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Tugas matematika 1 (smstr 2)
1.
Nama :Miza Pisari NPM
: 0031417 Kelas : 1 Elektronika A Tugas : MATEMATIKA Carilah nilai turunan dari fungsi berikut ini. 1. π¦ = π₯7 + 6π₯5 + 12π₯3 2. π¦ = π₯β5 + 2π₯β3 + 3π₯β7 3. π¦ = βπ₯2 + βπ₯73 4. π¦ = 2 β π₯ + 6 βπ₯3 5. π¦ = π₯2 .sin 2π₯ 6. π¦ = ( π₯2 + 1). cos3π₯ 7. π¦ = sin 5π₯ . tan 2π₯ 8. π¦ = cos3π₯ . cot 4π₯ 9. π¦ = sin 5π₯ cos 7π₯ 10. π¦ = cos 6π₯ tan 2π₯ Jawab: 1. π¦β² = 7π₯6 + 30π₯4 + 36π₯2 2. π¦β² = β5π₯β6 β6π₯β4 β21π₯β8 3. π¦ = π₯ 2 2 + π₯ 7 3 = π₯ + π₯ 7 3 π¦β² = 1 + 7 3 π₯ 7 3 β 3 3 = 1 + 7 3 π₯ 4 3 π¦β² = 1 + 7 3 β π₯43 4. π¦ = 2 π₯ 1 2 + 6 π₯ 3 2 = 2π₯ β 1 2 + 6π₯ β 3 2 π¦β² = 2. β 1 2 . π₯ β 1 2 β 2 2 + 6. β 3 2 . π₯ β 3 2 β 2 2 = βπ₯ β 3 2β9π₯ β 5 2 = β 1 π₯ 3 2 β 9 π₯ 5 2 π¦β² = β 1 βπ₯3 β 9 βπ₯5 5. π’ = π₯2 β π’β² = 2π₯ π£ = sin 2π₯ β π£β² = 2. cos π₯ π¦β² = π’β² . π£ + π’. π£β² = 2π₯. sin 2π₯ + π₯2 .2. cos π₯ π¦β² = 2π₯. sin2π₯ + 2π₯2 . cos π₯ 6. π’ = π₯2 + 1 β π’β² = 2π₯ π£ = cos3π₯ β π£β² = β3. sin 3π₯ π¦β² = π’β² . π£ + π’. π£β² = 2π₯. cos3π₯ + ( π₯2 + 1). β3. sin 3π₯ π¦β² = 2π₯. cos3π₯ + (β3π₯2 β 3).sin 3π₯ 7. π’ = sin 5π₯ β π’β² = 5. cos5π₯ π£ = tan 2π₯ β π£β² = 2. π ππ2 2π₯ π¦β² = π’β² . π£ + π’. π£β² = 5. cos5π₯ . tan2π₯ + sin 5π₯ . 2. π ππ2 2π₯ π¦β² = 5. cos5π₯ .tan 2π₯ + 2. sin 5π₯ . π ππ2 2π₯ 8. π’ = cos3π₯ β π’β² = β3.sin 3π₯ π£ = cot4π₯ β π£β² = β4. ππ π2 4π₯ π¦β² = π’β² . π£ + π’. π£β² = β3. sin 3π₯ . cot 4π₯ + cos3π₯ . β4. ππ π2 4π₯ π¦β² = β3. sin 3π₯ . cot 4π₯ β 4. cos3π₯ . ππ π2 4π₯ 9. π’ = sin 5π₯ β π’β² = 5. cos5π₯ π£ = cos7π₯ β π£β² = β7. sin 7π₯
2.
π¦β² = π’β² . π£ β
π’. π£β² π£2 = 5. cos5π₯ . cos7π₯ β sin 5π₯ . β7. sin 7π₯ (cos7π₯)2 π¦β² = 5. cos5π₯ .cos 7π₯ + 7. sin 5π₯ . sin 7π₯ πππ 27π₯ 10. π’ = cos6π₯ β π’β² = β6.sin 6π₯ π£ = tan 2π₯ β π£β² = 2. π ππ2 2π₯ π¦β² = π’β² . π£ β π’. π£β² π£2 = β6. sin 6π₯. tan 2π₯ β cos6π₯.2. π ππ2 2π₯ (tan 2π₯)2 π¦β² = β6. sin 6π₯. tan2π₯ β 2. cos6π₯. π ππ2 2π₯ π‘ππ2 2π₯
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