Unit 4: Contextual Applications
of Differentiation
单元 4：求导的
情境应用
Apply derivatives to real-world motion, related rates, linear approximation, and indeterminate limits — the bridge from technique to meaning.
将导数应用于现实世界中的运动、相关变化率、线性近似与不定式极限——从计算技巧通向意义理解的桥梁。
Interpreting the Meaning of the Derivative in Context
情境中导数（derivative）意义的解读
At its core, the derivative \(f'(x)\) measures the instantaneous rate of change of the output with respect to the input. In pure mathematics, that means the slope of the tangent line. In every applied problem, however, those abstract words acquire concrete meaning: dollars per year, litres per minute, degrees per hour, people per day. The ability to translate between "derivative" and "rate of change in context" is a skill tested relentlessly on the AP Exam.
从本质上讲，导数 \(f'(x)\) 度量的是输出对输入的瞬时变化率（instantaneous rate of change）。在纯数学中，这就是切线（tangent line）的斜率（slope）。但在每一道应用题中，这些抽象的字眼都会获得具体含义：每年多少美元、每分钟多少升、每小时多少度、每天多少人。能否在"导数"与"情境中的变化率"之间自如转换，是 AP 考试反复考查的核心技能。
When an exam question says, "Interpret \(f'(3) = -7\) in the context of the problem," it is asking for a complete sentence: At \(x = 3\), the [quantity described by \(f\)] is [decreasing / increasing] at a rate of [|value|] [units of \(f\) per unit of \(x\)]. You must include all four pieces: the specific input value, the direction of change, the magnitude, and the units.
当考题写"请在题意背景下解读 \(f'(3) = -7\)"时，它要求的是一句完整的句子：在 \(x = 3\) 时，[由 \(f\) 描述的量] 正以 [|数值|] [\(f\) 的单位 每 \(x\) 的单位] 的速率 [递减 / 递增]。四个要素缺一不可：具体的输入值、变化的方向、变化的大小，以及测量单位（units of measurement）。
decreasing）；导数为正 → 递增（increasing）。千万不要说"导数正在变化"——导数告诉你的是原始量是如何变化的。
velocity）为……"。速度、加速度（acceleration）和速率（speed）这几个词专用于运动问题。在非运动情境中，请用"变化率（rate of change）"，并让措辞与题目场景一致。
Worked Example — Interpreting a Derivative in Context例题 —— 在情境中解读一个导数
The population of a city, \(P(t)\), is measured in thousands of people, where \(t\) is years since 2010. We are told \(P'(8) = 2.4\).
某城市的人口 \(P(t)\) 以千人为单位计量，其中 \(t\) 表示自 2010 年起的年数。已知 \(P'(8) = 2.4\)。
Step 1 — Identify the variables and units
第 1 步 —— 辨明变量与单位
\(P(t)\) represents population, measured in thousands of people. The input \(t\) is measured in years since 2010.
\(P(t)\) 表示人口，以千人为单位。输入 \(t\) 以自 2010 年起的年数为单位。
\[ \text{Units of }P'(t)=\frac{\text{thousands of people}}{\text{year}} \]Step 2 — Evaluate the time in context
第 2 步 —— 把时间放回情境
Since \(t=8\), the calendar year is \(2010+8=2018\).
由 \(t=8\)，对应的公历年份为 \(2010+8=2018\)。
Step 3 — State the interpretation in a full sentence
第 3 步 —— 用一句完整的话作出解读
At \(t=8\) (that is, in 2018), the population of the city is increasing at a rate of \(2.4\) thousand people per year.
