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TUGAS MATEMATIKA 1
D
I
S
U
S
U
N
Oleh :
Nama : Muhammad Wahyu Utama
Prodi : Teknik Elektronika
Kelas : 1 EA
Semester : 2 (Genap)
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
TAHUN AJARAN 2014/2015
Kawasan Industri Air Kantung Sungailiat, Bangka 33211
Telp. (0717) 93586, Fax. (0717) 93585
Email : polman@polman-babel.ac.id
Carilah nilai turunan dari fungsi berikut ini.
1. 𝑦 = π‘₯7
+ 6π‘₯5
+ 12π‘₯3
2. 𝑦 = π‘₯βˆ’5
+ 2π‘₯βˆ’3
+ 3π‘₯βˆ’7
3. 𝑦 = √ π‘₯2 + √ π‘₯73
4. 𝑦 =
2
√π‘₯
+
6
√π‘₯3
5. 𝑦 = π‘₯2
.sin 2π‘₯
6. 𝑦 = ( π‘₯2
+ 1).cos 3π‘₯
7. 𝑦 = sin 5π‘₯ . tan2π‘₯
8. 𝑦 = cos 3π‘₯ . cot 4π‘₯
9. 𝑦 =
sin 5π‘₯
cos7π‘₯
10. 𝑦 =
cos6π‘₯
tan 2π‘₯
Jawaban :
1. 𝑦′
= 7π‘₯6
+ 30π‘₯4
+ 36π‘₯2
2. 𝑦′
= βˆ’5π‘₯βˆ’6
βˆ’6π‘₯βˆ’4
βˆ’21π‘₯βˆ’8
3. 𝑦 = π‘₯
2
2 + π‘₯
7
3 = π‘₯ + π‘₯
7
3
𝑦′
= 1 +
7
3
π‘₯
7
3
βˆ’
3
3 = 1 +
7
3
π‘₯
4
3
𝑦′ = 1 +
7
3
√ π‘₯43
4. 𝑦 =
2
π‘₯
1
2
+
6
π‘₯
3
2
= 2π‘₯βˆ’
1
2 + 6π‘₯βˆ’
3
2
𝑦′
= 2.βˆ’
1
2
. π‘₯βˆ’
1
2
βˆ’
2
2 + 6.βˆ’
3
2
. π‘₯βˆ’
3
2
βˆ’
2
2 = βˆ’π‘₯βˆ’
3
2βˆ’9π‘₯βˆ’
5
2 = βˆ’
1
π‘₯
3
2
βˆ’
9
π‘₯
5
2
𝑦′ = βˆ’
1
√ π‘₯3
βˆ’
9
√ π‘₯5
5. 𝑒 = π‘₯2
β†’ 𝑒′
= 2π‘₯
𝑣 = sin 2π‘₯ β†’ 𝑣′
= 2. cos π‘₯
𝑦′
= 𝑒′
. 𝑣 + 𝑒. 𝑣′
= 2π‘₯. sin2π‘₯ + π‘₯2
. 2.cos π‘₯
𝑦′ = 2π‘₯. sin 2π‘₯ + 2π‘₯2
.cos π‘₯
6. 𝑒 = π‘₯2
+ 1 β†’ 𝑒′
= 2π‘₯
𝑣 = cos 3π‘₯ β†’ 𝑣′
= βˆ’3.sin 3π‘₯
𝑦′
= 𝑒′
. 𝑣 + 𝑒. 𝑣′
= 2π‘₯. cos 3π‘₯ + ( π‘₯2
+ 1).βˆ’3. sin3π‘₯
𝑦′ = 2π‘₯. cos 3π‘₯ + (βˆ’3π‘₯2
βˆ’ 3).sin 3π‘₯
7. 𝑒 = sin 5π‘₯ β†’ 𝑒′
= 5.cos 5π‘₯
𝑣 = tan 2π‘₯ β†’ 𝑣′
= 2. 𝑠𝑒𝑐2
2π‘₯
𝑦′
= 𝑒′
. 𝑣 + 𝑒. 𝑣′
