AP Calculus BC

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 $\lim_{x\to\infty}\dfrac{3x^2-5}{2x^2+x}$. Divide top and bottom by the highest power, $x^2$: $\dfrac{3-5/x^2}{2+1/x}\to\dfrac{3-0}{2+0}=\dfrac{3}{2}$. Because the limit is a finite number, the line $y=\tfrac{3}{2}$ is a horizontal asymptote of the graph.

• 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 $g(x)=\dfrac{\sin x}{x}$ with the quotient rule $\left(\tfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}$: take $u=\sin x$, $v=x$, giving $g'(x)=\dfrac{x\cos x-\sin x}{x^2}$. Keep the order $u'v-uv'$ in the numerator — swapping the terms flips the sign and loses the mark.

• 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)$^5$" and 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 to $x^2+y^2=25$ at $(3,4)$. Here $\dfrac{dy}{dx}=-\dfrac{3}{4}$, so the tangent line is $y-4=-\tfrac{3}{4}(x-3)$ – perpendicular to the radius, as geometry predicts.

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. If $f(2)=5$ and $f'(2)=3$, and $g$ is the inverse of $f$, then $(5,2)$ lies on $g$ and $g'(5)=\dfrac{1}{f'(2)}=\dfrac{1}{3}$.

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.

Distinguish carefully between velocity (has direction) and speed (does not) – the exam tests this exact difference.

Worked example. A particle moves with $s(t)=t^3-6t^2+9t$. Then $v(t)=3(t-1)(t-3)$, so it is at rest at $t=1$ and $t=3$ and changes direction at each. At $t=2$, $v=-3<0$ and $a(2)=6(2)-12=0$; just after, $a>0$ while $v<0$, so the particle is slowing down there.

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).

Substituting numbers too early is the classic error: differentiate the general relationship first, then plug in.

Worked example. A spherical balloon's volume grows at $\dfrac{dV}{dt}=100\ \text{cm}^3/\text{s}$. From $V=\tfrac43\pi r^3$, differentiate first: $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$. At $r=5$, $100=4\pi(25)\dfrac{dr}{dt}$, so $\dfrac{dr}{dt}=\dfrac{1}{\pi}\approx0.32\ \text{cm/s}$.

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 $a$.

Over- or underestimate? The answer follows from concavity 凹凸性. If the graph is concave up near $a$ (it curves above its tangent), the tangent-line value is an underestimate 低估. If it is concave down, the tangent line lies above the curve, giving an overestimate 高估. Exam parts test this reasoning, so justify 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. $\displaystyle\lim_{x\to0}\frac{\sin x}{x}$ gives $\tfrac00$, so differentiate top and bottom: $\displaystyle\lim_{x\to0}\frac{\cos x}{1}=1$. And $\displaystyle\lim_{x\to0}\frac{e^{2x}-1}{x}$ is also $\tfrac00$; it becomes $\displaystyle\lim_{x\to0}\frac{2e^{2x}}{1}=2$.

• 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. For $f(x)=x^2$ on $[1,3]$ the average rate is $\dfrac{9-1}{2}=4$; setting $f'(c)=2c=4$ gives $c=2$, which lies 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. For $f(x)=x^3-3x^2$, $f'(x)=3x(x-2)$ is zero at $x=0,2$. Signs give $+,-,+$, so $x=0$ is a local maximum ($f=0$) and $x=2$ a local minimum ($f=-4$).

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. With $100\ \text{m}$ of fence for a rectangular pen against a wall (only three sides fenced), let the ends be $x$ and the far side $y=100-2x$. The area $A(x)=x(100-2x)=100x-2x^2$ has $A'(x)=100-4x=0$ at $x=25$; since $A''=-4<0$ this is the maximum, giving $y=50$ and $A=1250\ \text{m}^2$.

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.

Where a derivative measures a rate, an integral 积分 measures an accumulation 累积 – a total built up from a rate. If a rate of change acts over an interval, the area between its graph and the axis gives the net accumulated change. This "area = total change" idea is the foundation of integral calculus.

A Riemann sum 黎曼和 estimates the area under a curve by adding the areas of thin rectangles. Split $[a,b]$ into subintervals and use the function's height at the left endpoint, right endpoint, or midpoint of each. A trapezoidal sum 梯形法 uses trapezoids instead, averaging the two endpoint heights – usually more accurate. With more, thinner rectangles the estimate improves.