在 \(t=8\)（即 2018 年）时，该城市的人口正以 每年 \(2.4\) 千人 的速率递增。
Straight-Line Motion: Position, Velocity, & Acceleration
直线运动（straight-line motion）：位置、速度与加速度
Rectilinear (straight-line) motion is the most iconic application of the derivative. A particle moves along a line, and its position at time \(t\) is \(s(t)\) (or \(x(t)\)). Differentiation links position, velocity, and acceleration in a clean chain:
直线（直线方向）运动是导数最具代表性的应用。质点（particle）沿一条直线运动，时刻 \(t\) 时的位置（position）为 \(s(t)\)（或 \(x(t)\)）。求导（differentiate）把位置、速度（velocity）和加速度（acceleration）串成一条干净的链条：
Key Relationships
关键关系
| Quantity | Definition | Sign Meaning |
|---|---|---|
| Position \(s(t)\) | Location on the number line | Positive → right of origin; negative → left |
| Velocity \(v(t) = s'(t)\) | Rate of change of position | \(v > 0\) → moving right; \(v < 0\) → moving left |
| Speed \(|v(t)|\) | Magnitude of velocity | Always \(\geq 0\); how fast regardless of direction |
| Acceleration \(a(t) = v'(t)\) | Rate of change of velocity | Same sign as \(v\) → speeding up; opposite → slowing down |
| 量 | 定义 | 符号含义 |
|---|---|---|
| 位置 \(s(t)\) | 在数轴上的位置 | 正 → 原点右侧；负 → 原点左侧 |
| 速度 \(v(t) = s'(t)\) | 位置的变化率 | \(v > 0\) → 向右运动；\(v < 0\) → 向左运动 |
速率（speed） \(|v(t)|\) | 速度的大小 | 始终 \(\geq 0\)；不论方向，只看快慢 |
| 加速度 \(a(t) = v'(t)\) | 速度的变化率 | 与 \(v\) 同号 → 加速；异号 → 减速 |
direction）加速。
Worked Example — Analyzing Particle Motion例题 —— 分析质点运动（particle motion）
A particle moves along the \(x\)-axis with position \(s(t) = t^3 - 6t^2 + 9t + 2\) for \(t \geq 0\).
某质点沿 \(x\) 轴运动，当 \(t \geq 0\) 时其位置为 \(s(t) = t^3 - 6t^2 + 9t + 2\)。
Step 1 — Differentiate to find velocity and acceleration
第 1 步 —— 求导得到速度和加速度
\[ v(t)=s'(t)=3t^2-12t+9=3(t-1)(t-3) \] \[ a(t)=v'(t)=6t-12=6(t-2) \]Step 2 — Find when the particle is at rest
第 2 步 —— 判断质点何时静止
A particle is at rest when \(v(t)=0\).
当 \(v(t)=0\) 时质点处于静止。
\[ 3(t-1)(t-3)=0 \quad \Rightarrow \quad t=1,\;3 \]Step 3 — Analyze the sign of velocity
第 3 步 —— 分析速度的符号
Moving right.
向右运动。
Moving left.
向左运动。
So the particle changes direction at \(t=1\) and \(t=3\), because the sign of \(v\) changes at both times.
因此质点在 \(t=1\) 与 \(t=3\) 处改变方向，因为在这两个时刻 \(v\) 的符号都发生了变化。
Step 4 — Check speeding up or slowing down
第 4 步 —— 判断是加速还是减速
At \(t=2\):
当 \(t=2\) 时：
\[ v(2)=3(2-1)(2-3)=-3, \qquad a(2)=6(2-2)=0 \]Since acceleration is zero, the particle is neither speeding up nor slowing down at that instant.
由于加速度为零，质点在此瞬时既不加速也不减速。
At \(t=0.5\):
当 \(t=0.5\) 时：
\[ v(0.5)=3(-0.5)(-2.5)=3.75, \qquad a(0.5)=6(-1.5)=-9 \]Velocity is positive and acceleration is negative, so the signs are opposite. The particle is slowing down.
速度为正、加速度为负，二者异号。质点正在减速。
Rates of Change in Applied Contexts Other Than Motion
非运动情境中的变化率
The derivative as a rate of change extends far beyond moving particles. Any time a quantity depends on another quantity and you want to know how fast the first is changing with respect to the second, you are looking at a derivative. The underlying mathematical structure is identical to motion problems — only the vocabulary and units change.
