Tugas Matematika Dasar Prisma Wijayanto

1. MIRDA PRISMA WIJAYANTO
PHYSICS INTERNATIONAL EDUCATIONPROGRAM
NIM 120210152032
“ BASIC MATHEMATICS ASSIGNMENT “
We have a function F(x) = ( x – 3 )7 tan2 ( sin3 (2x-3)5 ), find :
a. F’(x)
b. F”(x)
c. Find the formula of tangent line at point (
3
2
, −
2187
128
)
d. Find the maximum and minimum local, maximum and minimum absolute, and the
inflection point on interval ( -π, π )
ANSWER :
 BASIC FORMULAS :
1. sin3
x = sinx.sinx.sinx
sin3
x’
= a. U’
V + V’
U
cosx.sinx + cosx.sinx = 2.sinx.cosx
b. U’
V + V’
U
cosx.sin2
x +2.sinx.cosx.sinx = 3. sin2
x.cosx
sin3
x’
= 3. sin2
x.cosx
2. tan2
x = tanx.tanx
tan2
x’
= U’
V + V’
U
= sec2
x.tanx+ sec2
x.tanx = 2.sec2
x.tanx
tan2
x’
= 2.sec2
x.tanx
3. sec2
x = secx.secx
sec2
x’
= U’
V + V’
U
= (secx.tanx)secx+(secx.tanx)secx = 2.sec2
x.tanx
sec2
x’
= 2.sec2
x.tanx
1. Finding F’
(x) = [( x – 3 )7
tan2
( sin3
(2x-3)5
)]’
a.) Derivation of ( sin3
(2x-3)5
)

2. ( sin3
(2x-3)5
)’
= 3.sin2
(2x-3).cos(2x-3).5(2x-3)4
b.) Derivation of tan2
( sin3
(2x-3)5
)
y = u2
u=tan v v= sin3
(2x-3)5
DxY = DuY . DvU . DxV
= 2sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].3sin2
(2x - 3)cos(2x - 3).5 (2x - 3)4
= 30 sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].sin2
(2x - 3)cos(2x - 3). (2x - 3)4
c.) Derivation of ( x – 3 )7
tan2
( sin3
(2x-3)5
)
U=( x – 3 )7
V= tan2
( sin3
(2x-3)5
)
U’=7( x – 3 )6
V’=30 sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].sin2
(2x - 3)cos(2x - 3). (2x - 3)4
U’
V + V’
U = 7( x – 3 )6
. tan2
( sin3
(2x-3)5
) + 30 sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].sin2
(2x -
3)cos(2x - 3). (2x - 3)4
. ( x – 3 )7
So,
F’
(x) = 7( x – 3 )6
. tan2
( sin3
(2x-3)5
) + 30 sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].sin2
(2x - 3)cos(2x
- 3). (2x - 3)4
. ( x – 3 )7
2. Finding F”
(x) = [7( x – 3 )6
. tan2
( sin3
(2x-3)5
) + 30 sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].sin2
(2x -
3)cos(2x - 3). (2x - 3)4
. ( x – 3 )7
]’
a.) Derivation of7( x – 3 )6
. tan2
( sin3
(2x-3)5
)
U=7( x – 3 )6
V= tan2
( sin3
(2x-3)5
)
U’=42( x – 3 )5
V’=30 sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].sin2
(2x - 3)cos(2x - 3). (2x - 3)4
U’
V + V’
U = 42( x – 3 )5
. tan2
( sin3
(2x-3)5
) + 30 sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].sin2
(2x -
3)cos(2x - 3). (2x - 3)4
. 7( x – 3 )6
b.) Derivation of(2x - 3)4
. ( x – 3 )7
U=(2x - 3)4
V=( x – 3 )7
U’=4(2x - 3)3
V’=7( x – 3 )6
U’
V + V’
U = 4(2x - 3)3
. ( x – 3 )7
+ 7( x – 3 )6
. (2x - 3)4
c.) Derivation ofcos(2x - 3). (2x - 3)4
. ( x – 3 )7
U= cos(2x - 3) V=(2x - 3)4
. ( x – 3 )7
U’
V + V’
U = -sin(2x - 3). (2x - 3)4
.(x - 3)7
+ 4(2x - 3)3
. ( x – 3 )7
+ 7( x – 3 )6
. (2x - 3)4
d.) Derivation ofsin2
(2x - 3)cos(2x - 3). (2x - 3)4
. ( x – 3 )7
U= sin2
(2x - 3) V= cos(2x - 3). (2x - 3)4
. ( x – 3 )7
U’
V + V’
U = [ 2sin(2x - 3).cos2
(2x - 3). (2x - 3)4
. ( x – 3 )7
] + [ -sin3
(2x - 3). (2x - 3)4
. ( x – 3 )7
+ 4sin2
(2x - 3)3
. ( x – 3 )7
+ 7sin2
(2x - 3)5
. ( x – 3 )6
]

