1. The document is a mathematics assignment from Devi Yunita, an electronics engineering student at Politeknik Manufaktur Negeri Bangka Belitung. 2. It contains 10 problems asking to find the derivatives of various functions. 3. Devi Yunita provides the detailed work and solutions for finding the derivatives of each function as requested in the assignment.

1. Tugas Matematika 1
D
I
S
U
S
U
N
Oleh :
Nama : Devi Yunita
Prodi : Teknik Elektronika
Kelas : 1 EA
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
TAHUN AJARAN 2014/2015

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
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

3. 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