导数作为变化率的应用范围远远超出运动中的质点。任何时候，只要一个量依赖于另一个量，而你想知道前者关于后者的变化快慢，看到的就是导数。其底层的数学结构与运动问题完全相同——只是用词和单位换了一套。
Common real-world contexts include population growth, the spread of a disease, the temperature of a cooling object, the rate at which a company's revenue changes with the number of units sold (marginal revenue), the rate water flows into or out of a tank, and the changing concentration of a chemical in a mixture. In every case, the derivative of the quantity function gives the instantaneous rate of change, and its units are always the output units divided by the input units.
常见的现实情境包括：人口增长（指数增长（exponential growth）的典型例子）、疾病传播、冷却物体的温度、公司收入随售出数量的变化（边际收入）、水流入或流出水箱的速率、混合物中化学物质浓度的变化。在所有这些情境中，量函数的导数给出瞬时变化率，单位始终是输出单位除以输入单位。
function）是什么？自变量是什么？导数在告诉我什么？回答好这三个问题，无论情境是什么，计算过程都是一样的。
average）变化率为 \(\frac{f(b) - f(a)}{b - a}\)——即割线（secant line）的斜率。在 \(x = a\) 处的瞬时（instantaneous）变化率为 \(f'(a)\)——即切线的斜率。考题经常考查你能否区分这两者，尤其当数据以表格形式给出时（此时你只能算出表中相邻点之间的平均变化率）。
Worked Example — Non-Motion Rate of Change例题 —— 非运动情境的变化率
The temperature of coffee, in °F, \(t\) minutes after being poured is modelled by \(T(t) = 70 + 130e^{-0.04t}\). Find and interpret \(T'(10)\).
咖啡倒出后 \(t\) 分钟时的温度（单位：°F）由模型 \(T(t) = 70 + 130e^{-0.04t}\) 给出。求 \(T'(10)\) 并加以解读。
Step 1 — Differentiate the model
第 1 步 —— 对模型求导
\[ T'(t)=130\bigl(-0.04\bigr)e^{-0.04t}=-5.2e^{-0.04t} \]The derivative has units of degrees Fahrenheit per minute.
该导数的单位为华氏度每分钟。
Step 2 — Evaluate at \(t=10\)
第 2 步 —— 代入 \(t=10\) 求值
\[ T'(10)=-5.2e^{-0.4}\approx -5.2(0.6703)\approx -3.486 \]Step 3 — Interpret the result
第 3 步 —— 解读结果
At 10 minutes after the coffee is poured, the temperature of the coffee is decreasing at a rate of about \(3.486\) °F per minute.
在咖啡倒出后第 10 分钟时，咖啡的温度正以约每分钟 \(3.486\) °F的速率递减。
Introduction to Related Rates
相关变化率（related rates）入门
In a related-rates problem, two or more quantities are changing with respect to time, and they are connected by an equation. You know the rate of change of one quantity and need to find the rate of change of another. The chain rule is the engine that makes this work: since every variable depends on \(t\) (time), differentiating the connecting equation with respect to \(t\) links the rates.
在相关变化率问题中，两个或多个量都随时间变化，并由一个方程相互联系。已知某个量的变化率，要求另一个量的变化率。能让这一切成立的引擎正是链式法则（chain rule）：由于每个变量都依赖于 \(t\)（时间），将这一联系方程对 \(t\) 求导，便把各个变化率连接了起来。
The Five-Step Framework
五步框架
Related-rates problems follow a consistent procedure. First, draw a picture and label all variables (not specific numbers, unless they are constants). Second, write an equation that relates the variables. Third, differentiate both sides with respect to \(t\), applying the chain rule to every variable. Fourth, substitute all known values at the specific instant. Fifth, solve for the unknown rate and interpret the result with units.
相关变化率问题遵循一套固定流程。第一，画图并标出所有变量（不要标具体数值，除非该数值是常量）。第二，写出方程把变量联系起来。第三，对两边关于 \(t\) 求导，对每一个变量都用链式法则。第四，代入在指定瞬时下所有已知值。第五，解出未知的变化率，并附上单位解读结果。
Solving Related Rates Problems
求解相关变化率问题
This topic puts the five-step framework from Topic 4.4 into practice. The key challenge is translating a word problem into a mathematical equation and then correctly applying implicit differentiation with respect to time. Let's work through the most common problem types.
