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POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Tugas 1 MTK2 Page 1
TUGAS 1
MATEMATIKA 2
D
I
S
U
S
U
N
Oleh :
Nama : Tia Lestari Dewi Yanti
NPM : 003 14 29
Prodi : Teknik Elektronika
Kelas : 1Elektronika A
Semester : 2 (Dua)
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Kawasan Industri Air Kantung Sungailiat, Bangka 33211
Telp. (0717) 93586, Fax. (0717) 93585
Email : polman@polman-babel.ac.id
Website : www.polman-babel.ac.id
TAHUN AJARAN 2014/2015
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Tugas 1 MTK2 Page 2
Carilah nilai turunan dari fungsi berikut ini !
1. 𝑦 = 𝑥7
+ 6𝑥5
+ 12𝑥3
2. 𝑦 = 𝑥−5
+ 2𝑥−3
+ 3𝑥−7
3. 𝑦 = √ 𝑥2 + √ 𝑥73
4. 𝑦 =
2
√𝑥
+
6
√𝑥3
5. 𝑦 = 𝑥2
.sin 2𝑥
6. 𝑦 = ( 𝑥2
+ 1).cos 3𝑥
7. 𝑦 = sin 5𝑥 . tan2𝑥
8. 𝑦 = cos 3𝑥 . cot 4𝑥
9. 𝑦 =
sin 5𝑥
cos7𝑥
10. 𝑦 =
cos6𝑥
tan 2𝑥
Jawaban :
1. 𝑦′
= 7𝑥6
+ 30𝑥4
+ 36𝑥2
2. 𝑦′
= −5𝑥−6
−6𝑥−4
−21𝑥−8
3. 𝑦 = 𝑥
2
2 + 𝑥
7
3 = 𝑥 + 𝑥
7
3
𝑦′
= 1 +
7
3
𝑥
7
3
−
3
3 = 1 +
7
3
𝑥
4
3
𝑦′ = 1 +
7
3
√ 𝑥43
4. 𝑦 =
2
𝑥
1
2
+
6
𝑥
3
2
= 2𝑥−
1
2 + 6𝑥−
3
2
𝑦′
= 2.−
1
2
. 𝑥−
1
2
−
2
2 + 6.−
3
2
. 𝑥−
3
2
−
2
2 = −𝑥−
3
2−9𝑥−
5
2 = −
1
𝑥
3
2
−
9
𝑥
5
2
𝑦′ = −
1
√ 𝑥3
−
9
√ 𝑥5
5. 𝑢 = 𝑥2
→ 𝑢′
= 2𝑥
𝑣 = sin 2𝑥 → 𝑣′
= 2. cos 𝑥
𝑦′
= 𝑢′
. 𝑣 + 𝑢. 𝑣′
= 2𝑥. sin2𝑥 + 𝑥2
. 2.cos 𝑥
𝑦′ = 2𝑥. sin 2𝑥 + 2𝑥2
.cos 𝑥
6. 𝑢 = 𝑥2
+ 1 → 𝑢′
= 2𝑥
𝑣 = cos 3𝑥 → 𝑣′
= −3.sin 3𝑥
𝑦′
= 𝑢′
. 𝑣 + 𝑢. 𝑣′
= 2𝑥. cos 3𝑥 + ( 𝑥2
+ 1).−3. sin3𝑥
𝑦′ = 2𝑥. cos 3𝑥 + (−3𝑥2
− 3).sin 3𝑥
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Tugas 1 MTK2 Page 3
