiable} \Rightarrow \text{Continuous}$$
(converse fails)（逆命题不成立）
Unit 2 — Practice Quiz
单元 2 ——综合小测
Test yourself across the full unit. Your score updates live in the navigation bar.
用本测验自检全单元掌握情况。分数会在导航栏实时更新。
Readiness Checklist
就绪自查清单
Click each item you've mastered. Aim for 100% before exam day.
点击每条已掌握的内容。考试前争取做到 100%。
- Compute the derivative from the limit definition用极限定义计算导数
- Distinguish average rate of change from instantaneous rate区分平均变化率与瞬时变化率
- Estimate derivatives numerically from tables and graphs通过数值表和图像估计导数
- Determine differentiability and its relationship to continuity判断可微性及其与连续性的关系
- Apply the Power Rule to polynomial and radical expressions对多项式与根式应用幂法则
- Use the constant multiple, sum, and difference rules使用常数倍、求和与求差法则
- Differentiate $\sin x$, $\cos x$, $e^x$, and $\ln x$对 $\sin x$、$\cos x$、$e^x$、$\ln x$ 求导
- Apply the Product Rule correctly and simplify正确应用乘积法则并化简
- Apply the Quotient Rule correctly and simplify正确应用商的法则并化简
- Differentiate $\tan x$, $\cot x$, $\sec x$, $\csc x$对 $\tan x$、$\cot x$、$\sec x$、$\csc x$ 求导
- 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
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。