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AP Calculus AB Study Guide

Comprehensive study guide for AP Calculus AB, aligned with the College Board Course and Exam Description. Covers all AB-only topics: limits, derivatives, integrals, and differential equations.

The limit of f(x)f(x) as xx approaches aa is LL if f(x)f(x) can be made arbitrarily close to LL by taking xx sufficiently close to aa (but not equal to aa):

limxaf(x)=L\lim_{x \to a} f(x) = L

A two-sided limit exists if and only if both one-sided limits exist and are equal:

limxaf(x)=L    limxaf(x)=limxa+f(x)=L\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L

If limxaf(x)\lim_{x \to a} f(x) and limxag(x)\lim_{x \to a} g(x) both exist, then:

LawExpression
Sumlim(f+g)=limf+limg\lim (f + g) = \lim f + \lim g
Differencelim(fg)=limflimg\lim (f - g) = \lim f - \lim g
Productlim(fg)=(limf)(limg)\lim (fg) = (\lim f)(\lim g)
Quotientlim(f/g)=limflimg\lim (f / g) = \dfrac{\lim f}{\lim g}, provided limg0\lim g \neq 0
Powerlimfn=(limf)n\lim f^n = (\lim f)^n
Constant multiplelimcf=climf\lim cf = c \lim f

If g(x)f(x)h(x)g(x) \leq f(x) \leq h(x) for all xx near aa (except possibly at aa), and limxag(x)=limxah(x)=L\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L, then limxaf(x)=L\lim_{x \to a} f(x) = L.

Commonly used to evaluate limx0sinxx=1\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1.

A function ff is continuous at x=ax = a if all three conditions are met:

  1. f(a)f(a) is defined
  2. limxaf(x)\displaystyle\lim_{x \to a} f(x) exists
  3. limxaf(x)=f(a)\displaystyle\lim_{x \to a} f(x) = f(a)

Types of discontinuity:

  • Removable: A hole in the graph (limit exists but function is undefined or unequal)
  • Jump: Left and right limits exist but are not equal
  • Infinite (essential): A vertical asymptote (function approaches ±\pm\infty)

If ff is continuous on [a,b][a, b] and NN is any value between f(a)f(a) and f(b)f(b), then there exists at least one c(a,b)c \in (a, b) such that f(c)=Nf(c) = N.

If limxaf(x)g(x)\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)} produces an indeterminate form 00\frac{0}{0} or ±±\frac{\pm\infty}{\pm\infty}, then:

limxaf(x)g(x)=limxaf(x)g(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

provided the limit on the right exists. May be applied repeatedly if necessary.

The derivative of ff at x=ax = a is the instantaneous rate of change:

f(a)=limh0f(a+h)f(a)hf'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}

Geometrically, f(a)f'(a) is the slope of the tangent line to the graph of ff at (a,f(a))(a, f(a)).

Power Rule:

ddxxn=nxn1\frac{d}{dx} x^n = nx^{n-1}

Product Rule:

ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x)\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)

Quotient Rule:

ddx[f(x)g(x)]=f(x)g(x)f(x)g(x)[g(x)]2\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}

Chain Rule:

ddxf(g(x))=f(g(x))g(x)\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)
Function f(x)f(x)Derivative f(x)f'(x)
cc (constant)00
xnx^nnxn1nx^{n-1}
sinx\sin xcosx\cos x
cosx\cos xsinx-\sin x
tanx\tan xsec2x\sec^2 x
secx\sec xsecxtanx\sec x \tan x
cscx\csc xcscxcotx-\csc x \cot x
cotx\cot xcsc2x-\csc^2 x
exe^xexe^x
lnx\ln x1x\dfrac{1}{x}
axa^xaxlnaa^x \ln a
logax\log_a x1xlna\dfrac{1}{x \ln a}
arcsinx\arcsin x11x2\dfrac{1}{\sqrt{1 - x^2}}
arccosx\arccos x11x2\dfrac{-1}{\sqrt{1 - x^2}}
arctanx\arctan x11+x2\dfrac{1}{1 + x^2}

When yy is defined implicitly as a function of xx, differentiate both sides with respect to xx, treating yy as a function of xx (using the chain rule where needed), then solve for dydx\dfrac{dy}{dx}.

