AP CALCULUS AB REVIEW SHEET
LIMITS LAWS
lim 𝑓(𝑥 ) = 𝐿 exists if an only
𝑥→𝑎
is lim− 𝑓 (𝑥 ) = lim+ 𝑓 (𝑥 ) = 𝐿
LIMITS
𝑥→𝑎
➢ lim
𝑥→0
➢ lim
𝑥→𝑎
sin 𝑐𝑥
𝑥
𝑥→0
sin 𝑥
➢ lim
𝑥→∞ 𝑥
sin2 𝑥
➢ lim
𝑥→0
= 1,
𝑐𝑥
cos 𝑥−1
CONTINUITY
A function is continuous at a if
lim− 𝑓 (𝑥 ) = lim+ 𝑓(𝑥 ) = 𝑓(𝑎)
L’HOPITALS RULE
If lim 𝑓(𝑥) = 0 and lim 𝑔(𝑥) = 0
TYPES OF DISCONTINUITY
𝑥→𝑐
𝑥→𝑎
𝑥→𝑎
𝑥→𝑐
𝑓′ (𝑥)
𝑓(𝑥)
then lim 𝑔(𝑥) = lim 𝑔′(𝑥) = 𝐿.
𝑥→𝑐
Infinte or
Asymptote
Point or
Removable
=0
𝑥→𝑐
or
lim 𝑓(𝑥) = ±∞ and lim 𝑔(𝑥) =
±∞
=0
=0
𝑥
𝑥→𝑐
DEFINITION OF THE
DERIVATIVE
𝑥→𝑐
Jump
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
𝑓 ′ (𝑥 ) = lim
𝑓(𝑥 ) − 𝑓(𝑎)
𝑥→𝑎
𝑥−𝑎
𝑓 ′ (𝑥 ) = lim
A derivative does not exist when…
… because the
left and right
hand derivative
are not equal
Corner
ADVANCED DERIVATIES
BASIC DERIVATIVES
CONSTANT:
Cusp
𝑑
𝑑𝑥
Discontinuity
[𝑐 ] = 0
𝑑
POWER RULE: 𝑑𝑥 [𝑥 𝑛 ] = 𝑛𝑥 𝑛−1
PRODUCT RULE
𝑑
[𝑓(𝑥) ∙ 𝑔(𝑥 )] = 𝑓(𝑥 )𝑔′ (𝑥 ) + 𝑔(𝑥 )𝑓 ′ (𝑥 )
𝑑𝑥
“1(d2)+2(d1)”
QUOTIENT RULE
𝑑 𝑓 (𝑥 )
𝑔(𝑥 )𝑓 ′ (𝑥 ) − 𝑓 (𝑥 )𝑔′(𝑥)
[
]=
[𝑔(𝑥 )]2
𝑑𝑥 𝑔(𝑥 )
Ho(dHi) – Hi(dHo)
"
"
HoHo
CHAIN RULE
𝑑
[𝑓(𝑔(𝑥))] = 𝑓′(𝑔(𝑥 )) ∙ 𝑔′(𝑥)
𝑑𝑥
INVERSE TRIG DERIVATIVES
𝑑
𝑢′
[sin−1 𝑢] =
𝑑𝑥
√1 − 𝑢2
𝑑
−𝑢′
−1
[cos 𝑢] =
𝑑𝑥
√1 − 𝑢2
𝑑
𝑢′
[tan−1 𝑢] =
𝑑𝑥
1 + 𝑢2
𝑑
−𝑢′
[cot −1 𝑢] =
𝑑𝑥
1 + 𝑢2
𝑑
𝑢′
[sec −1 𝑢] =
𝑑𝑥
|𝑢|√𝑢2 − 1
𝑑
−𝑢′
[csc −1 𝑢] =
𝑑𝑥
|𝑢|√𝑢2 − 1
… because
the derivative
is undefined.
