Your FREE AP Calculus AB Flashcards 2026 – 200+ Cards

The single value $f(x)$ approaches as $x \to a$. It can exist even when $f(a)$ is undefined.

(1) $f(a)$ is defined, (2) $\lim_{x \to a} f(x)$ exists, and (3) the limit equals $f(a)$.

The left-hand and right-hand limits are equal: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$.

$\lim_{x \to 0} \dfrac{\sin x}{x} = 1$ — a key special limit.

$\lim_{x \to 0} \dfrac{1 - \cos x}{x} = 0$.

A 'hole': the limit exists but doesn't equal $f(a)$ (or $f(a)$ is undefined). Often from a factor that cancels.

The left- and right-hand limits both exist but disagree, so the two-sided limit does not exist.

The function approaches $\pm\infty$ at the point — a vertical asymptote.

If $g(x) \le f(x) \le h(x)$ near $a$ and $\lim g = \lim h = L$, then $\lim f = L$.

Compare degrees: lower numerator → 0; equal → ratio of leading coefficients; higher numerator → $\pm\infty$.

If $f$ is continuous on $[a,b]$, it takes every value between $f(a)$ and $f(b)$; opposite signs force a root between.

When any of the three conditions fails — at a removable (hole), jump, or infinite discontinuity.

Use algebra first — factor and cancel, multiply by a conjugate, or simplify a complex fraction — then substitute.

$y = L$ is a horizontal asymptote if $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$.

$x = a$ is a vertical asymptote if $f(x) \to \pm\infty$ as $x \to a$ from one or both sides.

If $f$ is continuous at $a$, then $\lim_{x \to a} f(x) = f(a)$ — just plug in.

$\lim_{x \to a} c = c$ — a constant function's limit is the constant.

Every polynomial is continuous for all real $x$; rational functions are continuous except where the denominator is 0.

For large $|x|$, a polynomial behaves like its leading term, which sets the limits at $\pm\infty$.

$\lim_{x \to -\infty} e^x = 0$; and $\lim_{x \to \infty} e^x = \infty$.

Continuous at the boundary if the two pieces' one-sided limits match the defined value there.

A limit describes where the function is headed; it can exist even where $f$ is undefined or differs.

Limits distribute over sums, products, and quotients (denominator limit ≠ 0), so split a complex limit into parts.

Continuity is a required hypothesis for all four major theorems — always check it.

$\dfrac{d}{dx} x^n = n x^{n-1}$ for any real exponent $n$.

$f'(a) = \lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h}$ — the slope of the tangent line at $x = a$.

$\dfrac{f(a+h) - f(a)}{h}$ — the average rate of change (a secant slope). Its limit is the derivative.

The slope of the line tangent to the graph of $f$ at $x = a$ — the instantaneous rate of change.

$\dfrac{d}{dx} c = 0$ — the derivative of a constant is zero.

$\dfrac{d}{dx} [c f(x)] = c f'(x)$.

$(f \pm g)' = f' \pm g'$ — differentiate term by term.

$(f g)' = f' g + f g'$.

$\left( \dfrac{f}{g} \right)' = \dfrac{f' g - f g'}{g^2}$ — 'low d-high minus high d-low, over low squared.'

$\dfrac{d}{dx} e^x = e^x$ — its own derivative.

$\dfrac{d}{dx} \ln x = \dfrac{1}{x}$ (for $x > 0$).

$\dfrac{d}{dx} \sin x = \cos x$.

$\dfrac{d}{dx} \cos x = -\sin x$.

$\dfrac{d}{dx} \tan x = \sec^2 x$.

$\dfrac{d}{dx} \sec x = \sec x \tan x$.

$\dfrac{d}{dx} \cot x = -\csc^2 x$.

$\dfrac{d}{dx} \csc x = -\csc x \cot x$.

$\dfrac{d}{dx} a^x = a^x \ln a$ for a constant base $a > 0$.

$f''(x)$, the derivative of $f'$ — measures concavity and is acceleration in motion problems.

If $f$ is differentiable at $a$, it is continuous at $a$. The converse is false.

At a corner, a cusp, a vertical tangent, or any discontinuity — even if $f$ is continuous there.

Convert roots and reciprocals to powers first: $\sqrt{x} = x^{1/2}$, $\dfrac{1}{x^2} = x^{-2}$.

$\dfrac{d}{dx}(mx + b) = m$ — the slope is constant.

Use the difference quotient of two nearby points: $\dfrac{f(x_2) - f(x_1)}{x_2 - x_1}$.

$y = f(a) + f'(a)(x - a)$ — point-slope form using the derivative as slope.

The line perpendicular to the tangent at a point; its slope is $-\dfrac{1}{f'(a)}$.

