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TUGAS MATEMATIKA 1
Disusun Oleh
Nama : Achmad Trybuana Kurniasandy
Kelas : 1 EA
Jurusan : Teknik Elektronika
Semester : 2 ( DUA )
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
2014/2015
Kawasan Industri Air Kantung, Sungailiat-Bangka 33211
Telp : 0717-93586, 95252 Faks : 0717-93585
E-mail : polman@polman-babel.ac.id
Website : http://polman-babel.ac.id
Carilah nilai turunan dari fungsi berikut ini !
1. ๐’š = ๐’™ ๐Ÿ•
+ ๐Ÿ”๐’™ ๐Ÿ“
+ ๐Ÿ๐Ÿ๐’™ ๐Ÿ‘
Jawaban :
๐‘ฆโ€ฒ
= 7๐‘ฅ6
+ 30๐‘ฅ4
+ 36๐‘ฅ2
2. ๐’š = ๐’™โˆ’๐Ÿ“
+ ๐Ÿ๐’™โˆ’๐Ÿ‘
+ ๐Ÿ‘๐’™โˆ’๐Ÿ•
Jawaban :
๐‘ฆโ€ฒ
= โˆ’5๐‘ฅโˆ’6
โˆ’6๐‘ฅโˆ’4
โˆ’21๐‘ฅโˆ’8
3. ๐’š = โˆš๐’™ ๐Ÿ + โˆš ๐’™ ๐Ÿ•๐Ÿ‘
Jawaban :
๐‘ฆ = ๐‘ฅ
2
2 + ๐‘ฅ
7
3 = ๐‘ฅ + ๐‘ฅ
7
3
๐‘ฆโ€ฒ
= 1 +
7
3
๐‘ฅ
7
3
โˆ’
3
3 = 1 +
7
3
๐‘ฅ
4
3
๐‘ฆโ€ฒ = 1 +
7
3
โˆš ๐‘ฅ43
4. ๐’š =
๐Ÿ
โˆš๐’™
+
๐Ÿ”
โˆš ๐’™ ๐Ÿ‘
Jawaban :
๐‘ฆ =
2
๐‘ฅ
1
2
+
6
๐‘ฅ
3
2
= 2๐‘ฅโˆ’
1
2 + 6๐‘ฅโˆ’
3
2
๐‘ฆโ€ฒ
= 2.โˆ’
1
2
. ๐‘ฅโˆ’
1
2
โˆ’
2
2 + 6.โˆ’
3
2
. ๐‘ฅโˆ’
3
2
โˆ’
2
2 = โˆ’๐‘ฅโˆ’
3
2โˆ’9๐‘ฅโˆ’
5
2 = โˆ’
1
๐‘ฅ
3
2
โˆ’
9
๐‘ฅ
5
2
๐‘ฆโ€ฒ = โˆ’
1
โˆš ๐‘ฅ3
โˆ’
9
โˆš ๐‘ฅ5
5. ๐’š = ๐’™ ๐Ÿ
. ๐ฌ๐ข๐ง ๐Ÿ๐’™
Jawaban :
๐‘ข = ๐‘ฅ2
โ†’ ๐‘ขโ€ฒ
= 2๐‘ฅ
๐‘ฃ = sin 2๐‘ฅ โ†’ ๐‘ฃโ€ฒ
= 2. cos ๐‘ฅ
๐‘ฆโ€ฒ
= ๐‘ขโ€ฒ
. ๐‘ฃ + ๐‘ข. ๐‘ฃโ€ฒ
= 2๐‘ฅ. sin2๐‘ฅ + ๐‘ฅ2
. 2.cos ๐‘ฅ
๐‘ฆโ€ฒ = 2๐‘ฅ. sin 2๐‘ฅ + 2๐‘ฅ2
.cos ๐‘ฅ
6. ๐’š = ( ๐’™ ๐Ÿ
+ ๐Ÿ). ๐œ๐จ๐ฌ ๐Ÿ‘๐’™
Jawaban :
๐‘ข = ๐‘ฅ2
+ 1 โ†’ ๐‘ขโ€ฒ
= 2๐‘ฅ
