Generated by AskSia.ai — graphs, formulas, traps
f'(x) = limh→0 [f(x+h) − f(x)] / h| Rule | Form | Result |
|---|---|---|
| Power | d/dx[xⁿ] | n·x^(n−1) |
| Product | d/dx[fg] | f'g + fg' |
| Quotient | d/dx[f/g] | (f'g − fg')/g² |
| Chain | d/dx[f(g)] | f'(g)·g' |
(sin x)' = cos x (cos x)' = −sin x (eˣ)' = eˣ (ln x)' = 1/x (tan x)' = sec²xMVT: if f continuous on [a,b], differentiable on (a,b), ∃ c with f'(c) = (f(b) − f(a))/(b−a). EVT: continuous on [a,b] → attains absolute max + min.
L'Hôpital (only 0/0 or ∞/∞): lim f/g = lim f'/g'. Other indeterminates (0·∞, ∞−∞, 1^∞): rewrite to fit.
d/dx[sin(2x)] is 2 cos(2x), not cos(2x). The most common AP free-response loss: doing the outer derivative and forgetting to multiply by the derivative of what's inside. Mark every chain with an arrow.
dy/dx = (dy/dt)/(dx/dt) d²y/dx² = (d/dt[dy/dx])/(dx/dt)arc length: L = ∫√((dx/dt)² + (dy/dt)²) dt speed: |v(t)|| Convert | Formula |
|---|---|
| polar → Cartesian | x = r cos θ, y = r sin θ |
| Cartesian → polar | r² = x²+y², tan θ = y/x |
| Polar area | A = ½ ∫αβ r(θ)² dθ |
| Polar arc length | L = ∫√(r² + (dr/dθ)²) dθ |
Tangent line to polar curve: convert to parametric x(θ) = r(θ) cos θ, y(θ) = r(θ) sin θ, then dy/dx = (dy/dθ)/(dx/dθ).
Polar area is ½ ∫ r² dθ. Forgetting either the 1/2 or the square kills the entire problem. AP graders see this slip every year. Sketch the curve first to identify θ-bounds — sometimes a curve traces twice as θ goes 0 to 2π.
| Pattern | Method | Form |
|---|---|---|
| (stuff)ⁿ · stuff' | u-sub | u^(n+1)/(n+1) |
| poly · eˣ, poly · sin/cos, ln x | parts (LIATE) | uv − ∫v du |
| √(a²−x²), √(a²+x²), √(x²−a²) | trig sub | x = a sin/tan/sec θ |
| P(x)/Q(x), proper rational | partial fractions | decompose then integrate |
| improper (∞ limit) | limit of bounded | limb→∞ ∫ |
IBP: ∫ u dv = uv − ∫ v du | LIATE: Logs · Inverse trig · Algebraic · Trig · ExpBoomerang IBP: when ∫ u dv gives back the original integral on the right (∫ eˣ sin x), add to both sides, divide by 2.
If integrand is undefined at an endpoint OR limit is ±∞, the integral is improper. You must use a limit — not just plug in. Forgetting this on FRQ loses 2-3 points.
f(x) = Σn=0∞ f⁽ⁿ⁾(a) (x − a)ⁿ / n! (Maclaurin = Taylor at a = 0)| Function | Series | Converges |
|---|---|---|
| eˣ | 1 + x + x²/2! + x³/3! + … | all x |
| sin x | x − x³/3! + x⁵/5! − … | all x |
| cos x | 1 − x²/2! + x⁴/4! − … | all x |
| 1/(1−x) | 1 + x + x² + x³ + … | |x| < 1 |
| ln(1+x) | x − x²/2 + x³/3 − … | −1 < x ≤ 1 |
Lagrange remainder: |Rn(x)| ≤ M · |x − a|^(n+1) / (n+1)! M ≥ |f^(n+1)(c)|Trick: ∫ sin(x²) dx has no closed form — but substitute into sin series, integrate term-by-term, get ∫ x² − x⁶/6 + x¹⁰/120 − …
It's n!, not n. And it's (x − a)ⁿ, not just xⁿ — center matters. Off-by-one or off-by-center kills full credit on AP FRQ.
dy/dx = f(x) · g(y) ⇒ ∫ dy/g(y) = ∫ f(x) dx + CSteps: (1) separate variables to opposite sides. (2) integrate both sides. (3) apply initial condition to find C. (4) solve for y if possible.
y_{n+1} = y_n + h · f(x_n, y_n) — step forward by h. Underestimates when concave up, over when concave down.| Model | ODE | Solution |
|---|---|---|
| Exponential | dy/dt = k y | y = y₀ eᵏᵗ |
| Bounded (Newton's cooling) | dT/dt = k(T − T∞) | T = T∞ + (T₀−T∞)eᵏᵗ |
| Logistic | dy/dt = ky(1 − y/L) | y = L/(1 + Ae⁻ᵏᵗ) |
Logistic peak growth at y = L/2 (inflection). At carrying capacity L, growth → 0.
