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AP Calculus AB/BC Cheat Sheet


  • lim⁡x→af(x)=L\lim_{x \to a} f(x) = L means f(x)f(x) can be forced arbitrarily close to LL for xx near aa (with x≠ax \ne a).
  • Two-sided limit exists exactly when the one-sided limits agree:
lim⁡x→a−f(x)=lim⁡x→a+f(x)=L.\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L.
  • For polynomials and rational functions, direct substitution works when the denominator is nonzero.

If lim⁡x→af(x)=L\lim_{x \to a} f(x) = L and lim⁡x→ag(x)=M\lim_{x \to a} g(x) = M:

lim⁡(f±g)=L±M,lim⁡(fg)=LM,lim⁡fg=LM (M≠0),lim⁡fn=Ln.\lim (f \pm g) = L \pm M, \qquad \lim (fg) = LM, \qquad \lim \frac{f}{g} = \frac{L}{M}\ (M \ne 0), \qquad \lim f^n = L^n. 00,∞∞,0⋅∞,∞−∞,1∞,00,∞0.\frac{0}{0}, \quad \frac{\infty}{\infty}, \quad 0 \cdot \infty, \quad \infty - \infty, \quad 1^\infty, \quad 0^0, \quad \infty^0.

Techniques: factor and cancel, multiply by a conjugate, combine fractions, use a known trig limit, or divide by the dominant power of xx.

lim⁡x→0sin⁡xx=1,lim⁡x→0tan⁡xx=1,lim⁡x→01−cos⁡xx=0.\lim_{x \to 0} \frac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \frac{\tan x}{x} = 1, \qquad \lim_{x \to 0} \frac{1-\cos x}{x} = 0.

If g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) near aa and lim⁡x→ag=lim⁡x→ah=L\lim_{x \to a} g = \lim_{x \to a} h = L, then lim⁡x→af=L\lim_{x \to a} f = L.

  • degree top << degree bottom: limit is 00 (horizontal asymptote y=0y=0),
  • degrees equal: limit is the ratio of leading coefficients,
  • degree top >> degree bottom: no horizontal asymptote (possible slant asymptote).

All three must hold: f(a)f(a) exists, lim⁡x→af(x)\lim_{x \to a} f(x) exists, and lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a).

Discontinuity types: removable (hole), jump, infinite (vertical asymptote), oscillatory.

If ff is continuous on [a,b][a,b] and NN is between f(a)f(a) and f(b)f(b), then f(c)=Nf(c) = N for some c∈(a,b)c \in (a,b).


Unit 2: Differentiation: Definition and Fundamental Properties

Section titled “Unit 2: Differentiation: Definition and Fundamental Properties”
f′(x)=lim⁡h→0f(x+h)−f(x)h,f′(a)=lim⁡x→af(x)−f(a)x−a.f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}, \qquad f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}.

Interpretations: instantaneous rate of change, slope of the tangent line, limit of secant slopes.

  • Differentiable at aa implies continuous at aa; the converse is false (corner, cusp, vertical tangent, discontinuity).
ddx(c)=0,ddx(xn)=nxn−1,ddx[cf]=cf′.\frac{d}{dx}(c) = 0, \qquad \frac{d}{dx}(x^n) = nx^{n-1}, \qquad \frac{d}{dx}[cf] = cf'. ddx[f±g]=f′±g′,ddx[fg]=f′g+fg′.\frac{d}{dx}[f \pm g] = f' \pm g', \qquad \frac{d}{dx}[fg] = f'g + fg'. ddx[fg]=f′g−fg′g2,g≠0.\frac{d}{dx}\left[\frac{f}{g}\right] = \frac{f'g - fg'}{g^2}, \qquad g \ne 0. ddx(sin⁡x)=cos⁡x,ddx(cos⁡x)=−sin⁡x,ddx(tan⁡x)=sec⁡2x.\frac{d}{dx}(\sin x) = \cos x, \qquad \frac{d}{dx}(\cos x) = -\sin x, \qquad \frac{d}{dx}(\tan x) = \sec^2 x. ddx(sec⁡x)=sec⁡xtan⁡x,ddx(csc⁡x)=−csc⁡xcot⁡x,ddx(cot⁡x)=−csc⁡2x.\frac{d}{dx}(\sec x) = \sec x \tan x, \qquad \frac{d}{dx}(\csc x) = -\csc x \cot x, \qquad \frac{d}{dx}(\cot x) = -\csc^2 x. ddx(ex)=ex,ddx(ax)=axln⁡a,ddx(ln⁡x)=1x.\frac{d}{dx}(e^x) = e^x, \qquad \frac{d}{dx}(a^x) = a^x \ln a, \qquad \frac{d}{dx}(\ln x) = \frac{1}{x}.

