Tugas 1 MTK2
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This document contains a mathematics assignment with 10 problems asking to find the derivative of various functions. It provides the functions and asks the student named Kuntoro from the Polytechnic Manufacturing State of Bangka Belitung to find the derivatives. It then shows the student's work and answers for finding the derivatives of each function provided in the 10 problems.
Tugas 1 MTK2
- 1. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG Tugas 1 MTK2 Page 1 TUGAS 1 MATEMATIKA 2 D I S U S U N Oleh : Nama : Kuntoro NPM : 003 14 15 Prodi : Teknik Elektronika Kelas : 1E 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