AP 미적분학 AB

Calculus is the mathematics of change 变化 and of accumulation 累积. It answers two big questions: how fast is something changing right now, and how much has piled up so far? Unit 1 builds the one tool both questions rest on – the limit 极限.

Start with a puzzle. A car's speedometer reads $60$ km/h. What does that mean at a single instant 瞬间? Speed is distance over time. But at one instant no time passes and no distance is covered, so the fraction looks like $\tfrac{0}{0}$ – undefined.

• The average rate of change 平均变化率 uses a whole interval 区间: the change in one quantity divided by the change in another. It divides by zero, and so is undefined, when the change in the input would be zero.
• The instantaneous rate of change 瞬时变化率 is what we want at a point. It is the value the average rate approaches 趋近 as the interval shrinks toward zero length.

The clever move is not to plug in zero (undefined), but to watch what the average rate approaches as the interval gets smaller and smaller. That approaching value is a limit. So calculus lets us describe change at an instant – as a limit of average rates over ever-shorter intervals. This one idea powers the derivative 导数 (Unit 2) and, run in reverse, the integral 积分 (Unit 6). Everything else in this unit defines limits carefully and computes them reliably.

Given a function $f$, the limit of $f(x)$ as $x$ approaches $c$ is a real number $R$ if $f(x)$ can be made arbitrarily 任意地 close to $R$ by taking $x$ sufficiently 足够 close to $c$ – but not equal to $c$. We write

and read it: "the limit of $f(x)$, as $x$ approaches $c$, equals $R$."

The last words are the heart of a limit: it describes the behavior 行为 of $f$ near $c$, not the value at $c$. The function may be undefined at $c$, or defined but equal to something else – the limit does not care.

A limit can be shown in three ways: graphically 用图象, numerically 用数值 (a table), and analytically 用解析式 (algebra). Learning to move between these representations is a core skill.

(Note: the epsilon-delta definition of a limit is not tested on the AP Exam, so this handout does not use it.)

A graph is often the fastest way to read a limit. To find $\displaystyle \lim_{x \to c} f(x)$, run your finger along the curve toward $x = c$ from each side and ask: what height is the curve heading for?

• Trace from the left (inputs smaller than $c$): this gives the left-hand limit 左极限, $\displaystyle \lim_{x \to c^-} f(x)$.
• Trace from the right (inputs larger than $c$): this gives the right-hand limit 右极限, $\displaystyle \lim_{x \to c^+} f(x)$.
• These are the one-sided limits 单侧极限. If both head to the same height $R$, then the two-sided limit exists and $\displaystyle \lim_{x \to c} f(x) = R$.

Crucially, ignore the point itself. Graphs mark the difference between the limit and the value:

• An open circle 空心圆 marks a height the curve approaches but does not reach – a "hole" 空洞.
• A closed circle 实心圆 marks the actual value $f(c)$.

So a curve may approach $R = 3$ from both sides (limit is $3$) while a filled dot sits at height $5$ (value $f(c) = 5$). The limit is $3$; the two need not match.

A limit does not exist (often written DNE) when the two sides disagree (a jump 跳跃), when the function is unbounded 无界 (grows without limit), or when it oscillates 振荡 forever near $c$. For example:

Watch the scale 比例 of a graph: a zoomed-out picture can hide important behavior near a point, so confirm with algebra when you can.

When you have data or a formula but no picture, a table 表格 of values estimates a limit numerically. Choose inputs that creep toward $c$ from both sides and watch the outputs.

For example, to estimate $\displaystyle \lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ (which is $\tfrac{0}{0}$ at $x=2$):

$x$	$1.9$	$1.99$	$1.999$	$\to 2 \leftarrow$	$2.001$	$2.01$	$2.1$
$f(x)$	$3.9$	$3.99$	$3.999$	?	$4.001$	$4.01$	$4.1$

Both sides march toward $4$, so we estimate the limit is $4$. A table only suggests a value – it is a numerical estimate, not a proof.

Most limits are found analytically using limit theorems 极限定理. If $\lim_{x\to c} f(x)$ and $\lim_{x\to c} g(x)$ both exist, the limit of a combination is the same combination of the limits:

• Sum / difference: $\displaystyle \lim_{x\to c}\big[f(x)\pm g(x)\big] = \lim_{x\to c}f(x) \pm \lim_{x\to c}g(x)$
• Product: $\displaystyle \lim_{x\to c}\big[f(x)\,g(x)\big] = \lim_{x\to c}f(x)\cdot\lim_{x\to c}g(x)$
• Quotient: $\displaystyle \lim_{x\to c}\frac{f(x)}{g(x)} = \frac{\lim_{x\to c}f(x)}{\lim_{x\to c}g(x)}$, provided the bottom limit is not $0$.
• Composite 复合函数: if $g$ is continuous at $\lim_{x\to c} f(x)$, then $\displaystyle \lim_{x\to c} g\big(f(x)\big) = g\!\left(\lim_{x\to c} f(x)\right)$.

The practical rule: for a function built from polynomials, roots, and the like, first try direct substitution 直接代入 – put $x = c$ in. If you get a real number, that is the limit. One-sided limits obey the same theorems, read from one direction only.

