1. The document is a mathematics assignment from Barry Alkahfi, a student at POLMAN Bangka Belitung University. 2. The assignment contains 10 problems to derive the derivatives of various functions. 3. Barry provides the solutions and steps to derive each derivative function.

1. Tugas matematika 1
Disusun oleh:
Nama : Barry Alkahfi
Kelas : 1 EA
Prodi : Teknik Elektronika
Semester : 2(genap)
POLITEKNIK MANUFAKTUR (POLMAN) NEGERI
BANGKA BELITUNG
TAHUN 2014/2015
Industri Air Kantung Sungailiat 33211
Bangka Induk Propinsi Kepulauan Bangka Belitung
Telp : (0717) 431335 ext. 2281, 2126
Fax : (0717) 93585 email : polman@zircon.timah.co.id

2. Tugas 1
Carilahturunandarinilaifungsiberikut !!!
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 . tan 2𝑥
8. 𝑦 = cos 3𝑥 . cot 4𝑥
9. 𝑦 =
sin 5𝑥
cos 7𝑥
10. 𝑦 =
cos 6𝑥
tan 2𝑥

3. 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
𝑦′−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. sin 2𝑥 + 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.sin 3𝑥
𝑦′ = 2𝑥.cos 3𝑥 + (−3𝑥2
− 3). sin3𝑥

4. 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.sin 5𝑥 . 𝑠𝑒𝑐2
2𝑥
8. 𝑢 = cos 3𝑥 → 𝑢′
= −3.sin 3𝑥
𝑣 = cot4𝑥 → 𝑣′
= −4. 𝑐𝑠𝑐2
4𝑥
𝑦′
= 𝑢′
. 𝑣 + 𝑢. 𝑣′
= −3. sin3𝑥 . cot 4𝑥 + cos 3𝑥 . −4. 𝑐𝑠𝑐2
4𝑥
𝑦′
= −3.sin 3𝑥 . cot 4𝑥 − 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 − sin 5𝑥 . −7.sin 7𝑥
(cos 7𝑥)2
𝑦′
=
5.cos 5𝑥 . cos 7𝑥 + 7.sin 5𝑥 . sin 7𝑥
𝑐𝑜𝑠27𝑥
10. 𝑢 = cos 6𝑥 → 𝑢′
= −6.sin 6𝑥
𝑣 = tan 2𝑥 → 𝑣′
= 2. 𝑠𝑒𝑐2
2𝑥
𝑦′
=
𝑢′
. 𝑣 − 𝑢. 𝑣′
𝑣2
=
−6. sin6𝑥. tan 2𝑥 − cos 6𝑥. 2. 𝑠𝑒𝑐2
2𝑥
(tan 2𝑥)2
𝑦′
=
−6.sin 6𝑥. tan 2𝑥 − 2.cos 6𝑥. 𝑠𝑒𝑐2
2𝑥
𝑡𝑎𝑛22𝑥