๐‘ฃ = cos 3๐‘ฅ โ†’ ๐‘ฃโ€ฒ
= โˆ’3.sin 3๐‘ฅ
๐‘ฆโ€ฒ
= ๐‘ขโ€ฒ
. ๐‘ฃ + ๐‘ข. ๐‘ฃโ€ฒ
= 2๐‘ฅ. cos 3๐‘ฅ + ( ๐‘ฅ2
+ 1).โˆ’3. sin3๐‘ฅ
๐‘ฆโ€ฒ = 2๐‘ฅ. cos 3๐‘ฅ + (โˆ’3๐‘ฅ2
โˆ’ 3).sin 3๐‘ฅ
7. ๐’š = ๐ฌ๐ข๐ง ๐Ÿ“๐’™ . ๐ญ๐š๐ง ๐Ÿ๐’™
Jawaban :
๐‘ข = sin 5๐‘ฅ โ†’ ๐‘ขโ€ฒ
= 5.cos 5๐‘ฅ
๐‘ฃ = tan 2๐‘ฅ โ†’ ๐‘ฃโ€ฒ
= 2. ๐‘ ๐‘’๐‘2
2๐‘ฅ
๐‘ฆโ€ฒ
= ๐‘ขโ€ฒ
. ๐‘ฃ + ๐‘ข. ๐‘ฃโ€ฒ
= 5.cos 5๐‘ฅ . tan 2๐‘ฅ + sin 5๐‘ฅ . 2. ๐‘ ๐‘’๐‘2
2๐‘ฅ
๐‘ฆโ€ฒ
= 5.cos 5๐‘ฅ. tan 2๐‘ฅ + 2. sin5๐‘ฅ . ๐‘ ๐‘’๐‘2
2๐‘ฅ
8. ๐’š = ๐œ๐จ๐ฌ ๐Ÿ‘๐’™ . ๐œ๐จ๐ญ ๐Ÿ’๐’™
Jawaban :
๐‘ข = cos 3๐‘ฅ โ†’ ๐‘ขโ€ฒ
= โˆ’3. sin3๐‘ฅ
๐‘ฃ = cot4๐‘ฅ โ†’ ๐‘ฃโ€ฒ
= โˆ’4. ๐‘๐‘ ๐‘2
4๐‘ฅ
๐‘ฆโ€ฒ
= ๐‘ขโ€ฒ
. ๐‘ฃ + ๐‘ข. ๐‘ฃโ€ฒ
= โˆ’3.sin 3๐‘ฅ . cot4๐‘ฅ + cos 3๐‘ฅ .โˆ’4. ๐‘๐‘ ๐‘2
4๐‘ฅ
๐‘ฆโ€ฒ
= โˆ’3.sin 3๐‘ฅ . cot4๐‘ฅ โˆ’ 4. cos 3๐‘ฅ . ๐‘๐‘ ๐‘2
4๐‘ฅ
9. ๐’š =
๐ฌ๐ข๐ง ๐Ÿ“๐’™
๐œ๐จ๐ฌ ๐Ÿ•๐’™
Jawaban :
๐‘ข = sin 5๐‘ฅ โ†’ ๐‘ขโ€ฒ
= 5.cos 5๐‘ฅ
๐‘ฃ = cos 7๐‘ฅ โ†’ ๐‘ฃโ€ฒ
= โˆ’7.sin 7๐‘ฅ
๐‘ฆโ€ฒ
=
๐‘ขโ€ฒ
. ๐‘ฃ โˆ’ ๐‘ข. ๐‘ฃโ€ฒ
๐‘ฃ2
=
5.cos 5๐‘ฅ . cos 7๐‘ฅ โˆ’ sin5๐‘ฅ . โˆ’7.sin 7๐‘ฅ
(cos 7๐‘ฅ)2
๐‘ฆโ€ฒ
=
5.cos 5๐‘ฅ. cos 7๐‘ฅ + 7.sin 5๐‘ฅ . sin7๐‘ฅ
๐‘๐‘œ๐‘ 27๐‘ฅ
10. ๐’š =
๐œ๐จ๐ฌ ๐Ÿ”๐’™
๐ญ๐š๐ง ๐Ÿ๐’™
Jawaban :
๐‘ข = cos 6๐‘ฅ โ†’ ๐‘ขโ€ฒ
= โˆ’6. sin6๐‘ฅ
๐‘ฃ = tan 2๐‘ฅ โ†’ ๐‘ฃโ€ฒ
= 2. ๐‘ ๐‘’๐‘2
2๐‘ฅ
๐‘ฆโ€ฒ
=
๐‘ขโ€ฒ
. ๐‘ฃ โˆ’ ๐‘ข. ๐‘ฃโ€ฒ
๐‘ฃ2
=
โˆ’6.sin 6๐‘ฅ. tan2๐‘ฅ โˆ’ cos 6๐‘ฅ. 2. ๐‘ ๐‘’๐‘2
2๐‘ฅ
(tan 2๐‘ฅ)2
๐‘ฆโ€ฒ
=
โˆ’6.sin 6๐‘ฅ. tan 2๐‘ฅ โˆ’ 2.cos 6๐‘ฅ. ๐‘ ๐‘’๐‘2
2๐‘ฅ
๐‘ก๐‘Ž๐‘›22๐‘ฅ

