AP Calculas AB

What does lim f(x) tell us?

End behavior / Horiontal Asymptote

3 Reasons we don’t take a dirrivative

Discontinuity, Sharp Turn, Vertical Tangent

Know this statement if the prompt says f(x) is differentiable

f(x) is diffrientiable and since diffrientability implies continuity f(x) is continuous

dirrivative of sin(x)

cos(x)

dirrivative of cos(x)

-sin(x)

dirrivative of tan(x)

sec²x

dirrivative of sec(x)

sec(x) * tan(x)

dirrivative of csc(x)

-csc(x) * cot(x)

dirrivative of cot(x)

-csc²x

Particle is moving down

v(t) is less than 0

Particle is moving up

v(t) is greater than 0

Particle is stopped

v(t)=0

Particle is slowing down

v(t) and v’(t) have diffrent signs

The particle is speeding up

v(t) and v’(t) have the same sign

The particle’s aceleration is negative

v’(t) is less than 0

the particles aceleration is positive

v’(t) is greater than 0

the particle changes direction

v(t) changes signs

average velocity

change in position / change in time

average aceleration

change in velocity / change in time

f(x) is increasing

f’(x) is greater than 0

f(x) Is deacreasing

f’(x) is less than 0

f(x) has a relative maximum

f’(x) changes from - to +

f(x) has a relative minimum

f’(x) changed from + to -

f’(x) has an exstrema

f’(x) changes signs

f(x) is concave down

f’’(x) is less than 0

f(x) is concave down

f’’(x) is less than 0

f(x) has a critical point

f’(x) = 0 or DNE

f(x) has a point of inflection

f’’(x) changes signs and slope of f’(x) changes signs

degree on top is than less than degree on bottom as x approaches infinity

0

degree on top is greater than degree on bottom as x approaches infinity

DNE

Degree on top is Equal to degree on bottom

ratio of leading coefficients

Conditions for IVT

f is continuous on [A,B]

Conditions for MVT

f is continuous on [A,B] and diffrientiable on (A,B)

Graph of y= e^x

(0,1) and (1,e)

Graph of y= ln(x)

(1,0) and (e,1)

e =

2.72

e is between

1/3 < 1/e < 1/2

dirrivative of ln(u)

u’ / u

dirrivative of e^u

e^u * u’

dirrivative of a^u

a * u’ lna

intagrating a^x

a^x * 1/lna +c

intagrating e^x

e^x + c

intagrating 1/x

ln|x| + c

if f and g are inverses with (a,b) on f and (b,a) on g, f’(a)

1/g’(b)

change of bases log_ab

ln b / ln a

direct variation

y= kx

inverse variation

y = k/x

exponential growth

y= Ce^kt

intagrating sin(x)

-cos(x) + c

intagrating cos(x)

sin(x) + c

intagrating csc²(x)

-cot(x) + c

intagrating sec²(x)

tan(x)

intagrating sec(x) * tan(x)

sec(x) + c

intagrating csc(x) * tan(x)

sec(x) +c

intagrating csc(x) * cot(x)

-csc(x) + c

ln 1

0

e⁰

1

ln 0

DNE

intagrating cot(x)

ln|sin(x)|

intagrating sec(x)

ln|sec(x) + tan(x)| + c

intagrating csc(x)

-ln|csc(x) + cot(x)| + c

dirivative of arc sin(u)

u’ / squareroot 1 - u²

dirivative arc cos(u)

-u’ / squareroot 1 - u²

dirivative arc tan

u’ / 1+u²

dirivative arc cot(u)

-u’ / 1+u²

dirrivative of arc sec(u)

u’ / |u| square root u² -1

dirrivative of arc csc(u)

-u’ / |u| square root u² -1

integral of du/ square root a² - u²

arc sin(u/a) + c

Total DIstance

intagration of |v(t)|

intagration of tan(x)

-ln|cos(x)| + c

intagral of du / a² + u²

1/a arc tan u/a +c

intagration of du/ u * square root u² + a²

1/a arc sec |u| / a +c

finding area by disk

pie integral (radius)²

finiding area by washer

pie intragal of (outer)² - pie (inner)²

Cross Section: squares

intagral of (base)²

Cross Section: squares equilateral triangles

squareroot of 3 / 4 intagral of (base)²

Cross Section: isosceles right triangles

½ intagral of (base)²

cross section: semicircles

pie / 8 (base)²

Steps for solving a separable differential equations

seperate integrate solve

Rate Problems

initial + Rate in - Rate out