AP Calculus AB — Cheatsheet
Formulas, exam-day tips, and key terms on one page.
Formulas & relationships
Existence of a limit
lim(x→a) f(x) = L ⟺ lim(x→a⁻) f(x) = L AND lim(x→a⁺) f(x) = L
Both one-sided limits must exist and be equal. If they disagree, the two-sided limit does not exist.
Squeeze Theorem
If g(x) ≤ f(x) ≤ h(x) near a and lim(x→a) g(x) = lim(x→a) h(x) = L, then lim(x→a) f(x) = L
The classic use: since −x² ≤ x²·sin(1/x) ≤ x² and both bounds → 0, the middle → 0 as well.
A special trig limit
lim(x→0) sin(x)/x = 1
Proved with the Squeeze Theorem. Memorize it; it underlies the derivative of sin(x). Note x must be in radians.
End behavior of a rational function
lim(x→±∞) (aₙxⁿ + … )/(bₘxᵐ + … ) = 0 if n < m; aₙ/bₘ if n = m; ±∞ if n > m
Only the leading terms survive at infinity. Degrees equal ⇒ divide the leading coefficients.
Continuity at a point
f continuous at a ⟺ f(a) defined AND lim(x→a) f(x) exists AND lim(x→a) f(x) = f(a)
All three conditions. The IVT requires this at every point of a closed interval [a, b].
Definition of the derivative
f′(x) = lim(h→0) [f(x + h) − f(x)] / h
Equivalent form at a point: f′(a) = lim(x→a) [f(x) − f(a)]/(x − a). Both compute the tangent slope.
Power rule
d/dx[xⁿ] = n·xⁿ⁻¹
Works for every constant n. Special cases: d/dx[c] = 0 and d/dx[x] = 1.
Linearity of the derivative
d/dx[c·f(x)] = c·f′(x) and d/dx[f(x) ± g(x)] = f′(x) ± g′(x)
Constants pass through; sums and differences split apart. Differentiate a polynomial term by term.
Product rule
d/dx[f·g] = f′·g + f·g′
Derivative of first times second, plus first times derivative of second. Not f′·g′.
Quotient rule
d/dx[f/g] = (f′·g − f·g′) / g²
Low d-high minus high d-low, all over low squared. The order of subtraction is essential.
Chain rule
d/dx[f(g(x))] = f′(g(x)) · g′(x)
Outer derivative (inside untouched) times inner derivative. In Leibniz form, dy/dx = dy/du · du/dx.
Core trig derivatives
d/dx[sin x] = cos x · d/dx[cos x] = −sin x · d/dx[tan x] = sec²x
Also: d/dx[sec x] = sec x·tan x, d/dx[csc x] = −csc x·cot x, d/dx[cot x] = −csc²x.
Chain rule on y-terms
d/dx[yⁿ] = n·yⁿ⁻¹ · (dy/dx)
Treat y as a function of x. Any term containing y produces a dy/dx factor; terms in x alone do not.
Derivative of an inverse function
If g = f⁻¹, then g′(a) = 1 / f′(g(a))
The inverse’s slope at a is the reciprocal of f’s slope at the matching point b = g(a) where f(b) = a.
Key inverse trig derivatives
d/dx[arcsin x] = 1/√(1 − x²) · d/dx[arctan x] = 1/(1 + x²)
And d/dx[arccos x] = −1/√(1 − x²). Chain rule applies for arcsin(u): u′/√(1 − u²).
Second derivative and motion
v(t) = s′(t) · a(t) = s″(t) = v′(t)
Velocity is the first derivative of position; acceleration is the second derivative of position (the derivative of velocity).
Exponential and log derivatives
d/dx[eˣ] = eˣ · d/dx[ln x] = 1/x · d/dx[aˣ] = aˣ·ln a
With the chain rule: d/dx[e^u] = e^u·u′ and d/dx[ln u] = u′/u.
Differentiating a relationship in time
If V = (4/3)πr³, then dV/dt = 4πr² · (dr/dt)
Every variable carries a d/dt. The chain rule links dV/dt to dr/dt through the geometry.
Linear approximation
L(x) = f(a) + f′(a)·(x − a)
The tangent line at x = a. Use it to estimate f(x) for x near a. Concave up ⇒ underestimate; concave down ⇒ overestimate.
