A)169 B)144 C)121 D)100
⇣ Math
⇣ p ⌘
⇣ p ⌘
⇣ p ⌘
p ⌘ 1431
45
13
58
35
13
19
19
17
A) LAB
;
B)
;
C)
;
D)
;
session
12
2
2
2
2
2
2
2
2
⇣
⌘
⇣
⌘
⇣
⌘
⇣
⌘
1
5
1
5
1
3
1
3
p ;p
p ;p
A) p6 ; pe and
B) p7 ; pe and
e
e
6
7
⇣
⌘
⇣
⌘
⇣
⌘
⇣
⌘
1
3
1
3
1
5
1
5
p ;p
p ;p
C) p4 ; pe and
D) p8 ; pe and
e
e
4
8
Quiz 20
Example 1:
Use differentials to estimate the value (81.5) /4 sin °(31)
3
Example 2:
A)169 B)144 C)121 D)100
⇣ p ⌘
⇣ p ⌘
⇣ p ⌘
⇣ p ⌘
A) 13
; 245
B) 19
; 213
C) 19
; 258
D) 17
; 235
2
2
2
2
⇣
⌘
⇣
⌘
⇣
⌘
⇣
⌘
1
5
1
5
1
3
1
3
p ; p
p ; p
A) p6 ; pe and
B) p7 ; pe and
e
e
6
7
⇣
⌘
⇣
⌘
⇣
⌘
⇣
⌘
1
3
1
3
1
5
1
5
p ; p
p ; p
C) p4 ; pe and
D) p8 ; pe and
e
e
4
8
3
(81.5)
Use differentials to estimate the
value/4
sin °(31)
12
dy
Example 3:
Find the differential
(81.5)3/4 sin °(31)
dy
for
y = 5x3 cos (x)
y = 5x3 cos (x)
Example 4:
A spherical ball bearing will be coated by 0.03 cm of protective coating. If the radius of this ball bearing is 7
cm, approximately how much coating will be required? (use π≈3.14)
12
12
y = 5x cos (x)
1
10
Example 5:
y = 5x3 cos (x)
Give the derivative of
earcsin (5x)
1
10
at the point where x =
A)169 B)144 C)121 . D)100
⇣ p ⌘
⇣ p ⌘
⇣ p ⌘ earcsin⇣(5x) p ⌘
45
13
58
35
19
19
17
A) 13
;
B)
;
C)
;
D)
;
2
2
2
2
2
2
2
2
⇣
⌘
⇣
⌘
⇣
⌘
⇣
⌘
A)169 B)144 C)121 D)100
1
5
1
5
1
3
1
3
p ;p
B) p7 ; pe and
⇣ p ⌘
⇣ p ⌘ A) ⇣p6 ;ppe⌘ and⇣ pp6⌘; pe
e
7
45
13
58
35
19
19
17
A) 13
;
B)
;
C)
;
D)
;
⇣ 2 2 ⌘
⇣2 2
⌘
⇣
⌘
⇣
⌘
2
2
2
2
1
3
1
5
1
5
⇣
⌘
⇣
⌘ C) ⇣p1 ; p3 ⌘ and⇣
⌘
p ;p
p ;p
D) p8 ; pe and
e
e
41 p3e
43
8
p1 p5
p1 p5
p
p1 p
A) 6 ; e and
⇣
⌘
⇣
1
3
C) p4 ; pe and
6
;
B)
e
p1 ; p3
e
4
(81.5)3/4 sin °(31)
⌘
7
;
e
and
⇣ 3/4 ⌘
⇣
1
5
(81.5)
°(31)
D) p8 ; psin
and
e
7
;
e
p1 ; p5
e
8
⌘
dy
dy
A)169 B)144 C)121 D)100
⇣ p ⌘
⇣ p ⌘
⇣ p y⌘= 5x3 cos
⇣ (x)p ⌘
45
13
58
13
19
19
1
A) 2 ; 2
B) 2 ; 2
C) 2 ; 2 x =D) 17
; 235
2
10
1
⇣
⌘ x =⇣ 10
⌘
⇣
⌘
⇣
⌘
1
5
1
5
1
3
1
3
p ; p
p ; p
A) p6Question
; pe and#
B) p7 ; pe and
e
e
6
arcsin (5x) 7
e
⇣
⌘
⇣
⌘
⇣
⌘
⇣
⌘
1 p3
1 p3
1 p5
1 p5
p
p
p
p
C) 4Taking
; e and
; e,lnuse
; e lnand
; e 4.9
ln 5 ⇡ 1.609
4.9D)
differentials
to
5 estimate
⇡ 1.609
4
8
8 ln
y = 5x3 cos (x)
earcsin (5x)
81.5)3/4 sin °(31)
12
dy
x=
12
y = 5x3 cos (x)
1
10
earcsin (5x)
n 5 ⇡ 1.609 ln 4.9
Question #
Consider the function f (x) =
p
x . Approximate the change in f as x changes from 8 to 9
12
12
D)100 p
(x) ⇣= xp ⌘
⌘ fExample
⇣ p ⌘
6:
3
C) 19
; 258
D) 17
; 235
2
2
x 3
⌘ lim 2⇣
⌘
⇣
⌘
x!3 x
9
5
1
3
1
3
p
p ; p
B) p7 ; pe and
e
e
7
x
10
1
⌘ lim ⇣
⌘
⇣
⌘
3
1
5
1
5
x!0
x
p
p ; p
D) p8 ; pe and
e
e
8
3 + 3x 3ex
lim
x!0 7x(ex
1)
✓ ✓ ◆◆2x
11
lim cos
x!1
x
✓
◆
5
5
im
!0
x ln3(1 + x)
y = 5x cos (x)
6
lim (ex + 1) /x
x!1
(5x)
earcsin
12
f (x) =
p
x
x 3
x!3 x2 7: 9
Example
lim
10x 1
lim
x!0
x
3 + 3x 3ex
lim
x!0 7x(ex
1)
✓
✓ ◆◆2x
11
lim cos
x!1
x
✓
◆
5
5
im
!0
x ln (1 + x)
6
lim (ex + 1) /x
x!1
12
x
9
10x 1
lim 8:
Example
x!0
x
3 + 3x 3ex
lim
x!0 7x(ex
1)
✓ ✓ ◆◆2x
11
lim cos
x!1
x
✓
◆
5
5
lim
x!0
x ln (1 + x)
6
lim (ex + 1) /x
x!1
12
3 + 3x 3ex
lim
x!0 7x(e
Example
9: x 1)
✓ ✓ ◆◆2x
11
lim cos
x!1
x
✓
◆
5
5
lim
x!0
x ln (1 + x)
6
lim (ex + 1) /x
x!1
12
✓
✓
◆◆2x
11
lim cos
Example
10: x
x!1
✓
◆
5
5
lim
x!0 x
ln (1 + x)
6
lim (ex + 1) /x
x!1
12
✓
5
5
im
x!0 Example
x ln11:
(1 + x)
6
lim (ex + 1) /x
x!1
12
◆
x 3
lim
Question
x!3 x2 #9
9x 1
lim
x!0
5x
3 + 3x 3ex
lim
x!0 7x(ex
1)
✓ ✓ ◆◆2x
11
lim cos
x!1
x
✓
◆
5
5
lim
x!0
x ln (1 + x)
lim (ex + 1) /x
6
x!1
12
Question #
p
4 1 + x2
lim
x!1
3x2
5x 4 tan (x)
x!0 2 sin (x)
10x
lim
p
4 1 + x2
lim
x!1 # 3x2
Question
5x 4 tan (x)
x!0 2 sin (x)
10x
lim