Tugas 2 MTK2
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Tugas 2 MTK2
- 1. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG Tugas 2 MTK2 Page 1 TUGAS 2 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 2 MTK2 Page 2 Tentukanlah nilai ππ¦ ππ₯ dari fungsi berikut ini ! 1. π¦ = β π₯5 + 6π₯2 + 3 2. π¦ = β π₯4 + 6π₯ + 1 3 3. π¦ = β π₯2 β 5π₯ 5 4. π¦ = 1 βπ₯4+2π₯ 5. π¦ = 1 βπ₯2β6π₯ 3 6. π¦ = 1 βπ₯2β5π₯+2 5 7. π¦ = sin β π₯2 + 6π₯ 8. π¦ = cos β π₯3 + 2 3 9. π¦ = sin 1 βπ₯2+2 10. π¦ = cos 1 βπ₯2+6 3 Jawaban : 1. π¦ = β π₯5 + 6π₯2 + 3 Misal u = π₯5 + 6π₯2 + 3 , maka ππ’ ππ₯ = 5π₯4 + 12π₯ π¦ = β π’ = π’ 1 2 , maka ππ¦ ππ’ = 1 2 π’β 1 2 = 1 2 (π₯5 + 6π₯2 + 3)β 1 2 Maka ππ¦ ππ₯ = ππ¦ ππ’ . ππ’ ππ₯ = 1 2 (π₯5 + 6π₯2 + 3)β 1 2 . (5π₯4 + 12π₯) ππ¦ ππ₯ = 1 2 (5π₯4 + 12π₯) (π₯5 + 6π₯2 + 3) 1 2 = 1 2 (5π₯4 + 12π₯) β π₯5 + 6π₯2 + 3 2. π¦ = β π₯4 + 6π₯ + 1 3 Misal u = π₯4 + 6π₯ + 1 , maka ππ’ ππ₯ = 4π₯3 + 6 π¦ = β π’3 = π’ 1 3 , maka ππ¦ ππ’ = 1 3 π’β 2 3 = 1 3 (π₯4 + 6π₯ + 1)β 2 3 Maka ππ¦ ππ₯ = ππ¦ ππ’ . ππ’ ππ₯ = 1 3 (π₯4 + 6π₯ + 1)β 2 3 . (4π₯3 + 6) ππ¦ ππ₯ = 1 3 (4π₯3 + 6) (π₯4 + 6π₯ + 1) 2 3 = 1 3 (4π₯3 + 6) β(π₯4 + 6π₯ + 1)23 3. π¦ = β π₯2 β 5π₯ 5 Misal u = π₯2 β 5π₯ , maka ππ’ ππ₯ = 2π₯ β 5
- 3. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG Tugas 2 MTK2 Page 3 π¦ = β π’5 = π’ 1 5 , maka ππ¦ ππ’ = 1 5 π’β 4 5 = 1 5 (π₯2 β 5π₯)β 4 5 Maka ππ¦ ππ₯ = ππ¦ ππ’ . ππ’ ππ₯ = 1 5 (π₯2 β 5π₯)β 4 5 .(2π₯ β 5) ππ¦ ππ₯ = 1 5 (2π₯ β 5) (π₯2 β 5π₯) 4 5 = 1 5 (2π₯ β 5) β(π₯2 β 5π₯)45 4. π¦ = 1 βπ₯4+2π₯ = 1 (π₯4+2π₯) 1 2 = (π₯4 + 2π₯)β 1 2 Misal u = π₯4 + 2π₯ , maka ππ’ ππ₯ = 4π₯3 + 2 π¦ = π’β 1 2 , maka ππ¦ ππ’ = β 1 2 π’β 3 2 = β 1 2 (π₯4 + 2π₯)β 3 2 Maka ππ¦ ππ₯ = ππ¦ ππ’ . ππ’ ππ₯ = β 1 2 (π₯4 + 2π₯)β 3 2 .(4π₯3 + 2) ππ¦ ππ₯ = β 1 2 (4π₯3 + 2) (π₯4 + 2π₯) 3 2 = β2π₯3 β 1 β(π₯4 + 2π₯)3 5. π¦ = 1 βπ₯2β6π₯ 3 = 1 (π₯2β6π₯) 1 3 = (π₯2 β 6π₯)β 1 3 Misal u = π₯2 β 6π₯ , maka ππ’ ππ₯ = 2π₯ β 6 π¦ = π’β 1 3 , maka ππ¦ ππ’ = β 1 3 π’β 4 3 = β 1 3 (π₯2 β 6π₯)β 4 3 Maka ππ¦ ππ₯ = ππ¦ ππ’ . ππ’ ππ₯ = β 1 3 (π₯2 β 6π₯)β 4 3 .(2π₯ β 6) ππ¦ ππ₯ = β 1 3 . (2π₯ β 6) (π₯2 β 6π₯)β 4 3 = β 1 3 (2π₯ β 6) β(π₯2 β 6π₯)43
