This document contains a student's math homework assignment on calculating derivatives. It includes 10 problems asking the student to find the derivative of given functions. The student provides the step-by-step work and solutions for each problem. Some examples of functions differentiated are y=x7+6x5+12x3, y=sin5x tan2x, and y=cos6x/tan2x.

1. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Tugas 1 MTK2 Page 1
TUGAS 1
MATEMATIKA 2
D
I
S
U
S
U
N
Oleh :
Nama : Miza Pisari
NPM : 003 14 17
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