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AP Calculus

AP Calculus

memorize.aimemorize.ai (lvl 286)
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40 cards
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Section 1

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d/dx [arctanu] =

Front

1 / 40

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Last updated

6 years ago

Date created

Mar 1, 2020

Cards (40)

Section 1

(40 cards)

d/dx [arctanu] =

Front

u' / (1 + u^2)

Back

d/dx [cotx] =

Front

-csc^2 x

Back

∫ secxdx =

Front

ln /secx + tanx/ + C

Back

d/dx [cosx] =

Front

-sinx

Back

∫ x^n dx =

Front

x^(n+1) / (n+1) + C

Back

∫ sec^2 dx =

Front

tanx + C

Back

Product Rule

Front

y = uv y' = uv' + vu'

Back

∫ du/√a^2 - u^2

Front

arcsin u/a + C

Back

∫ cosxdx =

Front

sinx + C

Back

d/dx [arccosu] =

Front

- u'/√(1-u^2)

Back

∫ tanxdx =

Front

-ln /cosx/ + C

Back

y = e^u

Front

y' = e^u du/dx

Back

lim x->0 sinx/x =

Front

1

Back

lim x --> 0 1-cosx/x =

Front

0

Back

Quotient Rule

Front

y = u/v y' = (uv'-vu') / v²

Back

∫ du / a^2 + u^2 =

Front

1/a arctan u/a + C

Back

d/dx [arcsinu] =

Front

u'/√(1-u^2)

Back

y = a^x

Front

y' = a^x lna

Back

Area Between Two Curves

Front

∫ b on top [f(x) - g(x)] f(x) is top function g(x) is bottom function

Back

Rolle's Theorem

Front

Let f be continuous on [a,b] and differentiable on (a,b). If f(a)=f(b) then there is at least one c on (a,b) such that f'(c)=0

Back

d/dx [sinx] =

Front

cosx

Back

∫ a^u du

Front

1/lna a^u + C

Back

d/dx [C] =

Front

0

Back

∫ csc^2 x dx =

Front

-cotx + C

Back

d/dx [cscx] =

Front

-cotxcscx

Back

d/dx [secx] =

Front

secxtanx

Back

∫ cscxdx =

Front

- ln /cscx + cox/ + C

Back

∫ sinxdx =

Front

-cosx + C

Back

∫ cscxcotxdx =

Front

-cscx + C

Back

Volume

Front

π ∫ b on top [R(x) ^2 - r(x) ^2 ] dx

Back

∫ e^u du

Front

e^u + C

Back

Average Value

Front

( 1/(b-a) ) ∫ b on top ∫f(x)dx

Back

Conditions of Continuity

Front

I. f(c) must exist II. f(c) = lim x --> c must exist III. f(c) = lim x --> f(c)

Back

∫ du/u =

Front

ln /u/ + C

Back

∫ cotxdx =

Front

ln /sinx/ + C

Back

d/dx [x^n] =

Front

nx^n-1

Back

d/dx [tanx] =

Front

sec^2 x

Back

The Mean Value Theorem

Front

If f is continuous on [a,b] and differentiable on (a,b), then there exists a c on (a,b) such that f'(c)=( f(b)-f(a) )/ (b-a)

Back

∫ secxtanxdx =

Front

secx + C

Back

y = ln u

Front

y' = 1/u du/dx

Back