Exam skill: be able to compute left, right, midpoint, and trapezoidal estimates from a table or graph, and state whether each over- or under-estimates based on whether the function is increasing/decreasing or concave up/down.

Writing a Riemann sum with summation notation $\sum_{k=1}^{n} f(x_k)\,\Delta x$ and letting the number of rectangles grow without bound gives the exact area – the definite integral 定积分:

The integral is the limit of Riemann sums; $a$ and $b$ are the limits of integration.

An accumulation function 累积函数 $g(x)=\int_a^x f(t)\,dt$ gives the accumulated area from $a$ up to $x$. The Fundamental Theorem of Calculus (FTC) 微积分基本定理 says its derivative is the integrand:

Differentiation and integration are inverse operations. With a variable upper limit and the chain rule, $\dfrac{d}{dx}\int_a^{u(x)} f(t)\,dt=f(u(x))\,u'(x)$.

Because $g'(x)=f(x)$, the graph of $f$ tells you everything about $g$: $g$ increases where $f>0$, decreases where $f<0$, has extrema where $f$ crosses zero, and is concave up where $f$ is increasing. Reading these connections off a graph of $f$ is a classic free-response task.

Definite integrals obey useful rules: reversing the limits negates the value ($\int_b^a=-\int_a^b$), an integral over a zero-width interval is $0$, they add over adjacent intervals ($\int_a^c=\int_a^b+\int_b^c$), and constants factor out. Use these to combine or split given integral values.

The evaluation form of the FTC computes a definite integral from an antiderivative 原函数 $F$ (where $F'=f$):

So integrating a rate of change over $[a,b]$ gives the net change in the quantity – the single most-used result in the course.

Worked example. $\displaystyle\int_1^3 (2x+1)\,dx$: an antiderivative is $F(x)=x^2+x$, so the value is $F(3)-F(1)=12-2=10$.

An indefinite integral 不定积分 $\int f(x)\,dx=F(x)+C$ is the family of all antiderivatives (hence the constant of integration $C$). Reverse each derivative rule: the power rule becomes $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$ (for $n\neq-1$), with $\int \frac1x\,dx=\ln|x|+C$, and the antiderivatives of $e^x$, $\sin x$, $\cos x$, and $\sec^2 x$ come straight from their derivatives.

u-substitution 换元积分 reverses the chain rule: choose $u=g(x)$ so that $g'(x)$ also appears, turning $\int f(g(x))g'(x)\,dx$ into $\int f(u)\,du$. Remember to convert $dx$ to $du$ and, for a definite integral, either change the limits to $u$-values or convert back to $x$ at the end.

When a rational integrand is "top-heavy" (numerator degree $\ge$ denominator degree), long division rewrites it as a polynomial plus a proper fraction you can integrate. Completing the square in a denominator turns it into a form like $u^2+a^2$, leading to an arctangent antiderivative $\frac1a\arctan\frac{u}{a}+C$.

Integration by parts 分部积分 reverses the product rule:

Choose $u$ to simplify when differentiated and $dv$ to be easy to integrate (the LIATE guide: logs, inverse-trig, algebraic, trig, exponential). It handles products like $\int x e^x\,dx$ and $\int x\ln x\,dx$, sometimes applied twice.

Worked example. For $\int x e^x\,dx$, choose $u=x$ ($du=dx$) and $dv=e^x\,dx$ ($v=e^x$):

Partial fractions 部分分式 split a rational function with a factorable denominator into a sum of simpler fractions:

each of which integrates to a logarithm. This technique is essential for the logistic differential equation in the next unit.

An improper integral 反常积分 has an infinite limit of integration or an infinite discontinuity in the integrand. Evaluate it as a limit: $\int_a^\infty f\,dx=\lim_{b\to\infty}\int_a^b f\,dx$. If the limit is a finite number the integral converges 收敛; otherwise it diverges 发散.

Worked example. $\displaystyle\int_1^\infty \frac{1}{x^2}\,dx=\lim_{b\to\infty}\left[-\frac1x\right]_1^b=\lim_{b\to\infty}\left(1-\frac1b\right)=1$, so it converges to $1$. By contrast $\int_1^\infty \frac1x\,dx$ gives $\lim_{b\to\infty}\ln b=\infty$ and diverges – the same integrand-shape can go either way.

The BC exam expects you to recognize which method fits: basic rules, substitution (a chain-rule pattern), by parts (a product), partial fractions (a factorable rational), or long division/completing the square. Being able to look at an integral and pick the right tool quickly is itself a tested skill.

• 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.