At x=ax=a, tangent slope is f′(a)f'(a), tangent line is y−f(a)=f′(a)(x−a)y - f(a) = f'(a)(x-a), and the normal slope is −1/f′(a)-1/f'(a) when f′(a)≠0f'(a) \ne 0.

f′′(x)f''(x) measures concavity (or acceleration). For position s(t)s(t): velocity v(t)=s′(t)v(t) = s'(t), acceleration a(t)=v′(t)=s′′(t)a(t) = v'(t) = s''(t), speed is ∣v(t)∣\lvert v(t) \rvert.

Near x=ax=a: f(x)≈f(a)+f′(a)(x−a)f(x) \approx f(a) + f'(a)(x-a).


Unit 3: Differentiation: Composite, Implicit, and Inverse Differentiation

Section titled “Unit 3: Differentiation: Composite, Implicit, and Inverse Differentiation”

If y=f(g(x))y = f(g(x)), then dydx=f′(g(x))g′(x)\dfrac{dy}{dx} = f'(g(x))g'(x), equivalently dydx=dydu⋅dudx\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}.

Differentiate both sides with respect to xx, multiplying by dydx\dfrac{dy}{dx} each time a derivative hits a yy term, then solve for dydx\dfrac{dy}{dx}.

If b=f(a)b = f(a) and f′(a)≠0f'(a) \ne 0:

(f−1)′(b)=1f′(a),(f−1)′(x)=1f′(f−1(x)).(f^{-1})'(b) = \frac{1}{f'(a)}, \qquad (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}. ddx(arcsin⁡x)=11−x2,ddx(arccos⁡x)=−11−x2,ddx(arctan⁡x)=11+x2.\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1-x^2}}, \qquad \frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}. ddxeu=euu′,ddxln⁡∣u∣=u′u.\frac{d}{dx} e^{u} = e^{u}u', \qquad \frac{d}{dx} \ln\lvert u \rvert = \frac{u'}{u}.

Logarithmic differentiation: take ln⁡\ln of both sides first when the variable is in both base and exponent (e.g. y=xxy = x^x).

  1. Draw and label a diagram.
  2. Write an equation relating the variables.
  3. Differentiate implicitly with respect to time.
  4. Substitute the requested instant (not before).
  5. Keep units consistent.

Unit 4: Contextual Applications of Differentiation

Section titled “Unit 4: Contextual Applications of Differentiation”

Q′(t)Q'(t) is the instantaneous rate of change of QQ, with units of QQ per unit of tt. Always interpret both sign and units.

v(t)=s′(t),a(t)=v′(t)=s′′(t),speed=∣v(t)∣.v(t) = s'(t), \qquad a(t) = v'(t) = s''(t), \qquad \text{speed} = \lvert v(t) \rvert.

Speed increases when vv and aa have the same sign; speed decreases when they have opposite signs.

V′(t)=(rate in)−(rate out).V'(t) = (\text{rate in}) - (\text{rate out}). L(x)=f(a)+f′(a)(x−a),dy=f′(x) dx,Δf≈f′(a) Δx.L(x) = f(a) + f'(a)(x-a), \qquad dy = f'(x)\,dx, \qquad \Delta f \approx f'(a)\,\Delta x.

Profit P(x)=R(x)−C(x)P(x) = R(x) - C(x); marginal cost/revenue/profit are C′(x)C'(x), R′(x)R'(x), P′(x)P'(x).

A complete interpretation names the quantity, the input value, the direction (sign), and the units, e.g. “at t=5t=5 the population is increasing at 4040 fish per year.”


Unit 5: Analytical Applications of Differentiation

Section titled “Unit 5: Analytical Applications of Differentiation”

x=cx=c is critical if f′(c)=0f'(c) = 0 or f′(c)f'(c) does not exist, with cc in the domain.