本节把小节 4.4 的五步框架付诸实战。关键挑战在于：把文字题翻译成数学方程，然后正确地对时间做隐式求导。下面我们逐一处理最常见的几类题型。
Expanding circle: \(A = \pi r^2\) → \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\)
Expanding sphere: \(V = \frac{4}{3}\pi r^3\) → \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)
Right triangle (ladder): \(x^2 + y^2 = L^2\) → \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\)
Cone filling: \(V = \frac{1}{3}\pi r^2 h\) (may need to eliminate \(r\) using similar triangles)
扩张的圆： \(A = \pi r^2\) → \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\)
膨胀的球： \(V = \frac{4}{3}\pi r^3\) → \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)
直角三角形（梯子）： \(x^2 + y^2 = L^2\) → \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\)
圆锥注水： \(V = \frac{1}{3}\pi r^2 h\) （可能需要用相似三角形消去 \(r\)）
Worked Example — Expanding Circle例题 —— 扩张的圆
A stone dropped in a pond creates a circular ripple whose radius increases at \(3\) cm/s. How fast is the area increasing when the radius is \(10\) cm?
一颗石子落入池塘，激起一圈圆形涟漪，其半径以 \(3\) cm/s 的速率增大。当半径为 \(10\) cm 时，圆面积增大的速率是多少？
Step 1 — Write the relationship
第 1 步 —— 写出关系式
\[ A=\pi r^2 \]Step 2 — Differentiate with respect to time
第 2 步 —— 关于时间求导
\[ \frac{dA}{dt}=2\pi r\frac{dr}{dt} \]Step 3 — Substitute the known values
第 3 步 —— 代入已知值
\[ \frac{dA}{dt}=2\pi(10)(3)=60\pi \]Step 4 — State the answer with units
第 4 步 —— 给出带单位的答案
When the radius is \(10\) cm, the area is increasing at
当半径为 \(10\) cm 时，面积增大的速率为
\[ \frac{dA}{dt}=60\pi\text{ cm}^2/\text{s}\approx 188.5\text{ cm}^2/\text{s} \]Worked Example — Sliding Ladder例题 —— 滑动的梯子
A 13-foot ladder leans against a wall. The base slides away at \(2\) ft/s. How fast is the top sliding down when the base is \(5\) feet from the wall?
一架 13 英尺长的梯子靠在墙上。梯脚以 \(2\) ft/s 的速率向外滑动。当梯脚距墙 \(5\) 英尺时，梯顶向下滑动的速率是多少？
Step 1 — Relate the variables
第 1 步 —— 写出变量关系
Let \(x\) be the distance of the base from the wall and \(y\) be the height of the top on the wall. Then
设 \(x\) 为梯脚到墙的距离，\(y\) 为梯顶在墙上的高度，则
\[ x^2+y^2=13^2=169 \]Step 2 — Find \(y\) at the instant of interest
第 2 步 —— 求指定瞬时的 \(y\)
When \(x=5\),
当 \(x=5\) 时，
\[ 5^2+y^2=169 \Rightarrow 25+y^2=169 \Rightarrow y^2=144 \Rightarrow y=12 \]Step 3 — Differentiate with respect to time
第 3 步 —— 关于时间求导
\[ 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 \]Step 4 — Substitute known values and solve
第 4 步 —— 代入已知值并求解
\[ 2(5)(2)+2(12)\frac{dy}{dt}=0 \] \[ 20+24\frac{dy}{dt}=0 \] \[ \frac{dy}{dt}=-\frac{20}{24}=-\frac{5}{6} \]Step 5 — Interpret the sign
第 5 步 —— 解读符号
The negative sign means the height is decreasing, so the top of the ladder is sliding down at
负号表示高度在减小，因此梯顶正在以下列速率向下滑动：
\[ \left|\frac{dy}{dt}\right|=\frac{5}{6}\text{ ft/s} \]Approximating Values Using Local Linearity & Linearization
用局部线性与线性近似（linear approximation）估算函数值
Near the point of tangency, a differentiable function looks almost like its tangent line. This is the principle of local linearity: if you zoom in closely enough around a point, the curve becomes indistinguishable from a straight line. We exploit this by using the tangent line to approximate function values that would otherwise be difficult to calculate.
