Equations of note
1.5
a 4 −b 4 =(a−b)(a+b)(a2 +b2 )
{
|x|= x if x≥0
−x if x <0
}
sin θ
=1 ; lim sin θ=θ
θ
θ⇒0
θ⇒0
lim
|x−a|<k o r −k <( x−a)<k
|x−a|≤k o r −k≤( x−a)≤k
cos θ−1
1−cos θ
=0 o r lim
=0
θ
θ
θ⇒0
θ⇒0
lim
1.7
1.10
f ( x +Δ x)−f ( x) Δ x f ( x +h)−f ( x)
=
;
Δx
Δy
h
lim
1.9
|x|= √ x 2= x if x≥0
−x if x <0
3
3
2
2
a −b =(a−b)(a +ab+b )
1
1
=lim =0 ; use to solve when
2
going towards ∞
x ⇒∞ x
x ⇒∞ x
{
}
|x 2|= √ x 4 = x 2 if x 2≥0
Riley’s Notes
The main takeaway from this unit is learning how derivations work, and building a solid
understanding of limit definitions. For studying, I would recommend simply just using
the above equations in as many practice questions as possible – they become easier to
understand when you actually apply the theory within this unit.
As of Calculus 12, most of the theory stuff doesn’t really matter – instead just opt for
remembering the limit definition, absolute value rules, limits, and working with infinity
as a limit.
If this package has been printed out, feel free to add whatever equations you feel are
important below.
Equations of Note
2.1
V prism= A base∗h
d n
( x )=n∗x n−1 - Power rule
dx
V cube =x 3
2.2
du
dⱴ
–u
d u
dx
dx
( )=
- Quotient rule
2
dx ⱴ
ⱴ
ⱴ
d n
du
- Power rule extension
(u )=nu(n−1) ∗
dx
dx
d
dⱴ
du
- Product rule
(uⱴ)=u +ⱴ
dx
dx
dx
2.3
p
q
d
p
u = u
dx
q
p
−1
q
V cylinder =π r 2 h
1
V pyramid = A base∗h
3
V rectprism =lwh
1
V cone = π r 2 h
3
ie, if r and h change in the volume of a cylinder,
dv
dv
Δ v≈ ∗ dr + ∗ dh
dh
dh
if more variables needed, treat this similar to the
product rule extention
2.5
dy dy du - Chain Rule
= ∗
dx du dx
(aka setting y=a+cun , u= A +Cx N )
∗
du
- Fraction Rule
dx
2.6
2.4
odd (−θ)=−odd (θ) −even(θ)=even(θ)
f ( x)≈ L( x)=f (a)+f ’(a)( x−a) -
2.7
Linearization of f(x) at a.
d
sin x=cos x
dx
d
cos x=−sin x
dx
d
tan x=sec 2 x
dx
(1+ x)k ≈1+kx - Approx. Power rule
df =f ’( x)dx - Replace f with any function,
such as volume, x with any unit, such as
radius.
d
csc x=−csc x cot x
dx
d
sec x=sec x tan x
dx
d
cot x=−csc 2 x
dx
All functions starting with “c” (ie cos x)
have a negative derivative.
Riley’s Notes
Wow, what a unit. Much more to remember here, but these are significantly easier to
wrap your head around (except 2.4, nobody likes 2.4). Remember your power rule
extension, product rule, all of 2.4 and the chain rule – and you’re good as gold!
Equations of Note
2.9
3.1
dy dy du dⱴ
= ∗
∗
Chain Rule extension
dx du dⱴ dx
g ' (f ( x))=
3.2
↕ Inverse functions
(inverses – aka arc[functions] - sin−1 x=arcsin x)
dx
1
=
dy dy / dx
y=sin−1 x
iff
x=sin y
and
−1
x=cos y and
−1
x=tan y and
y=cos x
y=tan x
−1
y=csc x
x=csc y and
y=sec−1 x
x=sec y and
y=cot−1 x
x=cot y and
π
π
≤ y≤
2
2
0≤ y≤π
π
π
− < y<
2
2
π
π
y≠0 , − ≤ y≤
2
2
π
y≠ ,
0≤ y≤π
2
0< y < π
−
Step-by-step (generally)
Given a function of y=f ( x),
To determine the equation of
the inverse of g( x) (aka f −1 ( x))
1. Switch x|y
2. Solve for y
3. Determine the derivative
1
1
π
sec−1 x = cos−1 ( ) csc−1 x = sin−1 ( ) cot−1 x = −tan−1 x
x
x
2
Derivatives of inverse trig functions
d
1
du
(sin−1 u)=
∗
2
dx
√ 1−u dx
d
1
du
(tan−1 u)=
∗
2
dx
dx
1+u
d
1
du
(sec−1 u)=
∗
2
dx
|u|√ u −1 dx
if |u|>1
Memorize these guys!
3.3 (log natural derivative)
Product
ln ab=ln a+ln b
Power
m
ln x =m ln x
Quotient
a
ln =ln a – ln b
b
Root
normal logs if need be)
of g( x).
d
1
du
(cos−1 u)=−
∗
←
2
dx
√ 1−u dx
←
d
1
du
(cot−1 u)=−
∗
←
dx
1+u2 dx
d
1
du
(csc−1 u)=−
∗
2
dx
|u|√ u −1 dx
if |u|>1
3.3 (log rules)
(you can replace ln with
1
f ' ( x)
d
1 du
(ln u)= ∗
u>0
dx
u dx
1
n
ln √ x=ln x 1/n= ln x
n
More on next page ↓
3.3 (Log Natural Differentiation)
3.4 Exponential Properties
Process:
From PC 12 → y=c x
c>0 , c≠1
1. Write the equation in the form y=f ( x).
Base
→ y=e x
2. Take the natural log of both sides.
Inverse
→ y=log e x=ln x
3. Use the properties of logs to simplify the
expression.
4. Differentiate the equation implicitly with
respect to x.
1. ln e x =x ; x ϵ R
2. e ln x =x ; x >0
3. e( x+ y)=e x ∗ e y
4. e kx =(e x )k
5. Solve for y’.
5. e( x− y)=
3.4 General log derivatives
ex
ey
1
6. e−x = x
e
d
1
du
(log a u)=
∗
dx
u ln a dx
d x
(e )=e x
dx
d u
du
(e )=e u ∗
dx
dx
3.5
a x =e x ln a
y=log a x=ln x=
ln x
ln a
y=log a u=
ln u
ln a
Riley’s Notes
Yeah, there’s starting to be a bit of a scaling requirement of what you should remember,
but most of these (log/exponent laws) are from last year – and the rest builds ontop of
what we already knew. Just remember the new stuff (derivatives and oddball rules) and
this unit will be a wrap.