Increasing / decreasing and the First Derivative Test

Section titled “Increasing / decreasing and the First Derivative Test”
  • f′>0f' > 0: ff increasing; f′<0f' < 0: ff decreasing.
  • f′f' goes ++ to −- at cc: local max; −- to ++: local min; no sign change: neither.
  • f′′>0f'' > 0: concave up; f′′<0f'' < 0: concave down; inflection point where concavity changes.
  • If f′(c)=0f'(c) = 0: f′′(c)>0f''(c) > 0 gives a local min, f′′(c)<0f''(c) < 0 gives a local max, f′′(c)=0f''(c) = 0 is inconclusive.

Evaluate ff at all interior critical points and at both endpoints aa and bb, then compare values.

If ff is continuous on [a,b][a,b] and differentiable on (a,b)(a,b), some c∈(a,b)c \in (a,b) satisfies

f′(c)=f(b)−f(a)b−a.f'(c) = \frac{f(b)-f(a)}{b-a}.

Rolle’s Theorem is the case f(a)=f(b)f(a) = f(b).

For 00\dfrac{0}{0} or ∞∞\dfrac{\infty}{\infty} only:

lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x).\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}.

Identify the quantity, write it as a one-variable function, set the feasible domain, find critical points, then test candidates.

xn+1=xn−f(xn)f′(xn).x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.

Unit 6: Integration and Accumulation of Change

Section titled “Unit 6: Integration and Accumulation of Change”
∫xn dx=xn+1n+1+C (n≠−1),∫1x dx=ln⁡∣x∣+C,∫ex dx=ex+C.\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \ (n \ne -1), \qquad \int \frac{1}{x}\,dx = \ln\lvert x \rvert + C, \qquad \int e^x\,dx = e^x + C. ∫cos⁡x dx=sin⁡x+C,∫sin⁡x dx=−cos⁡x+C,∫sec⁡2x dx=tan⁡x+C.\int \cos x\,dx = \sin x + C, \qquad \int \sin x\,dx = -\cos x + C, \qquad \int \sec^2 x\,dx = \tan x + C. ∫11+x2 dx=arctan⁡x+C,∫11−x2 dx=arcsin⁡x+C.\int \frac{1}{1+x^2}\,dx = \arctan x + C, \qquad \int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C. ∑i=1nf(xi∗) Δx,Δx=b−an.\sum_{i=1}^n f(x_i^*)\,\Delta x, \qquad \Delta x = \frac{b-a}{n}.

Left/right/midpoint sums; if ff is increasing, a left sum underestimates and a right sum overestimates (reverse if decreasing).

∫abf(x) dx=lim⁡n→∞∑i=1nf(xi∗) Δx.\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*)\,\Delta x.

Gives signed area / net accumulation / total change of a rate.

∫abf(x) dx=F(b)−F(a)where F′=f.\int_a^b f(x)\,dx = F(b) - F(a) \quad \text{where } F' = f.

If g(x)=∫axf(t) dtg(x) = \displaystyle\int_a^x f(t)\,dt, then g′(x)=f(x)g'(x) = f(x). Chain-rule form:

ddx∫u(x)v(x)f(t) dt=f(v(x))v′(x)−f(u(x))u′(x).\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\,dt = f(v(x))v'(x) - f(u(x))u'(x).

With u=g(x)u = g(x), du=g′(x) dxdu = g'(x)\,dx:

∫f(g(x))g′(x) dx=∫f(u) du.\int f(g(x))g'(x)\,dx = \int f(u)\,du. favg=1b−a∫abf(x) dx.f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx. ∫abf(x) dx≈Δx2[y0+2y1+2y2+⋯+2yn−1+yn].\int_a^b f(x)\,dx \approx \frac{\Delta x}{2}\left[y_0 + 2y_1 + 2y_2 + \cdots + 2y_{n-1} + y_n\right].

A general solution carries a constant of integration (a family of curves); an initial condition pins it to a particular solution.

Slope fields draw dy/dxdy/dx at many points. Euler’s method (step size hh, with dy/dx=f(x,y)dy/dx = f(x,y)):

xn+1=xn+h,yn+1=yn+h f(xn,yn).x_{n+1} = x_n + h, \qquad y_{n+1} = y_n + h\,f(x_n,y_n).

