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1.
TUGAS 2 MATEMATIKA 2 Nama
: Cinjy Saylendra Monica Prodi : Teknik Elektronika Kelas : 1E A Semester : 2 (Dua) Tahun Ajaran : 2014/2015 Soal : 1. ๐ฆ = โ ๐ฅ5 + 6๐ฅ2 + 3 2. ๐ฆ = โ ๐ฅ4 + 6๐ฅ + 1 3 3. ๐ฆ = โ ๐ฅ2 โ 5๐ฅ 5 4. ๐ฆ = 1 โ๐ฅ4+2๐ฅ 5. ๐ฆ = 1 โ๐ฅ2โ6๐ฅ 3 6. ๐ฆ = 1 โ๐ฅ2โ5๐ฅ+2 5 7. ๐ฆ = sin โ ๐ฅ2 + 6๐ฅ 8. ๐ฆ = cos โ ๐ฅ3 + 2 3 9. ๐ฆ = sin 1 โ๐ฅ2+2 10. ๐ฆ = cos 1 โ๐ฅ2+6 3 Jawaban : 1. ๐ฆ = โ ๐ฅ5 + 6๐ฅ2 + 3 u = ๐ฅ5 + 6๐ฅ2 + 3 , maka ๐๐ข ๐๐ฅ = 5๐ฅ4 + 12๐ฅ ๐ฆ = โ ๐ข = ๐ข 1 2 , maka ๐๐ฆ ๐๐ข = 1 2 ๐ขโ 1 2 = 1 2 ( ๐ฅ5 + 6๐ฅ2 + 3)โ 1 2 ๐๐ฆ ๐๐ฅ = ๐๐ฆ ๐๐ข . ๐๐ข ๐๐ฅ = 1 2 (๐ฅ5 + 6๐ฅ2 + 3)โ 1 2 . (5๐ฅ4 + 12๐ฅ)
2.
๐๐ฆ ๐๐ฅ = 1 2 (5๐ฅ4 + 12๐ฅ) (๐ฅ5 +
6๐ฅ2 + 3) 1 2 = 1 2 (5๐ฅ4 + 12๐ฅ) โ ๐ฅ5 + 6๐ฅ2 + 3 2. ๐ฆ = โ ๐ฅ4 + 6๐ฅ + 1 3 u = ๐ฅ4 + 6๐ฅ + 1 , maka ๐๐ข ๐๐ฅ = 4๐ฅ3 + 6 ๐ฆ = โ ๐ข3 = ๐ข 1 3 , maka ๐๐ฆ ๐๐ข = 1 3 ๐ขโ 2 3 = 1 3 (๐ฅ4 + 6๐ฅ + 1)โ 2 3 ๐๐ฆ ๐๐ฅ = ๐๐ฆ ๐๐ข . ๐๐ข ๐๐ฅ = 1 3 (๐ฅ4 + 6๐ฅ + 1)โ 2 3 . (4๐ฅ3 + 6) ๐๐ฆ ๐๐ฅ = 1 3 (4๐ฅ3 + 6) (๐ฅ4 + 6๐ฅ + 1) 2 3 = 1 3 (4๐ฅ3 + 6) โ(๐ฅ4 + 6๐ฅ + 1)23 3. ๐ฆ = โ ๐ฅ2 โ 5๐ฅ 5 u = ๐ฅ2 โ 5๐ฅ , maka ๐๐ข ๐๐ฅ = 2๐ฅ โ 5 ๐ฆ = โ ๐ข5 = ๐ข 1 5 , maka ๐๐ฆ ๐๐ข = 1 5 ๐ขโ 4 5 = 1 5 (๐ฅ2 โ 5๐ฅ)โ 4 5 ๐๐ฆ ๐๐ฅ = ๐๐ฆ ๐๐ข . ๐๐ข ๐๐ฅ = 1 5 (๐ฅ2 โ 5๐ฅ)โ 4 5 . (2๐ฅ โ 5) ๐๐ฆ ๐๐ฅ = 1 5 (2๐ฅ โ 5) (๐ฅ2 โ 5๐ฅ) 4 5 = 1 5 (2๐ฅ โ 5) โ(๐ฅ2 โ 5๐ฅ)45 4. ๐ฆ = 1 โ๐ฅ4+2๐ฅ = 1 (๐ฅ4+2๐ฅ) 1 2 = (๐ฅ4 + 2๐ฅ)โ 1 2 u = ๐ฅ4 + 2๐ฅ , maka ๐๐ข ๐๐ฅ = 4๐ฅ3 + 2 ๐ฆ = ๐ขโ 1 2 , maka ๐๐ฆ ๐๐ข = โ 1 2 ๐ขโ 3 2 = โ 1 2 ( ๐ฅ4 + 2๐ฅ)โ 3 2 ๐๐ฆ ๐๐ฅ = ๐๐ฆ ๐๐ข . ๐๐ข ๐๐ฅ = โ 1 2 (๐ฅ4 + 2๐ฅ)โ 3 2 . (4๐ฅ3 + 2)
3.