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Tugas Matematika 1

  • 1. TUGAS MATEMATIKA 1 Disusun Oleh Nama : Achmad Trybuana Kurniasandy Kelas : 1 EA Jurusan : Teknik Elektronika Semester : 2 ( DUA ) POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG 2014/2015 Kawasan Industri Air Kantung, Sungailiat-Bangka 33211 Telp : 0717-93586, 95252 Faks : 0717-93585 E-mail : polman@polman-babel.ac.id Website : http://polman-babel.ac.id
  • 2. Carilah nilai turunan dari fungsi berikut ini ! 1. ๐’š = ๐’™ ๐Ÿ• + ๐Ÿ”๐’™ ๐Ÿ“ + ๐Ÿ๐Ÿ๐’™ ๐Ÿ‘ Jawaban : ๐‘ฆโ€ฒ = 7๐‘ฅ6 + 30๐‘ฅ4 + 36๐‘ฅ2 2. ๐’š = ๐’™โˆ’๐Ÿ“ + ๐Ÿ๐’™โˆ’๐Ÿ‘ + ๐Ÿ‘๐’™โˆ’๐Ÿ• Jawaban : ๐‘ฆโ€ฒ = โˆ’5๐‘ฅโˆ’6 โˆ’6๐‘ฅโˆ’4 โˆ’21๐‘ฅโˆ’8 3. ๐’š = โˆš๐’™ ๐Ÿ + โˆš ๐’™ ๐Ÿ•๐Ÿ‘ Jawaban : ๐‘ฆ = ๐‘ฅ 2 2 + ๐‘ฅ 7 3 = ๐‘ฅ + ๐‘ฅ 7 3 ๐‘ฆโ€ฒ = 1 + 7 3 ๐‘ฅ 7 3 โˆ’ 3 3 = 1 + 7 3 ๐‘ฅ 4 3 ๐‘ฆโ€ฒ = 1 + 7 3 โˆš ๐‘ฅ43 4. ๐’š = ๐Ÿ โˆš๐’™ + ๐Ÿ” โˆš ๐’™ ๐Ÿ‘ Jawaban : ๐‘ฆ = 2 ๐‘ฅ 1 2 + 6 ๐‘ฅ 3 2 = 2๐‘ฅโˆ’ 1 2 + 6๐‘ฅโˆ’ 3 2 ๐‘ฆโ€ฒ = 2.โˆ’ 1 2 . ๐‘ฅโˆ’ 1 2 โˆ’ 2 2 + 6.โˆ’ 3 2 . ๐‘ฅโˆ’ 3 2 โˆ’ 2 2 = โˆ’๐‘ฅโˆ’ 3 2โˆ’9๐‘ฅโˆ’ 5 2 = โˆ’ 1 ๐‘ฅ 3 2 โˆ’ 9 ๐‘ฅ 5 2 ๐‘ฆโ€ฒ = โˆ’ 1 โˆš ๐‘ฅ3 โˆ’ 9 โˆš ๐‘ฅ5
  • 3. 5. ๐’š = ๐’™ ๐Ÿ . ๐ฌ๐ข๐ง ๐Ÿ๐’™ Jawaban : ๐‘ข = ๐‘ฅ2 โ†’ ๐‘ขโ€ฒ = 2๐‘ฅ ๐‘ฃ = sin 2๐‘ฅ โ†’ ๐‘ฃโ€ฒ = 2. cos ๐‘ฅ ๐‘ฆโ€ฒ = ๐‘ขโ€ฒ . ๐‘ฃ + ๐‘ข. ๐‘ฃโ€ฒ = 2๐‘ฅ. sin2๐‘ฅ + ๐‘ฅ2 . 