AP Calculus BC — Cheatsheet
Formulas, exam-day tips, and key terms on one page.
Formulas & relationships
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. If they differ, the two-sided limit does not exist.
Two limits worth memorizing
lim(x→0) sin(x)/x = 1 · lim(x→0) (1 − cos x)/x = 0
These special trig limits appear constantly; direct substitution gives 0/0, but the true limits are 1 and 0.
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 parts are required. The equality in the third part is the one students most often skip.
End behavior of rational functions
deg(bottom) > deg(top): limit 0 · equal degrees: ratio of leading coefficients · deg(top) > deg(bottom): ±∞
These three cases decide the horizontal asymptote (or its absence) as x → ±∞.
Intermediate Value Theorem
f continuous on [a, b] and N between f(a) and f(b) ⟹ ∃ c in (a, b) with f(c) = N
An existence theorem: it promises a c exists but gives no formula for it.
Limit definition of the derivative
f′(x) = lim(h→0) [f(x + h) − f(x)] / h
The difference quotient [f(x+h) − f(x)]/h is the secant slope; its limit as h→0 is the tangent slope.
Product and quotient rules
(fg)′ = f′g + fg′ · (f/g)′ = (f′g − fg′) / g²
The quotient rule numerator is "low d-high minus high d-low"; the order of subtraction matters.
The chain rule
d/dx [ f(g(x)) ] = f′(g(x)) · g′(x)
Derivative of the outer (with the inner left inside) times the derivative of the inner. In Leibniz form: dy/dx = (dy/du)(du/dx).
Differentiating a y-term
d/dx [ y³ ] = 3y² · (dy/dx) · d/dx [ x·y ] = y + x·(dy/dx)
A y raised to a power picks up dy/dx by the chain rule; a product of x and y needs the product rule too.
Derivative of an inverse function
If g = f⁻¹ and f(a) = b, then g′(b) = 1 / f′(a) = 1 / f′(g(b)).
The slope of the inverse at b is the reciprocal of the slope of f at the matching input a.
Inverse trig derivatives
d/dx[arcsin x] = 1/√(1 − x²) · d/dx[arctan x] = 1/(1 + x²) · d/dx[arcsec x] = 1/(|x|√(x² − 1))
arccos, arccot, and arccsc are just the negatives of arcsin, arctan, and arcsec respectively.
Differentiating a volume relation in time
V = (4/3)πr³ ⟹ dV/dt = 4πr² · (dr/dt)
Every geometric formula becomes a rate equation once differentiated with respect to t via the chain rule.
First-derivative test for a maximum
If f′ changes + → − at c, then f has a local maximum at c.
A sign change from positive to negative means the function rises then falls — a peak. The reverse (− → +) is a local minimum.
Mean Value Theorem
f′(c) = [ f(b) − f(a) ] / (b − a) for some c in (a, b)
The right side is the secant slope (average rate); f′(c) is the tangent slope (instantaneous rate). MVT guarantees they match at some c.
The motion chain and distance
v(t) = x′(t) · a(t) = v′(t) · total distance = ∫ₐᵇ |v(t)| dt
Displacement is ∫v dt (signed); total distance is the integral of |v| (unsigned), splitting at direction changes.
FTC Part 1 (with the chain rule)
d/dx ∫ₐ^{u(x)} f(t) dt = f(u(x)) · u′(x)
When the upper limit is a function u(x) rather than plain x, multiply by u′(x). This chain-rule twist is heavily tested.
FTC Part 2 (evaluation)
∫ₐᵇ f(x) dx = F(b) − F(a), where F′ = f
Any antiderivative works — the constant of integration cancels in the subtraction.
Integration by parts
∫ u dv = u·v − ∫ v du
You pick u (to differentiate) and dv (to integrate); then du = u′ dx and v = ∫dv.
Decomposition over distinct linear factors
P(x) / [(x − a)(x − b)] = A/(x − a) + B/(x − b)
One unknown constant per distinct linear factor. Solve for A and B by clearing denominators and substituting convenient x-values.