Vertical Tangent
TRIG DERIVATIVES
𝑑
[sin 𝑢] = cos 𝑢 𝑑𝑢
𝑑𝑥
𝑑
[cos 𝑢] = − sin 𝑢 𝑑𝑢
𝑑𝑥
𝑑
[tan 𝑢] = sec 2 𝑢 𝑑𝑢
𝑑𝑥
𝑑
[cot 𝑢] = − csc 2 𝑢 𝑑𝑢
𝑑𝑥
𝑑
[sec 𝑢] = sec 𝑢 tan 𝑢 𝑑𝑢
𝑑𝑥
𝑑
[csc 𝑢] = − csc 𝑢 cot 𝑢 𝑑𝑢
𝑑𝑥
LOG, EXPONENTIAL, & INVERSES
𝑑 𝑔(𝑥)
[𝑒
] = 𝑒 𝑔(𝑥) ∙ 𝑔′ (𝑥 )
𝑑𝑥
𝑑
𝑔′(𝑥)
[ln 𝑔(𝑥)] =
𝑑𝑥
𝑔(𝑥)
𝑑 𝑔(𝑥)
[𝑎
] = ln 𝑎 ∙ 𝑎 𝑔(𝑥) ∙ 𝑔′(𝑥)
𝑑𝑥
𝑑
𝑔′(𝑥)
[log 𝑎 𝑔(𝑥)] =
𝑑𝑥
ln 𝑎 𝑔(𝑥)
𝑑 −1
1
[𝑓 (𝑥)] = ′ −1
𝑑𝑥
𝑓 [𝑓 (𝑥 )]
𝑑
𝑓(𝑥)
[|𝑓(𝑥 )|] =
𝑓′(𝑥)
𝑑𝑥
|𝑓(𝑥 )|
THEOREMS
MEAN VALUE THM
If f is continuous on
[𝑎, 𝑏] and differentiable
on (𝑎, 𝑏), then there
exists a c on the interval
(𝑎, 𝑏) such that
𝑓(𝑏) − 𝑓(𝑎)
𝑓 ′ (𝑐 ) =
𝑏−𝑎
ROLLE’S THM
If f is continuous on [𝑎, 𝑏],
differentiable on (𝑎, 𝑏),
and 𝑓 (𝑎) = 𝑓(𝑏), then
there exists a c on the
interval (𝑎, 𝑏) such that
𝑓(𝑏) − 𝑓(𝑎)
𝑓 ′ (𝑐 ) =
=0
𝑏−𝑎
(
(
GRAPH OF 𝒇 𝒙), 𝒇′ 𝒙), & 𝒇′′(𝒙)
Connections between 𝑓′(𝑥) and 𝑓(𝑥)
SPEED
NET & TOTAL DISTANCE
EXTREME VALUE THM
If f is continuous on [𝑎, 𝑏],
then f has an absolute
maximum and an absolute
minimum value at 𝑥 =
𝑎, 𝑥 = 𝑏, or when 𝑓 ′ (𝑥 ) =
0 or 𝑓 ′ (𝑥) is undefined.
Connections between 𝑓′′(𝑥) and 𝑓(𝑥)
If 𝑓 ′ (𝑥 ) …
Then 𝑓 (𝑥) …
If 𝑓 ′′ (𝑥 ) …
Then 𝑓 (𝑥) …
… is 0 or undefined
…has a potential
… is 0 or
…has a potential point
at 𝑥 = 𝑎
relative extrema
undefined at 𝑥 = 𝑎 of inflection
… is positive
… is increasing
… is positive
… is concave up
… is negative
… is decreasing
… in negative
… is concave down
… changes from
… 𝑓(𝑎) is a relative
… changes from
… (𝑎, 𝑓(𝑎) is a point of
positive to negative
maximum
signs
inflection
… changes from
… 𝑓(𝑎) is a relative
negative to positive
minimum
nd
2 Derivative Test:
𝑓 ′ (𝑎) = 0 or undefined 𝑓 ′′ (𝑥 ) < 0 𝑎 is a relative maximum since f is concave down
𝑓 ′ (𝑎) = 0 or undefined 𝑓 ′′ (𝑥 ) > 0 𝑎 is a relative minimum since f is concave up
Speed = |𝑣(𝑡)|
FTC
INTERMEDIATE VALUE
THM
If f is continuous on [𝑎, 𝑏]
and there is a value of k
between 𝑓(𝑎) and 𝑓(𝑏),
then there exists at least
on value of c on (𝑎, 𝑏)
such that 𝑓(𝑐 ) = 𝑘.
Increasing
• Velocity and Acceleration
have the same sign or
direction
Net Distance:
The distance from where the object begins
and where it ends.
Decreasing
• Velocity and Acceleration
have opposite sign or
direction
Total Distance:
The sum of all distances moved in any
direction.
On the interval [a, b], where 𝑝(𝑡) is the position function…
… the net distance is
I. |𝑝(𝑎) − 𝑝(𝑏)|
𝑏
II. ∫𝑎 𝑣(𝑡) 𝑑𝑡
… the total distance is
I. |𝑝(𝑎) − 𝑝(𝑐 )| + |𝑝(𝑐 ) − 𝑝(𝑏)|, where c is
where the object changes direction.
𝑏
II. ∫𝑎 |𝑣(𝑡)| 𝑑𝑡
Fundamental Theorem of Calculus Part 1
If 𝑓(𝑥) is continuous on [a, b] and 𝐹(𝑥) is the
anti-derivative of 𝑓(𝑥), then
Fundamental Theorem of Calculus Part 2
𝑏
∫ 𝑓 (𝑥 )𝑑𝑥 = 𝐹 (𝑏) − 𝐹(𝑎)
𝑔(𝑥)
If 𝐹 (𝑥 ) = ∫𝑎 𝑓 (𝑡)𝑑𝑡, then
𝐹 ′ (𝑥 ) = 𝑓(𝑔(𝑥 )) ∙ 𝑔′(𝑥).
TANGENT
𝑎
Tangent Line at 𝑥 = 𝑎:
𝑦 − 𝑓 (𝑎) = 𝑓 ′ (𝑎)(𝑥 − 𝑎)
Normal Line at 𝑥 = 𝑎:
𝑦 − 𝑓 (𝑎 ) = −
1
(𝑥 − 𝑎 )
𝑓′(𝑎)