在切点附近，可微函数看起来几乎与它的切线（tangent line）完全一样。这就是局部线性原理：把一个点附近放得足够大，曲线就变得与一条直线无法区分。我们利用这一点，用切线来近似（approximation）那些原本难以直接计算的函数值。
tangent line approximation）（线性近似）Why would you use an approximation instead of computing \(f(x)\) exactly? On the AP Exam, you might be given a function defined by a differential equation or a table, where an exact formula is unavailable. The tangent line approximation lets you estimate nearby values using only a single known point and the derivative there.
为什么要用近似而不直接精确计算 \(f(x)\)？在 AP 考试中，函数可能由微分方程或表格定义，根本拿不到显式公式。切线近似允许你只用一个已知点和那里的导数值，就估出附近的函数值。
Over- and Underestimates
高估与低估
The tangent line approximation is not always on the same side of the curve. Whether it overestimates or underestimates depends on the concavity of \(f\) near the point of tangency.
切线近似并不总在曲线的同一侧。它是高估还是低估，取决于 \(f\) 在切点附近的凹凸性（concavity）。
concave up）（\(f'' > 0\)），切线位于曲线下方，因此线性近似是低估。若 \(f\) 在 \(a\) 附近下凹（concave down）（\(f'' < 0\)），切线位于曲线上方，因此线性近似是高估。可以这样记忆：上凹像一只碗——任意点处的切线都贴在碗底之下。
Worked Example — Approximating \(\sqrt{26}\)例题 —— 近似 \(\sqrt{26}\)
Use the linearization of \(f(x) = \sqrt{x}\) at \(a = 25\) to approximate \(\sqrt{26}\).
用 \(f(x) = \sqrt{x}\) 在 \(a = 25\) 处的线性近似来估算 \(\sqrt{26}\)。
Step 1 — Compute \(f(a)\) and \(f'(a)\)
第 1 步 —— 求 \(f(a)\) 与 \(f'(a)\)
\[ f(25)=\sqrt{25}=5 \] \[ f'(x)=\frac{1}{2\sqrt{x}} \quad \Rightarrow \quad f'(25)=\frac{1}{2\cdot 5}=\frac{1}{10} \]Step 2 — Build the linearization
第 2 步 —— 构造线性近似
\[ L(x)=f(a)+f'(a)(x-a)=5+\frac{1}{10}(x-25) \]Step 3 — Approximate the target value
第 3 步 —— 近似目标值
\[ L(26)=5+\frac{1}{10}(26-25)=5+0.1=5.1 \]Step 4 — Check whether the estimate is high or low
第 4 步 —— 判断估计偏高还是偏低
\[ f''(x)=-\frac{1}{4x^{3/2}}<0 \]Because \(f\) is concave down near \(x=25\), the tangent line lies above the curve. Therefore, \(5.1\) is an overestimate of \(\sqrt{26}\).
因为 \(f\) 在 \(x=25\) 附近下凹，切线位于曲线之上。因此 \(5.1\) 是 \(\sqrt{26}\) 的高估。
Using L'Hôpital's Rule for Indeterminate Forms
用洛必达法则（L'Hôpital's rule）处理不定式（indeterminate form）
Sometimes direct substitution into a limit yields \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) — forms that are called indeterminate because they do not have a definite value without further analysis. L'Hôpital's Rule provides an elegant resolution: under these conditions, the limit of the ratio equals the limit of the ratio of the derivatives.