Improper integral over an infinite interval
∫ₐ^∞ f(x) dx = lim(b→∞) ∫ₐᵇ f(x) dx
Convert the infinite bound to a finite b, integrate, then let b → ∞. A finite result means convergence.
What a slope field encodes
At each point (x, y), segment slope = dy/dx = f(x, y)
The field is the geometry of the equation; a solution curve threads through it tangent to every segment.
Exponential growth/decay model
dy/dt = k·y ⟹ y = y₀·e^{kt}
The classic separable equation: rate proportional to amount. k > 0 is growth, k < 0 is decay; y₀ is the initial amount.
Euler’s method update
x_{n+1} = x_n + Δx · y_{n+1} = y_n + f(x_n, y_n)·Δx
The new y is the old y plus slope times step size. "Rise = slope × run" applied one step at a time.
Logistic differential equation
dP/dt = kP(1 − P/M)
M is the carrying capacity. Growth rate is near-exponential for small P and falls to 0 as P → M.
Area between curves
A = ∫ₐᵇ [ f(x) − g(x) ] dx, with f on top and g on the bottom
Integrate top minus bottom. If integrating in y instead, it becomes right minus left: ∫ [x_right − x_left] dy.
Average value of a function
f_avg = (1/(b − a)) · ∫ₐᵇ f(x) dx
Integrate, then divide by the length of the interval. Do not confuse with average rate of change, [f(b) − f(a)]/(b − a).
Disk and washer methods
disk: V = π ∫ₐᵇ [R(x)]² dx · washer: V = π ∫ₐᵇ ([R(x)]² − [r(x)]²) dx
R is the outer radius (to the far edge), r the inner radius (to the hole). Use a washer whenever the region does not touch the axis.
Arc length of y = f(x)
L = ∫ₐᵇ √( 1 + (dy/dx)² ) dx
Comes from the Pythagorean length √(dx² + dy²) of each infinitesimal piece, factored to pull out dx.
Parametric first derivative
dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0
The slope of the curve is the ratio of the two t-derivatives — a direct consequence of the chain rule.
Area enclosed by a polar curve
A = ½ ∫_α^β [ r(θ) ]² dθ
The ½r² comes from the area of a circular sector. Integrate over the θ-interval that traces the region exactly once.
Speed as the magnitude of velocity
speed = |v(t)| = √( (dx/dt)² + (dy/dt)² )
Speed is a scalar: the length of the velocity vector. Velocity carries direction; speed is just how fast.
Parametric arc length
L = ∫ₐᵇ √( (dx/dt)² + (dy/dt)² ) dt
Identical to the integral of speed over time, so arc length = total distance traveled by the particle.
Two benchmark series
geometric Σ arⁿ: converges ⟺ |r| < 1, to a/(1 − r) · p-series Σ 1/nᵖ: converges ⟺ p > 1
These two are the reference points most comparison and ratio arguments lean on.
Taylor series centered at a
f(x) = Σ (n=0 to ∞) [ fⁿ(a) / n! ] (x − a)ⁿ = f(a) + f′(a)(x−a) + [f″(a)/2!](x−a)² + …
The nth coefficient is fⁿ(a)/n!. Setting a = 0 gives the Maclaurin series.
Radius of convergence via the ratio test
lim |cₙ₊₁(x − a)ⁿ⁺¹ / cₙ(x − a)ⁿ| < 1 ⟹ |x − a| < R
Apply the ratio test, solve the resulting inequality for |x − a|; the bound is R. Then test x = a ± R by hand.
Alternating series error bound
| S − Sₙ | ≤ | a_{n+1} |
For a convergent alternating series, the error after n terms is no larger than the absolute value of the first omitted term.
Lagrange error bound
| Rₙ(x) | ≤ ( M / (n+1)! ) · | x − a |ⁿ⁺¹
M is a bound on |f^(n+1)| on the interval between a and x. Controls the error of the nth-degree Taylor polynomial.
On the exam
- On the AP exam, always try substitution first and read the form. Continuous-looking value → done. 0/0 → simplify (factor/rationalize/special limit). Nonzero/0 → infinite limit (asymptote). Naming the form tells you the technique.