Direct substitution sometimes gives the indeterminate form 未定式 $\tfrac{0}{0}$. This does not mean the limit fails – it means you must rewrite the function into an equivalent form 等价形式 that removes the trouble, then substitute. Three standard moves:

• Factor and cancel 因式分解并约分 a rational function 有理函数. Example: $\displaystyle \lim_{x\to 2}\frac{x^2-4}{x-2} = \lim_{x\to 2}\frac{(x-2)(x+2)}{x-2} = \lim_{x\to 2}(x+2) = 4$.
• Multiply by the conjugate 共轭 to simplify a radical 根式. Example: $\displaystyle \lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \lim_{x\to 0}\frac{x}{x\big(\sqrt{x+1}+1\big)} = \frac{1}{2}$.
• Use alternate forms of trigonometric functions (identities) to simplify.

The cancelled factor is why the original graph had a hole: the two functions agree everywhere except at $x=c$, so they share the same limit there.

This is a skill topic, not new content: choose the right tool for the limit in front of you.

1. Try direct substitution first. A real answer means you are done.
2. Getting $\tfrac{0}{0}$? Rewrite – factor and cancel, or use the conjugate, or a trig identity – then substitute.
3. A non-zero number over $0$ (like $\tfrac{5}{0}$)? The limit is infinite or DNE – check the sign from each side (see vertical asymptotes below).
4. As $x\to\pm\infty$? Compare the dominant 主导 terms (see limits at infinity).
5. Trapped between two functions? The squeeze theorem may apply.

The squeeze theorem 夹逼定理 (also called the sandwich theorem) finds a limit by trapping the function between two others. If $g(x) \le f(x) \le h(x)$ near $c$, and

then $f$ is squeezed to the same place: $\displaystyle \lim_{x\to c} f(x) = L$.

The two famous results proved this way, both used throughout calculus, are:

Another skill topic: the same limit lives in a graph, a table, and an algebraic form, and you should be able to translate between them. A graph shows the shape and any holes or jumps; a table gives numerical evidence; algebra gives an exact value and a reason. Strong answers use one representation to confirm another.

A function is discontinuous 间断 at $c$ when its graph "breaks" there. There are three types:

• Removable discontinuity 可去间断 – a single hole. The two-sided limit exists, but the point is missing or misplaced.
• Jump discontinuity 跳跃间断 – the two one-sided limits exist but disagree, so the curve jumps.
• Infinite discontinuity 无穷间断 – the function blows up to $\pm\infty$ at a vertical asymptote 垂直渐近线.

Continuity is defined by a three-part test. A function $f$ is continuous 连续 at $x=c$ exactly when all three hold:

In words: the point is there, the limit is there, and the two agree. If any one fails, $f$ is discontinuous at $c$. This test is the backbone of nearly every continuity question, so learn it as a checklist.

A function is continuous on an interval 在区间上连续 if it is continuous at every point of that interval. You rarely check point by point, because whole families are continuous on their domains:

Polynomial, rational, power, exponential 指数, logarithmic 对数, and trigonometric 三角 functions are continuous at every point of their domains.

So a rational function is continuous everywhere except where its denominator is zero; $\ln x$ is continuous for $x>0$; and so on. Knowing this lets you declare continuity quickly and correctly.

If the limit exists at a hole, the discontinuity is removable: redefine the function at that one point to equal the limit, and the graph is repaired. Formally, set the missing value to $\displaystyle \lim_{x\to c} f(x)$.

For a piecewise-defined function 分段函数, continuity at a boundary $x=c$ needs the two pieces to meet: the left piece's value, the right piece's value, and $f(c)$ must all be equal. This is a common exam setup – you solve for a parameter 参数 (an unknown constant) that makes the pieces match:

The idea of a limit extends to infinite limits 无穷极限. When a function grows without bound near $x=c$, we write $\lim_{x\to c} f(x) = \pm\infty$. This describes a vertical asymptote at $x=c$: the graph hugs the vertical line $x=c$ and shoots off toward $\pm\infty$.

This happens where a non-zero number is divided by something approaching $0$, such as at a zero of a denominator that does not cancel. Always check each side separately – the two sides can shoot opposite ways (one to $+\infty$, one to $-\infty$).

We can also let the input grow: limits at infinity 无穷远处的极限 describe the end behavior 末端行为 of a function as $x\to\pm\infty$. If the outputs settle toward a finite value $L$, then $y=L$ is a horizontal asymptote 水平渐近线.

For a rational function, compare the degrees 次数 of the top and bottom:

• top degree < bottom degree $\Rightarrow$ limit is $0$ (asymptote $y=0$);
• top degree = bottom degree $\Rightarrow$ limit is the ratio of the leading coefficients 首项系数之比;
• top degree > bottom degree $\Rightarrow$ the function is unbounded (no horizontal asymptote).

More generally, we compare the relative magnitudes 相对大小 (relative growth rates) of functions: far out, an exponential beats any polynomial, and a polynomial beats any logarithm. On the exam, "as $t\to\infty$, which quantity is larger/where does the rate settle?" is answered with a limit at infinity.

The Intermediate Value Theorem 介值定理 is an existence theorem 存在性定理 – it guarantees a value exists without telling you where:

If $f$ is continuous on the closed interval $[a,b]$, and $d$ is any number between $f(a)$ and $f(b)$, then there is at least one number $c$ in $(a,b)$ with $f(c)=d$.

An unbroken curve cannot skip a height between its endpoints – it must pass through every one.

Exam skill – how to justify with the IVT. These questions appear almost every year (for example, "Must there be a value $c$ with $R(c)=155$?" or "Is there a time when $r'(t)=-6$?"). A full-credit justification has three moves:

1. State continuity. Say the function is continuous on $[a,b]$ (often because it is differentiable, or given continuous).
2. Show $d$ is trapped. Compute the two endpoint values and show the target $d$ lies between them, e.g. $f(a) < d < f(b)$.
3. Conclude by name. "By the Intermediate Value Theorem, there is a $c$ in $(a,b)$ with $f(c)=d$."

