Unlike AP Physics or AP Statistics, the AP Calculus exam provides no formula sheet: every derivative rule, theorem statement, and integral form below has to come from memory. This page collects what is worth knowing cold for AP Calculus AB, organized by family and annotated with what each fact is for. These are the standard results, not an official College Board document; confirm the exam's exact provided materials and calculator policy on AP Central. Taking BC? The BC reference carries everything here plus the series, parametric, polar, and integration-technique facts. Every entry links to the unit where Mistake Master teaches and drills it.
Units 1-2
Limits and Continuity
What a limit claims, when a function is continuous, and how the derivative is born as a limit. Where it slips: answering with the function's value at a point when the question asks for the limit, and citing the IVT without first stating that the function is continuous.
Both forms appear on the exam, usually in disguise: a limit shaped like either one IS a derivative, and recognizing which function and which point beats grinding out the algebra.
Three requirements: the limit exists (left agrees with right), the function value is defined, and the two match. A jump, a hole, and a relocated point each break a different one.
$$ f \text{ continuous on } [a,b] \Rightarrow f \text{ attains every value between } f(a) \text{ and } f(b) $$
Intermediate Value Theorem
Guarantees a value is reached somewhere in the interval; it never locates it. State continuity on the closed interval before invoking it, since that hypothesis is the whole theorem.
Differentiability is the stronger condition. The absolute value function is continuous everywhere yet not differentiable at zero: corners, cusps, and vertical tangents all survive continuity.
The differentiation toolkit, from the power rule to L'Hospital. Where it slips: the forgotten inner derivative in the chain rule, and the order of the quotient rule's numerator.
$$ \dfrac{d}{dx}\,x^n = n\,x^{\,n-1} $$
Power rule
Works for every real exponent. Rewrite roots and reciprocals as powers before differentiating; the square root of x is x to the one half.
Differentiate the outer function at the inner one, then multiply by the inner derivative. That trailing factor is the single most dropped item in the course.
Also: secant gives secant times tangent, cotangent gives negative cosecant squared, cosecant gives negative cosecant times cotangent. Every co-function carries the minus sign.
The power rule needs a constant exponent, so a to the x is not x times a to the x minus 1. For other log bases, divide: log base a of x differentiates to 1 over x ln a.
Evaluate the original derivative at the inverse's output: the x whose image is b, not at b itself. Reading the wrong input is where nearly every miss on this rule comes from.
$$ \lim \dfrac{f(x)}{g(x)} = \lim \dfrac{f'(x)}{g'(x)} \quad \text{only for } \tfrac{0}{0} \text{ or } \tfrac{\infty}{\infty} $$
L'Hospital's rule
Verify and state the indeterminate form before applying it; on the FRQ that verification earns the point. Differentiate the top and bottom separately, never with the quotient rule.
The guarantee theorems: each trades hypotheses for a conclusion. Where it slips: quoting the conclusion without checking, or writing down, the hypotheses that earn it.
$$ f'(c) = \dfrac{f(b) - f(a)}{b - a} \; \text{ for some } c \text{ in } (a,b) $$
Mean Value Theorem
Somewhere, the instantaneous rate equals the average rate. Requires continuity on the closed interval and differentiability on the open interval; both belong in the justification.
$$ f(a) = f(b) \Rightarrow f'(c) = 0 \; \text{ for some } c \text{ in } (a,b) $$
Rolle's Theorem
The MVT with equal endpoint values, so the guaranteed slope is zero. Same hypotheses: continuous on the closed interval, differentiable on the open one.
$$ \text{compare } f \text{ at critical points and endpoints} $$
Candidates test
On a closed interval, absolute extrema live only at critical points or endpoints. Evaluate the function itself at each candidate and compare; the derivative's sign is not the finish line.
Antiderivatives, substitution, and the theorem that ties accumulation to rate. Where it slips: bounds left in terms of x after a u-substitution, and the dropped absolute value on the log.
Raise the exponent, then divide by the new exponent. The excluded case is exactly the one with its own rule: the antiderivative of 1 over x is ln of the absolute value of x, plus C.
$$ \int e^x \, dx = e^x + C, \quad \int \cos x \, dx = \sin x + C, \quad \int \sin x \, dx = -\cos x + C $$
Core antiderivatives
Also: secant squared integrates to tangent. The minus sign lives on the integral of sine, not of cosine; differentiating your answer takes five seconds and catches the swap.
The chain rule run in reverse: the integrand must contain the inner function's derivative, up to a constant. On a definite integral, convert the bounds to u-values too.
The derivative of an accumulation function is the integrand, evaluated at the upper limit. If the upper limit is g(x), the chain rule appends a factor of g'(x).
Any antiderivative F evaluates the definite integral: upper minus lower. Net change of F equals the integral of its rate, which is how most applied FRQ parts are set up.
$$ \int_a^b f = -\int_b^a f, \qquad \int_a^b f = \int_a^c f + \int_c^b f $$
Definite integral properties
Reversing the bounds flips the sign; the interval splits at any point between. Constants factor out and sums split term by term, but products and quotients do not.
Area, volume, and motion. Where it slips: squaring the difference of the radii instead of subtracting their squares, and reporting displacement when the question asks for total distance.
$$ A = \int_a^b \big[ f(x) - g(x) \big] \, dx $$
Area between curves
Top minus bottom when integrating in x; right minus left when integrating in y. If the curves cross, split the integral at the intersection so the difference stays positive.
Each slice is a solid circle whose radius runs from the axis of revolution to the curve. Revolving around a shifted axis changes the radius, not the formula.
Integrate the area of the slice. The region gives each slice's base length; feed that length into the shape's own area formula, halving the base first for a semicircle's radius.
Velocity is the derivative of position, acceleration of velocity. Speed is the absolute value of velocity, and a particle speeds up exactly when velocity and acceleration share a sign.
Integrating velocity gives net change in position; integrating speed gives ground covered. They differ whenever the particle turns around, which is exactly when exams ask.
Separating variables and the one growth model AB owns. Where it slips: applying the initial condition after sloppy exponentiation instead of right after integrating, and losing the constant inside the exponent.
Move every y to one side and every x to the other before integrating; a stray factor left behind poisons both integrals. One constant of integration, applied immediately, is enough.
$$ \dfrac{dy}{dt} = ky \;\Rightarrow\; y = y_0\,e^{kt} $$
Exponential growth and decay
Rate proportional to amount forces this solution: initial value y-zero, growth for positive k, decay for negative k. Recognize the differential equation and skip the re-derivation.
Knowing the formula is not the same as not slipping on it
Most AP Calculus points are lost to a small set of predictable reasoning errors, not to forgotten formulas. Mistake Master diagnoses which ones are costing you points, then drills only those.