$f'(x), f''(x), f'''(x), f^{(4)}(x), \dots$ — repeatedly differentiate.

$\dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2\sqrt{x}}$.

$\dfrac{d}{dx} \dfrac{1}{x} = -\dfrac{1}{x^2}$.

Average = secant slope over $[a,b]$; instantaneous = tangent slope = derivative at a point.

Constants factor straight through: $(cf)' = c f'$ and $\int cf\,dx = c\int f\,dx$.

$f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$ gives the derivative as a function.

$\dfrac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$ — outer derivative at the inner, times the inner derivative.

Derivative of the outside (leaving the inside alone) times the derivative of the inside.

$\cos(g(x)) \cdot g'(x)$ — e.g. $\dfrac{d}{dx}\sin(4x) = 4\cos(4x)$.

$e^{g(x)} \cdot g'(x)$ — e.g. $\dfrac{d}{dx} e^{3x^2} = 6x\,e^{3x^2}$.

$\dfrac{g'(x)}{g(x)}$ — the inner derivative over the inner function.

$n[g(x)]^{n-1} \cdot g'(x)$ — the power rule with the chain rule.

Differentiate both sides in $x$, attaching $\dfrac{dy}{dx}$ to every $y$-term, then solve for $\dfrac{dy}{dx}$.

Because $y$ is treated as a function of $x$, so the chain rule applies whenever you differentiate a $y$-term.

If $g = f^{-1}$, then $g'(x) = \dfrac{1}{f'(g(x))}$.

$\dfrac{d}{dx} \arcsin x = \dfrac{1}{\sqrt{1 - x^2}}$.

$\dfrac{d}{dx} \arccos x = -\dfrac{1}{\sqrt{1 - x^2}}$.

$\dfrac{d}{dx} \arctan x = \dfrac{1}{1 + x^2}$.

$-2x \sin(x^2)$ by the chain rule (outer $-\sin$, inner $2x$).

$\dfrac{x}{\sqrt{x^2 + 1}}$ — chain rule on $(x^2+1)^{1/2}$.

$3x^2 \sec^2(x^3)$ by the chain rule.

$10(2x + 1)^4$ — bring down 5, reduce power, times inner derivative 2.

Find $\dfrac{dy}{dx}$ implicitly, then substitute the point's coordinates to get the tangent slope.

For $f(g(h(x)))$, differentiate outermost to innermost, multiplying all the inner derivatives.

Take $\ln$ of both sides before differentiating — useful for products, quotients, or variable exponents.

$\dfrac{1}{x}$ — since $\dfrac{5}{5x} = \dfrac{1}{x}$.

If $h(x) = f(g(x))$, then $h'(x) = f'(g(x)) g'(x)$; read the needed values from the table.

Differentiate implicitly, substitute the point for the slope, then write the tangent line in point-slope form.

$a^{g(x)} \ln a \cdot g'(x)$ — exponential chain rule with base $a$.

Differentiate $\dfrac{dy}{dx}$ again implicitly, substituting the first derivative where $\dfrac{dy}{dx}$ appears.

Two quantities linked by an equation change with time; differentiate the equation in $t$ and solve for the unknown rate.

Write the relating equation, differentiate with respect to $t$, THEN substitute the known values and rates.

$L(x) = f(a) + f'(a)(x - a)$ estimates $f$ near $a$.

Only when the limit is the indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ — verify first.

For an indeterminate $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, $\lim \dfrac{f}{g} = \lim \dfrac{f'}{g'}$.

$v(t) = s'(t)$ — velocity is the derivative of position.

$a(t) = v'(t) = s''(t)$ — acceleration is the second derivative of position.

The particle is at rest when $v(t) = 0$; these are candidate times for a change of direction.

Speeding up when $v$ and $a$ have the SAME sign; slowing down when their signs DIFFER.

Speed is $|v(t)|$ (always ≥ 0); velocity carries a sign giving direction.

Occurs where $v(t)$ changes sign (passes through 0), not merely where $v = 0$.

From $A = \pi r^2$, $\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$.

From $V = \tfrac{4}{3}\pi r^3$, $\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}$.

From $V = s^3$, $\dfrac{dV}{dt} = 3s^2 \dfrac{ds}{dt}$.

Use $x^2 + y^2 = L^2$; differentiate to get $2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt} = 0$.

Applying it to a limit that ISN'T indeterminate (e.g. a 0/3 form) gives a wrong answer.

$dy = f'(x)\,dx$ — the change along the tangent line, used for linear approximation.

A tangent-line estimate overestimates when the graph is concave down and underestimates when concave up.

Net change uses signed velocity; total distance uses $|v|$ (split where $v$ changes sign).

Any 'how fast is __ changing' question is a derivative with respect to time.