Skipping the continuity statement, or not showing $d$ is between the endpoints, loses the point – the theorem requires both conditions.

Worked example. Evaluate $\lim_{x\to 3}\dfrac{x^2-9}{x-3}$. Direct substitution gives $\tfrac{0}{0}$ (indeterminate), so factor: $\dfrac{(x-3)(x+3)}{x-3}=x+3$ for $x\ne 3$. Then $\lim_{x\to 3}(x+3)=6$. The graph has a removable discontinuity (a hole) at $x=3$ — the limit exists even though the function is undefined there.

• A limit describes what $f(x)$ approaches, which need not equal $f(a)$ — the two-sided limit exists only if both sides agree.
• Try direct substitution first; for a $\tfrac00$ form, factor and cancel or rationalise before substituting.
• A function is continuous at $a$ when the limit exists, $f(a)$ is defined, and they are equal.
• Use the Intermediate Value Theorem to guarantee a root: a continuous function that changes sign on $[a,b]$ takes every value between.
• Read horizontal asymptotes from end behaviour (limits at $\pm\infty$) and vertical asymptotes where the denominator (not the numerator) is zero.

Unit 1 built the limit. Unit 2 uses it to define the derivative 导数 – the exact rate of change at a point.

Over an interval, the average rate of change is a difference quotient 差商. Two equivalent forms appear:

The first uses a step of size $h$ from $a$; the second uses two points $x$ and $a$. Both compute $\dfrac{\text{change in output}}{\text{change in input}}$ over the interval.

The instantaneous 瞬时 rate of change at $x=a$ is what the difference quotient approaches as the interval shrinks to zero. This limit is the derivative at $a$, written $f'(a)$:

provided the limit exists.

Let the point $a$ vary and the derivative becomes a new function:

This is the definition of the derivative (sometimes called differentiating "by first principles" 用定义求导). Its value at each $x$ is the instantaneous rate of change there.

Common notations 记号 for the derivative of $y=f(x)$ are:

The derivative can be represented graphically, numerically, analytically, and verbally – be ready to move between them.

Geometric meaning. The derivative at a point is the slope 斜率 of the tangent line 切线 to the graph there. So the tangent line at $x=a$ passes through $\big(a, f(a)\big)$ with slope $f'(a)$:

Writing this line is a routine exam task, so keep the point-slope form ready.

You do not always have a formula. When a function is given by a table 表格 or a graph, estimate the derivative $f'(a)$ with a difference quotient over a small interval around $a$. A table with values on both sides of $a$ gives the best estimate:

Technology (a calculator) can also estimate a derivative at a point.

Exam skill (appears almost every year). Questions such as "Approximate $M'(7.5)$ using the average rate of change of $M$ over the interval $5 \le t \le 10$" ask for exactly this difference quotient. Show the setup:

Full credit needs the numbers plugged in and the correct units 单位 (output units per input unit), since these come from real-world models.

Differentiability is stronger than continuity. The key relationship:

If $f$ is differentiable 可导 at a point, then $f$ is continuous 连续 there.

So differentiability implies continuity. The reverse is false: a continuous function can fail to be differentiable. Two ways this happens:

• A corner 尖点: the left and right difference-quotient limits disagree, as with $f(x)=|x|$ at $x=0$.
• A vertical tangent 垂直切线: the slope is infinite (no real number), as with $f(x)=\sqrt[3]{x}$ at $x=0$.

Also, a point outside the domain of $f$ cannot be in the domain of $f'$. Use the contrapositive on the exam: if $f$ is not continuous at $a$, then $f$ is not differentiable at $a$.

From here we use rules instead of the limit definition each time. The power rule 幂法则 handles any power of $x$:

It works for whole-number powers, negative powers ($\tfrac{1}{x}=x^{-1}$), and roots ($\sqrt{x}=x^{1/2}$) – rewrite as a power first, then apply the rule.

These rules let you differentiate term by term:

• Constant: $\dfrac{d}{dx}\,k = 0$ (a constant does not change).
• Constant multiple 常数倍: $\dfrac{d}{dx}\big[k\,f(x)\big] = k\,f'(x)$.
• Sum / difference: $\dfrac{d}{dx}\big[f(x)\pm g(x)\big] = f'(x)\pm g'(x)$.

Combined with the power rule, they differentiate any polynomial 多项式 term by term. Example:

Learn these four building-block derivatives by heart:

Note the minus sign on the derivative of cosine, and that $e^{x}$ is its own derivative.

A limit that is really a derivative (LIM-3.A.1). Sometimes a limit is secretly the definition of a known derivative. If you recognize

for a function $f$ whose derivative you know, just evaluate $f'(a)$. For example, $\displaystyle \lim_{h\to 0}\frac{\sin\!\big(\tfrac{\pi}{2}+h\big)-1}{h} = \left.\frac{d}{dx}\sin x\right|_{x=\pi/2} = \cos\tfrac{\pi}{2} = 0$.

A product of two functions is not differentiated by multiplying the derivatives. Use the product rule 乘积法则:

"Derivative of the first times the second, plus the first times the derivative of the second." Example: Exam questions often build a new function from given pieces, e.g. $k'(x) = \big(f(x)\big)^2 g(x)$, and ask you to combine rules while reading values from a table.

