AP Calculus AB: Stuff You Must Know Cold

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A comprehensive reference set of vocabulary terms, theorems, and formulas adapted from the 'Stuff You Must Know Cold' handout for AP Calculus AB exam preparation.

Last updated 7:15 PM on 6/2/26
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30 Terms

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Power Rule

ddx(xn)=nxn1\frac{d}{dx}(x^n) = nx^{n-1}

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Derivative of tan(x)\tan(x)

ddx(tan(x))=sec2(x)\frac{d}{dx}(\tan(x)) = \sec^2(x)

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Chain Rule

ddx[f(u)]=f(u)dudx\frac{d}{dx}[f(u)] = f'(u)\frac{du}{dx} or dydx=dydududx\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}

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Product Rule

ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} or uv+vuuv' + vu'

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Quotient Rule

ddx(uv)=vdudxudvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} or vuuvv2\frac{vu' - uv'}{v^2}

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Intermediate Value Theorem (IVT)

If the function f(x)f(x) is continuous on [a,b][a, b] and yy is a number between f(a)f(a) and f(b)f(b), then there exists at least one number x=cx = c in the open interval (a,b)(a, b) such that f(c)=yf(c) = y.

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Mean Value Theorem (MVT)

If the function f(x)f(x) is continuous on [a,b][a, b] AND the first derivative exists on the interval (a,b)(a, b), then there is at least one number x=cx = c in (a,b)(a, b) such that f(c)=f(b)f(a)baf'(c) = \frac{f(b) - f(a)}{b - a}.

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Rolle’s Theorem

If the function f(x)f(x) is continuous on [a,b][a, b], the first derivative exists on (a,b)(a, b), and f(a)=f(b)f(a) = f(b), then there is at least one number x=cx = c in (a,b)(a, b) such that f(c)=0f'(c) = 0.

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Extreme Value Theorem (EVT)

If the function f(x)f(x) is continuous on [a,b][a, b], then the function is guaranteed to have an absolute maximum and an absolute minimum on the interval.

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Critical Point

A point where dydx=0\frac{dy}{dx} = 0 or is undefined; endpoints must also be considered for absolute extrema.

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Point of Inflection

A point where concavity changes and d2ydx2\frac{d^2y}{dx^2} changes sign (goes from positive to negative or vice versa).

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Derivative of an Inverse Function

If ff has an inverse function gg, then g(x)=1f(g(x))g'(x) = \frac{1}{f'(g(x))}.

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Average Rate of Change (ARoC)

msec=f(b)f(a)bam_{sec} = \frac{f(b) - f(a)}{b - a}

14
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First Derivative Test: Relative Maximum

Occurs if f(x)=0f'(x) = 0 or DNE and the sign of f(x)f'(x) changes from positive (+) to negative (-).

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Horizontal Asymptotes Cases

  1. Numerator exponent < denominator: y=0y = 0. 2. Numerator exponent > denominator: DNE. 3. Numerator exponent = denominator: y=aby = \frac{a}{b} (leading coefficients quotient).
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Fundamental Theorem of Calculus (FTC)

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

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Speed

The absolute value of velocity: v(t)|v(t)|.

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Displacement

t0tfv(t)dt\int_{t_0}^{t_f} v(t)\,dt

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Total Distance

Integral of v(t)|v(t)| over the time interval.

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Average Velocity

ΔxΔt=final positioninitial positiontotal time\frac{\Delta x}{\Delta t} = \frac{\text{final position} - \text{initial position}}{\text{total time}}

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Exponential Growth and Decay

Uses y=Cekty = Ce^{kt}. Occurs when the rate of change of yy is proportional to yy (y=kyy' = ky).

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Mean Value Theorem for Integrals (Average Value)

If f(x)f(x) is continuous on [a,b][a, b], then there exists x=cx = c such that favg=1baabf(x)dxf_{avg} = \frac{1}{b - a} \int_a^b f(x)\,dx.

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Trapezoidal Rule (Even Intervals)

abf(x)dxba2n[y0+2y1+2y2++2yn1+yn]\int_a^b f(x)\,dx \approx \frac{b - a}{2n} [y_0 + 2y_1 + 2y_2 + \dots + 2y_{n-1} + y_n]

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Area Between Two Curves (Slices \perp to x-axis)

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

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Volume by Disk Method (About x-axis)

V=πab[R(x)]2dxV = \pi \int_a^b [R(x)]^2\,dx

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Volume by Washer Method (About x-axis)

V=πab([R(x)]2[r(x)]2)dxV = \pi \int_a^b ([R(x)]^2 - [r(x)]^2)\,dx

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Cross Section Volume: Equilateral Triangles

V=34ab(base)2dxV = \frac{\sqrt{3}}{4} \int_a^b (base)^2\,dx

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Basic Integral: duu\int \frac{du}{u}

lnu+C\ln|u| + C

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Derivative of sin1(ua)\sin^{-1}\left(\frac{u}{a}\right)

ddx[sin1(ua)]=1a2u2dudx\frac{d}{dx} \left[ \sin^{-1}\left(\frac{u}{a}\right) \right] = \frac{1}{\sqrt{a^2 - u^2}} \frac{du}{dx}

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Derivative of tan1(ua)\tan^{-1}\left(\frac{u}{a}\right)

ddx[tan1(ua)]=aa2+u2dudx\frac{d}{dx} \left[ \tan^{-1}\left(\frac{u}{a}\right) \right] = \frac{a}{a^2 + u^2} \frac{du}{dx}