L’Hôpital’s Rule
If lim f(x)/g(x) is 0/0 or ∞/∞, then lim f(x)/g(x) = lim f′(x)/g′(x)
Differentiate numerator and denominator separately — not with the quotient rule. Re-check the form after each application.
Motion relationships
v(t) = s′(t) · a(t) = v′(t) · speed = |v(t)|
Speeding up ⇔ v and a share a sign. Direction changes where v = 0 and switches sign.
Mean Value Theorem
f′(c) = [f(b) − f(a)] / (b − a) for some c in (a, b)
Requires f continuous on [a, b] and differentiable on (a, b). The right side is the average rate of change (secant slope).
First Derivative Test
f′ changes + to − at c ⇒ local max; f′ changes − to + at c ⇒ local min
No sign change ⇒ not an extremum. Critical points are where f′ = 0 or f′ is undefined (and c is in the domain).
Concavity and the Second Derivative Test
f″ > 0 ⇒ concave up (local min if f′ = 0); f″ < 0 ⇒ concave down (local max if f′ = 0)
Inflection point where f″ changes sign. If f″(c) = 0 at a critical point, the test is inconclusive.
Optimization strategy
objective Q(x) ← substitute constraint → optimize Q′(x) = 0 → test candidates and endpoints
A closed-interval extremum occurs at a critical point or an endpoint. Always justify which candidate wins.
Riemann sum and the integral
∫[a to b] f(x) dx = lim(n→∞) Σ f(xᵢ*)·Δx, where Δx = (b − a)/n
A left sum takes xᵢ* at left endpoints, a right sum at right endpoints. The limit is the exact net signed area.
Reverse power rule
∫xⁿ dx = xⁿ⁺¹/(n + 1) + C (n ≠ −1)
For n = −1, ∫(1/x) dx = ln|x| + C. Also ∫eˣ dx = eˣ + C and ∫cos x dx = sin x + C.
Fundamental Theorem of Calculus (Part 2)
∫[a to b] f(x) dx = F(b) − F(a), where F′ = f
Find an antiderivative, evaluate at the upper limit minus the lower limit. No + C needed for a definite integral.
U-substitution
∫ f(g(x))·g′(x) dx = ∫ f(u) du, where u = g(x), du = g′(x) dx
For definite integrals, either back-substitute to x, or change the limits: x = a, b become u = g(a), g(b).
Fundamental Theorem of Calculus (Part 1)
d/dx ∫[a to x] f(t) dt = f(x); with a variable limit, d/dx ∫[a to u(x)] f(t) dt = f(u(x))·u′(x)
Differentiating an accumulation function returns the integrand at the top limit, times the derivative of that limit.
Slope at a point
At (x₀, y₀), the segment slope is dy/dx evaluated at (x₀, y₀)
Horizontal segments where dy/dx = 0; steeper segments where |dy/dx| is large. Slope depends only on x ⇒ identical columns.
Separation of variables
dy/dx = g(x)·h(y) → ∫ (1/h(y)) dy = ∫ g(x) dx → solve for y, apply the initial condition
One constant C after integrating. Substitute the initial condition to find C, then solve for y.
Exponential growth/decay model
dy/dt = ky ⟹ y = y₀·e^(kt)
y₀ is the initial amount, k the relative rate. k > 0 grows, k < 0 decays. Half-life and doubling time follow from setting y = y₀/2 or 2y₀.
Solving for a characteristic time
Doubling: 2y₀ = y₀·e^(kt) ⟹ t = (ln 2)/k; Half-life: t = (ln 2)/|k|
The initial amount y₀ always cancels, so these times depend only on the rate constant k.
Area between curves
A = ∫[a to b] (top − bottom) dx or A = ∫[c to d] (right − left) dy
Always the larger function minus the smaller. Limits come from the intersection points of the curves.
Volume methods
Disk: V = π∫[a to b] R² dx · Washer: V = π∫[a to b] (R² − r²) dx · Cross sections: V = ∫[a to b] A(x) dx
Washers subtract the squares of the radii. For cross sections, integrate the area formula of the given shape.
Average value of a function
f_avg = (1/(b − a)) ∫[a to b] f(x) dx
The integral of f divided by the interval length. Different from the average rate of change [f(b) − f(a)]/(b − a).
