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>
Mathematics
>
Differential Rules and Applications in Calculus Explained
Differential Rules and Applications in Calculus Explained
School
University of Alberta
*
*We aren't endorsed by this school
Course
MATH 100
Subject
Mathematics
Date
Dec 12, 2024
Pages
12
Uploaded by MagistrateProton22759
Chapter
3
Differential
Rules
&
Applications
of
Differentiation
CRITICAL
NUMBERS
OF
THE
FUNCTION
1)
f(x)
:
2x-3x2
5
.
1
g(x)
=
Ex
9
)
f(x)
=
4x2
- 9xx
(2x
+
3
f(x)
=
2
6x
g(x)
=
x
f'(x)
=
12x2
18x-12
let
/
'(x)
=
0
g'(x)
=
-
(
'(x)
=
G(dX2
-
3x
2)
2
6X
=
0
let
f'(x)
=
0
=
le
+
q'(x)
=
0
0
=
6(2x2
3x
2)
:
X
=
0
=
6(2x2
-
4x
+
x
2)
-
4
.
1
4
X
=
t
6
.
)
g(x)
:
(x
+
11
0
=
6(2x(X
2)
+
1(x
2))
4
+
1
3
2
)
((x)
=
5
+
8X
q'(x)
=
2x
+
1
=
0
X
=
1/2
ā
'(X)
=
8
X
-
2
=
j
X
=
2
let
f(x)
=
0
letq'(x)
=
0
:.
NONE
8
:
X
+
1
10)
S(t)
=
2t"
+
3t2
-
6t
+
4
3
)
(
(x)
=
X"
3x
+
1
X
=
-
1
s'(t)
=
6t2
+
6t-6
f(x)
=
3x2
3
s'(l)
=
6(
+
2
+
t
-
1)
let
f(x)
=
0
M
)
(
(x)
=
516x
2x
let
s'(t)
=
0
0
:
3x2
3
l
'(X)
=
6
-
6x2
0
:
6
It
It-17
=
X
=
I
X
=
b
=
4a)
2a
X
=
=1
8)
/
(H)
=
273
+
372
+
6t
+
4
i
f
(t)
=
Gt2
+
6t
+
6
4
.
)
(
(t)
=
+
3
+
6t2
+
3t
-1
f'(t)
=
G(
+
2
+
t
+
1)
f'(l)
:
3t2
+
12t
+
3
let
fll)
:
0
let
f'(t)
=
0
:
32
+
12t
+
3
11
)
((t)
:
+
414t3
+
at
Ja
!
"I
:
4516213t
+
11
X
=
b
=
m
4a
s
(t)
=
4
+
3
+
12
+
2
+
47
x
=
-
12
!
4(3)(3)
X
=
UNDEFINED
s(t)
:
0
2(3)
X
=
0
X
:
Elit
:
X
=
NONE
*
12
)
fil
15
.
7
VIX1
:
F
V(X)
=
X(x
2)
1)
ru
+
1) I
v'(x)
=
(X
2)
+
(x
t
-
(x
2i)
flo
:
v'(x)
=
(x
-
2)
+ 2
v'(X)
=
2(X
2)
+
X
I'Ir)
:
L(X-2)'la
let
!
'
(r)
=
0
v'(x)
=
2x
-
4
+
x
G(X
2)1
0
=
p
+
1
,
v2
=
1
r
=
If
~
'(x)
=
V'(x)
=
0
v'(X)
:
GNE
13
.
)
f(0)
:
sir
128)
0
:
3x
4
X
=
2
+
'(0)
:
<
/sin
1281)
Ā·
Cos
(20)
:
2
x
=
%'
10)
:
4
sinGE
COS2t
-(indo
Ā·
16
)
T(X1
=
x 2(2x-17
"
i'(X)
=
<x(2x
-
1)
+
(x
-
2. (ax-
17."
.