The second derivative f(x)=d2ydx2f''(x) = \dfrac{d^2y}{dx^2} gives the rate of change of the first derivative. The nn-th derivative is denoted f(n)(x)f^{(n)}(x).

If ff is continuous on [a,b][a, b] and differentiable on (a,b)(a, b), then there exists at least one c(a,b)c \in (a, b) such that:

f(c)=f(b)f(a)baf'(c) = \frac{f(b) - f(a)}{b - a}

Increasing, Decreasing, and Critical Points

Section titled “Increasing, Decreasing, and Critical Points”
  • ff is increasing on an interval if f(x)>0f'(x) > 0 for all xx in that interval
  • ff is decreasing on an interval if f(x)<0f'(x) < 0 for all xx in that interval
  • A critical number is an interior point cc of the domain where f(c)=0f'(c) = 0 or f(c)f'(c) does not exist
  • ff is concave up where f(x)>0f''(x) > 0 (graph curves upward)
  • ff is concave down where f(x)<0f''(x) < 0 (graph curves downward)
  • An inflection point occurs where ff'' changes sign (concavity changes)

First Derivative Test: At a critical point cc:

  • ff' changes from positive to negative \Rightarrow local maximum
  • ff' changes from negative to positive \Rightarrow local minimum

Second Derivative Test: At a critical point cc where f(c)=0f'(c) = 0:

  • f(c)>0f''(c) > 0 \Rightarrow local minimum
  • f(c)<0f''(c) < 0 \Rightarrow local maximum
  • f(c)=0f''(c) = 0 \Rightarrow inconclusive

A global maximum/minimum is the absolute largest/smallest value of ff on its entire domain. Check endpoints, critical points, and any discontinuities.

  1. Identify the quantity to be optimised and write it as a function of one variable
  2. Determine the feasible domain
  3. Find critical numbers by setting the derivative to zero
  4. Evaluate the function at critical numbers and endpoints
  5. Identify the optimum value
  1. Identify all quantities that change with time
  2. Write an equation relating the quantities
  3. Differentiate both sides with respect to time tt (chain rule)
  4. Substitute known values and solve for the unknown rate

Procedure:

  1. Find domain, intercepts, and symmetry
  2. Identify asymptotes (vertical, horizontal, slant)
  3. Find the first derivative. Determine increasing/decreasing intervals and local extrema
  4. Find the second derivative. Determine concavity and inflection points
  5. Sketch the graph using all gathered information

FF is an antiderivative of ff if F(x)=f(x)F'(x) = f(x). The general antiderivative is:

f(x)dx=F(x)+C\int f(x)\, dx = F(x) + C

where CC is the constant of integration.

The definite integral is defined as the limit of Riemann sums:

abf(x)dx=limni=1nf(xi)Δx\int_a^b f(x)\, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x

where Δx=ban\Delta x = \dfrac{b - a}{n} and xix_i^* is a sample point in the ii-th subinterval.

  • Left sum: xi=xi1x_i^* = x_{i-1}
  • Right sum: xi=xix_i^* = x_i
  • Midpoint sum: xi=xi1+xi2x_i^* = \dfrac{x_{i-1} + x_i}{2}
  • Trapezoidal rule: abf(x)dxΔx2[f(a)+2i=1n1f(xi)+f(b)]\displaystyle\int_a^b f(x)\, dx \approx \frac{\Delta x}{2}\left[f(a) + 2\sum_{i=1}^{n-1}f(x_i) + f(b)\right]

Properties:

  • abf(x)dx=baf(x)dx\displaystyle\int_a^b f(x)\, dx = -\int_b^a f(x)\, dx
  • aaf(x)dx=0\displaystyle\int_a^a f(x)\, dx = 0
  • ab[f(x)±g(x)]dx=abf(x)dx±abg(x)dx\displaystyle\int_a^b [f(x) \pm g(x)]\, dx = \int_a^b f(x)\, dx \pm \int_a^b g(x)\, dx
  • abcf(x)dx=cabf(x)dx\displaystyle\int_a^b cf(x)\, dx = c\int_a^b f(x)\, dx
  • Additivity: abf(x)dx=acf(x)dx+cbf(x)dx\displaystyle\int_a^b f(x)\, dx = \int_a^c f(x)\, dx + \int_c^b f(x)\, dx