For a quotient, use the quotient rule 商法则:

"Bottom times derivative of top, minus top times derivative of bottom, all over bottom squared." The order matters because of the minus sign, so write the numerator carefully. Example:

The remaining trigonometric derivatives are not memorized separately – you rewrite them with identities 恒等式 and apply the quotient (or product) rule. For instance, $\tan x = \dfrac{\sin x}{\cos x}$, so the quotient rule gives

The same method (writing $\cot x=\tfrac{\cos x}{\sin x}$, $\sec x=\tfrac{1}{\cos x}$, $\csc x=\tfrac{1}{\sin x}$) gives $-\csc^2 x$, $\sec x\tan x$, and $-\csc x\cot x$.

Higher-order derivatives. Differentiating $f'$ again gives the second derivative 二阶导数 $f''(x)$ (or $\tfrac{d^2y}{dx^2}$) – the rate of change of the rate of change. An exam part like "Find $k''(3)$" just means differentiate twice, then substitute. You can also estimate a second derivative from a table by applying the average-rate-of-change method to the $f'$ values.

Worked example. Differentiate $f(x)=x^2\sin x$ with the product rule $(uv)'=u'v+uv'$: take $u=x^2$ ($u'=2x$) and $v=\sin x$ ($v'=\cos x$), giving $f'(x)=2x\sin x+x^2\cos x$. Always read the structure first — a product needs the product rule, not the power rule applied to each factor separately.

• The derivative is the slope of the tangent — the limit of the secant slope $\tfrac{f(x+h)-f(x)}{h}$ as $h\to0$.
• Memorise the rules: power, product, quotient, and the derivatives of $\sin$, $\cos$, $e^x$, and $\ln x$.
• Differentiability implies continuity, but not the reverse (a corner or cusp is continuous yet not differentiable).
• Distinguish average rate of change (secant slope over an interval) from instantaneous rate (the derivative at a point).
• Give a tangent-line equation as $y-f(a)=f'(a)(x-a)$.

Unit 2 differentiated single functions. Unit 3 differentiates functions built inside other functions. The chain rule 链式法则 differentiates a composite function 复合函数 $f\big(g(x)\big)$:

"Derivative of the outer function (leaving the inside alone), times the derivative of the inside." The inner derivative $g'(x)$ is the piece students forget, so always ask "what is the inside, and what is its derivative?" Example:

In Leibniz notation, with $y=f(u)$ and $u=g(x)$, the rule reads $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$ – the intermediate $du$ appears to "cancel." Exam questions often give a table for $f$, $g$, $f'$, $g'$ and ask for $h'(a)$ where $h(x)=f\big(g(x)\big)$; evaluate $f'\big(g(a)\big)\cdot g'(a)$ by reading values.

Worked example. Differentiate $h(x)=(2x^2+1)^5$. The outer function is "(something)$^5$", the inner is $2x^2+1$:

Some curves are defined implicitly – by an equation in $x$ and $y$ that is not solved for $y$, such as $x^2+y^2=25$. Implicit differentiation 隐函数求导 finds $\dfrac{dy}{dx}$ without solving for $y$ first. It is just the chain rule, treating $y$ as a function of $x$.

The method: differentiate both sides with respect to $x$; every time you differentiate a $y$-term, multiply by $\dfrac{dy}{dx}$ (the chain rule); then solve algebraically for $\dfrac{dy}{dx}$. For $x^2+y^2=25$:

Worked example. Find the tangent line to the circle $x^2+y^2=25$ at the point $(3,4)$. From above $\dfrac{dy}{dx}=-\dfrac{x}{y}=-\dfrac{3}{4}$ there, so the tangent is

Notice the tangent is perpendicular to the radius, as geometry promises – a good sanity check.

Exam skill – "Show that $\dfrac{dy}{dx}=\ldots$". This exact prompt appears most years (e.g. "Show that $\dfrac{dy}{dx}=\dfrac{2y}{y^2-2x}$"). Because the target is given, you must show every algebra step cleanly: differentiate both sides, use the product/chain rules on mixed $xy$ terms, collect all $\dfrac{dy}{dx}$ terms on one side, factor, and divide. A correct final line that skips the algebra earns little. Follow-up parts then ask for a tangent line, or where the tangent is horizontal ($\tfrac{dy}{dx}=0$, so the numerator is $0$) or vertical (the denominator is $0$).

If $g$ is the inverse function 反函数 of $f$ (so $f(g(x))=x$), the chain rule links their derivatives:

In words: the derivative of the inverse at a point is the reciprocal 倒数 of the derivative of the original function at the matching point. A common exam setup gives a table and a point $(a,b)$ on $f$ (so $(b,a)$ is on $g$), then asks for $g'(b)=\dfrac{1}{f'(a)}$.

Worked example. Suppose $f(2)=5$ and $f'(2)=3$, and $g$ is the inverse of $f$. Because $(2,5)$ lies on $f$, the point $(5,2)$ lies on $g$, and

The graphs of $f$ and $g$ are mirror images across $y=x$, so their slopes at matching points are reciprocals.

The same idea gives the derivatives of the inverse trigonometric functions 反三角函数. The three you should know:

Combine these with the chain rule when the input is itself a function, e.g. $\dfrac{d}{dx}\arctan(3x)=\dfrac{3}{1+9x^2}$.

A skill topic: real derivatives mix several rules, so read the structure of the expression first, from the outside in.