2]
:
<(sin40)
T'(x)
2x
(2x
1)
%
+
4x2
12
+
+
3(2X
17"
i'(x)
=
6x(2x
1)
+
4x
14
)
g(8)
=
O
sint
3(2X
1)"
q
10)
=
1
+
cost
i'(x)
-
x
let
gf)
=
g
O
=
It
cost
Cost
=
1
i'(x)
:
16x"
Ex
-
=
T
3(2x
-
1)
+
'
(x)
=
DNE
X
=
i
+
hi
,
ne
I
i'(x)
=
2x(8x
3)
3(2X-1)"3
X
=
-
It
I'l
=
0
X
=
0020
Mean
Value
Theorem
19
.
1
f(x)
=
1
X2
,
10
,
37
1
.
)
((x1
=
x2-4x
+
5
,
(1
,
5]
Man
=
f (3)-f(0)
3 -
f
Man
=
((b)-fla)
b
a
Man
:
1-1312-11-01
=
/(5)
fill
3
5
-
1
10
-
2
=
52
415)
+
5
[12-4(1)
+
5]
:
19
=
4
f'(x)
=
2x
f'(c
=
2
=
25-2015
1
+
4
5
4
-
2
=
E
f(x)
=
GX
-
4
=
8/4
=
2
c
=
2
2
f'()
=
2c
4
=
G
2)
-
4
=
h
2
=
2
+
4
ā„
18
)
f(x)
:
x 3-2X
+
1
(-2
,
3)
f'(l)
=
3x2-
2
Man
=
f(3)-f(
-
2)
f'
(c)
=
3c2
-
2
=
5
3
(2)
32
=
5
+
2
Max
=
33
2(3771
-
((-2)"
2(
2)
+
1)
E
5
:
Fle
May
=
27 -4
+
1
-
(
8
+
4
+
1)
5
Man
=
27
6
+
1
+
8
4
-1
=
25
5
21
)
f(x)
:
+
,
11
,
2)
20
.
1
f(x)
=
2x3
+
x
=
-
X
1
,
(0
,
2]
Man
=
f(27-f(0)
Man
:
Ef()
2
0
Man
=
((2)3
+
1213-2-1-1-17
Ā·
It
:
+=
2
May
=
16
+
4
3
+
1
=
a
2
f(x)
=
x(0)
=
f'(x)
=
6x2
+
2x
-
1
f'
(l)
=
6c=
+
2.
1
=
&
t'(c)
:
t
=
62
+
2c
q
-
1
=
0
Gc2
+
2c-10
=
0
2)
(3x2
+
c
5)
=
0
=
-1
:
c
="
c
=
b
=
b2
Ha
11
,
27
:
cre
2a
=
=
IIF
41311
+
5)
2(3)
"
24
)
62
.
)
f(x)
=
x
,
(1
,
4)
fis
cont
(2
.
51
12f'(x)14(2
,
5)
Man
=
f
(4)
fill
show
that
3[f(5)-f(e)
<
12
3
-
Man
=
Fr
hypothesis
of
Mean
Value
Theorem
!
'
(2)
=
/(5)
(12)
l'x1
=
I'll
test
5-2
:
f'(c)
=
1151
-
f(a)
=
f(x)
3
:
f(d)-(15)-f(2)
1
fla)
q
=
c
3
23
.
7
/x)
:
It
ETF
;
(2
,
97
1
:
(
(2)
Man
=
f(q)-f(2)
31
(15) - f(2)112
M
man
Let
mas
=
i
I'(x)
=
1.
(x-17
2
ā
'Ill
=
31c-1)
:
12-12"
:
-
(1
:
c
=
(1)2
+
1
INCREASING
/DECREASING
26
)
f(x)
=
X5
+
4x
4
25
.
)
f(x)
=
x3
+
2x2
- X
+
1
f'(x)
=
3x2
+
4x
-
1
f(x)
=
5x"
+
12x2
let
!
