Unlike AP Physics or AP Statistics, the AP Calculus exam provides no formula sheet: every derivative rule, theorem statement, integral form, and series fact below has to come from memory. This page collects what is worth knowing cold for AP Calculus BC, organized by family, with the BC-only material tagged so AB review stays separable. 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 AB? The AB reference is this page without the BC-only layer. 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.
The three techniques BC adds to the antidifferentiation toolkit. Where it slips: choosing u and dv so the new integral is harder than the old one, and treating a divergent improper integral as if it had a value.
$$ \int u \, dv = uv - \int v \, du $$
Integration by parts BC only
The product rule run in reverse. Pick u to get simpler when differentiated (logs and inverse trig first) and dv to be something you can integrate; the goal is an easier integral, not just a different one.
A rational function with a factorable denominator splits into pieces that each integrate to a logarithm. BC keeps to distinct linear factors, so this template is the whole technique.
Write the limit explicitly; the integral converges only if the limit is finite. The same move handles an integrand that blows up at an endpoint of the interval.
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.
Square the derivative before adding the 1, and keep everything under the root. The parametric version in Unit 9 uses the same idea with both coordinate derivatives.
Separating variables, plus the two BC additions: Euler's method and the logistic model. Where it slips: applying the initial condition after sloppy exponentiation instead of right after integrating, and reading the logistic equation's fastest growth at the carrying capacity instead of half of it.
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.
Step along tangent lines: each new y is the old y plus slope times step, with the slope recomputed from the differential equation at each new point, never reused.
$$ \dfrac{dy}{dt} = ky \left( 1 - \dfrac{y}{L} \right) $$
Logistic model BC only
Solutions approach the carrying capacity L without crossing it, and growth is fastest when y equals L over 2. Both facts are read off the equation, no solving required.
Motion in the plane and curves that are not functions of x. Where it slips: flipping the quotient in dy/dx, and dropping the one half in polar area.
$$ \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} $$
Parametric slope BC only
The y-derivative goes on top, and the expression is defined where dx/dt is not zero. A vertical tangent appears where dx/dt vanishes while dy/dt does not.
Distance traveled is the integral of speed: the same integrand as the speed formula. This one formula answers both arc length and total distance questions in Unit 9.
$$ A = \dfrac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta $$
Polar area BC only
The one half is part of the formula, not optional. Between two curves, use one half the integral of outer r squared minus inner r squared over the shared angle interval.
$$ x = r\cos\theta, \quad y = r\sin\theta, \quad \dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta} $$
Polar slope BC only
Convert to x and y first, then take the parametric quotient with theta as the parameter. The derivative dr over d-theta alone is not a slope on the graph.
Convergence tests, error bounds, and the Taylor machinery. Where it slips: using the nth term test to prove convergence, and skipping the endpoint check on an interval of convergence.
Converges exactly when the ratio's absolute value is below 1, to first term over 1 minus r. The a in the numerator is the first term actually present, whatever the starting index.
A divergence test only. Terms going to zero proves nothing: the harmonic series is the standing counterexample, terms shrinking to zero while the sum diverges.
Know each by name; the FRQ asks you to cite the one you used. Ratio test: below 1 converges absolutely, above 1 diverges, exactly 1 says nothing and another test must finish the job.
For an alternating series whose terms decrease in magnitude to zero, the error after n terms is at most the magnitude of the first omitted term. It applies only to alternating series.
Centered at a; Maclaurin means a equals 0. The kth coefficient is the kth derivative at the center divided by k factorial, and that division is the most dropped piece.
$$ e^x = \sum_{n=0}^{\infty} \dfrac{x^n}{n!}, \quad \sin x = \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}, \quad \cos x = \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n}}{(2n)!} $$
Standard Maclaurin series BC only
Sine takes the odd powers, cosine the even; all three converge for every x. The fourth to memorize: 1 over 1 minus x is the sum of x to the n, valid only for absolute value of x below 1.
M is a bound on the absolute value of the next derivative, the (n+1)st, on the interval between a and x. It bounds the error of any Taylor polynomial, alternating or not.
The ratio test applied to a power series yields the radius. The endpoints always come back inconclusive from the ratio test, so each must be tested separately with another test.
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.
