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Tugas matematika 2 (semester 2)
- 1. Nama :Nikmah Utami
NPM : 0031421
Kelas : 1 Elektronika A
Tugas : MATEMATIKA
Tentukanlah nilai
ππ¦
ππ₯
dari fungsi berikut ini !
1. π¦ = β π₯5 + 6π₯2 + 3
2. π¦ = β π₯4 + 6π₯ + 1
3
3. π¦ = β π₯2 β 5π₯
5
4. π¦ =
1
βπ₯4+2π₯
5. π¦ =
1
βπ₯2β6π₯
3
6. π¦ =
1
βπ₯2β5π₯+2
5
7. π¦ = sin β π₯2 + 6π₯
8. π¦ = cos β π₯3 + 2
3
9. π¦ = sin
1
βπ₯2+2
10. π¦ = cos
1
βπ₯2+6
3
Jawaban :
1. π¦ = β π₯5 + 6π₯2 + 3
Misal u = π₯5
+ 6π₯2
+ 3 , maka
ππ’
ππ₯
= 5π₯4
+ 12π₯
π¦ = β π’ = π’
1
2 , maka
ππ¦
ππ’
=
1
2
π’β
1
2 =
1
2
(π₯5
+ 6π₯2
+ 3)β
1
2
Maka
ππ¦
ππ₯
=
ππ¦
ππ’
.
ππ’
ππ₯
=
1
2
(π₯5
+ 6π₯2
+ 3)β
1
2 . (5π₯4
+ 12π₯)
ππ¦
ππ₯
=
1
2
(5π₯4
+ 12π₯)
(π₯5 + 6π₯2 + 3)
1
2
=
1
2
(5π₯4
+ 12π₯)
β π₯5 + 6π₯2 + 3
2. π¦ = β π₯4 + 6π₯ + 1
3
Misal u = π₯4
+ 6π₯ + 1 , maka
ππ’
ππ₯
= 4π₯3
+ 6
π¦ = β π’3
= π’
1
3 , maka
ππ¦
ππ’
=
1
3
π’β
2
3 =
1
3
(π₯4
+ 6π₯ + 1)β
2
3
Maka
ππ¦
ππ₯
=
ππ¦
ππ’
.
ππ’
ππ₯
=
1
3
(π₯4
+ 6π₯ + 1)β
2
3 . (4π₯3
+ 6)
- 2. ππ¦
ππ₯
=
1
3
(4π₯3
+ 6)
(π₯4 + 6π₯ + 1)
2
3
=
1
3
(4π₯3
+ 6)
β(π₯4 + 6π₯ + 1)23
3. π¦ = β π₯2 β 5π₯
5
Misal u = π₯2
β 5π₯ , maka
ππ’
ππ₯
= 2π₯ β 5
π¦ = β π’5
= π’
1
5 , maka
ππ¦
ππ’
=
1
5
π’β
4
5 =
1
5
(π₯2
β 5π₯)β
4
5
Maka
ππ¦
ππ₯
=
ππ¦
ππ’
.
ππ’
ππ₯
=
1
5
(π₯2
β 5π₯)β
4
5 .(2π₯ β 5)
ππ¦
ππ₯
=
1
5
(2π₯ β 5)
(π₯2 β 5π₯)
4
5
=
1
5
(2π₯ β 5)
β(π₯2 β 5π₯)45
4. π¦ =
1
βπ₯4+2π₯
=
1
(π₯4+2π₯)
1
2
= (π₯4
+ 2π₯)β
1
2
Misal u = π₯4
+ 2π₯ , maka
ππ’
ππ₯
= 4π₯3
+ 2
π¦ = π’β
1
2 , maka
ππ¦
ππ’
= β
1
2
π’β
3
2 = β
1
2
(π₯4
+ 2π₯)β
3
2
Maka
ππ¦
ππ₯
=
ππ¦
ππ’
.
ππ’
ππ₯
= β
1
2
(π₯4
+ 2π₯)β
3
2 .(4π₯3
+ 2)
ππ¦
ππ₯
=
β
1
2
(4π₯3
+ 2)
(π₯4 + 2π₯)
3
2
=
β2π₯3
β 1
β(π₯4 + 2π₯)3
- 3. 5. π¦ =
1
βπ₯2β6π₯
3 =
1
(π₯2β6π₯)
1
3
= (π₯2
β 6π₯)β
1
3
Misal u = π₯2
β 6π₯ , maka
ππ’
ππ₯
= 2π₯ β 6
π¦ = π’β
1
3 , maka
ππ¦
ππ’
= β
1
3
π’β
4
3 = β
1
3
(π₯2
β 6π₯)β
4
3
Maka
ππ¦
ππ₯
=
ππ¦
ππ’
.
