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.
Differentiability at a seam
f differentiable at c ⟺ f continuous at c AND lim(x→c⁻) f′(x) = lim(x→c⁺) f′(x)
Both conditions, in that order. Matching slopes alone is not enough — a function can have equal one-sided slopes across a jump.
Estimating a derivative from a table
f′(a) ≈ [f(a + h) − f(a − h)] / (2h) · one-sided: [f(b) − f(a)] / (b − a)
Symmetric when values bracket the point; the divisor is always the distance between the two x-values used.
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.
Chain rule, three layers
d/dx f(g(h(x))) = f′(g(h(x))) · g′(h(x)) · h′(x)
One factor per layer. The count of factors should equal the count of function layers — a quick check that nothing was dropped.
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.
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.
Linearization at x = a
L(x) = f(a) + f′(a)(x − a) · f(x) ≈ L(x) for x near a
Concave up on the interval ⇒ L underestimates. Concave down ⇒ L overestimates.
L’Hôpital’s Rule
If lim f/g is 0/0 or ∞/∞, then lim f(x)/g(x) = lim f′(x)/g′(x)
Check the form on every pass, including after each application. Stop as soon as the form is no longer indeterminate.
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.
First Derivative Test at a critical point c
f′: + → − ⇒ relative max · f′: − → + ⇒ relative min · no sign change ⇒ neither
The conclusion comes from the sign change of f′ across c, not from the value of f at c.
Concavity and the Second Derivative Test
f″ > 0 ⇒ concave up ⇒ relative min at a critical point · f″ < 0 ⇒ concave down ⇒ relative max · f″ = 0 ⇒ inconclusive
Inflection points need a sign change in f″, which f″ = 0 alone does not guarantee.
Translating between f, f′ and f″
f ↑ ⟺ f′ > 0 · f rel. extremum ⟺ f′ changes sign · f concave up ⟺ f′ ↑ ⟺ f″ > 0 · f inflection ⟺ f′ has an extremum
Every question about f from a graph of f′ is one of these four rows.
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.
Trapezoidal sum, unequal widths
Σ [(f(xᵢ) + f(xᵢ₊₁))/2] · (xᵢ₊₁ − xᵢ) · equal widths: (Δx/2)[f(x₀) + 2f(x₁) + … + 2f(xₙ₋₁) + f(xₙ)]
The h/2 shortcut with doubled interior terms is valid only when every subinterval has the same width.
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.
Second Fundamental Theorem, with the chain rule
d/dx ∫ₐˣ f(t) dt = f(x) · d/dx ∫ₐ^(u(x)) f(t) dt = f(u(x)) · u′(x)
g″(x) = f′(x), so the concavity of the accumulation function is the increase or decrease of the integrand.
Definite integral by substitution
∫ₐᵇ f(g(x))·g′(x) dx = ∫_(g(a))^(g(b)) f(u) du
The new limits are the old ones passed through g. Once the limits are converted, never substitute back.
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.
The exponential model
dy/dt = ky, y(0) = y₀ ⇒ y = y₀e^(kt)
Growth for k > 0, decay for k < 0. Contrast with the logistic dy/dt = ky(L − y), whose growth levels off at the carrying capacity L.
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.
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).
Motion and net change
displacement = ∫ₐᵇ v(t) dt · total distance = ∫ₐᵇ |v(t)| dt · F(b) = F(a) + ∫ₐᵇ F′(t) dt
Position at a later time needs an initial position; the integral supplies only the change.
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.
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.
Parametric derivatives
dy/dx = (dy/dt)/(dx/dt) · d²y/dx² = [d/dt (dy/dx)] / (dx/dt)
Differentiate the slope expression with respect to t, then divide by dx/dt once more. Never divide the second t-derivatives.
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.
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.
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.
Polar area and slope
A = (1/2)∫ᵅᵝ r² dθ · between curves: (1/2)∫ᵅᵝ (r_outer² − r_inner²) dθ · dy/dx = (dy/dθ)/(dx/dθ)
Square first, then subtract. (r₁ − r₂)² is not r₁² − r₂².
Geometric series and the nth-term test
Σ arⁿ = a/(1 − r) for |r| < 1, divergent otherwise · lim aₙ ≠ 0 ⇒ Σaₙ diverges
aₙ → 0 does NOT imply convergence. The nth-term test proves divergence only.
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.
The tests, with their conditions
p-series: Σ1/n^p converges ⟺ p > 1 · Ratio: lim|aₙ₊₁/aₙ| = L, converges if L < 1, diverges if L > 1, inconclusive if L = 1 · Alternating: terms decreasing in magnitude AND → 0
The ratio test at L = 1 says nothing at all — that is exactly where p-series live, and where a comparison is needed instead.
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.
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.
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.
The four required Maclaurin series
e^x = Σ xⁿ/n! (all x) · sin x = Σ (−1)ⁿx^(2n+1)/(2n+1)! (all x) · cos x = Σ (−1)ⁿx^(2n)/(2n)! (all x) · 1/(1−x) = Σ xⁿ (|x| < 1)
sin has odd powers and cos has even ones — checking the parity of the exponents catches a swapped pair instantly.
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.
- When a free-response question hands you a piecewise function with unknown constants and says "differentiable," it is asking for two equations, not one. Write the continuity equation and the slope equation, then solve the system.
- Table questions almost always ask you to justify the estimate. "f′(5) ≈ [f(7) − f(3)]/(7 − 3) = 3" earns the point; the bare number 3 often does not. Show the quotient you used.
- 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.
- Nested chain rule questions are frequently multiple choice with all the right pieces and one missing factor among the distractors. Count your factors against the number of layers before choosing.
- 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.