7. 𝑢 = sin 5𝑥 → 𝑢′
= 5.cos 5𝑥
𝑣 = tan 2𝑥 → 𝑣′
= 2. 𝑠𝑒𝑐2
2𝑥
𝑦′
= 𝑢′
. 𝑣 + 𝑢. 𝑣′
= 5.cos 5𝑥 . tan 2𝑥 + sin 5𝑥 . 2. 𝑠𝑒𝑐2
2𝑥
𝑦′
= 5.cos 5𝑥. tan 2𝑥 + 2. sin5𝑥 . 𝑠𝑒𝑐2
2𝑥
8. 𝑢 = cos 3𝑥 → 𝑢′
= −3. sin3𝑥
𝑣 = cot4𝑥 → 𝑣′
= −4. 𝑐𝑠𝑐2
4𝑥
𝑦′
= 𝑢′
. 𝑣 + 𝑢. 𝑣′
= −3.sin 3𝑥 . cot4𝑥 + cos 3𝑥 .−4. 𝑐𝑠𝑐2
4𝑥
𝑦′
= −3.sin 3𝑥 . cot4𝑥 − 4. cos 3𝑥 . 𝑐𝑠𝑐2
4𝑥
9. 𝑢 = sin 5𝑥 → 𝑢′
= 5.cos 5𝑥
𝑣 = cos 7𝑥 → 𝑣′
= −7.sin 7𝑥
𝑦′
=
𝑢′
. 𝑣 − 𝑢. 𝑣′
𝑣2
=
5.cos 5𝑥 . cos 7𝑥 − sin5𝑥 . −7.sin 7𝑥
(cos 7𝑥)2
𝑦′
=
5.cos 5𝑥. cos 7𝑥 + 7.sin 5𝑥 . sin7𝑥
𝑐𝑜𝑠27𝑥
10. 𝑢 = cos 6𝑥 → 𝑢′
= −6. sin6𝑥
𝑣 = tan 2𝑥 → 𝑣′
= 2. 𝑠𝑒𝑐2
2𝑥
𝑦′
=
𝑢′
. 𝑣 − 𝑢. 𝑣′
𝑣2
=
−6.sin 6𝑥. tan2𝑥 − cos 6𝑥. 2. 𝑠𝑒𝑐2
2𝑥
(tan 2𝑥)2
𝑦′
=
−6.sin 6𝑥. tan 2𝑥 − 2.cos 6𝑥. 𝑠𝑒𝑐2
2𝑥
𝑡𝑎𝑛22𝑥

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POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG Tugas 1 MTK2

  • 1. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG Tugas 1 MTK2 Page 1 TUGAS 1 MATEMATIKA 2 D I S U S U N Oleh : Nama : Tia Lestari Dewi Yanti NPM : 003 14 29 Prodi : Teknik Elektronika Kelas : 1Elektronika A Semester : 2 (Dua) POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG Kawasan Industri Air Kantung Sungailiat, Bangka 33211 Telp. (0717) 93586, Fax. (0717) 93585 Email : polman@polman-babel.ac.id Website : www.polman-babel.ac.id TAHUN AJARAN 2014/2015