有时直接代入求极限会得到 \(\frac{0}{0}\) 或 \(\frac{\infty}{\infty}\)——这类形式称为不定式，因为不做进一步分析无法确定其值。洛必达法则给出了一个优雅的解决方法：在这些条件下，比值的极限等于导数之比的极限（limit）。
quotient rule）。不要把 \(\frac{f}{g}\) 当作一个整体分式来求导。
Worked Example — \(\displaystyle\lim_{x \to 0}\frac{\sin x}{x}\)例题 —— \(\displaystyle\lim_{x \to 0}\frac{\sin x}{x}\)
Step 1 — Verify the indeterminate form
第 1 步 —— 验证不定式形式
\[ \lim_{x\to 0}\frac{\sin x}{x} \to \frac{\sin 0}{0}=\frac{0}{0} \]Since the form is \(\frac{0}{0}\), L'Hôpital's Rule applies.
由于形式为 \(\frac{0}{0}\)，洛必达法则适用。
Step 2 — Differentiate numerator and denominator separately
第 2 步 —— 分别对分子、分母求导
\[ \lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\cos x}{1} \]Step 3 — Evaluate the simplified limit
第 3 步 —— 求出简化后的极限
\[ \lim_{x\to 0}\frac{\cos x}{1}=\frac{\cos 0}{1}=1 \]Worked Example — Applying L'Hôpital's Rule Twice例题 —— 两次应用洛必达法则
Find \(\displaystyle\lim_{x \to 0}\frac{1 - \cos x}{x^2}\).
求 \(\displaystyle\lim_{x \to 0}\frac{1 - \cos x}{x^2}\)。
Step 1 — Check the original form
第 1 步 —— 检查原始形式
\[ \lim_{x\to 0}\frac{1-\cos x}{x^2}\to \frac{1-\cos 0}{0^2}=\frac{0}{0} \]So L'Hôpital's Rule applies.
因此洛必达法则适用。
Step 2 — Apply L'Hôpital's Rule once
第 2 步 —— 应用一次洛必达法则
\[ \lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\frac{\sin x}{2x} \]Step 3 — Check again
第 3 步 —— 再次检查
\[ \lim_{x\to 0}\frac{\sin x}{2x}\to \frac{0}{0} \]The result is still indeterminate, so apply L'Hôpital's Rule a second time.
结果仍为不定式，因此再次应用洛必达法则。
Step 4 — Apply L'Hôpital's Rule again and evaluate
第 4 步 —— 再次应用洛必达法则并求值
\[ \lim_{x\to 0}\frac{\sin x}{2x}=\lim_{x\to 0}\frac{\cos x}{2}=\frac{\cos 0}{2}=\frac{1}{2} \]Worked Example — Rewriting \(0 \cdot \infty\) for L'Hôpital (\(\displaystyle\lim_{x \to 0^+}\, x \ln x\))例题 —— 把 \(0 \cdot \infty\) 改写后再用洛必达（\(\displaystyle\lim_{x \to 0^+}\, x \ln x\)）
L'Hôpital does not apply directly to a \(0 \cdot \infty\) form. The strategy is to move one factor to the denominator so the limit becomes \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then apply L'Hôpital.
洛必达法则不能直接用于 \(0 \cdot \infty\) 形式。策略是把其中一个因子移到分母，使极限变成 \(\frac{0}{0}\) 或 \(\frac{\infty}{\infty}\)，再用洛必达。
Step 1 — Identify the indeterminate form
第 1 步 —— 辨明不定式形式
As \(x \to 0^+\): \(x \to 0\) and \(\ln x \to -\infty\). So the form is \(0 \cdot (-\infty)\). This is indeterminate, but L'Hôpital does not apply yet — we need a quotient.