= 5.cos 5π‘₯ . tan 2π‘₯ + sin 5π‘₯ . 2. 𝑠𝑒𝑐2
2π‘₯
𝑦′
= 5.cos 5π‘₯. tan 2π‘₯ + 2. sin5π‘₯ . 𝑠𝑒𝑐2
2π‘₯
8. 𝑒 = cos 3π‘₯ β†’ 𝑒′
= βˆ’3. sin3π‘₯
𝑣 = cot4π‘₯ β†’ 𝑣′
= βˆ’4. 𝑐𝑠𝑐2
4π‘₯
𝑦′
= 𝑒′
. 𝑣 + 𝑒. 𝑣′
= βˆ’3.sin 3π‘₯ . cot4π‘₯ + cos 3π‘₯ .βˆ’4. 𝑐𝑠𝑐2
4π‘₯
𝑦′
= βˆ’3.sin 3π‘₯ . cot4π‘₯ βˆ’ 4. cos 3π‘₯ . 𝑐𝑠𝑐2
4π‘₯
9. 𝑒 = sin 5π‘₯ β†’ 𝑒′
= 5.cos 5π‘₯
𝑣 = cos 7π‘₯ β†’ 𝑣′
= βˆ’7.sin 7π‘₯
𝑦′
=
𝑒′
. 𝑣 βˆ’ 𝑒. 𝑣′
𝑣2
=
5.cos 5π‘₯ . cos 7π‘₯ βˆ’ sin5π‘₯ . βˆ’7.sin 7π‘₯
(cos 7π‘₯)2
𝑦′
=
5.cos 5π‘₯. cos 7π‘₯ + 7.sin 5π‘₯ . sin7π‘₯
π‘π‘œπ‘ 27π‘₯
10. 𝑒 = cos 6π‘₯ β†’ 𝑒′
= βˆ’6. sin6π‘₯
𝑣 = tan 2π‘₯ β†’ 𝑣′
= 2. 𝑠𝑒𝑐2
2π‘₯
𝑦′
=
𝑒′
. 𝑣 βˆ’ 𝑒. 𝑣′
𝑣2
=
βˆ’6.sin 6π‘₯. tan2π‘₯ βˆ’ cos 6π‘₯. 2. 𝑠𝑒𝑐2
2π‘₯
(tan 2π‘₯)2
𝑦′
=
βˆ’6.sin 6π‘₯. tan 2π‘₯ βˆ’ 2.cos 6π‘₯. 𝑠𝑒𝑐2
2π‘₯
π‘‘π‘Žπ‘›22π‘₯

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Wahyu1

  • 1. TUGAS MATEMATIKA 1 D I S U S U N Oleh : Nama : Muhammad Wahyu Utama Prodi : Teknik Elektronika Kelas : 1 EA Semester : 2 (Genap) POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG TAHUN AJARAN 2014/2015 Kawasan Industri Air Kantung Sungailiat, Bangka 33211 Telp. (0717) 93586, Fax. (0717) 93585 Email : polman@polman-babel.ac.id
  • 2. Carilah nilai turunan dari fungsi berikut ini. 1. 𝑦 = π‘₯7 + 6π‘₯5 + 12π‘₯3 2. 𝑦 = π‘₯βˆ’5 + 2π‘₯βˆ’3 + 3π‘₯βˆ’7 3. 𝑦 = √ π‘₯2 + √ π‘₯73 4. 𝑦 = 2 √π‘₯ + 6 √π‘₯3 5. 𝑦 = π‘₯2 .sin 2π‘₯ 6. 𝑦 = ( π‘₯2 + 1).cos 3π‘₯ 7. 𝑦 = sin 5π‘₯ . tan2π‘₯ 8. 𝑦 = cos 3π‘₯ . cot 4π‘₯ 9. 𝑦 = sin 5π‘₯ cos7π‘₯ 10. 𝑦 = cos6π‘₯ tan 2π‘₯ Jawaban : 1. 𝑦′ = 7π‘₯6 + 30π‘₯4 + 36π‘₯2 2. 