After integrating both sides, write +C immediately. Then apply initial condition to find C. Skipping +C gives the wrong family of solutions and the IC step yields nonsense.
Disk: V = π ∫ [R(x)]² dx Washer: V = π ∫ ([Rout]² − [Rin]²) dxShell: V = 2π ∫ x · h(x) dx Cross-section: V = ∫ A(x) dx| Use | When |
|---|---|
| Disk in x | rotate y=f(x) around x-axis |
| Disk in y | rotate x=g(y) around y-axis |
| Shell in x | rotate y=f(x) around y-axis |
| Cross-sections | cross-section A(x) given (square, semicircle, etc.) |
L = ∫ √(1 + (f')²) dx. Or for parametric: ∫ √((dx/dt)² + (dy/dt)²) dt.f̄ = (1/(b−a)) ∫ab f dx. AP FRQ near-guarantee.Accumulation: if f'(x) is given, then f(b) − f(a) = ∫ab f'(x) dx (FTC). Used for distance, work, total change.
Washer method = π ∫ ([Rout]² − [Rin]²) dx. NOT π ∫ (Rout − Rin)². Algebraically very different. Geometrically: difference of disk areas, not area of difference.
| Test | Use when | Result |
|---|---|---|
| nth term | always first | an ↛ 0 ⇒ diverges |
| Geometric | an = ar^n | conv ⇔ |r| < 1 |
| p-series | 1/n^p | conv ⇔ p > 1 |
| Integral | an = f(n), f decreasing | matches ∫ f |
| Comparison / Limit comp | vs known series | same fate |
| Ratio | factorials, n!, x^n | L < 1 conv |
| Root | (stuff)^n | L < 1 conv |
| Alternating | (−1)^n an, an ↘ 0 | conv |
Geometric sum: Σ ar^n = a/(1 − r) for |r| < 1Power series: Σ cn(x − a)ⁿ has radius R from ratio test. Converges absolutely for |x − a| < R, diverges for > R, must check x = a±R individually.
If an → 0, the series might still diverge (harmonic). nth term only proves divergence, never convergence. Many AP students confidently misapply this both ways.
| Question says… | Use § from | Move |
|---|---|---|
| 'find f'(x)' / 'differentiate' | § ① | power · product · quotient · chain |
| 'how fast' / 2 vars in time | § ① | related rates (5-step recipe) |
| 0/0 or ∞/∞ in limit | § ① | L'Hôpital |
| 'show ∃ c with…' | § ① | MVT / Rolle's / IVT |
| (stuff)ⁿ · stuff' integrand | § ② | u-sub |
| poly · eˣ, ln x, poly · sin/cos | § ② | IBP (LIATE) |
| √(a²±x²), √(x²−a²) | § ② | trig sub |
| P(x)/Q(x), proper rational | § ② | partial fractions |
| integrand / limit blows up | § ② | improper — write as limit |
| 'volume rotated' | § ③ | disk · washer · shell · cross-section |
| 'arc length' | § ③ | ∫√(1+(f')²) |
| 'average value of f' | § ③ | (1/(b−a)) ∫f |
| Σ converges? | § ④ | nth term FIRST, then test ladder |
| (−1)ⁿ in series | § ④ | alternating series test + error bound |
| 'radius of convergence' | § ⑤ | ratio test in (x−a), check endpoints |
| 'Maclaurin of g(x)' | § ⑤ | substitute / differentiate / multiply known series |
| 'estimate f(x)' near nice a | § ⑤ | Taylor at a, Lagrange remainder |
| x = f(t), y = g(t) | § ⑥ | parametric: dy/dx = (dy/dt)/(dx/dt) |
| r = f(θ), 'area inside' | § ⑥ | ½ ∫ r² dθ |
| r(t) particle, 'speed at' | § ⑥ | |v(t)| = √((dx/dt)²+(dy/dt)²) |
| dy/dx = f(x)·g(y) | § ⑦ | separate, integrate, +C, IC |
| 'slope field' / 'Euler's' | § ⑦ | follow slopes / step y_{n+1}=y_n+hf |
| 'population' / 'cooling' / 'logistic' | § ⑦ | recognize the model, pick solution form |
If you've spent 5 min on one method and the integral isn't simplifying, switch. AP problems are designed for one specific tool. Going down the wrong path is the #1 time sink — don't be stubborn.
Volumes are cubic units. Velocity vs speed (signed vs |·|). +C on indefinite, NOT definite. Always restate what you computed at the end so partial credit isn't lost to ambiguity.