If dydx=g(x)h(y)\dfrac{dy}{dx} = g(x)h(y), rewrite as 1h(y) dy=g(x) dx\dfrac{1}{h(y)}\,dy = g(x)\,dx and integrate both sides.

dydt=ky⟹y=Cekt.\frac{dy}{dt} = ky \quad \Longrightarrow \quad y = Ce^{kt}.

CC is the initial amount; k>0k>0 grows, k<0k<0 decays.

dydt=ky(1−yL).\frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right).

Carrying capacity LL; equilibria at y=0y=0 and y=Ly=L; growth is fastest at y=L/2y = L/2.


If f(x)≥g(x)f(x) \ge g(x) on [a,b][a,b]:

∫ab[f(x)−g(x)] dx,or with dy:∫cd[xright(y)−xleft(y)] dy.\int_a^b [f(x)-g(x)]\,dx, \qquad \text{or with } dy: \int_c^d [x_{\text{right}}(y) - x_{\text{left}}(y)]\,dy.

Split at intersection points when the curves cross.

net change=∫abR(t) dt,displacement=∫abv(t) dt,distance=∫ab∣v(t)∣ dt.\text{net change} = \int_a^b R(t)\,dt, \qquad \text{displacement} = \int_a^b v(t)\,dt, \qquad \text{distance} = \int_a^b \lvert v(t) \rvert\,dt.

Split the distance integral at sign changes of v(t)v(t).

V=∫abA(x) dx.V = \int_a^b A(x)\,dx.

Common cross-section areas: square A=s2A = s^2, semicircle A=12πr2A = \tfrac12 \pi r^2, equilateral triangle A=34s2A = \tfrac{\sqrt 3}{4}s^2.

V=π∫ab[R(x)]2 dx,V=π∫ab([R(x)]2−[r(x)]2) dx.V = \pi \int_a^b [R(x)]^2\,dx, \qquad V = \pi \int_a^b \left([R(x)]^2 - [r(x)]^2\right)\,dx. V=2π∫ab(radius)(height) dx.V = 2\pi \int_a^b (\text{radius})(\text{height})\,dx. L=∫ab1+[f′(x)]2 dx.L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx.

Evaluate as a limit, e.g. ∫1∞1x2 dx=lim⁡b→∞∫1b1x2 dx=1\displaystyle\int_1^\infty \frac{1}{x^2}\,dx = \lim_{b \to \infty} \int_1^b \frac{1}{x^2}\,dx = 1. A finite limit means it converges.


Unit 9: Parametric, Polar, and Vector-Valued Functions (BC-only)

Section titled “Unit 9: Parametric, Polar, and Vector-Valued Functions (BC-only)”

For x=f(t)x = f(t), y=g(t)y = g(t) with dx/dt≠0dx/dt \ne 0:

dydx=dy/dtdx/dt,d2ydx2=ddt(dydx)dx/dt.\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{dx/dt}.

Horizontal tangent: dy/dt=0dy/dt = 0, dx/dt≠0dx/dt \ne 0. Vertical tangent: dx/dt=0dx/dt = 0, dy/dt≠0dy/dt \ne 0.

speed=[x′(t)]2+[y′(t)]2,L=∫ab[x′(t)]2+[y′(t)]2 dt.\text{speed} = \sqrt{[x'(t)]^2 + [y'(t)]^2}, \qquad L = \int_a^b \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt. x=rcos⁡θ,y=rsin⁡θ,r2=x2+y2.x = r\cos\theta, \qquad y = r\sin\theta, \qquad r^2 = x^2 + y^2.

Polar slope, for r=f(θ)r = f(\theta):

dydx=r′(θ)sin⁡θ+r(θ)cos⁡θr′(θ)cos⁡θ−r(θ)sin⁡θ.\frac{dy}{dx} = \frac{r'(\theta)\sin\theta + r(\theta)\cos\theta}{r'(\theta)\cos\theta - r(\theta)\sin\theta}. A=12∫ab[r(θ)]2 dθ,L=∫ab[r(θ)]2+[r′(θ)]2 dθ.A = \frac12 \int_a^b [r(\theta)]^2\,d\theta, \qquad L = \int_a^b \sqrt{[r(\theta)]^2 + [r'(\theta)]^2}\,d\theta.