- When justifying continuity on the AP exam, cite all three conditions by name. Writing "f(a) is defined, the limit exists, and they are equal" earns the justification point that a bare numerical answer does not.
- For limits at infinity of rational functions, memorize the three-case degree rule — it turns most such problems into a one-line answer without any algebra. Reserve the divide-by-highest-power method for showing work when the exam demands justification.
- The IVT is an *existence* theorem — it never locates or counts solutions. If a prompt asks "how many" or "find the value," the IVT alone is insufficient; you would need additional tools like a sign analysis or the Mean Value Theorem.
- When a limit is written as lim(h→0) [f(a+h) − f(a)]/h, recognize it as the *definition of a derivative at a point*, f′(a). AP loves disguising a derivative as a raw limit — identifying it lets you skip the algebra and just differentiate.
- Chain rule questions escalate by nesting. For sin³(2x) = (sin(2x))³ you apply the rule twice: 3(sin 2x)² · cos(2x) · 2. Count the layers before you start so you know how many inner-derivative factors to expect.
- Implicit results usually contain both x and y — that is correct, not incomplete. To report a numerical slope you must plug in a full point (x, y) on the curve. Free-response graders expect the point substituted, not just the symbolic derivative.
- Inverse-function problems on the AP exam almost always give you a table or a point pairing. The move is mechanical: find where f equals the target output, evaluate f′ there, take the reciprocal. Practice it until it is automatic.
- Related-rates free-response answers must carry units and be evaluated at the stated instant. Write the differentiated equation, then substitute — showing the un-substituted derivative first is what earns the setup points even if arithmetic slips.
- On free response you must *justify* that your answer is a max or min — cite a sign change in f′ (first-derivative test), the sign of f′′ (second-derivative test), or the candidates comparison. An unjustified extremum loses the justification point.
- MVT justifications must state both hypotheses ("f is continuous on [a,b] and differentiable on (a,b)") before invoking the conclusion. Many free-response points hinge on that explicit verification, not on solving for c.
- Read motion questions carefully: "how far from start" (displacement, signed) and "total distance traveled" (unsigned) are different. For total distance, always locate where v = 0, split the interval, and sum the absolute values of the pieces.
- FTC Part 1 questions on the AP exam almost always disguise a chain-rule step by using a non-trivial upper limit. Before answering, check whether the upper limit is plain x or a function of x — the latter needs the extra u′(x) factor.
- Some integrals like ∫eˣ·sin(x) dx cycle back to the original after two rounds of parts. When that happens, set the integral equal to I, solve the algebraic equation for I, and divide — do not keep integrating forever.
- The fastest way to find the constants is the "cover-up" method: to get the constant over (x − a), cover that factor and substitute x = a into what remains. It computes each numerator in one step, ideal under time pressure.
- Improper integrals also arise from a vertical asymptote *inside* or at an endpoint of the interval, such as ∫₀¹ 1/√x dx. Split at the discontinuity and take a one-sided limit there — do not treat it as an ordinary definite integral.
- When a free-response question shows a slope field, you can sketch a particular solution without solving the equation — just start at the given point and follow the segments. Full credit comes from a curve tangent to the field, passing through the initial condition.
- On free response you must show the separation and the integration of both sides, then apply the initial condition explicitly. A bare final formula, even if correct, typically earns only partial credit — the graders reward the separated equation and the antiderivatives.
- Organize Euler’s method in a small table with columns for x, y, slope f(x,y), and Δy = slope·Δx. The table structure prevents arithmetic slips across steps and is exactly what free-response graders expect to see.
- You can often answer logistic questions without solving the differential equation. Read M directly from the equation as the P that zeroes the (1 − P/M) factor, then use the two facts: fastest growth at M/2, and the long-run limit is M.
- When the region is easier to slice horizontally — for instance bounded by curves given as x = h(y) — integrate with respect to y using right minus left. Choosing the orientation that avoids splitting the region saves time on free response.
- Read the wording precisely: "average value of the velocity" means integrate v and divide by the time interval, while "average acceleration" means [v(b) − v(a)]/(b − a). One is an integral average, the other an endpoint average — pick the right formula.
