AP Calc AB Exam Things to Know Cold

y=f(x) must be continuous where?

critical point: dy/dx=0 or undefined or endpoints

local minimum

dy/dx goes (-,0,+) or (-,undef.,+) or 2nd derivative >0

local maximum

dy/dx changes from + to 0 to - or + to undef. to - or 2nd derivative < 0

point of inflection

concavity changes, 2nd derivative goes from (+,0,-), (-,0,+), (+, undef, -) or (-, undef., +)

d/dx x^n

nx^n-1

d/dx sinx

cosx

d/dx cosx

-sinx

d/dx tanx

sec^2x

d/dx cotx

-csc^2x

d/dx secx

secxtanx

d/dx cscx

-cscxcotx

d/dx lnu

1/u du/dx

d/dx e^u

e^u du/dx

d/dx sin^-1u

1/√(1-u^2) du/dx

d/dx cos^-1x

1/√(1-x^2)

d/dx tan^-1x

1/(1+x^2)

d/dx cot^-1x

-1/(1+x^2)

d/dx sec^-1x

1/(|x| √(x^2-1))

d/dx csc^-1x

-1/(|x| √(x^2-1))

d/dx a^x

a^x ln(a)

d/dx loga(x)

1/xlna

ftc part 2

∫ab f(x) dx = F(b) - F(a)

ftc part 1

F(x) = ∫[a to x] f(t) dt, then F'(x) = f(x)

ivt

if f(x) is continuous on [a,b] and y is a number between f(a) and f(b), there is at least one number x=c in the open interval (a,b) where f(c)=y

mvt

if the function f(x) is continuous on [a,b] and the first derivative exists on the interval (a,b) then there's at least 1 number x=c in (a,b) where f'(c)=f(b)-f(a)/b-a

trapezoidal rule

∫ (a to b) = ((b-a)/2n)(f(x)+2f(x)+2f(x)+f(x))

mvt average value

if the function f(x) is continuous on [a,b] and the first derivative exists on the interval (a,b) then there's at least 1 number x=c in (a,b) where f'(c)= (a∫b f(b)-f(a))/b-a

disk method

V = pi ∫ R(x)^2 dx

washer method

V= pi ∫ [R(x)^2] - [r(x)^2]

general volume equation

v = a∫b A(x) dx

velocity

d/dt of position

acceleration

d/dt of velocity

displacement

t0∫tf vdt

distance

t0∫tf |v| dt

average velocity

final position - initial position / total time

∫x^n dx

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

∫ sinx dx

-cosx + c

∫tanx dx

ln |secx| + c

evt

every continuous function on a closed interval has a highest y value and lowest y value

∫secx dx

ln |secx+tanx| + c

∫ sec^2x dx

tanx + c

∫secxtanx

secx + c

∫ e^x dx

e^x + C

∫ 1/x dx

∫ ln|x| +c

∫cosx dx

sinx +c

∫cotxdx

ln|sinx+c|

∫cscx dx

ln|cscx-cotx| +c

∫csc^2x dx

-cotx + c

∫cscxcotxdx

-cscx+c

f(x) is increasing

f’(x) >0

f(x) is decreasing

f’(x) <0

concave up

f’’(x)>0

concave down

f’’(x)<0

point of inflection

f’’(x) changes sign