• Is it a sum? Differentiate term by term.
• A product or quotient? Apply that rule, and expect to use the chain rule inside.
• A composite (something inside something)? Chain rule.
• Given implicitly? Implicit differentiation.

Name the outermost operation, apply its rule, and recurse inward. Neatness prevents the sign and bookkeeping errors that cost marks.

Differentiating $f'$ produces the second derivative 二阶导数 $f''$; repeating gives higher-order derivatives. The notations:

The second derivative measures how the slope is changing; it drives concavity 凹凸性 and acceleration in later units. To find $f''$ implicitly, differentiate the expression for $\dfrac{dy}{dx}$ again (with the quotient and chain rules), then substitute $\dfrac{dy}{dx}$ back in.

• Use the chain rule for composite functions — differentiate the outside, then multiply by the derivative of the inside (the most-forgotten factor).
• For implicit differentiation, differentiate both sides with respect to $x$ and attach $\tfrac{dy}{dx}$ each time $y$ is differentiated, then solve.
• Get the second derivative by differentiating twice (velocity → acceleration).
• Combine rules carefully in layered expressions (chain inside product, etc.).
• The inverse function's graph is the reflection in $y=x$; its slope is the reciprocal of the original's at the matching point.

Once you can compute derivatives, you use them to describe the real world. The derivative $f'(x)$ is the instantaneous rate of change of $f$ with respect to its input. Reading and reporting this rate correctly is a graded skill.

Units matter. The unit of $f'(x)$ is the unit of $f$ divided by the unit of $x$. If $C(t)$ is a number of acres and $t$ is in weeks, then $C'(t)$ is in acres per week. On the exam, "Using correct units, interpret the meaning of $g'(140)$" wants a full sentence: the value, the quantity, the rate word "per", and the moment. For example: "$g'(140)=2.3$ means that at $x=140$, the quantity is increasing at about $2.3$ units per unit of $x$."

For a particle moving on a line, three functions of time are linked by differentiation:

• position 位置 $s(t)$;
• velocity 速度 $v(t)=s'(t)$ – signed; its sign gives direction;
• acceleration 加速度 $a(t)=v'(t)=s''(t)$.

Key readings (frequent exam parts):

• The particle is at rest 静止 when $v(t)=0$.
• It moves right/up when $v(t)>0$ and left/down when $v(t)<0$; it changes direction where $v$ changes sign.
• Speed 速率 is $|v(t)|$. Speed is increasing when $v$ and $a$ have the same sign (the particle is speeding up), and decreasing when they have opposite signs.

Worked example. A particle moves with $s(t)=t^3-6t^2+9t$. Then $v(t)=3t^2-12t+9=3(t-1)(t-3)$, so it is at rest at $t=1$ and $t=3$, and changes direction at each. At $t=2$, $v(2)=3(1)(-1)=-3<0$ (moving left) and $a(2)=6(2)-12=0$; just after, $a>0$ while $v<0$, so it is slowing down there. Distinguish carefully between velocity (has direction) and speed (does not) – the exam tests this exact difference.

The same derivative idea models any changing quantity: a draining tank, a spreading population, a cooling cup. Whenever a problem says "the rate at which...", it is describing a derivative. Read the units to know which quantity's rate you have, then interpret in context.

In a related rates 相关变化率 problem, several quantities change together over time, and you know some rates but want another. The engine is the chain rule: differentiate a relationship with respect to time $t$. Every variable becomes a function of $t$, so each derivative picks up a "$\,/\,dt$" factor. Product and quotient rules may also be needed.

A reliable procedure – and a full-credit template on the exam:

1. Name the variables and write down the given rates and the unknown rate (e.g. "$\dfrac{dh}{dt}=-2$ cm/day, find $\dfrac{dV}{dt}$").
2. Write an equation relating the quantities (often a geometric or volume formula).
3. Differentiate both sides with respect to $t$ (chain rule) – before substituting numbers.
4. Substitute the known values at the instant of interest, and solve for the unknown rate.
5. State the answer with units and the correct sign (a decreasing quantity has a negative rate).

Worked example. Air is pumped into a spherical balloon so its volume grows at $\dfrac{dV}{dt}=100\ \text{cm}^3/\text{s}$. How fast is the radius growing when $r=5\ \text{cm}$? Start from $V=\tfrac43\pi r^3$ and differentiate with respect to $t$ first:

Substituting $r=5$ too early is the classic error – differentiate the general relationship, then plug in.

Near a point of tangency, a smooth curve looks like its tangent line – this is local linearity 局部线性. So the tangent line gives a linear approximation 线性近似 (linearization) of the function near that point:

Use it to estimate $f$ at an $x$ close to $a$.

Worked example. Estimate $\sqrt{4.1}$. Take $f(x)=\sqrt{x}$ and $a=4$: $f(4)=2$ and $f'(x)=\dfrac{1}{2\sqrt{x}}$ so $f'(4)=\dfrac14$. Then $L(4.1)=2+\tfrac14(4.1-4)=2.025$ (the true value is $2.0248\ldots$). Because $\sqrt{x}$ is concave down, the tangent lies above the curve, so this is a slight overestimate 高估 – justify over/under with the sign of $f''$.

When direct substitution in a quotient of limits gives the indeterminate form 未定式 $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, you may use L'Hospital's Rule 洛必达法则:

provided the right-hand limit exists. Differentiate the top and bottom separately (this is not the quotient rule), then try the limit again. First confirm the form really is $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ – applying the rule to any other form is a mistake.

