Studied by 0 peopled/dx cotu
-u' csc^2(u)
d/dx [f(x)/g(x)] product rule
f(x)g'(x) + g(x)f'(x)
d/dx [f(x)/g(x)] quotient rule
g(x)f'(x)-f(x)g'(x)/g(x)^2
PVA
Position is x(t)
x'(t) = v(t)
v'(t) = a(t)
particle farthest left/down
Minimum of x(t)
particle farthest right/up
Maximum of x(t)
particle is at rest when
v(t)=0
speed increases when
v(t) and a(t) have the same sign
average velocity (when given position function)
[x(b) - x(a)]/ b-a
average velocity (when given velocity function)
1/b-a ∫b a [v(t) dt]
total displacement
∫b a v(t) dt
total distance (when given velocity)
∫ b a | v(t) | d(t)
d/dx [f(g(x)
f'(g(x))g'(x)
if g(x) is the inverse of f(x)
g'(b)= 1/f'(a)
d/dx (sin-1 u)
u'/sqrt(1-u^2)
d/dx (cos-1 u)
1/√(1-u^2) * -u'
d/dx (tan-1 u)
u'/(1+u^2)
d/dx (sq.rt u)
1 / 2[sq.rt u]
d/dx (a^u)
ln(a)a^u du/dx
d/dx (e^u)
e^u u'
graph of y = e^x
________|> (up)
d/dx loga(u)
u'/u * 1/lna
d/dx (ln u)
u'/u
y=lnx
_________|> (down)
if f(x) is on a closed interval [a,b], to find the absolute extrema, use the..
candidate's test (test the end points and critical numbers)
critical values occur when
f'(x) = 0 or f'(x) = DNE
f(x) increases/decreases
when f'(x) is (+) or (-)
f(x) has a relative minimum where
f'(x) changes from (-) to (+)
f(x) has a relative maximum where
f'(x) changes from (+) to (-)
May 13, 2024
I will make a 5 on the AP Calc test because success is where preparation and oppurtunity meet
My calculator should be on...
radian mode
I will round final frq answers to...
thousandths place
I WILL NOT LEAVE ANY FRQS
BLANK!!
I will keep trank of my own time
30 non cal. MCQ=1 hour
15 calc. MCQ=45 min
2 calc. FRQ=30 min
4 non calc. FRQ= 1 hour
I will go through all of the MCQ that I am confident about first and..
Save the hard ones for last
d/dx a^n
anx ^n-1
d/dx sin(u)
u'cos u
d/dx cos(u)
-u' sin u
d/dx tan u
u' sec^2 u
d/dx csc u
-u' csc u cot u
d/dx sec u
u' sec u tan u
f(x) has a POI where
f’(x) has a rel. max or min OR f’’(x) changes signs
f(x) is concave up
when f’(x) is increasing OR when f’’(x) is positive
f(x) is concave down
when f’(x) is decreasing OR when f’’(x) is negative
tangent lines
overapprox concave down f(x) & underapprox concave up f(x)
use MVT if f(x) is
-continous on (a,b)
-differentiable on (a,b)
then f’(c ) = (f(b)-f(a))/b-a
∫axn^ndx
(a(x^n+1 )/n+1) +c
intergration using area
∫b a f(x)dx= (area above x-axis) - (above below x-axis)
left riemann sums
overapprox decreasing f(x) & underapprox increasing f(x)
right riemann sums
overapprox increasing f(x) & underapprox decreasing f(x)
midpoint riemann sums
overapprox concave down f(x) & underapprox concave up f(x)
area of trapezoid
½ (h)(b1+b2)
trapezoid rule
overapprox. concave up f(x) & underapprox concave down f(x)
properties of definite intergrals
∫b a kf(x)dx = k∫b a f(x)dx
∫a a f(x)=0
∫b a (f(x) +_ g(x)) dx= ∫b a f(x)dx +_ ∫b a g(x)dx
∫a b f(x) dx= -∫b a f(x)dx
∫sin(u)du
-1/u’ cosu +c
∫cos(u)du
1/u’ sinu + c
∫sec²(u) du
1/u’ tanu +c
∫csc²(u)du
-1/u’ cotu+c
∫sec(u)tan(u)du
1/u’ secu+c
∫csc(u)cot(u)du
-1/u’ cscu+c
intergration by substitution
∫f(g)x))g’(x)dx =∫f(u)du =F(u) +c
u=g(x) & u’=g’(x)dx
∫tan(u)du
-ln|cos(u)|+c OR ln|sec(u)|+c
∫cot(u)du
ln|sinu| +c
∫sec(u)du
ln|sec(u) + tan(u)| +c
∫csc(u)du
-ln|cscu + cotu| +c
∫e^kx dx
1/k e^kx +c
Note
Flashcard