Motion by integration
displacement = ∫[a to b] v(t) dt · total distance = ∫[a to b] |v(t)| dt · s(t) = s(a) + ∫[a to t] v(τ) dτ
Velocity integrates to displacement; speed |v| integrates to total distance. An initial condition pins down the position.
On the exam
- For piecewise functions, always compute the two one-sided limits separately using the correct branch, then compare. Writing "lim⁻ = lim⁺ = L, therefore the limit exists" is exactly the justification AP readers want to see.
- Reach for the Squeeze Theorem whenever you see a bounded factor (like sin or cos of something) multiplied by a factor going to 0. Bound the trig part between −1 and 1, multiply through, and let the outer limits pinch the answer.
- A horizontal asymptote describes end behavior *only* — a graph is allowed to cross its horizontal asymptote at finite x-values. Do not claim a curve "can never touch" y = L; that restriction applies to vertical asymptotes, not horizontal ones.
- A full-credit IVT argument has three moving parts: state that f is continuous on [a, b], compute both endpoint values to show the target lies between them, then conclude "by the IVT" a c exists. Skipping the continuity sentence loses the point even when the arithmetic is right.
- When a problem says "using the definition of the derivative", you must show the limit of the difference quotient — the shortcut rules earn no credit on that specific prompt. Look for the h → 0 phrasing as your cue.
- Watch the sign and the new exponent carefully with negative powers. d/dx[x⁻²] = −2x⁻³ — the coefficient is negative *and* the exponent becomes more negative. Rushing this step is a frequent multiple-choice trap.
- Before reaching for the quotient rule, ask whether the denominator is a single term you can divide out. Rewriting (x³ + x)/x as x² + 1 turns a quotient-rule problem into a one-line power-rule problem and removes a chance to slip on signs.
- Think of the chain rule as peeling an onion: differentiate the outermost layer first, keep the inside intact, then multiply by the derivative of the inside, repeating for nested compositions. Track every layer and you will never drop a factor.
- To find a tangent slope on an implicit curve you need both coordinates, because dy/dx typically depends on x *and* y. Differentiate, solve for dy/dx, then substitute the full point — not just the x-value.
- The formula g′(a) = 1/f′(g(a)) shows up on the AP exam most often in table form: you are handed values of f and f′ and asked for the inverse’s derivative. Identify the matching point from the table, then take the reciprocal.
- On motion problems, read the question’s verb carefully: "how fast" usually means velocity (first derivative), while "how is the velocity changing" means acceleration (second derivative). Matching the derivative order to the phrasing is worth easy points.
- When several rules stack, name the outermost operation before computing anything. "This is a product, whose second factor is a composite" gives you a plan and keeps the chain-rule factors from getting lost in a long expression.
- AP related-rates responses must show the differentiated equation with the d/dt factors before any numbers appear. Readers look for that general relationship; a correct final number with no setup earns little credit.
- To justify over- versus underestimate, cite concavity explicitly: "f″ < 0 on the interval, so the graph is concave down and the tangent line overestimates." A bare "overestimate" with no reason to concavity does not earn the justification point.
- Always write the phrase "form is 0/0" (or ∞/∞) before applying L’Hôpital on free response. Stating the indeterminate form is part of a valid justification; jumping straight to differentiating the pieces can cost the point even with a correct answer.
- For "speeding up or slowing down", always compare the signs of v and a at that instant. Same sign ⇒ speeding up; opposite signs ⇒ slowing down. The sign of acceleration by itself tells you nothing about speed.
- To use the MVT on the exam, first state that f is continuous on [a, b] and differentiable on (a, b), then compute the average rate and set f′(c) equal to it. Solving for c without confirming the hypotheses loses justification credit.
- A critical point is only a *candidate* for an extremum — you must show a sign change in f′ to confirm it. Where f′ = 0 but does not change sign (as at x = 0 for y = x³), there is no local max or min.
- When the Second Derivative Test returns f″(c) = 0, do not guess — state that the test is inconclusive and switch to the First Derivative Test, examining the sign change of f′ around c to classify the point.
- Optimization free-response points hinge on justification: after finding a critical point, state a reason it gives the extreme value (sign of f′ or f″), and on a closed domain compare it against the endpoint values. A number with no justification rarely earns full credit.
- AP problems often give data in a table and ask for a specific Riemann sum. Read carefully whether left, right, or midpoint endpoints are required, and note that subintervals may have *unequal* widths — compute each rectangle’s width separately then.