2.cos ๐‘ฅ ๐‘ฆโ€ฒ = 2๐‘ฅ. sin 2๐‘ฅ + 2๐‘ฅ2 .cos ๐‘ฅ 6. ๐’š = ( ๐’™ ๐Ÿ + ๐Ÿ). ๐œ๐จ๐ฌ ๐Ÿ‘๐’™ Jawaban : ๐‘ข = ๐‘ฅ2 + 1 โ†’ ๐‘ขโ€ฒ = 2๐‘ฅ ๐‘ฃ = cos 3๐‘ฅ โ†’ ๐‘ฃโ€ฒ = โˆ’3.sin 3๐‘ฅ ๐‘ฆโ€ฒ = ๐‘ขโ€ฒ . ๐‘ฃ + ๐‘ข. ๐‘ฃโ€ฒ = 2๐‘ฅ. cos 3๐‘ฅ + ( ๐‘ฅ2 + 1).โˆ’3. sin3๐‘ฅ ๐‘ฆโ€ฒ = 2๐‘ฅ. cos 3๐‘ฅ + (โˆ’3๐‘ฅ2 โˆ’ 3).sin 3๐‘ฅ 7. ๐’š = ๐ฌ๐ข๐ง ๐Ÿ“๐’™ . ๐ญ๐š๐ง ๐Ÿ๐’™ Jawaban : ๐‘ข = sin 5๐‘ฅ โ†’ ๐‘ขโ€ฒ = 5.cos 5๐‘ฅ ๐‘ฃ = tan 2๐‘ฅ โ†’ ๐‘ฃโ€ฒ = 2. ๐‘ ๐‘’๐‘2 2๐‘ฅ ๐‘ฆโ€ฒ = ๐‘ขโ€ฒ . ๐‘ฃ + ๐‘ข. ๐‘ฃโ€ฒ = 5.cos 5๐‘ฅ . tan 2๐‘ฅ + sin 5๐‘ฅ . 2. ๐‘ ๐‘’๐‘2 2๐‘ฅ ๐‘ฆโ€ฒ = 5.cos 5๐‘ฅ. tan 2๐‘ฅ + 2. sin5๐‘ฅ . ๐‘ ๐‘’๐‘2 2๐‘ฅ 8. ๐’š = ๐œ๐จ๐ฌ ๐Ÿ‘๐’™ . ๐œ๐จ๐ญ ๐Ÿ’๐’™ Jawaban : ๐‘ข = cos 3๐‘ฅ โ†’ ๐‘ขโ€ฒ = โˆ’3. sin3๐‘ฅ ๐‘ฃ = cot4๐‘ฅ โ†’ ๐‘ฃโ€ฒ = โˆ’4. ๐‘๐‘ ๐‘2 4๐‘ฅ ๐‘ฆโ€ฒ = ๐‘ขโ€ฒ . ๐‘ฃ + ๐‘ข. ๐‘ฃโ€ฒ = โˆ’3.sin 3๐‘ฅ . cot4๐‘ฅ + cos 3๐‘ฅ .โˆ’4. ๐‘๐‘ ๐‘2 4๐‘ฅ
  • 4. ๐‘ฆโ€ฒ = โˆ’3.sin 3๐‘ฅ . cot4๐‘ฅ โˆ’ 4. cos 3๐‘ฅ . ๐‘๐‘ ๐‘2 4๐‘ฅ 9. ๐’š = ๐ฌ๐ข๐ง ๐Ÿ“๐’™ ๐œ๐จ๐ฌ ๐Ÿ•๐’™ Jawaban : ๐‘ข = sin 5๐‘ฅ โ†’ ๐‘ขโ€ฒ = 5.cos 5๐‘ฅ ๐‘ฃ = cos 7๐‘ฅ โ†’ ๐‘ฃโ€ฒ = โˆ’7.sin 7๐‘ฅ ๐‘ฆโ€ฒ = ๐‘ขโ€ฒ . ๐‘ฃ โˆ’ ๐‘ข. ๐‘ฃโ€ฒ ๐‘ฃ2 = 5.cos 5๐‘ฅ . cos 7๐‘ฅ โˆ’ sin5๐‘ฅ . โˆ’7.sin 7๐‘ฅ (cos 7๐‘ฅ)2 ๐‘ฆโ€ฒ = 5.cos 5๐‘ฅ. cos 7๐‘ฅ + 7.sin 5๐‘ฅ . sin7๐‘ฅ ๐‘๐‘œ๐‘ 27๐‘ฅ 10. ๐’š = ๐œ๐จ๐ฌ ๐Ÿ”๐’™ ๐ญ๐š๐ง ๐Ÿ๐’™ Jawaban : ๐‘ข = cos 6๐‘ฅ โ†’ ๐‘ขโ€ฒ = โˆ’6. sin6๐‘ฅ ๐‘ฃ = tan 2๐‘ฅ โ†’ ๐‘ฃโ€ฒ = 2. ๐‘ ๐‘’๐‘2 2๐‘ฅ ๐‘ฆโ€ฒ = ๐‘ขโ€ฒ . ๐‘ฃ โˆ’ ๐‘ข. ๐‘ฃโ€ฒ ๐‘ฃ2 = โˆ’6.sin 6๐‘ฅ. tan2๐‘ฅ โˆ’ cos 6๐‘ฅ. 2. ๐‘ ๐‘’๐‘2 2๐‘ฅ (tan 2๐‘ฅ)2 ๐‘ฆโ€ฒ = โˆ’6.sin 6๐‘ฅ. tan 2๐‘ฅ โˆ’ 2.cos 6๐‘ฅ. ๐‘ ๐‘’๐‘2 2๐‘ฅ ๐‘ก๐‘Ž๐‘›22๐‘ฅ