ππ’
ππ₯
= β
1
3
(π₯2
β 6π₯)β
4
3 .(2π₯ β 6)
ππ¦
ππ₯
=
β
1
3
. (2π₯ β 6)
(π₯2 β 6π₯)β
4
3
=
β
1
3
(2π₯ β 6)
β(π₯2 β 6π₯)43
6. π¦ =
1
βπ₯2β5π₯+2
5 =
1
(π₯2β5π₯+2)
1
5
= (π₯2
β 5π₯ + 2)β
1
5
Misal u = π₯2
β 5π₯ + 2 , maka
ππ’
ππ₯
= 2π₯ β 5
π¦ = π’β
1
5 , maka
ππ¦
ππ’
= β
1
5
π’β
6
5 = β
1
5
(π₯2
β 5π₯ + 2)β
6
5
Maka
ππ¦
ππ₯
=
ππ¦
ππ’
.
ππ’
ππ₯
= β
1
5
(π₯2
β 5π₯ + 2)β
6
5 .(2π₯ β 5)
ππ¦
ππ₯
=
β
1
5
. (2π₯ β 5)
(π₯2 β 5π₯ + 2)
6
5
=
β
1
5
(2π₯ β 5)
β(π₯2 β 5π₯ + 2)65
7. π¦ = sin β π₯2 + 6π₯
Misal u = π₯2
+ 6π₯ , maka
ππ’
ππ₯
= 2π₯ + 6
π£ = β π’ = π’
1
2 , maka
ππ£
ππ’
=
1
2
π’β
1
2 =
1
2
(π₯2
+ 6π₯ )β
1
2
π¦ = sin π£ , maka
ππ¦
ππ£
= cos π£ = cos β π’ = cos β π₯2 + 6π₯
- 4. Maka
ππ¦
ππ₯
=
ππ¦
ππ£
.
ππ£
ππ’
.
ππ’
ππ₯
= cos β π₯2 + 6π₯ .
1
2
(π₯2
+ 6π₯ )β
1
2 .(2π₯ + 6)
ππ¦
ππ₯
=
1
2
. (2π₯ + 6) .cos β π₯2 + 6π₯
(π₯2 + 6π₯ )
1
2
=
( π₯ + 3) .cos β π₯2 + 6π₯
β π₯2 + 6π₯
8. π¦ = cos β π₯3 + 2
3
Misal u = π₯3
+ 2 , maka
ππ’
ππ₯
= 3π₯2
π£ = β π’3
= π’
1
3 , maka
ππ£
ππ’
=
1
3
π’β
2
3 =
1
3
(π₯3
+ 2)β
2
3
π¦ = cos π£ , maka
ππ¦
ππ£
= βsin π£ = βsin β π’3
= βsin β π₯3 + 2
3
Maka
ππ¦
ππ₯
=
ππ¦
ππ£
.
ππ£
ππ’
.
ππ’
ππ₯
= βsin β π₯3 + 2
3
.
1
3
(π₯3
+ 2)β
2
3 .3π₯2
ππ¦
ππ₯
=
1
3
. 3π₯2
. βsin β π₯3 + 2
3
(π₯3 + 2)
2
3
=
π₯2
. βsin β π₯3 + 2
3
β(π₯3 + 2)23
9. π¦ = sin
1
βπ₯2+2
= sin
1
(π₯2+2)
1
2
= sin(π₯2
+ 2)β
1
2
Misal u = π₯2
+ 2 , maka
ππ’
ππ₯
= 2π₯
π£ = π’β
1
2 , maka
ππ£
ππ’
= β
1
2
π’β
3
2 = β
1
2
(π₯2
+ 2)β
3
2
π¦ = sin π£ , maka
ππ¦
ππ£
= cos π£ = cos π’β
1
2 = cos(π₯2
+ 2)β
1
2
Maka
ππ¦
ππ₯
=
ππ¦
ππ£
.
ππ£
ππ’
.
ππ’
ππ₯
= cos(π₯2
+ 2)β
1
2 . β
1
2
(π₯2
+ 2)β
3
2 .2π₯
- 5. ππ¦
ππ₯
=
β
1
2
. 2π₯ . cos(π₯2
+ 2)β
1
2
(π₯2 + 2)
3
2
=
βπ₯ . cos(π₯2
+ 2)β
1
2
β(π₯2 + 2)3
10. π¦ = cos
1
βπ₯2+6
3 = cos
1
(π₯2+6)
1
3
= cos(π₯2
+ 6)β
1
3
Misal u = π₯2
+ 6 , maka
ππ’
ππ₯
= 2π₯
π£ = π’β
1
3 , maka
ππ£
ππ’
= β
1
3
π’β
4
3 = β
1
3
(π₯2
+ 6)β
4
3
π¦ = cos π£ , maka
ππ¦
ππ£
= βsin π£ = βsin π’β
1
3 = βsin(π₯2
+ 6)β
1
3
Maka
ππ¦
ππ₯
=
ππ¦
ππ£
.
ππ£
ππ’
.
ππ’
ππ₯
= βsin(π₯2
+ 6)β
1
3 . β
1
3
(π₯2
+ 6)β
4
3 .2π₯
ππ¦
ππ₯
=
β
1
3
. 2π₯ . βsin(π₯2
+ 6)β
1
3
(π₯2 + 6)
4
3
=
1
3
. 2π₯ . sin(π₯2
+ 6)β
1
3
β(π₯2 + 6)43