Best near the point of tangency; error grows as you move away and as curvature increases.

Differentiate position → velocity → acceleration; integrate acceleration → velocity → position.

Units of $f'(x)$ are units of $f$ per unit of $x$ (e.g. meters per second).

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, some $c$ gives $f'(c) = \dfrac{f(b) - f(a)}{b - a}$.

A point where $f'(x) = 0$ or $f'(x)$ is undefined — a candidate for a relative extremum.

$f$ is increasing on that interval (graph rising).

$f$ is decreasing on that interval (graph falling).

At a critical point: $f'$ changing $+ \to -$ is a relative max; $- \to +$ is a relative min.

At a critical point: $f'' > 0$ ⟹ relative min; $f'' < 0$ ⟹ relative max; $f'' = 0$ ⟹ inconclusive.

Where $f''(x) > 0$ — the graph cups upward (∪); $f'$ is increasing.

Where $f''(x) < 0$ — the graph caps downward (∩); $f'$ is decreasing.

A point where concavity changes — where $f''(x)$ changes sign.

A function continuous on a closed, bounded interval attains both an absolute max and an absolute min there.

Evaluate $f$ at all critical points AND both endpoints; the largest/smallest values are the absolute extrema.

$f$ continuous on the closed $[a,b]$ AND differentiable on the open $(a,b)$.

A special MVT: if also $f(a) = f(b)$, then some $c$ in $(a,b)$ has $f'(c) = 0$.

Express the quantity as a function of one variable, differentiate, find critical points, and test them (and endpoints).

If $f''(c) = 0$ at the critical point, it's inconclusive — use the First Derivative Test instead.

A point higher (max) or lower (min) than all nearby points; occurs only at critical points.

Possible — e.g. $f(x) = x^3$ has $f'(0) = 0$ but no extremum there (sign of $f'$ doesn't change).

Sign-analyze $f'$: positive intervals are increasing, negative intervals are decreasing.

Sign-analyze $f''$: positive intervals are concave up, negative intervals are concave down.

Where $f'$ is zero/undefined are extrema candidates; where $f'$ is large the graph is steep.

For absolute extrema on a closed interval, the only candidates are critical points and endpoints.

A max of $f$ sits where $f' = 0$ and $f'' < 0$; an inflection of $f$ sits where $f'' = 0$ changes sign.

Somewhere the instantaneous rate (tangent slope) equals the average rate (secant slope) over the interval.

Both give critical points; the second includes corners, cusps, and vertical tangents.

Local extrema are highest/lowest nearby; global (absolute) extrema are highest/lowest over the whole interval.

A concave-up graph lies above its tangent lines; a concave-down graph lies below them.

The MVT guarantees the instantaneous rate equals the average rate at some interior point $c$.

Use $f'$ sign for increase/decrease, $f' = 0$ for extrema, and $f''$ sign for concavity to draw $f$.

$\int_a^b f(x)\,dx = F(b) - F(a)$, where $F$ is any antiderivative of $f$.

$\int_a^b f(x)\,dx$ is the signed area between $f$ and the $x$-axis from $a$ to $b$.

An area approximation using rectangles whose heights come from the function at left, right, or midpoints.

For an increasing function, the left sum underestimates and the right sum overestimates the true area.

$\dfrac{d}{dx} \int_a^x f(t)\,dt = f(x)$ — differentiating an accumulation function returns the integrand.

$\int_a^b f(x)\,dx = F(b) - F(a)$, where $F$ is any antiderivative of $f$.

A function $F$ with $F' = f$. The indefinite integral $\int f\,dx = F(x) + C$.

$\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ for $n \ne -1$.

$\int \dfrac{1}{x}\,dx = \ln|x| + C$.

$\int e^x\,dx = e^x + C$.

$\int \cos x\,dx = \sin x + C$.

$\int \sin x\,dx = -\cos x + C$.

$\int \sec^2 x\,dx = \tan x + C$.

Reverse the chain rule: let $u$ be an inner function, compute $du$, and rewrite the integral in $u$.

Either change the limits to $u$-values or convert back to $x$ before evaluating.

$g(x) = \int_a^x f(t)\,dt$ stores total change; by FTC Part 1, $g'(x) = f(x)$.

$\int_a^b f'(x)\,dx = f(b) - f(a)$ — the integral of a rate gives the net change.

$\int_a^b f\,dx = -\int_b^a f\,dx$, and $\int_a^a f\,dx = 0$.

$\int_a^c f\,dx = \int_a^b f\,dx + \int_b^c f\,dx$.

$\int c f(x)\,dx = c \int f(x)\,dx$ — pull constants out.

Uses the function value at each subinterval's midpoint; often more accurate than left or right sums.

Averages left and right sums; for a concave-up function it overestimates the integral.