当 \(x \to 0^+\) 时：\(x \to 0\) 且 \(\ln x \to -\infty\)。因此形式为 \(0 \cdot (-\infty)\)。这是不定式，但此时洛必达法则还不适用——我们需要一个商。
Step 2 — Rewrite as \(\frac{\infty}{\infty}\)
第 2 步 —— 改写为 \(\frac{\infty}{\infty}\)
Move \(x\) into the denominator as \(\dfrac{1}{1/x}\):
把 \(x\) 写成 \(\dfrac{1}{1/x}\) 移入分母：
\[ x \ln x \;=\; \dfrac{\ln x}{\,1/x\,} \]As \(x \to 0^+\): \(\ln x \to -\infty\) and \(\dfrac{1}{x} \to +\infty\). The new form is \(\dfrac{-\infty}{+\infty}\) — now L'Hôpital applies.
当 \(x \to 0^+\) 时：\(\ln x \to -\infty\) 且 \(\dfrac{1}{x} \to +\infty\)。新的形式为 \(\dfrac{-\infty}{+\infty}\)——现在可以用洛必达了。
Step 3 — Apply L'Hôpital's Rule
第 3 步 —— 应用洛必达法则
\[ \lim_{x\to 0^+}\frac{\ln x}{1/x} \;=\; \lim_{x\to 0^+}\frac{\,1/x\,}{-1/x^{2}} \;=\; \lim_{x\to 0^+}\frac{1}{x}\cdot\bigl(-x^{2}\bigr) \;=\; \lim_{x\to 0^+}\bigl(-x\bigr) \]Step 4 — Evaluate
第 4 步 —— 求值
\[ \lim_{x\to 0^+}\bigl(-x\bigr) \;=\; 0 \]Therefore \(\displaystyle\lim_{x\to 0^+}x\ln x = 0\).
因此 \(\displaystyle\lim_{x\to 0^+}x\ln x = 0\)。
Exam Strategy
应试策略
MCQ — Contextual Interpretation
Multiple-choice questions love to present four interpretations of a derivative and ask which one is correct. Look for the answer that includes: (1) a specific time/input, (2) "increasing" or "decreasing", (3) the correct magnitude, and (4) correct units. Wrong answers often swap units, confuse total change with rate, or omit "per unit."
选择题 —— 情境解读
选择题最爱给出一个导数的四种解读，问哪一项正确。要找包含以下四个要素的选项：(1) 具体的时刻/输入值，(2) "递增"或"递减"，(3) 正确的数值大小，(4) 正确的单位。错误选项常常会换错单位、把总变化与变化率混淆，或漏掉"每单位"。
FRQ — Related Rates
Related-rates FRQs are scored on process as much as the final answer. Show: the equation you start with, the derivative step, the substitution step, and the interpretation. If you skip straight to the numerical answer, you may only earn 1 of 3–4 available points.
自由回答 —— 相关变化率
相关变化率的自由回答题，过程的分值不亚于最终答案。要写明：所用的方程、求导的步骤、代入的步骤，以及最终的解读。如果直接跳到数值答案，3–4 分中可能只能拿到 1 分。
Table Problems
When a function is given by a table and you're asked for a rate of change at a specific point, use the average rate of change over the smallest interval containing that point. If asked to approximate using linearization, use the tangent line built from nearby table values.
表格题
当函数以表格形式给出，题目要求某点处的变化率时，使用包含该点的最小区间上的平均变化率。如要求用线性近似来估算，则使用由邻近表格值构造的切线。
L'Hôpital's — Must-Show Work
On FRQs, you must explicitly state that the limit yields $0/0$ or $\infty/\infty$ before applying L'Hôpital's Rule. Writing "by L'Hôpital's Rule" without verification loses points. Also remember: differentiate the numerator and denominator separately — do not use the quotient rule on the fraction.
洛必达 —— 必须写过程
在自由回答题中，应用洛必达法则之前必须明确说明极限为 $0/0$ 或 $\infty/\infty$。只写"由洛必达法则"而不验证会扣分。另外，要分别对分子、分母求导——不要把这个分式当作整体用商的法则。
Common Mistakes
常见错误
1. Missing units in interpretations. "The population is increasing at 2.4" is incomplete. It must be "2.4 thousand people per year."
2. Confusing speed with velocity. Speed \(= |v(t)|\) and is always non-negative. Saying "the speed is \(-5\)" is meaningless.