3. e.) Derivation of tan[sin3
(2x-3)5
].sin2
(2x - 3)cos(2x - 3). (2x - 3)4
.( x – 3 )7
U= tan[sin3
(2x-3)5
] V= sin2
(2x - 3)cos(2x - 3). (2x - 3)4
. ( x – 3 )7
U’
V + V’
U = [ sec2
(sin3
(2x - 3)5
).3 sin2
(2x - 3)3
. cos(2x - 3).5(2x - 3)4
].[ sin2
(2x - 3)cos(2x - 3).
(2x - 3)4
. ( x – 3 )7
] + [ 2sin(2x - 3).cos2
(2x - 3). (2x - 3)4
. ( x – 3 )7
+ (-sin3
(2x - 3)5
). ( x – 3 )7
+ 4sin2
(2x - 3)3
. ( x – 3 )7
+ 7sin2
(2x - 3)5
. ( x – 3 )6
].[ tan(sin3
(2x-3)5
) ]
f.) Derivation of30 sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].sin2
(2x - 3)cos(2x - 3). (2x - 3)4
. ( x – 3 )7
U=30 sec2
[sin3
(2x-3)5
] V= tan[sin3
(2x-3)5
].sin2
(2x - 3)cos(2x - 3). (2x - 3)4
. ( x – 3 )7
U’
V + V’
U = ( 60 sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].3 sin2
(2x - 3)cos(2x - 3).5(2x - 3)4
).(
tan[sin3
(2x-3)5
].sin2
(2x – 3).cos(2x - 3)4
. ( x – 3 )7
) + ( sec2
(sin3
(2x - 3)5
).3 sin2
(2x - 3)3
. cos(2x -
3).5(2x - 3)4
].[ sin2
(2x - 3)cos(2x - 3). (2x - 3)4
. ( x – 3 )7
] + [ 2sin(2x - 3).cos2
(2x - 3). (2x -
3)4
. ( x – 3 )7
+ (-sin3
(2x - 3)5
). ( x – 3 )7
+ 4sin2
(2x - 3)3
. ( x – 3 )7
+ 7sin2
(2x - 3)5
. ( x – 3 )6
).(
30 sec2
[sin3
(2x-3)5
] )
So,
F”
(x) = (42( x – 3 )5
. tan2
( sin3
(2x-3)5
) + 30 sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].sin2
(2x -
3)cos(2x - 3). (2x - 3)4
.7( x – 3 )6
) + ( 60 sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].3 sin2
(2x - 3)cos(2x
- 3).5(2x - 3)4
).( tan[sin3
(2x-3)5
].sin2
(2x – 3).cos(2x - 3)4
.( x – 3 )7
) + ( sec2
(sin3
(2x - 3)5
).3
sin2
(2x - 3)3
. cos(2x - 3).5(2x - 3)4
].[ sin2
(2x - 3)cos(2x - 3). (2x - 3)4
. ( x – 3 )7
] + [ 2sin(2x -
3).cos2
(2x - 3). (2x - 3)4
. ( x – 3 )7
+ (-sin3
(2x - 3)5
).( x – 3 )7
+ 4sin2
(2x - 3)3
. ( x – 3 )7
+ 7sin2
(2x - 3)5
. ( x – 3 )6
).( 30 sec2
[sin3
(2x-3)5
] )
3. Finding the Formula of tangent line at the point (
3
2
, −
2187
128
)
 m = F’
(x)
m = 7( x – 3 )6
. tan2
( sin3
(2x-3)5
) + 30 sec2
[sin3
(2x-3)5
].tan[sin3
(2x-3)5
].sin2
(2x - 3)cos(2x - 3).
(2x - 3)4
. ( x – 3 )7
we can substitute x for
3
2
, so we got :
m = 7(
3
2
– 3 )6
. tan2
( sin3
(2
3
2
-3)5
) + 30 sec2
[sin3
(2
3
2
-3)5
].tan[sin3
(2
3
2
-3)5
].sin2
(2
3
2
- 3)cos(2
3
2
- 3).
(2
3
2
- 3)4
. (
3
2
– 3 )7
m = 0
 y – b = m ( x – a )
y – (−
2187
128
) = 0
y = −
𝟐𝟏𝟖𝟕
𝟏𝟐𝟖
= −𝟏𝟕
𝟏𝟏
𝟏𝟐𝟖

