Tugas 1 MTK2 punya gue, lo mau apa??

1. Nama : Siti Fatimah
NPM : 0031427
Kelas : 1 EA
Mata Kuliah : Matematika2
Carilahnilai turunandari fungsi berikutini.
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). cos3𝑥
7 . 𝑦 = sin 5𝑥.tan 2𝑥
8 . 𝑦 = cos3𝑥.cot4𝑥
9 . 𝑦 =
sin5𝑥
cos7𝑥
10 . 𝑦 =
cos 6𝑥
tan 2𝑥
Penyelesaian:
1 . 𝑦 = 𝑥7 + 6𝑥5 + 12𝑥3
𝑦′ = 7𝑥6 + 30𝑥4 + 36𝑥2
2 . 𝑦 = 𝑥−5 + 2𝑥−3 + 3𝑥−7
𝑦′ = −5𝑥−6 − 6𝑥−4 − 21𝑥−8
3 . 𝑦 = √𝑥2 + √𝑥73
𝑦 = 𝑥 + 𝑥
7
3
𝑦′ = 1 +
7
3
𝑥
7
3
−3

2. 𝑦′ = 1 +
7
3
𝑥
4
3
𝑦′ = 1 +
7
3
√𝑥43
4 . 𝑦 =
2
√ 𝑥
+
6
√𝑥3
𝑦 =
2
𝑥
1
2
+
6
𝑥
3
2
𝑦 = 2𝑥−
1
2 + 6𝑥−
3
2
𝑦′ = −𝑥−
3
2 − 9𝑥−
5
2
𝑦′ = −
1
𝑥
3
2
−
9
𝑥
5
2
𝑦′ = −
1
√𝑥3
−
9
√ 𝑥5
5 . 𝑦 = 𝑥2.sin 2𝑥
Misal 𝑢 = 𝑥2 → 𝑢′ = 2𝑥
𝑣 = sin 2𝑥 → 𝑣′ = 2. cos 𝑥
𝑦′ = 𝑢′ 𝑣 + 𝑢𝑣′
𝑦′ = 2𝑥. sin 2𝑥 + 𝑥2.2cos 𝑥
𝑦′ = 2𝑥. sin 2𝑥 + 2𝑥2.cos 𝑥
6 . 𝑦 = ( 𝑥2 + 1). cos3𝑥
Misal 𝑢 = 𝑥2 + 1 → 𝑢′ = 2𝑥
𝑣 = cos3𝑥 → 𝑣′ = −3.sin 3𝑥
𝑦′ = 𝑢′ 𝑣 + 𝑢𝑣′
𝑦′ = 2𝑥. cos 3𝑥 + ( 𝑥2 + 1). −3.sin 3𝑥
𝑦′ = 2𝑥. cos3𝑥 + (−3𝑥2 − 3).sin 3𝑥
7 . 𝑦 = sin 5𝑥.tan 2𝑥
Misal 𝑢 = sin 5𝑥 → 𝑢′ = 5. cos5𝑥
𝑣 = tan 2𝑥 → 𝑣′ = 2. 𝑠𝑒𝑐2 2𝑥

3. 𝑦′ = 𝑢′ 𝑣 + 𝑢𝑣′
𝑦′ = 5. cos5𝑥.tan 2𝑥 + sin 5𝑥. 2. 𝑠𝑒𝑐2 2𝑥
𝑦′ = 5. cos5𝑥.tan 2𝑥 + 2. sin 5𝑥. 𝑠𝑒𝑐2 2𝑥
8 . 𝑦 = cos3𝑥.cot4𝑥
Misal 𝑢 = cos3𝑥 → 𝑢′ = −3. sin 3𝑥
𝑣 = cot4𝑥 → 𝑣′ = −4. 𝑐𝑜𝑠𝑒𝑐 2 4𝑥
𝑦′ = 𝑢′ 𝑣 + 𝑢𝑣′
𝑦′ = −3. sin 3𝑥. cot4𝑥 + cos3𝑥 . − 4. 𝑐𝑜𝑠𝑒𝑐 2 4𝑥
𝑦′ = −3. sin 3𝑥 . cot4𝑥 + 4. cos3𝑥 . 𝑐𝑜𝑠𝑒𝑐 2 4𝑥
9 . 𝑦 =
sin5𝑥
cos7𝑥
Misal 𝑢 = sin 5𝑥 → 𝑢′ = 5.cos5𝑥
𝑣 = cos7𝑥 → 𝑣′ = −7. sin 7𝑥
𝑦′ =
𝑢′. 𝑣 − 𝑢. 𝑣′
𝑣2
𝑦′ =
5.cos5𝑥 .cos7𝑥 − sin 5𝑥 .−7.sin 7𝑥
(cos7𝑥)2
𝑦′ =
5.cos5𝑥 .cos7𝑥 + 7.sin 5𝑥.sin 7𝑥
𝑐𝑜𝑠27𝑥
10 . 𝑦 =
cos 6𝑥
tan 2𝑥
Misal 𝑢 = cos6𝑥 → 𝑢′ = −6 .sin 6𝑥
𝑣 = tan2𝑥 → 𝑣′ = 2. 𝑠𝑒𝑐22𝑥
𝑦′ =
𝑢′. 𝑣 − 𝑢. 𝑣′
𝑣2
𝑦′ =
−6 .sin 6𝑥 . tan2𝑥 − cos6𝑥.2. 𝑠𝑒𝑐22𝑥
(tan 2𝑥)2
𝑦′ =
−6 .sin 6𝑥 . tan2𝑥 − 2 .cos6𝑥. 𝑠𝑒𝑐22𝑥
𝑡𝑎𝑛2 2𝑥