Worked example. Evaluate $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}$. Substituting gives $\tfrac00$, so differentiate top and bottom: $\displaystyle\lim_{x\to 0}\frac{\cos x}{1}=\cos 0=1$ – confirming the famous limit from Unit 1.

• In motion problems: velocity is the derivative of position, acceleration the derivative of velocity; speed increases when velocity and acceleration share a sign.
• For related rates, differentiate the relating equation with respect to time, then substitute the given values last.
• Use the tangent line for a linear approximation near a known point; it is accurate only close by.
• Read the sign of a rate: positive means the quantities move together, negative means opposite.
• Always state units and interpret the answer in context.

The Mean Value Theorem 中值定理 (MVT) links the average rate of change to an instantaneous one:

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is at least one point $c$ in $(a,b)$ where

$$f'(c) = \frac{f(b)-f(a)}{b-a}.$$

In words: somewhere inside the interval, the instantaneous rate equals the average rate. Geometrically, some tangent line is parallel to the line joining the endpoints.

Exam skill. Like the IVT, the MVT is an existence theorem, and questions ask you to justify. Full credit needs: (1) state $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$; (2) compute the average rate $\frac{f(b)-f(a)}{b-a}$; (3) conclude "by the MVT there is a $c$ in $(a,b)$ with $f'(c)$ equal to that value." Both hypotheses must be named.

Worked example. Verify the MVT for $f(x)=x^2$ on $[1,3]$. It is a polynomial, so continuous and differentiable everywhere. The average rate is $\dfrac{f(3)-f(1)}{3-1}=\dfrac{9-1}{2}=4$. Setting $f'(c)=2c=4$ gives $c=2$, which does lie in $(1,3)$ – the guaranteed point.

The Extreme Value Theorem 极值定理 (EVT) guarantees extremes exist: a function continuous on a closed interval $[a,b]$ attains both an absolute maximum and an absolute minimum on it.

A critical point 临界点 is an interior point where $f'(x)=0$ or $f'(x)$ does not exist. All local (relative) extrema 局部极值 occur at critical points – but not every critical point is an extremum. So critical points are the candidates; you must test each.

The first derivative tells you where $f$ rises or falls:

• $f'(x) > 0$ on an interval $\Rightarrow$ $f$ is increasing 递增 there;
• $f'(x) < 0$ $\Rightarrow$ $f$ is decreasing 递减.

On the exam, "find the intervals where $f$ is increasing" means: find the critical points, then test the sign of $f'$ between them, and justify with the sign of $f'$ (a stated reason, not just an interval).

To classify a critical point $x=c$ as a local max, local min, or neither, check how $f'$ changes sign there:

• $f'$ changes $+$ to $-$ at $c$ $\Rightarrow$ local maximum 极大值;
• $f'$ changes $-$ to $+$ at $c$ $\Rightarrow$ local minimum 极小值;
• $f'$ does not change sign $\Rightarrow$ neither.

Always state the sign change as your justification.

Worked example. Classify the critical points of $f(x)=x^3-3x^2$. Here $f'(x)=3x^2-6x=3x(x-2)$, zero at $x=0$ and $x=2$. Testing signs: $f'>0$ for $x<0$, $f'<0$ on $(0,2)$, and $f'>0$ for $x>2$. So $f'$ turns $+\to-$ at $x=0$ (a local maximum, $f(0)=0$) and $-\to+$ at $x=2$ (a local minimum, $f(2)=-4$).

On a closed interval, absolute (global) extrema occur only at critical points or endpoints. The candidates test:

1. List all critical points in $[a,b]$ and the two endpoints.
2. Evaluate $f$ at each candidate.
3. The largest output is the absolute maximum; the smallest is the absolute minimum.

Show the table of values – the comparison is the argument.

The second derivative describes bending:

• $f'' > 0$ $\Rightarrow$ $f$ is concave up 上凹 (curving like a cup; $f'$ is increasing);
• $f'' < 0$ $\Rightarrow$ $f$ is concave down 下凹 ($f'$ is decreasing).

A point of inflection 拐点 is where concavity changes, i.e. where $f''$ changes sign (not merely where $f''=0$). Report its $x$-coordinate and justify with the sign change of $f''$.

An alternative way to classify a critical point $c$ where $f'(c)=0$:

• $f''(c) > 0$ $\Rightarrow$ concave up $\Rightarrow$ local minimum;
• $f''(c) < 0$ $\Rightarrow$ concave down $\Rightarrow$ local maximum;
• $f''(c) = 0$ $\Rightarrow$ the test is inconclusive – fall back on the first derivative test.

Special case: if a continuous function has only one critical point on an interval and it is a local extremum, that point is also the absolute extremum there.

Key features of $f$, $f'$, and $f''$ mirror each other. To sketch or read graphs:

• $f$ increasing $\Leftrightarrow$ $f'$ above the axis; $f$ has a local max $\Leftrightarrow$ $f'$ crosses from $+$ to $-$.
• $f$ concave up $\Leftrightarrow$ $f'$ increasing $\Leftrightarrow$ $f''$ above the axis; $f$ has an inflection point $\Leftrightarrow$ $f'$ has a local extremum $\Leftrightarrow$ $f''$ crosses zero.

This is the skill of reading one graph to describe another. A very common exam setup gives the graph of $f'$ and asks about $f$: where is $f$ increasing (where $f'>0$), where are $f$'s extrema (where $f'$ crosses zero, with a sign change), where is $f$ concave up (where $f'$ is increasing). Answer questions about $f$ using the height and slope of the $f'$ graph.