Part 1: If ff is continuous on [a,b][a, b], then the function F(x)=axf(t)dtF(x) = \int_a^x f(t)\, dt is differentiable and:

F(x)=f(x)F'(x) = f(x)

More generally, if G(x)=u(x)v(x)f(t)dtG(x) = \int_{u(x)}^{v(x)} f(t)\, dt, then:

G(x)=f(v(x))v(x)f(u(x))u(x)G'(x) = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x)

Part 2: If ff is continuous on [a,b][a, b] and FF is any antiderivative of ff, then:

abf(x)dx=F(b)F(a)\int_a^b f(x)\, dx = F(b) - F(a)

For integrals of the form f(g(x))g(x)dx\int f(g(x))g'(x)\, dx:

  1. Let u=g(x)u = g(x), then du=g(x)dxdu = g'(x)\, dx
  2. Rewrite the integral entirely in terms of uu
  3. Evaluate the integral
  4. Substitute back x=g1(u)x = g^{-1}(u)

For definite integrals, transform the limits: when x=ax = a, u=g(a)u = g(a); when x=bx = b, u=g(b)u = g(b).

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

A=ab[f(x)g(x)]dxA = \int_a^b [f(x) - g(x)]\, dx

If curves cross, split the integral at intersection points and take absolute values.

The average value of ff on [a,b][a, b] is:

favg=1baabf(x)dxf_{\text{avg}} = \frac{1}{b - a}\int_a^b f(x)\, dx

A first-order separable differential equation has the form:

dydx=g(x)h(y)\frac{dy}{dx} = g(x)h(y)

Solve by separating variables:

dyh(y)=g(x)dx    dyh(y)=g(x)dx\frac{dy}{h(y)} = g(x)\, dx \implies \int \frac{dy}{h(y)} = \int g(x)\, dx

A slope field (direction field) is a graphical representation of a first-order differential equation dydx=f(x,y)\dfrac{dy}{dx} = f(x, y). At each point (x,y)(x, y) on a grid, a short line segment is drawn with slope f(x,y)f(x, y). Solutions to the differential equation are curves that are tangent to the line segments at every point.

For a quantity yy that changes at a rate proportional to itself:

dydt=ky    y(t)=y0ekt\frac{dy}{dt} = ky \implies y(t) = y_0 e^{kt}
  • k>0k > 0: exponential growth
  • k<0k < 0: exponential decay