'
(x)
=
0
let
f(x)
=
0
5x*
+
12x2
=
0
0
=
3x2
+
4x
-
1
x
(5x21(2)
=
0
Ā·
x
=
-
undefine
:
f
is
increasing
On
all
#S
2) 7
f(x)
=
Glanx-lan
=
X
f(x)
=
2secix-(2-tanx
Ā·
secix]
=
Isec
*
x(1-tanx]
interval
f(x)
inc/des
f
(x)
=
0
1- tanx
=
0
1
=
tanx
x
=
Ļ
(4
+
In
10
.
En
inc
CON
(A vI
+
Y
interval
f"(X)
12)
dec
28
)
y
=
6x2
2x3
xY
7- 0
,
1
62)
CD
y
=
12x
6x2
-
4X
(1
62
,
0
62)
+
w
lat
,
)
inc
y"
=
12
12x
12x2
10
62
,
2)
CD
curve
up
a
let
y"
=
0
intervals
ā
(-2)
=
3(2)2
+
4(2)
1
0
=
12-12x-
12x2
=
12
8
1
=
3
0
:
12
(1
X
-
x2)
I
.
0
:
-12(x21x-1)
+
107
:
-
*:
11
,
1
4111(1)
X
=
- 1
618033989
:
INC
:
1-c
,
-2- m)
&
(2
,
a)
2(1)
X
=
0
618033989
3
I
DEC
:
1-2
-
Els
,
-2
+
/
**
(
-
115/a
29
)
y
=
X
(1
+
x)"
y"
=
(1
+
x 7
"2
(8
+
8x
=
2x
+
Rx
-
(2x
(x))
d
4)
.
2xIt
4)
((tx)3
y
=
2x(1
+
x)
X2
911
+
x7
(01x
A s
2(1
+
x)"
(1
+
x)
%::
YX
x)
1"
OXT
y)
=
4X
+
4x2
X2
2(1
+
x)3/
g'
:
el
E
y"
=
((1
+
x(
3.
(4
+
6x)
(6 --
-
(1
+
x)x
.
(4x
+
3xz)]
:
(2
(1
+
x)3/2)2
ele
for
function
f(x]
y"
=
(8
+
12x)(1
+
x)
*
-
(((2x
+
axz(1
+
x)"]
Y
(1
+
x)3
y"
=
(8
+
12x)(1
+
x73/
(12x19x2)(1
+
x)"2
4(1
+
x)3
y"
=
(1
+
x)'
/18
+
12x)
(1
+
x)
12x-ax2)
4(1
+
x)3
30
.
)
y
:
x)e
interval
y"
(
-
20
.
-1)
-
CD
y
=
(1
+
x)
(20(1
+
x)
Ā·
x)
(-1
,
2)
=>
LD
(1
+
x)Y
(2
,
c)
ā
Ch
y
=
(
+
2x
+
x
=
-
X
-
2x2
curve
up
on
interval
(1
+
x)4
(G
,
c)
9
:
y"
=
(11x)"
-
2x
-
(4(1
+
x)3(1
x2)]
(1
+
x)
:
y"
=
2x((
+
x)"
-
((
+
x(3(4
4x4]
(1
+
x)
:
y"
=
(1
+
x))
2x(1
+
x)
-
4
+
4x
*
]
(1
+
x)
y"
=
-
2x
-
2x2
4
+
4xz
(1
+
x)5
y"
:
2x2-2x-
4
y"
:
(1
+
x)5
y"
=
0(x
2)(X
+
1)
X
=
2
X
=
+
31
)
y
=
X
intervals
y1
cu/cp
x2
-
3
(
-
0
,
3)
ā”
1
-
3
,
0)
=>
y
=
(x2
3)(3xx
(2x
-
x3]
10
.
3)
#
13
,
c)
#
(x2-3)2
y
=
3xY
9x2
2x4
(x2
3)2
y
=
X*
-
9x2
(x
=
3)2
y"
=
(x2
=
3)2
Ā·
(4x
-
18x)
-
(2(x2-3)
.