- 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.
- A complete over/under justification names three things: the sign of f″, the concavity that follows, and the position of the tangent line relative to the curve. Write all three.
- BC uses L’Hôpital far beyond limit questions: it is how you show a series term goes to zero, and how improper integrals are evaluated at a bound. The form-checking habit pays off in Units 6 and 10 as much as here.
- 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.
- Free-response justifications are graded on the sign change. Write "f′ changes from positive to negative at x = 1, so f has a relative maximum there" — the phrase "changes from … to …" is what the rubric looks for.
- The exam asks "for what values of x does the graph of f have a point of inflection? Justify." A justification that stops at "f″(x) = 0 there" is incomplete every time — say that f″ changes sign.
- When the stem says "the graph of f′ is shown," write "f′" in the margin next to the picture before reading the questions. Half the lost points in this unit come from answering as though the picture were f.
- 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.
- Riemann-sum questions on the calculator-free section reward writing the sum before evaluating it: the setup line carries its own point, and an arithmetic slip afterward costs less than a missing setup.
- 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.
- The graph-of-f accumulation question appears on nearly every exam and carries several points at once: values of g, an extremum with justification, concavity, and often a tangent line to g. Getting g′ = f stated early sets up all of them.
- BC Unit 6 is the largest unit on the exam, and much of its weight is technique selection rather than execution. When practicing, say which tool you are choosing and why before doing any algebra.
- 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, write the differential equation and the initial condition as a labeled pair before solving. Both are separately scored, and a correct equation salvages partial credit even when the integration goes wrong.
- 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.
- 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.
- Read the verb. "How far did it travel" is total distance; "what is its position" and "how far from its starting point" are displacement questions. The integrals differ by absolute-value bars and the answers differ by a lot.
- 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.
- 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."
- Parametric second-derivative questions are usually multiple choice and usually include the ratio-of-second-derivatives distractor. Compute dy/dx first, write it down, then differentiate that written expression — the discipline prevents the shortcut.
- 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.
- 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.
- 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.
- Polar free-response questions almost always award a point for correct limits separate from the integrand. Even if the algebra stalls, write the fully set-up integral with its bounds.
- Writing "aₙ → 0, therefore the series converges" is scored as a wrong statement even when the series does happen to converge. The rubric penalizes the reasoning, not just the conclusion.
- 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.
- Free-response convergence questions want the test named and its hypotheses verified. "By the alternating series test, since 1/√(n+3) is decreasing and tends to 0" earns credit; "it converges by AST" alone often does not.
- 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!.
- 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.
- 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.
- Series free-response questions are usually four parts that all rest on one expansion: write the series, differentiate or integrate it, bound the error, then state the interval. Getting the first part right carries the rest, so spend the time 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.
- Budget series practice heavily: Unit 10 is the largest BC-only block, and one full FRQ is a near-guaranteed Taylor/Maclaurin question — memorize the eˣ, sin x, cos x, and 1/(1 − x) series cold.
- When naming a convergence test on an FRQ, state its hypotheses explicitly (e.g. “terms are positive, decreasing, and approach 0”) — the justification point is for the conditions, not the test’s name.
- For parametric and polar questions, write the general formula (dy/dx = (dy/dt)/(dx/dt), A = (1/2)∫r² dθ) before substituting; the setup line earns points and prevents dropped factors like the 1/2.
- Distance traveled is the integral of speed √((dx/dt)² + (dy/dt)²) — never integrate the velocity components separately for distance, which gives displacement instead.
- On error-bound parts, identify which bound is being requested: alternating series bound needs only the next term, while the Lagrange bound needs a maximum of the next derivative.
- AB material is still most of the BC exam — do not let limits, the FTC, and separable differential equations get rusty while drilling series.
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. Choose u by LIATE — logarithmic, inverse trig, algebraic, trigonometric, exponential — taking the earliest type as u.
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 bounds the (n+1)th derivative on the interval. State what M is and why.
L’Hôpital’s rule (when it applies) — Only for 0/0 or ∞/∞ forms: replace f/g by f′/g′ and re-take the limit. Always confirm the indeterminate form first.
Partial fraction decomposition — For proper rational functions with factorable denominators: split into A/(x − a) + B/(x − b), solve for constants, integrate to logarithms.
Improper integral — An integral with an infinite limit or an unbounded integrand; evaluate as a limit. ∫₁^∞ 1/xᵖ converges iff p > 1; ∫₀¹ 1/xᵖ converges iff p < 1.
Arc length (function form) — L = ∫ₐᵇ √(1 + (f′(x))²) dx.
Euler’s method — Repeated tangent-line steps: yₙ₊₁ = yₙ + Δx·f(xₙ, yₙ), always using the slope at the current point.
Logistic differential equation — dP/dt = kP(1 − P/L). Growth is nearly exponential when P is small and slows to zero as P approaches the carrying capacity L.
Parametric derivative — dy/dx = (dy/dt)/(dx/dt), defined wherever dx/dt ≠ 0. Not the derivative of y with respect to t.
Parametric arc length / distance traveled — L = ∫√((dx/dt)² + (dy/dt)²) dt — the integral of speed.
Speed of a vector-valued particle — |v(t)| = √((x′)² + (y′)²); speed is a scalar, direction lives in the velocity vector.
Polar area formula — A = (1/2)∫ r² dθ over the sweep of the region — do not forget the 1/2.
Polar to rectangular slope — dy/dx = (dy/dθ)/(dx/dθ) with x = r cos θ, y = r sin θ (product rule in θ since r depends on θ).
Geometric series — Σarⁿ converges to a/(1 − r) when |r| < 1 and diverges otherwise. The only common series whose exact sum is easy to state.