3. Substituting before differentiating in related rates. Plugging in numbers before taking \(\frac{d}{dt}\) kills the problem. Variables first, numbers after.
4. Applying L'Hôpital's without verifying the form. If the limit is not \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), L'Hôpital's Rule does not apply. Applying it anyway produces a wrong answer.
5. Using the quotient rule instead of L'Hôpital's. L'Hôpital's says differentiate top and bottom separately. It is not the derivative of the fraction.
6. Forgetting concavity when assessing over/underestimate. Without knowing whether \(f\) is concave up or down, you cannot say whether a linearization is too high or too low.
7. Saying "the derivative is changing" instead of "the quantity is changing." The derivative describes how the original quantity changes. Say "the temperature is decreasing," not "the derivative is decreasing."
1. 解读时漏写单位。"人口正以 2.4 增长"是不完整的。必须写成"每年 2.4 千人"。
2. 把速率与速度混为一谈。速率 \(= |v(t)|\)，始终非负。说"速率为 \(-5\)"毫无意义。
3. 相关变化率里在求导前就代数。在 \(\frac{d}{dt}\) 之前代入数值会把问题做坏。先保留变量，后代数。
4. 不验证形式就用洛必达。如果极限不是 \(\frac{0}{0}\) 或 \(\frac{\infty}{\infty}\)，洛必达法则不适用。强行使用会得到错误答案。
5. 用商的法则代替洛必达。洛必达要求分子分母分别求导，不是把整个分式作为商求导。
6. 判断高估/低估时忘记凹凸性。不知道 \(f\) 是上凹还是下凹，就无法判断线性近似偏高还是偏低。
7. 说"导数在变化"而不是"该量在变化"。导数描述原始量是如何变化的。要说"温度在降低"，而不是"导数在减小"。
Flashcards
闪卡
Click any card to flip it.
点击任意卡片可翻转。
underestimate低估
\[\lim\frac{f}{g} = \lim\frac{f'}{g'}\]
overestimate高估
Unit Quiz
单元小测
Readiness Checklist
备考清单
Click each item you've mastered. Aim for 100% before exam day.
点击你已掌握的每一项。考试之前争取做到 100%。
- Interpret the derivative as a rate of change in a real-world context在现实情境中把导数解读为变化率
- Analyze straight-line motion using position, velocity, and acceleration用位置、速度和加速度分析直线运动
- Determine when an object is speeding up vs. slowing down判断物体何时加速、何时减速
- Use units correctly when interpreting derivatives in context在情境中解读导数时正确使用单位
- Set up related-rates equations from a geometric relationship由几何关系建立相关变化率方程
- Differentiate both sides of an implicit equation with respect to time对隐式方程两边关于时间求导
- Solve classic related-rates problems (ladder, cone, shadow, balloon)求解经典相关变化率问题（梯子、圆锥、影子、气球）
- Construct a linearization $L(x) = f(a) + f'(a)(x-a)$构造线性近似 $L(x) = f(a) + f'(a)(x-a)$
- Use linear approximation to estimate function values near a point用线性近似估算某点附近的函数值
- Recognize when approximation over- or underestimates (concavity)识别近似何时高估或低估（凹凸性）
- Identify indeterminate forms $0/0$ and $\infty/\infty$辨识不定式 $0/0$ 与 $\infty/\infty$
- Apply L'Hôpital's Rule with correct justification of hypotheses在正确验证前提下应用洛必达法则
- Rewrite a $0 \cdot \infty$ form as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before applying L'Hôpital (e.g., $x \ln x$)在使用洛必达前把 $0 \cdot \infty$ 改写为 $\frac{0}{0}$ 或 $\frac{\infty}{\infty}$（如 $x \ln x$）
- Select the appropriate tool (derivative rule, related rates, L'Hôpital) per problem针对每道题选择合适的工具（求导法则、相关变化率、洛必达）
AP-Style Practice Problems
AP 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 评分标准设计的选择题与简答题。为冲刺高分准备；建议在完成上方笔记和小测之后再进入。