Tugas 1 mtk2 yori
- 1. TUGAS 1 MATEMATIKA 2 Disusun Oleh : Nama : Dita Yoriza Prodi : Teknik Elektronika Kelas : 1E A Semester : 2 (Genap) 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 Carilah nilai turunan dari fungsi berikut ini !
- 2. 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
- 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 = −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
- 4. 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