- For cross-section problems, first decide whether the region’s height gives the *side* or the *diameter* of the shape. Semicircle and some square-on-the-diagonal setups use the height as a diameter, which changes the area formula — misreading this is a common lost point.
- Most arc-length integrals cannot be evaluated by hand, so on the calculator section, set up ∫√(1 + (dy/dx)²) dx precisely and let the calculator finish. The setup — correct derivative, correct bounds — is what the rubric rewards.
- Keep the pattern straight: dy/dt = 0 → horizontal tangent, dx/dt = 0 → vertical tangent. Because the slope is a fraction (dy/dt)/(dx/dt), the numerator controls "flat" and the denominator controls "steep/vertical."
- The number-one polar-area error is writing (R − r)² instead of (R² − r²). Square each radius separately, then subtract. When in doubt, compute the outer area and inner area as two integrals and subtract the results.
- For planar motion, keep three vectors straight: position ⟨x, y⟩, velocity ⟨x′, y′⟩, acceleration ⟨x″, y″⟩. "Speed" and "total distance" both use the magnitude √((x′)² + (y′)²) — speed at an instant, distance as its integral over time.
- Parametric arc length and "total distance traveled" are the same integral, ∫√((dx/dt)² + (dy/dt)²) dt. If a free-response question asks for the distance a particle travels in the plane, this is the formula — not the magnitude of the displacement vector.
- Match the test to the terms: factorials or nth powers → ratio test; a clean 1/nᵖ → p-series; a constant ratio between terms → geometric. Recognizing the *form* quickly is the skill that makes the series free-response manageable.
- Coefficient and derivative are linked by a factorial: the coefficient of (x − a)ⁿ is fⁿ(a)/n!, so fⁿ(a) = n! × (that coefficient). AP loves asking for a high derivative at the center — read off the coefficient and multiply by n!.
- A complete interval-of-convergence answer has three parts: apply the ratio test for R, then test each endpoint separately, then state the interval with the correct brackets. Free-response rubrics award a point specifically for the endpoint analysis.
- Choose the bound that fits: if the series alternates with decreasing terms, use the simple "first omitted term" bound. Otherwise reach for Lagrange, and take M as the maximum of the (n+1)th derivative on the relevant interval — often just its largest possible value there.
How to get a 5
- On FRQs, show the definite-integral setup (limits + integrand) before evaluating — setup points are earned even if arithmetic slips.
- For series, name the test and confirm its conditions explicitly (e.g., "terms decreasing and → 0"); unjustified convergence claims lose points.
- When a problem says "justify," reference a sign change, the sign of a derivative, or a named theorem — never just restate the answer.
- Keep answers exact (fractions, π, e) unless told to round, and store intermediate values in the calculator to protect the accuracy point.
Key terms
Derivative of tan x — sec²x
Derivative of arctan x — 1/(1 + x²)
∫ 1/(a² + x²) dx — (1/a) arctan(x/a) + C
Integration by parts — ∫ u dv = uv − ∫ v du
Maclaurin series for eˣ — Σ xⁿ/n! = 1 + x + x²/2! + x³/3! + …
Maclaurin series for sin x — Σ (−1)ⁿ x²ⁿ⁺¹/(2n+1)! = x − x³/3! + x⁵/5! − …
Arc length of y = f(x) on [a,b] — ∫ₐᵇ √(1 + (f'(x))²) dx
Area inside polar curve r = f(θ) — (1/2) ∫_α^β [f(θ)]² dθ
Parametric dy/dx — (dy/dt)/(dx/dt), provided dx/dt ≠ 0
p-series convergence — Σ 1/nᵖ converges if p > 1; diverges if p ≤ 1.
Logistic equation & carrying capacity — dP/dt = kP(1 − P/L); grows fastest at P = L/2, limit L.
Lagrange error bound — |Rₙ(x)| ≤ M·|x − a|ⁿ⁺¹/(n+1)!, where M ≥ |f⁽ⁿ⁺¹⁾| on the interval.