4. 4. a.) Finding the maximum local
for the maximum local, F’(x) > 0
 Suppose that F’(x) = 0
7( x – 3 )6. tan2 ( sin3 (2x-3)5 ) + 30 sec2[sin3 (2x-3)5].tan[sin3 (2x-3)5 ].sin2(2x - 3)cos(2x -
3). (2x - 3)4. ( x – 3 )7 = 0
We can consider to say that if F’(x) = 0, so sin3 (2x-3)5 = 0, because all elements have sin3
(2x-3)5
sin3 (2x-3)5 = 0 sin3 (2x-3)5 = 0
sin3 (2x-3)5 = sin 0o sin3 (2x-3)5 = sin 180o
(2x-3)5 = 0 (2x-3)5 = 180
x =
𝟑
𝟐
x = 2,91261725
 The maximum local, x > 2,91261725
b.) Finding the minimum local
For the minimum local, F’(x) < 0
 The minimum local, x <
𝟑
𝟐
c.) Finding the Maximum and Minimum Absolute
For the maximum absolute, F”(x) < 0
For the minimum absolute, F”(x) > 0
 To find the maximum or minimum, we can substitute x =
3
2
and x = 2,91261725 to F”(x)
1. For x =
3
2
F”
(x) = (42(
3
2
– 3 )5
. tan2
( sin3
(2
3
2
-3)5
) + 30 sec2
[sin3
(2
3
2
-3)5
].tan[sin3
(2
3
2
-3)5
].sin2
(2
3
2
- 3)cos(2
3
2
-
3). (2
3
2
- 3)4
. 7(
3
2
– 3 )6
) + ( 60 sec2
[sin3
(2
3
2
-3)5
].tan[sin3
(2
3
2
-3)5
].3 sin2
(2
3
2
- 3)cos(2
3
2
-
3).5(2
3
2
- 3)4
).( tan[sin3
(2
3
2
-3)5
].sin2
(2
3
2
– 3).cos(2
3
2
- 3)4
. (
3
2
– 3 )7
) + ( sec2
(sin3
(2
3
2
- 3)5
).3
sin2
(2
3
2
- 3)3
. cos(2
3
2
- 3).5(2
3
2
- 3)4
].[ sin2
(2
3
2
- 3)cos(2
3
2
- 3). (2
3
2
- 3)4
. (
3
2
– 3 )7
] + [
2sin(2
3
2
- 3).cos2
(2
3
2
- 3). (2
3
2
- 3)4
. (
3
2
– 3 )7
+ (-sin3
(2
3
2
- 3)5
). (
3
2
– 3 )7
+ 4sin2
(2
3
2
- 3)3
.
(
3
2
– 3 )7
+ 7sin2
(2
3
2
- 3)5
. (
3
2
– 3 )6
).( 30 sec2
[sin3
(2
3
2
-3)5
] )
F”
(x) = 0

5. 2. For x = 2,91261725
F”
(x) = (42(2,91261725– 3 )5
. tan2
( sin3
(2. 2,91261725 - 3)5
) + 30 sec2
[sin3
(2. 2,91261725 -
3)5
].tan[sin3
(2. 2,91261725 - 3)5
].sin2
(2.2,91261725 - 3)cos(2. 2,91261725 - 3).
(2. 2,91261725- 3)4
. 7(2,91261725 – 3 )6
) + ( 60 sec2
[sin3
(2. 2,91261725-3)5
].tan[sin3
(2. 2,91261725 - 3)5
].3 sin2
(2. 2,91261725 - 3)cos(2. 2,91261725 - 3).5(2. 2,91261725 -
3)4
).( tan[sin3
(2. 2,91261725 - 3)5
].sin2
(2.2,91261725– 3).cos(2. 2,91261725 - 3)4
.
(2,91261725 – 3 )7
) + ( sec2
(sin3
(2. 2,91261725 - 3)5
).3 sin2
(2. 2,91261725 - 3)3
.
cos(2.2,91261725- 3).5(2. 2,91261725- 3)4
].[ sin2
(2.2,91261725-
3)cos(2.2,91261725 - 3). (2. 2,91261725- 3)4
. (2,91261725 – 3 )7
] + [ 2sin(2.
2,91261725 - 3).cos2
(2. 2,91261725 - 3). (2. 2,91261725- 3)4
. (2,91261725 – 3 )7
+ (-
sin3
(2. 2,91261725 - 3)5
). (2,91261725 – 3 )7
+ 4sin2
(2. 2,91261725 - 3)3
.
(2,91261725 – 3 )7
+ 7sin2
(2. 2,91261725 - 3)5
. (2,91261725 – 3 )6
).( 30 sec2
[sin3
(2.
2,91261725 - 3)5
] )
F”
(x) = 0
 Both the values of F”(x) are 0, so it means :
The maximum and minimum absolute are both nothing
d.) Finding the unflection point on interval ( -π, π )
The properties of Inflection point is F”(x) = 0. We had calculate before in exercise 4.c to
find the values of x for F”(x) = 0. So, the inflection point are :
a. (
𝟑
𝟐
, 𝟎 )
b. ( 2,91261725 , 0 )