Half-life: t1/2=ln2kt_{1/2} = \dfrac{\ln 2}{|k|}

ddxxn=nxn1,ddxex=ex,ddxlnx=1x\frac{d}{dx} x^n = nx^{n-1}, \quad \frac{d}{dx} e^x = e^x, \quad \frac{d}{dx} \ln x = \frac{1}{x} ddx[fg]=fg+fg,ddx[fg]=fgfgg2\frac{d}{dx}[fg] = f'g + fg', \quad \frac{d}{dx}\left[\frac{f}{g}\right] = \frac{f'g - fg'}{g^2} ddxf(g(x))=f(g(x))g(x)\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)xndx=xn+1n+1+C(n1)\int x^n\, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1) 1xdx=lnx+C\int \frac{1}{x}\, dx = \ln|x| + C exdx=ex+C,axdx=axlna+C\int e^x\, dx = e^x + C, \quad \int a^x\, dx = \frac{a^x}{\ln a} + C sinxdx=cosx+C,cosxdx=sinx+C\int \sin x\, dx = -\cos x + C, \quad \int \cos x\, dx = \sin x + C sec2xdx=tanx+C,csc2xdx=cotx+C\int \sec^2 x\, dx = \tan x + C, \quad \int \csc^2 x\, dx = -\cot x + CIVT: f(c)=N for some c(a,b) if f continuous on [a,b]\text{IVT: } f(c) = N \text{ for some } c \in (a,b) \text{ if } f \text{ continuous on } [a,b] MVT: f(c)=f(b)f(a)ba for some c(a,b)\text{MVT: } f'(c) = \frac{f(b)-f(a)}{b-a} \text{ for some } c \in (a,b) FTC 1: ddxaxf(t)dt=f(x)\text{FTC 1: } \frac{d}{dx}\int_a^x f(t)\, dt = f(x) FTC 2: abf(x)dx=F(b)F(a)\text{FTC 2: } \int_a^b f(x)\, dx = F(b) - F(a)
  1. Show all working. The AP exam awards partial credit for correct intermediate steps even when the final answer is wrong. Write out every step evidently.
  2. Justify your answers. On free-response questions, explicitly state which theorem, test, or rule you are applying (e.g., “by the Intermediate Value Theorem” or “by the Second Derivative Test”).
  3. Check your calculator’s mode. Ensure radians mode is selected before evaluating trigonometric expressions. This is one of the most common sources of error.
  4. Master u-substitution. Many integration problems on the AP exam can be solved with a well-chosen substitution. Practise identifying the inner function and its derivative.
  5. Verify answers graphically. On the calculator-active section, use your calculator to sketch graphs and check that your analytical results (extrema, inflection points, intercepts) match.
  6. Do not leave blanks. Even if you cannot complete a problem, write down relevant formulas, diagrams, or reasoning — partial credit may be awarded.
  7. Time management. Spend roughly 15 minutes per free-response question. If stuck, move on and return later.
  1. Forgetting the chain rule. When differentiating composite functions (e.g., sin(3x2)\sin(3x^2)), students often omit the derivative of the inner function. Always ask: is there an inner function?
  2. Sign errors in the quotient rule. The correct order is numerator-derivative times denominator minus numerator times denominator-derivative: fgfgg2\dfrac{f'g - fg'}{g^2}. Getting this backwards changes the sign.
  3. Dropping the constant of integration. When finding antiderivatives, always include +C+ C. On free-response questions, the constant is essential for solving initial value problems.
  4. Confusing average rate of change with instantaneous rate of change. Average rate of change is f(b)f(a)ba\dfrac{f(b) - f(a)}{b - a} (a slope of a secant line); instantaneous rate of change is f(a)f'(a) (slope of a tangent line).
  5. Misapplying L’Hopital’s rule. Only use L’Hopital’s rule for indeterminate forms 00\frac{0}{0} or ±±\frac{\pm\infty}{\pm\infty}. Always verify the indeterminate form before differentiating.
  6. Incorrect limits in u-substitution. When using substitution on a definite integral, either transform the limits of integration to uu-values or substitute back to xx before evaluating. Do not mix old and new limits.
  7. Confusing the MVT with the IVT. The MVT guarantees a point where the instantaneous rate of change equals the average rate of change (f(c)f'(c)). The IVT guarantees a point where the function takes on a specific value (f(c)=Nf(c) = N). Know the difference.
TopicKey Ideas
Limits and ContinuityEvaluating limits, limit laws, Squeeze theorem, IVT, L’Hopital’s rule, continuity conditions
DifferentiationPower/product/quotient/chain rules, implicit differentiation, derivatives of all standard functions
Applications of DerivativesMVT, increasing/decreasing, concavity, extrema, optimisation, related rates, curve sketching
IntegrationAntiderivatives, Riemann sums, FTC parts 1 and 2, u-substitution, area between curves
Differential EquationsSeparable equations, slope fields, exponential growth/decay

The AP Calculus AB exam tests your ability to apply these concepts in both multiple-choice and free-response formats. Focus on understanding why each rule works — the AP exam rewards conceptual understanding as much as mechanical computation. Practise past papers under timed conditions and review every mistake to build confidence and accuracy.

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

  • Confusing terminology or concepts that appear similar but have distinct meanings.
  • Overlooking key assumptions or boundary conditions that limit applicability.