Optimization 最优化 uses the derivative to find the largest or smallest value of a quantity on an interval. It is the candidates/derivative-test machinery applied to a real goal.

A dependable procedure:

1. Write the quantity to optimize as a function of one variable (use a constraint equation to eliminate extras).
2. State the interval of allowed inputs.
3. Find critical points ($f'=0$ or undefined) and test them (first- or second-derivative test, or candidates test if the interval is closed).
4. Answer the question asked, with units and interpretation in context – the maximum area, the minimum cost, etc.

Worked example. A farmer has $100\ \text{m}$ of fence for a rectangular pen against a wall, so only three sides need fencing. Maximize the area. Let the two ends be $x$ and the side parallel to the wall be $y$; the constraint is $2x+y=100$, so $y=100-2x$. The area is

Since $A''=-4<0$ this is a maximum, giving $y=50$ and $A=25\times50=1250\ \text{m}^2$.

All of this extends to implicitly defined relations. A critical point of an implicit relation is where $\dfrac{dy}{dx}=0$ (horizontal tangent) or is undefined (vertical tangent). Because $\dfrac{dy}{dx}$ is usually a relation in $x$ and $y$, and the second derivative involves $x$, $y$, and $\dfrac{dy}{dx}$, substitute your first-derivative expression back in when finding $\dfrac{d^2y}{dx^2}$, then reason about concavity from its sign.

• $f'>0$ means increasing, $f'<0$ decreasing; candidates for extrema are where $f'=0$ or is undefined.
• Classify a critical point with the first-derivative sign change or the second-derivative test ($f''>0$ minimum, $f''<0$ maximum).
• $f''>0$ is concave up, $f''<0$ concave down; a point of inflection is where concavity changes ($f''$ changes sign).
• For an absolute extremum on a closed interval, also check the endpoints.
• Justify every conclusion by citing the sign of $f'$ or $f''$ — the exam demands the reasoning, not just the answer.

Differentiation found rates. Integration runs the idea in reverse: given a rate of change, it finds the accumulated change 累积变化. The key picture: the area between the graph of a rate function and the $x$-axis gives the total accumulation.

• If the rate is positive over an interval, the accumulated change is positive; if negative, negative. Area below the axis counts as negative.
• Simple regions (triangles, rectangles) can be found with geometry 几何.
• Units: the area's unit is the rate's unit times the input's unit. A rate in vehicles-per-hour times hours gives vehicles.

When exact area is hard, approximate it with a Riemann sum 黎曼和 – split the interval into subintervals and add up rectangle (or trapezoid) areas. The four standard estimates:

• Left Riemann sum – rectangle height from the left endpoint of each subinterval.
• Right Riemann sum – height from the right endpoint.
• Midpoint Riemann sum – height from the midpoint 中点.
• Trapezoidal sum 梯形法 – average the two endpoint heights (a trapezoid).

Subintervals may be uniform (equal width) or nonuniform – read widths from the table.

Over- or underestimate? Judge from the behavior of the function: for an increasing function, a left sum underestimates and a right sum overestimates; a trapezoidal sum overestimates when the function is concave up and underestimates when concave down. Exam parts ask you to state which and why.

Worked example. A table gives $f(0)=3$, $f(2)=5$, $f(4)=8$, $f(6)=9$. Estimate $\int_0^6 f(x)\,dx$ with a right Riemann sum of three equal subintervals ($\Delta x=2$). Use the right endpoint of each strip:

Since $f$ is increasing, this right sum is an overestimate; the left sum $2(3+5+8)=32$ would be an underestimate.

As the subinterval widths shrink to zero, the Riemann sum approaches an exact value – the definite integral 定积分:

Here $\Delta x_i$ is the width of the $i$th subinterval and $x_i^{*}$ a point inside it. So a definite integral is the limit of a Riemann sum, and you should be able to translate each into the other.

A definite integral with a variable upper limit defines a new accumulation function 累积函数. The Fundamental Theorem of Calculus 微积分基本定理 (first part) says differentiation undoes this accumulation: if $f$ is continuous, then

So if $g(x)=\int_a^x f(t)\,dt$, then $g'(x)=f(x)$ and $g''(x)=f'(x)$. This is the engine behind the very common "let $g(x)=\int_a^x f(t)\,dt$" questions.

Because $g'(x)=f(x)$, everything from Unit 5 applies to an accumulation function using the graph of $f$:

• $g$ is increasing where $f>0$ and decreasing where $f<0$;
• $g$ has a local extremum where $f$ crosses zero (with a sign change);
• $g$ is concave up where $f$ is increasing; inflection points of $g$ occur where $f$ has a local extremum.

To get a value of $g$, compute the signed area: $g(x)=\int_a^x f(t)\,dt$, adding areas above the axis and subtracting areas below.

These properties simplify computation and appear constantly:

The last (splitting at an interior point $c$) lets you build a total integral from pieces read off a graph.

The second part of the Fundamental Theorem evaluates a definite integral using an antiderivative 原函数. If $F'=f$, then

So: find any antiderivative $F$, then subtract its values at the two limits. This is how most exact integrals are computed. It also gives the net change view: $\int_a^b g'(t)\,dt = g(b)-g(a)$, so a starting value plus accumulated change gives a later value, e.g. $g(5)=g(0)+\int_0^5 g'(t)\,dt$.