Area where $f < 0$ counts as negative in a definite integral.

$\int a^x\,dx = \dfrac{a^x}{\ln a} + C$ for base $a > 0,\ a \ne 1$.

$\int \dfrac{1}{1 + x^2}\,dx = \arctan x + C$.

$\dfrac{d}{dx} \int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)$ — chain rule on the limit.

Antiderivatives differ by a constant, so the general antiderivative includes an arbitrary $+ C$.

Unlike an indefinite integral (a family of functions), $\int_a^b f\,dx$ is a single value.

As the number of subintervals → ∞ (width → 0), the Riemann sum approaches $\int_a^b f\,dx$.

$\int k\,dx = kx + C$.

$u = x^2 + 1$, $du = 2x\,dx$ ⟹ $\int u^4\,du = \tfrac{1}{5}(x^2+1)^5 + C$.

Equals 0 by symmetry; an even function gives $2\int_0^a f\,dx$.

Use left, right, midpoint, or trapezoidal sums with the given values to approximate $\int f\,dx$.

Units of $\int f\,dx$ are units of $f$ times units of $x$ (e.g. gallons when $f$ is gal/min).

An equation relating a function to its derivative(s), e.g. $\dfrac{dy}{dx} = ky$.

Can be rewritten so all $y$-terms (with $dy$) are on one side and all $x$-terms (with $dx$) on the other.

Separate variables, integrate both sides (add $C$), then use an initial condition to find $C$.

$\dfrac{dy}{dt} = ky$ has solution $y = y_0 e^{kt}$; growth if $k > 0$, decay if $k < 0$.

A grid of short segments with slope $\dfrac{dy}{dx} = f(x,y)$ at each point, picturing solution curves.

A solution curve follows the nearby segment directions; segments are horizontal where $\dfrac{dy}{dx} = 0$.

The general solution has the $+C$; a particular solution fixes $C$ using an initial condition.

A given point $y(x_0) = y_0$ used to solve for the constant of integration.

From $y = y_0 e^{kt}$: doubling time solves $2 = e^{kt}$; half-life solves $\tfrac{1}{2} = e^{kt}$.

When solving for $y$ from $y^2 = \dots$, the initial condition decides whether to take the + or − root.

Where $|f(x,y)|$ is large; horizontal where $f(x,y) = 0$ and vertical-ish where it's undefined.

Substitute $y$ and its derivative back into the differential equation; both sides must match.

A quantity whose rate of change is proportional to its current amount — exponential growth or decay.

In a slope field, distinct solution curves never intersect (each point has one slope).

Rewrite as $\dfrac{1}{h(y)}\,dy = g(x)\,dx$, then integrate both sides.

Analytic solution methods on AP Calculus AB are limited to separable first-order equations.

$\int_a^b [f(x) - g(x)]\,dx$ where $f$ is the top curve and $g$ the bottom — always top minus bottom.

Find the intersection points; they are the limits $a$ and $b$ of integration.

$\dfrac{1}{b - a} \int_a^b f(x)\,dx$ — the constant height of an equal-area rectangle on $[a,b]$.

$\int_{t_1}^{t_2} v(t)\,dt$ — the net change in position (signed).

$\int_{t_1}^{t_2} |v(t)|\,dt$ — split the integral where the velocity changes sign.

$V = \pi \int R^2\,dx$ — use when the region touches the axis of rotation (no gap).

$V = \pi \int (R^2 - r^2)\,dx$ — use when a gap leaves a hollow core; $R$ outer, $r$ inner radius.

$V = \int A(x)\,dx$ — integrate the area of one cross section (square, triangle, semicircle) along the axis.

If the side length is $s(x)$, the cross-section area is $s(x)^2$; integrate that along the base.

$\int_a^b R(t)\,dt$ gives the total accumulated amount when $R$ is a rate of change.

$s(t) = s(t_0) + \int_{t_0}^{t} v(\tau)\,d\tau$ — add the net change to the starting position.

Subtracting bottom from top reversed flips the sign — area must be positive, so order matters.

Split the integral at each crossing so 'top minus bottom' stays correct on each piece.

Average VALUE integrates the function; average RATE OF CHANGE is the secant slope $\dfrac{f(b)-f(a)}{b-a}$.

When revolving $y = f(x)$ about the $x$-axis, the disk radius is $R = f(x)$, so area $= \pi f(x)^2$.

Revolve about the $y$-axis (or integrate vertical extent) by writing radii as functions of $y$ and $\int \dots dy$.

A continuous $f$ attains its average value at some $c$ in $[a,b]$.

Area between curves is the limit of summing thin vertical strips of height (top − bottom).

When displacement $\int v\,dt = 0$ over the interval, the particle ends where it began.