- 4. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG Tugas 2 MTK2 Page 4 6. π¦ = 1 βπ₯2β5π₯+2 5 = 1 (π₯2β5π₯+2) 1 5 = (π₯2 β 5π₯ + 2)β 1 5 Misal u = π₯2 β 5π₯ + 2 , maka ππ’ ππ₯ = 2π₯ β 5 π¦ = π’β 1 5 , maka ππ¦ ππ’ = β 1 5 π’β 6 5 = β 1 5 (π₯2 β 5π₯ + 2)β 6 5 Maka ππ¦ ππ₯ = ππ¦ ππ’ . ππ’ ππ₯ = β 1 5 (π₯2 β 5π₯ + 2)β 6 5 .(2π₯ β 5) ππ¦ ππ₯ = β 1 5 . (2π₯ β 5) (π₯2 β 5π₯ + 2) 6 5 = β 1 5 (2π₯ β 5) β(π₯2 β 5π₯ + 2)65 7. π¦ = sin β π₯2 + 6π₯ Misal u = π₯2 + 6π₯ , maka ππ’ ππ₯ = 2π₯ + 6 π£ = β π’ = π’ 1 2 , maka ππ£ ππ’ = 1 2 π’β 1 2 = 1 2 (π₯2 + 6π₯ )β 1 2 π¦ = sin π£ , maka ππ¦ ππ£ = cos π£ = cos β π’ = cos β π₯2 + 6π₯ Maka ππ¦ ππ₯ = ππ¦ ππ£ . ππ£ ππ’ . ππ’ ππ₯ = cos β π₯2 + 6π₯ . 1 2 (π₯2 + 6π₯ )β 1 2 .(2π₯ + 6) ππ¦ ππ₯ = 1 2 . (2π₯ + 6) .cos β π₯2 + 6π₯ (π₯2 + 6π₯ ) 1 2 = ( π₯ + 3) .cos β π₯2 + 6π₯ β π₯2 + 6π₯ 8. π¦ = cos β π₯3 + 2 3 Misal u = π₯3 + 2 , maka ππ’ ππ₯ = 3π₯2 π£ = β π’3 = π’ 1 3 , maka ππ£ ππ’ = 1 3 π’β 2 3 = 1 3 (π₯3 + 2)β 2 3 π¦ = cos π£ , maka ππ¦ ππ£ = βsin π£ = βsin β π’3 = βsin β π₯3 + 2 3 Maka ππ¦ ππ₯ = ππ¦ ππ£ . ππ£ ππ’ . ππ’ ππ₯ = βsin β π₯3 + 2 3 . 1 3 (π₯3 + 2)β 2 3 .3π₯2
- 5. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG Tugas 2 MTK2 Page 5 ππ¦ ππ₯ = 1 3 . 3π₯2 . βsin β π₯3 + 2 3 (π₯3 + 2) 2 3 = π₯2 . βsin β π₯3 + 2 3 β(π₯3 + 2)23 9. π¦ = sin 1 βπ₯2+2 = sin 1 (π₯2+2) 1 2 = sin(π₯2 + 2)β 1 2 Misal u = π₯2 + 2 , maka ππ’ ππ₯ = 2π₯ π£ = π’β 1 2 , maka ππ£ ππ’ = β 1 2 π’β 3 2 = β 1 2 (π₯2 + 2)β 3 2 π¦ = sin π£ , maka ππ¦ ππ£ = cos π£ = cos π’β 1 2 = cos(π₯2 + 2)β 1 2 Maka ππ¦ ππ₯ = ππ¦ ππ£ . ππ£ ππ’ . ππ’ ππ₯ = cos(π₯2 + 2)β 1 2 . β 1 2 (π₯2 + 2)β 3 2 .2π₯ ππ¦ ππ₯ = β 1 2 . 2π₯ . cos(π₯2 + 2)β 1 2 (π₯2 + 2) 3 2 = βπ₯ . cos(π₯2 + 2)β 1 2 β(π₯2 + 2)3 10. π¦ = cos 1 βπ₯2+6 3 = cos 1 (π₯2+6) 1 3 = cos(π₯2 + 6)β 1 3 Misal u = π₯2 + 6 , maka ππ’ ππ₯ = 2π₯ π£ = π’β 1 3 , maka ππ£ ππ’ = β 1 3 π’β 4 3 = β 1 3 (π₯2 + 6)β 4 3 π¦ = cos π£ , maka ππ¦ ππ£ = βsin π£ = βsin π’β 1 3 = βsin(π₯2 + 6)β 1 3 Maka ππ¦ ππ₯ = ππ¦ ππ£ . ππ£ ππ’ . ππ’ ππ₯ = βsin(π₯2 + 6)β 1 3 . β 1 3 (π₯2 + 6)β 4 3 .2π₯ ππ¦ ππ₯ = β 1 3 . 2π₯ . βsin(π₯2 + 6)β 1 3 (π₯2 + 6) 4 3 = 1 3 . 2π₯ . sin(π₯2 + 6)β 1 3 β(π₯2 + 6)43