Worked example. Evaluate $\int_1^3 (2x+1)\,dx$. An antiderivative is $F(x)=x^2+x$, so

An indefinite integral 不定积分 is the family of all antiderivatives, written with a constant of integration 积分常数:

Reverse each derivative rule to build the basic antiderivatives:

$u$-substitution 换元积分法 reverses the chain rule. Choose an inside function $u=g(x)$, so $du=g'(x)\,dx$, and rewrite the integral entirely in $u$:

Look for a function and its derivative both present. For a definite integral, either change the limits to $u$-values or convert back to $x$ before substituting the original limits.

Worked example. Evaluate $\int 2x\cos(x^2)\,dx$. The inside function is $u=x^2$, whose derivative $2x\,dx=du$ is present, so

Spotting that $2x$ is exactly $\dfrac{du}{dx}$ is the whole trick.

Two algebraic set-up moves let more integrals fit the basic forms: polynomial long division 多项式长除法 when the top degree is $\ge$ the bottom degree of a rational function, and completing the square 配方法 to turn a quadratic denominator into a form that integrates to an arctangent or logarithm.

A skill topic: match the integral to a method. Try a basic antiderivative first; look for a $u$-substitution (an inside function whose derivative is also present); use algebra (division, completing the square, splitting a fraction) to reshape the integrand into a standard form. Naming the structure first prevents wasted effort.

• Integration is antidifferentiation; use the power rule $\int x^n\,dx=\tfrac{x^{n+1}}{n+1}+C$ and don't forget the $+C$.
• The Fundamental Theorem links the two: $\int_a^b f'(x)\,dx=f(b)-f(a)$, and $\tfrac{d}{dx}\int_a^x f(t)\,dt=f(x)$.
• Approximate a definite integral with Riemann sums or the trapezoidal rule from a table of values.
• A definite integral is a signed area (below the axis counts negative); split at sign changes for total area.
• Use u-substitution and remember to change the limits (or back-substitute) accordingly.

A differential equation 微分方程 is an equation that relates a function to its own derivatives. It describes a situation by its rate of change. For example, "the rate of change of a quantity is proportional to its size" becomes

Learning to translate a sentence about a rate into a differential equation is the first skill of this unit.

A solution 解 of a differential equation is a function that makes it true. You can verify a proposed solution by differentiating it and substituting into the equation: if both sides match, it is a solution. Note a differential equation usually has infinitely many solutions – a whole family – differing by a constant.

A slope field 斜率场 draws the differential equation as a grid of short segments: at each point $(x,y)$ the segment has slope $\dfrac{dy}{dx}$ evaluated there. To sketch one, plug several points into the right-hand side and draw a small segment with that slope at each. The picture shows the shape of the solution curves without solving.

Exam skill. A common part gives a portion of a slope field and asks you to sketch the particular solution through a given point – start at the point and follow the segments, staying tangent to them.

Solutions to a differential equation are functions or families of functions. Read a slope field to reason about behavior: where the segments are flat ($\tfrac{dy}{dx}=0$) the solution is momentarily level; where they steepen the solution rises or falls faster; horizontal rows of equal slope suggest the rate depends only on $y$ (or only on $x$).

Many exam differential equations are solved by separation of variables 分离变量法. If $\dfrac{dy}{dx}$ factors into a function of $x$ times a function of $y$, move all $y$'s to one side and all $x$'s to the other, then integrate both sides:

Add the constant of integration once (on the $x$-side). This gives the general solution 通解.

Worked example. Solve $\dfrac{dy}{dx}=xy$ with the initial condition $y(0)=2$. Separate and integrate:

Applying $y(0)=2$ gives $A=2$, so the particular solution is $y=2e^{x^2/2}$.

An initial condition 初始条件 – a known point $(x_0,y_0)$ – pins down one curve from the family. Substitute it to solve for the constant $C$; the result is the particular solution 特解. There is exactly one solution through a given point. Watch for domain restrictions 定义域限制: keep the branch that contains the initial point (for example, the correct sign of a square root).

The most important model is exponential growth and decay 指数增长与衰减. The equation $\dfrac{dy}{dt}=ky$ (rate proportional to size) has the solution:

where $y_0$ is the value at $t=0$. Here $k>0$ gives growth and $k<0$ gives decay. You can derive this by separation of variables ($\int \tfrac{dy}{y} = \int k\,dt$), and it models processes like population growth, radioactive decay, and Newton's law of cooling. A follow-up part may ask for $\dfrac{d^2y}{dt^2}$: differentiate $\dfrac{dy}{dt}=ky$ again (using $\dfrac{dy}{dt}=ky$) to express it in terms of $y$.

Worked example. A sample decays by $\dfrac{dy}{dt}=-0.10\,y$ (in years) from $y_0=50\ \text{g}$. The solution is $y=50e^{-0.10t}$, so after $10$ years $y=50e^{-1}=18.4\ \text{g}$. Its half-life solves $25=50e^{-0.10t}$, i.e. $e^{-0.10t}=\tfrac12$, giving $t=\dfrac{\ln 2}{0.10}=6.9\ \text{years}$ – independent of the starting amount.

• Solve a separable equation by getting all $y$ on one side and all $x$ on the other, then integrating both sides (add $+C$ once).
• Use the initial condition to find $C$ (a particular solution).
• Sketch or read a slope field: the little segments show $\tfrac{dy}{dx}$ at each point, and a solution curve follows them.
• Recognise exponential models $\tfrac{dy}{dt}=ky\Rightarrow y=Ce^{kt}$ (growth/decay).
• A differential equation gives the slope — you must integrate to recover the function.