  • 2. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG Tugas 1 MTK2 Page 2 Carilah nilai turunan dari fungsi berikut ini ! 1. 𝑦 = 𝑥7 + 6𝑥5 + 12𝑥3 2. 𝑦 = 𝑥−5 + 2𝑥−3 + 3𝑥−7 3. 𝑦 = √ 𝑥2 + √ 𝑥73 4. 𝑦 = 2 √𝑥 + 6 √𝑥3 5. 𝑦 = 𝑥2 .sin 2𝑥 6. 𝑦 = ( 𝑥2 + 1).cos 3𝑥 7. 𝑦 = sin 5𝑥 . tan2𝑥 8. 𝑦 = cos 3𝑥 . cot 4𝑥 9. 𝑦 = sin 5𝑥 cos7𝑥 10. 𝑦 = cos6𝑥 tan 2𝑥 Jawaban : 1. 𝑦′ = 7𝑥6 + 30𝑥4 + 36𝑥2 2. 𝑦′ = −5𝑥−6 −6𝑥−4 −21𝑥−8 3. 𝑦 = 𝑥 2 2 + 𝑥 7 3 = 𝑥 + 𝑥 7 3 𝑦′ = 1 + 7 3 𝑥 7 3 − 3 3 = 1 + 7 3 𝑥 4 3 𝑦′ = 1 + 7 3 √ 𝑥43 4. 𝑦 = 2 𝑥 1 2 + 6 𝑥 3 2 = 2𝑥− 1 2 + 6𝑥− 3 2 𝑦′ = 2.− 1 2 . 𝑥− 1 2 − 2 2 + 6.− 3 2 . 𝑥− 3 2 − 2 2 = −𝑥− 3 2−9𝑥− 5 2 = − 1 𝑥 3 2 − 9 𝑥 5 2 𝑦′ = − 1 √ 𝑥3 − 9 √ 𝑥5 5. 𝑢 = 𝑥2 → 𝑢′ = 2𝑥 𝑣 = sin 2𝑥 → 𝑣′ = 2. cos 𝑥 𝑦′ = 𝑢′ . 𝑣 + 𝑢. 𝑣′ = 2𝑥. sin2𝑥 + 𝑥2 . 2.cos 𝑥 𝑦′ = 2𝑥. sin 2𝑥 + 2𝑥2 .cos 𝑥 6. 𝑢 = 𝑥2 + 1 → 𝑢′ = 2𝑥 𝑣 = cos 3𝑥 → 𝑣′ = −3.sin 3𝑥 𝑦′ = 𝑢′ . 𝑣 + 𝑢. 𝑣′ = 2𝑥. cos 3𝑥 + ( 𝑥2 + 1).−3. sin3𝑥 𝑦′ = 2𝑥. cos 3𝑥 + (−3𝑥2 − 3).sin 3𝑥
  • 3. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG Tugas 1 MTK2 Page 3 7. 𝑢 = sin 5𝑥 → 𝑢′ = 5.cos 5𝑥 𝑣 = tan 2𝑥 → 𝑣′ = 2. 𝑠𝑒𝑐2 2𝑥 𝑦′ = 𝑢′ . 𝑣 + 𝑢. 𝑣′ = 5.cos 5𝑥 . tan 2𝑥 + sin 5𝑥 . 2. 𝑠𝑒𝑐2 2𝑥 𝑦′ = 5.cos 5𝑥. tan 2𝑥 + 2. sin5𝑥 . 𝑠𝑒𝑐2 2𝑥 8. 𝑢 = cos 3𝑥 → 𝑢′ = −3. sin3𝑥 𝑣 = cot4𝑥 → 𝑣′ = −4. 𝑐𝑠𝑐2 4𝑥 𝑦′ = 𝑢′ . 𝑣 + 𝑢. 𝑣′ = −3.sin 3𝑥 . cot4𝑥 + cos 3𝑥 .−4. 𝑐𝑠𝑐2 4𝑥 𝑦′ = −3.sin 3𝑥 . cot4𝑥 − 4. cos 3𝑥 . 𝑐𝑠𝑐2 4𝑥 9. 𝑢 = sin 5𝑥 → 𝑢′ = 5.cos 5𝑥 𝑣 = cos 7𝑥 → 𝑣′ = −7.sin 7𝑥 𝑦′ = 𝑢′ . 𝑣 − 𝑢. 𝑣′ 𝑣2 = 5.cos 5𝑥 . cos 7𝑥 − sin5𝑥 . −7.sin 7𝑥 (cos 7𝑥)2 𝑦′ = 5.cos 5𝑥. cos 7𝑥 + 7.sin 5𝑥 . sin7𝑥 𝑐𝑜𝑠27𝑥 10. 𝑢 = cos 6𝑥 → 𝑢′ = −6. sin6𝑥 𝑣 = tan 2𝑥 → 𝑣′ = 2. 𝑠𝑒𝑐2 2𝑥 𝑦′ = 𝑢′ . 𝑣 − 𝑢. 𝑣′ 𝑣2 = −6.sin 6𝑥. tan2𝑥 − cos 6𝑥. 2. 𝑠𝑒𝑐2 2𝑥 (tan 2𝑥)2 𝑦′ = −6.sin 6𝑥. tan 2𝑥 − 2.cos 6𝑥. 𝑠𝑒𝑐2 2𝑥 𝑡𝑎𝑛22𝑥