For r(t)=⟨x(t),y(t)⟩\mathbf{r}(t) = \langle x(t), y(t) \rangle: velocity r′(t)\mathbf{r}'(t), acceleration r′′(t)\mathbf{r}''(t), speed ∣r′(t)∣\lvert \mathbf{r}'(t) \rvert. Differentiate and integrate component by component.


Unit 10: Infinite Sums and Series (BC-only)

Section titled “Unit 10: Infinite Sums and Series (BC-only)”
∑n=0∞arn=a1−rwhen ∣r∣<1 (diverges otherwise).\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \quad \text{when } \lvert r \rvert < 1 \ \text{(diverges otherwise)}. ∑n=1∞1np converges iff p>1;∑n=1∞1n diverges.\sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges iff } p > 1; \qquad \sum_{n=1}^{\infty} \frac{1}{n} \text{ diverges}.
  • nth-term test: if lim⁡n→∞an≠0\lim_{n \to \infty} a_n \ne 0, the series diverges (can only prove divergence).
  • Integral test: positive, continuous, decreasing ff with f(n)=anf(n) = a_n; series and ∫f\int f share fate.
  • Direct comparison and limit comparison (with 0<lim⁡an/bn<∞0 < \lim a_n/b_n < \infty).
  • Alternating series test: ∑(−1)nbn\sum (-1)^n b_n converges if bnb_n decreases and bn→0b_n \to 0.
  • Ratio test: L=lim⁡∣an+1an∣L = \lim \left\lvert \dfrac{a_{n+1}}{a_n} \right\rvert; root test: L=lim⁡∣an∣nL = \lim \sqrt[n]{\lvert a_n \rvert}. L<1L < 1 converges, L>1L > 1 diverges, L=1L = 1 inconclusive.

∑∣an∣\sum \lvert a_n \rvert converges: absolute (implies convergence). ∑an\sum a_n converges but ∑∣an∣\sum \lvert a_n \rvert diverges: conditional.

For ∑an(x−c)n\displaystyle\sum a_n (x-c)^n, there is a radius RR: converges for ∣x−c∣<R\lvert x-c \rvert < R, diverges for ∣x−c∣>R\lvert x-c \rvert > R, and endpoints must be tested separately. Use the ratio test to find RR.

f(x)=∑n=0∞f(n)(c)n!(x−c)n(Maclaurin: c=0).f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n \quad (\text{Maclaurin: } c=0).

Core series to memorize:

11−x=∑n=0∞xn (∣x∣<1),ex=∑n=0∞xnn!.\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \ (\lvert x \rvert < 1), \qquad e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}. sin⁡x=∑n=0∞(−1)nx2n+1(2n+1)!,cos⁡x=∑n=0∞(−1)nx2n(2n)!.\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}, \qquad \cos x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}.

Alternating series remainder: ∣Rn∣≤bn+1\lvert R_n \rvert \le b_{n+1} (first omitted term). Lagrange error bound:

∣Rn(x)∣≤M(n+1)!∣x−c∣n+1,\lvert R_n(x) \rvert \le \frac{M}{(n+1)!}\lvert x-c \rvert^{n+1},

where MM bounds the next derivative between cc and xx.


  1. Doing limit operations on forms that are not actually indeterminate (e.g. ∞/3\infty/3).
  2. Forgetting the chain-rule factor in differentiation or the reverse factor in u-substitution.
  3. Treating yy as a constant during implicit differentiation, or dropping dy/dxdy/dx.
  4. Calling every critical point an extremum without a sign-change or second-derivative check.
  5. Using L’Hopital when the form is not 0/00/0 or ∞/∞\infty/\infty.
  6. Forgetting +C+C on indefinite integrals.
  7. Reporting velocity when the question asks for speed, or displacement when it asks for total distance.
  8. Substituting the instant before differentiating in related-rates problems.
  9. (BC) Stopping after the radius of convergence without testing endpoints.
  10. (BC) Treating an→0a_n \to 0 as proof of convergence.

  1. Identify which unit/tool the problem belongs to before computing.
  2. Check whether direct substitution, a derivative rule, or an integral technique applies.
  3. Track units, and interpret the sign of any rate.
  4. For applications, draw the picture (slice, diagram, slope field) first.
  5. State theorems’ hypotheses (continuity, differentiability) when justifying.
  6. Confirm the answer’s magnitude and sign make sense in context.

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