2x(x
*
-
9x1]
(x2
- 3)4
y"
=
(x2
3)2(4x3
-
18x)
-
4x(x2
3)(X"
9x
?
)
(x2-3)4
y"
=
(x2 -
3)((x2 -
3)(4x3
-
18x)
-
4x(X
*
-
9x2)]
(x2
-
3)4
y"
=
(x
2
3)(#5
18x3
12x2
-
54x
4x5
+
36x3]
(x2
-
3)4
y"
=
Gx3
54X
=
bx(x2
-
9)
(x
?
- 3)3
(x2 -
3)3
y"
=
0X
=
0
,
3
,
3
CURVE
SKETCHING
E
.
7
INC/DEC
32
)
y
=
1
+
x
y
=
Itx2
1
X2
1-x2
A
)
DOMAIN
y
=
(1
-
xz)(2x)
(
2x(1
+
x2)]
11-x2)2
Domain
y
=
X F
11
,
XER
y
=
2x
2x3
)
2x
2x3)
(1
-
x2)2
B
.
)
INTERCEPT
X
int
y1
=
Gx
-
x
+
2x
+
2x3
let
y
=
0
no
xint
0
:
1
+
x2
(l-xe)2
let
X
=
0
:
Ho
,
yin
:
ye
X
=
I
intervals
y'
INC/DEC
c
)
Symmetry
=
f(-x)
=
f(x)
:
lo
ā
Co
,
1
t
INC
symmetric
about
the
yaxis
(1
,
a)
t
inc
w
D
)
ASIMPTOTES
F
.
)
local
min
In
m
10
,
1)
HA
y
=
lim
1
+
x2-
a
lim
=
A
X
+
1
-
1
-
x 2
X
=
1
lim
6)
concavity
5
:
y"
=
11-X12.4-
(2
/1-X4-2x
.
4x]
(1
-
y2)4
y"
=
4(1
x 2)2
1
-
16x2
(1
X)
(l-X2)4
y"
=
4(1
x22
+
16x2
11-x2
(l-x2)"
y"
:
4(1
X2(1
x"
+
4x)
11-x2)
Y
y"
=
4)c
xi
+
4x)
(l-x2)3
y"
=
H
4x2
+
164
(l
y2)3
x
=
=
16
!
4(4-4)
21
4)
y"-
40
7
hyperbola
ye-X
=
=
H
49
.
71"(x)
=
1
closest
to
(2
.
0
f(x)
=
X
+
C
41
)
53
.
)
[4
)
f(x)
=
+x
+
(x
+
4
INTRO
TO
INTEGRATION
50
.
)f"(x)
=
sinx
ANTIDERIVATIVES
f(x)
=
cosX
+
C
find
f(x)
f(x)
:
-
Sinx
+
(x
+
D
47
.
)
+"(x)
=
X*
+
x3
51)
&
"(x)
=
24X
f(x)
=
x
+
f"(x)
=
24(t)x
+
2
((x)
=
!
(4)x"
+
+(f)x5
+
xx
+
Df(x)
=
12x
+
e
f(x)
=
(x"
+
2x
+
D
f(x)
=
+
(x
+
D
f(x)
=
x
+
Exit
52
.
)g(X)
:
E
+
z
=
2
48
)f"
(x)
=
GOx"
45x2
f(x)
=
y
+
f"(x)
:
2x
+
=
E
f(x)
=
10x5
15
x
f(x)
=
2
.
Ext
+
ex
f(x)
=
12x5
-
15x3
+
C
f(x)
=
1x
+
ex
f(x)
=
12x"
Ex
+
ex
f(x)
:
.
x
+
Ex
+
Dx
f(x)
=
2x"
-
15x"
+
(x
+
b
f(x)
:
<Ex
+
Bx