𝑦′ = βˆ’5π‘₯βˆ’6 βˆ’6π‘₯βˆ’4 βˆ’21π‘₯βˆ’8 3. 𝑦 = π‘₯ 2 2 + π‘₯ 7 3 = π‘₯ + π‘₯ 7 3 𝑦′ = 1 + 7 3 π‘₯ 7 3 βˆ’ 3 3 = 1 + 7 3 π‘₯ 4 3 𝑦′ = 1 + 7 3 √ π‘₯43 4. 𝑦 = 2 π‘₯ 1 2 + 6 π‘₯ 3 2 = 2π‘₯βˆ’ 1 2 + 6π‘₯βˆ’ 3 2 𝑦′ = 2.βˆ’ 1 2 . π‘₯βˆ’ 1 2 βˆ’ 2 2 + 6.βˆ’ 3 2 . π‘₯βˆ’ 3 2 βˆ’ 2 2 = βˆ’π‘₯βˆ’ 3 2βˆ’9π‘₯βˆ’ 5 2 = βˆ’ 1 π‘₯ 3 2 βˆ’ 9 π‘₯ 5 2 𝑦′ = βˆ’ 1 √ π‘₯3 βˆ’ 9 √ π‘₯5 5. 𝑒 = π‘₯2 β†’ 𝑒′ = 2π‘₯ 𝑣 = sin 2π‘₯ β†’ 𝑣′ = 2. cos π‘₯ 𝑦′ = 𝑒′ . 𝑣 + 𝑒. 𝑣′ = 2π‘₯. sin2π‘₯ + π‘₯2 . 2.cos π‘₯ 𝑦′ = 2π‘₯. sin 2π‘₯ + 2π‘₯2 .cos π‘₯ 6. 𝑒 = π‘₯2 + 1 β†’ 𝑒′ = 2π‘₯ 𝑣 = cos 3π‘₯ β†’ 𝑣′ = βˆ’3.sin 3π‘₯ 𝑦′ = 𝑒′ . 𝑣 + 𝑒. 𝑣′ = 2π‘₯. cos 3π‘₯ + ( π‘₯2 + 1).βˆ’3. sin3π‘₯ 𝑦′ = 2π‘₯. cos 3π‘₯ + (βˆ’3π‘₯2 βˆ’ 3).sin 3π‘₯ 7. 𝑒 = sin 5π‘₯ β†’ 𝑒′ = 5.cos 5π‘₯ 𝑣 = tan 2π‘₯ β†’ 𝑣′ = 2. 𝑠𝑒𝑐2 2π‘₯ 𝑦′ = 𝑒′ . 𝑣 + 𝑒. 𝑣′ = 5.cos 5π‘₯ . tan 2π‘₯ + sin 5π‘₯ . 2. 𝑠𝑒𝑐2 2π‘₯ 𝑦′ = 5.cos 5π‘₯. tan 2π‘₯ + 2. sin5π‘₯ . 𝑠𝑒𝑐2 2π‘₯ 8. 𝑒 = cos 3π‘₯ β†’ 𝑒′ = βˆ’3. sin3π‘₯ 𝑣 = cot4π‘₯ β†’ 𝑣′ = βˆ’4. 𝑐𝑠𝑐2 4π‘₯ 𝑦′ = 𝑒′ . 𝑣 + 𝑒. 𝑣′ = βˆ’3.sin 3π‘₯ . cot4π‘₯ + cos 3π‘₯ .βˆ’4. 𝑐𝑠𝑐2 4π‘₯ 𝑦′ = βˆ’3.sin 3π‘₯ . cot4π‘₯ βˆ’ 4. cos 3π‘₯ . 𝑐𝑠𝑐2 4π‘₯
  • 3. 9. 𝑒 = sin 5π‘₯ β†’ 𝑒′ = 5.cos 5π‘₯ 𝑣 = cos 7π‘₯ β†’ 𝑣′ = βˆ’7.sin 7π‘₯ 𝑦′ = 𝑒′ . 𝑣 βˆ’ 𝑒. 𝑣′ 𝑣2 = 5.cos 5π‘₯ . cos 7π‘₯ βˆ’ sin5π‘₯ . βˆ’7.sin 7π‘₯ (cos 7π‘₯)2 𝑦′ = 5.cos 5π‘₯. cos 7π‘₯ + 7.sin 5π‘₯ . sin7π‘₯ π‘π‘œπ‘ 27π‘₯ 10. 𝑒 = cos 6π‘₯ β†’ 𝑒′ = βˆ’6. sin6π‘₯ 𝑣 = tan 2π‘₯ β†’ 𝑣′ = 2. 𝑠𝑒𝑐2 2π‘₯ 𝑦′ = 𝑒′ . 𝑣 βˆ’ 𝑒. 𝑣′ 𝑣2 = βˆ’6.sin 6π‘₯. tan2π‘₯ βˆ’ cos 6π‘₯. 2. 𝑠𝑒𝑐2 2π‘₯ (tan 2π‘₯)2 𝑦′ = βˆ’6.sin 6π‘₯. tan 2π‘₯ βˆ’ 2.cos 6π‘₯. 𝑠𝑒𝑐2 2π‘₯ π‘‘π‘Žπ‘›22π‘₯