Calculus AB

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].
Existence of a two-sided 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. The value f(a) is irrelevant to whether the limit exists.
Continuity at a point
f continuous at a ⟺ f(a) defined, lim(x→a) f(x) exists, and lim(x→a) f(x) = f(a)
Three separate conditions. Name the one that fails when explaining a discontinuity.
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.
One-sided derivatives
f′(a) exists ⟺ lim(h→0⁻) [f(a+h) − f(a)]/h = lim(h→0⁺) [f(a+h) − f(a)]/h, both finite
Both one-sided difference quotients must exist, be finite, and agree. A corner fails the agreement; a cusp or vertical tangent fails finiteness.
Symmetric difference quotient
f′(a) ≈ [f(a + h) − f(a − h)] / (2h)
Uses points on both sides of a, so it is generally more accurate than a one-sided quotient with the same h. Note the divisor is 2h, not h.
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.
Implicit differentiation, worked shape
From x² + y² = 25: 2x + 2y·(dy/dx) = 0 ⟹ dy/dx = −x/y
Every y term contributes dy/dx by the chain rule. Collect those terms, factor, and divide.
Inverse function derivative
If g = f⁻¹, then g′(b) = 1 / f′(a) where f(a) = b
Reciprocal of the derivative of f, evaluated at the ORIGINAL point a, not at b. Find a first.
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.
Units of a derivative
units of dy/dx = (units of y) / (units of x)
Reading the notation literally gives the units. A second derivative has units of y per unit of x squared — for example feet per second per second.
Speeding up and slowing down
speeding up ⟺ v(t)·a(t) > 0 slowing down ⟺ v(t)·a(t) < 0
Same signs means speeding up, regardless of direction of travel. Note this is a product test — it works without deciding which sign each has.
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.
Candidates Test
On [a, b] with f continuous: compare f(a), f(b), and f(c) at every critical c in (a, b)
Evaluate f, not f′. The largest value is the absolute maximum; the smallest is the absolute minimum.
Second derivative test
If f′(c) = 0 and f″(c) > 0, f has a local minimum at c; if f′(c) = 0 and f″(c) < 0, a local maximum
If f″(c) = 0 the test is inconclusive — fall back to the first derivative test, which always works.
Justification template
[which derivative] [does what — sign or sign change] [where], so [conclusion about f]
Three elements then the conclusion. Omitting any of the three is how correct work loses the justification point.
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.
Trapezoidal contribution on one subinterval
area ≈ [(f(x₀) + f(x₁)) / 2] · (x₁ − x₀)
Average of the two heights times the width of THAT subinterval. Sum over subintervals; widths need not be equal.
FTC Part 2 with chain rule
d/dx ∫ₐ^(u(x)) f(t)dt = f(u(x)) · u′(x)
Substitute the upper limit into the integrand, then multiply by its derivative. A variable lower limit contributes the same thing with a minus sign.
Integral of a constant
∫ₐᵇ c dx = c(b − a)
A constant contributes its value times the interval width — the area of a rectangle. Forgetting this term is the most common error in recombination problems.
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.
Exponential model from a proportionality
dy/dt = ky ⟹ y = y₀e^(kt)
A rate proportional to the amount present. k > 0 gives growth, k < 0 decay. This is the one solution worth memorizing rather than re-deriving.
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.
Area with horizontal slices
A = ∫_c^d [x_right(y) − x_left(y)] dy
Right minus left, both as functions of y, with limits that are y-values. The mirror of top minus bottom with dx.
Amount from a rate
f(b) = f(a) + ∫ₐᵇ f′(t) dt
Initial amount plus net change. The integral alone gives the change, never the amount.
Volume by cross sections
V = ∫ₐᵇ A(x) dx, with A(x) built from s = f(x) − g(x)
Square: s². Equilateral triangle: (√3/4)s². Semicircle on diameter s: (π/8)s². Disk of radius r: πr².

On the exam

How to get a 5

Key terms

Limit definition of the derivativef′(a) = lim(h→0) (f(a + h) − f(a))/h — the slope of the tangent line, an instantaneous rate of change.
Power, product, and quotient rulesPower: d/dx[xⁿ] = nxⁿ⁻¹. Product: (fg)' = f'g + fg'. Quotient: (f/g)' = (f'g − fg')/g².
Chain ruled/dx[f(g(x))] = f′(g(x))·g′(x). Differentiate the outside leaving the inside alone, then multiply by the derivative of the inside.
Key derivativesd/dx[sin x] = cos x; [cos x] = −sin x; [tan x] = sec²x; [eˣ] = eˣ; [ln x] = 1/x.
Fundamental Theorem of CalculusPart 2: ∫ₐᵇ f dx = F(b) − F(a). Part 1: d/dx ∫ₐˣ f(t)dt = f(x).
Mean Value TheoremIf f is continuous on [a, b] and differentiable on (a, b), some c in (a, b) has f′(c) = [f(b) − f(a)]/(b − a). Both hypotheses must be stated.
Intermediate Value TheoremIf f is CONTINUOUS on [a, b] and k lies between f(a) and f(b), some c in (a, b) has f(c) = k. Continuity on the closed interval must be stated.
Motion relationshipsv = x'; a = v'. Displacement = ∫v dt; total distance = ∫|v| dt. Speeding up when v and a share a sign.
Average value of a functionf_avg = (1/(b − a))∫ₐᵇ f(x)dx. Not the average of the endpoints, and not the average of the derivative.
L'Hôpital's RuleFor 0/0 or ∞/∞ only, lim f/g = lim f′/g′. Verify the indeterminate form first, and differentiate numerator and denominator separately — not as a quotient.
Disk/washer volume (about x-axis)Disk: V = π∫ R² dx. Washer: V = π∫ (R² − r²) dx (outer² − inner²).
Continuity at x = af(a) exists, lim(x→a) f(x) exists, and they are equal.
Formal idea of a limitlim(x→a) f(x) = L means f(x) gets arbitrarily close to L as x approaches a from both sides — the value f(a) itself is irrelevant.
Conditions for continuity at x = af(a) exists, lim(x→a) f(x) exists, and the two are equal. All three must hold.
The two special trig limitslim(x→0) sin(x)/x = 1 and lim(x→0) (1 − cos x)/x = 0.
Differentiability vs. continuityDifferentiable implies continuous, but not conversely — corners, cusps, and vertical tangents are continuous yet not differentiable.
Product ruled/dx[uv] = u′v + uv′. The derivative of a product is never the product of the derivatives.
Quotient ruled/dx[u/v] = (u′v − uv′)/v². Order matters in the numerator; reversing it flips the sign.
Implicit differentiationDifferentiate both sides with respect to x, applying the chain rule to every y term so each contributes dy/dx, then solve for dy/dx.
First derivative testf′ changing from positive to negative gives a local maximum, negative to positive a local minimum. No sign change means neither.
Second derivative meaningf″ > 0: concave up; f″ < 0: concave down; f″ changes sign: point of inflection.
Candidates test (absolute extrema)On a closed interval, evaluate f at every critical point and both endpoints; compare values.
Speeding up vs. slowing downA particle speeds up when velocity and acceleration share a sign, and slows down when their signs differ.
L’Hôpital’s ruleFor limits of form 0/0 or ∞/∞: the limit equals lim f′(x)/g′(x), provided that limit exists. Verify the form first.