- Verify any antiderivative by differentiating it back — if F′ returns the integrand, F is correct. This quick check catches reverse-power-rule slips (like forgetting to divide by the new exponent) before they cost points.
- On the exam, converting the limits to u-values is usually faster and less error-prone than back-substituting. Write the new limits explicitly next to the integral so you evaluate the u-antiderivative at the correct endpoints.
- For "how much is there at time b" problems, use final value = initial value + ∫[a to b] rate dt. Integrating the rate gives only the *change*; you must add the starting amount to get the total. This structure appears on nearly every AP integral free-response.
- When asked to sketch a solution curve on a given slope field, start at the initial point and keep your curve tangent to the nearby segments, never crossing them at odd angles. Follow the flow smoothly in both directions from the starting point.
- Free-response separable-equation problems award points for the separated equation, the antiderivatives of both sides, and finding C from the initial condition. Show each stage explicitly — a correct final formula with no work in between leaves points on the table.
- For half-life or doubling-time questions, set y = y₀/2 (or 2y₀) in y = y₀·e^(kt); the y₀ cancels, leaving an equation you solve with a natural log. The half-life depends only on k, never on the starting amount.
- When a problem gives a data point to find k, substitute it into y = y₀·e^(kt), isolate the exponential, and take a natural log — do not try to guess k. Carry enough decimal places through so the final rounded answer stays accurate.
- Before integrating, sketch the two curves and their intersections to see which is on top and whether they cross inside the interval. That quick picture prevents both limit errors and top/bottom mix-ups that graders penalize.
- Identify the cross-section shape before writing the integral: disks and washers bring a π (they are circles), but squares, triangles, and rectangles do not. Slapping a π on a non-circular cross section is a classic AP volume error.
- When a free-response question asks for the "average value" of a quantity, use the integral formula and divide by the interval length. If it asks for the "average rate of change", use the endpoint difference over the length instead — the two prompts want different computations.
- Watch the wording: "displacement" or "position change" means ∫v dt, while "total distance traveled" means ∫|v| dt with the interval split at sign changes. And "where is the particle" always needs the initial position added to the integral.
How to get a 5
- On calculator-active FRQs, write the setup integral/equation before using the calculator, and store answers to 3 decimals — bare answers lose points.
- Justify with the specific theorem or sign analysis: extrema cite f' sign change, concavity cites f'', and cite continuity/differentiability for MVT/IVT.
- Include units and interpret in context; remember total distance uses ∫|v| dt while displacement uses ∫v dt — a very common trap.
- Know FTC accumulation functions cold: g(x) = ∫ₐˣ f(t)dt problems appear yearly — g' = f and g'' = f' drive the reasoning.
Key terms
Limit definition of the derivative — f'(x) = lim(h→0) [f(x+h) − f(x)]/h.
Power, product, and quotient rules — Power: d/dx[xⁿ] = nxⁿ⁻¹. Product: (fg)' = f'g + fg'. Quotient: (f/g)' = (f'g − fg')/g².
Chain rule — d/dx[f(g(x))] = f'(g(x))·g'(x).
Key derivatives — d/dx[sin x] = cos x; [cos x] = −sin x; [tan x] = sec²x; [eˣ] = eˣ; [ln x] = 1/x.
Fundamental Theorem of Calculus — Part 2: ∫ₐᵇ f dx = F(b) − F(a). Part 1: d/dx ∫ₐˣ f(t)dt = f(x).
Mean Value Theorem — If f is continuous on [a,b] and differentiable on (a,b), some c has f'(c) = [f(b)−f(a)]/(b−a).
Intermediate Value Theorem — If f is continuous on [a,b] and N is between f(a) and f(b), some c in (a,b) has f(c) = N.
Motion relationships — v = x'; a = v'. Displacement = ∫v dt; total distance = ∫|v| dt. Speeding up when v and a share a sign.
Average value of a function — f_avg = (1/(b−a)) ∫ₐᵇ f(x) dx.
L'Hôpital's Rule — If lim f/g is 0/0 or ∞/∞, then lim f/g = lim f'/g'.
Disk/washer volume (about x-axis) — Disk: V = π∫ R² dx. Washer: V = π∫ (R² − r²) dx (outer² − inner²).
Continuity at x = a — f(a) exists, lim(x→a) f(x) exists, and they are equal.