๐๐ฆ ๐๐ฅ = โ 1 2 (4๐ฅ3 + 2) (๐ฅ4 +
2๐ฅ) 3 2 = โ2๐ฅ3 โ 1 โ(๐ฅ4 + 2๐ฅ)3 5. ๐ฆ = 1 โ๐ฅ2โ6๐ฅ 3 = 1 (๐ฅ2โ6๐ฅ) 1 3 = (๐ฅ2 โ 6๐ฅ)โ 1 3 u = ๐ฅ2 โ 6๐ฅ , maka ๐๐ข ๐๐ฅ = 2๐ฅ โ 6 ๐ฆ = ๐ขโ 1 3 , maka ๐๐ฆ ๐๐ข = โ 1 3 ๐ขโ 4 3 = โ 1 3 (๐ฅ2 โ 6๐ฅ)โ 4 3 ๐๐ฆ ๐๐ฅ = ๐๐ฆ ๐๐ข . ๐๐ข ๐๐ฅ = โ 1 3 (๐ฅ2 โ 6๐ฅ)โ 4 3 . (2๐ฅ โ 6) ๐๐ฆ ๐๐ฅ = โ 1 3 . (2๐ฅ โ 6) (๐ฅ2 โ 6๐ฅ)โ 4 3 = โ 1 3 (2๐ฅ โ 6) โ(๐ฅ2 โ 6๐ฅ)43 6. ๐ฆ = 1 โ๐ฅ2โ5๐ฅ+2 5 = 1 (๐ฅ2โ5๐ฅ+2) 1 5 = (๐ฅ2 โ 5๐ฅ + 2)โ 1 5 u = ๐ฅ2 โ 5๐ฅ + 2 , maka ๐๐ข ๐๐ฅ = 2๐ฅ โ 5 ๐ฆ = ๐ขโ 1 5 , maka ๐๐ฆ ๐๐ข = โ 1 5 ๐ขโ 6 5 = โ 1 5 (๐ฅ2 โ 5๐ฅ + 2)โ 6 5 ๐๐ฆ ๐๐ฅ = ๐๐ฆ ๐๐ข . ๐๐ข ๐๐ฅ = โ 1 5 (๐ฅ2 โ 5๐ฅ + 2)โ 6 5 .(2๐ฅ โ 5) ๐๐ฆ ๐๐ฅ = โ 1 5 . (2๐ฅ โ 5) (๐ฅ2 โ 5๐ฅ + 2) 6 5 = โ 1 5 (2๐ฅ โ 5) โ(๐ฅ2 โ 5๐ฅ + 2)65 7. ๐ฆ = sin โ ๐ฅ2 + 6๐ฅ u = ๐ฅ2 + 6๐ฅ , maka ๐๐ข ๐๐ฅ = 2๐ฅ + 6
4.
๐ฃ = โ
๐ข = ๐ข 1 2 , maka ๐๐ฃ ๐๐ข = 1 2 ๐ขโ 1 2 = 1 2 (๐ฅ2 + 6๐ฅ )โ 1 2 ๐ฆ = sin ๐ฃ , maka ๐๐ฆ ๐๐ฃ = cos ๐ฃ = cos โ ๐ข = cos โ ๐ฅ2 + 6๐ฅ ๐๐ฆ ๐๐ฅ = ๐๐ฆ ๐๐ฃ . ๐๐ฃ ๐๐ข . ๐๐ข ๐๐ฅ = cos โ ๐ฅ2 + 6๐ฅ . 1 2 (๐ฅ2 + 6๐ฅ )โ 1 2 .(2๐ฅ + 6) ๐๐ฆ ๐๐ฅ = 1 2 . (2๐ฅ + 6) .cos โ ๐ฅ2 + 6๐ฅ (๐ฅ2 + 6๐ฅ ) 1 2 = ( ๐ฅ + 3) .cos โ ๐ฅ2 + 6๐ฅ โ ๐ฅ2 + 6๐ฅ 8. ๐ฆ = cos โ ๐ฅ3 + 2 3 u = ๐ฅ3 + 2 , maka ๐๐ข ๐๐ฅ = 3๐ฅ2 ๐ฃ = โ ๐ข3 = ๐ข 1 3 , maka ๐๐ฃ ๐๐ข = 1 3 ๐ขโ 2 3 = 1 3 (๐ฅ3 + 2)โ 2 3 ๐ฆ = cos ๐ฃ , maka ๐๐ฆ ๐๐ฃ = โsin ๐ฃ = โsin โ ๐ข3 = โsin โ ๐ฅ3 + 2 3 ๐๐ฆ ๐๐ฅ = ๐๐ฆ ๐๐ฃ . ๐๐ฃ ๐๐ข . ๐๐ข ๐๐ฅ = โsin โ ๐ฅ3 + 2 3 . 1 3 (๐ฅ3 + 2)โ 2 3 .3๐ฅ2 ๐๐ฆ ๐๐ฅ = 1 3 . 3๐ฅ2 . โsin โ ๐ฅ3 + 2 3 (๐ฅ3 + 2) 2 3 = ๐ฅ2 . โsin โ ๐ฅ3 + 2 3 โ(๐ฅ3 + 2)23 9. ๐ฆ = sin 1 โ๐ฅ2+2 = sin 1 (๐ฅ2+2) 1 2 = sin(๐ฅ2 + 2)โ 1 2 u = ๐ฅ2 + 2 , maka ๐๐ข ๐๐ฅ = 2๐ฅ ๐ฃ = ๐ขโ 1 2 , maka ๐๐ฃ ๐๐ข = โ 1 2 ๐ขโ 3 2 = โ 1 2 (๐ฅ2 + 2)โ 3 2 ๐ฆ = sin ๐ฃ , maka ๐๐ฆ ๐๐ฃ = cos ๐ฃ = cos ๐ขโ 1 2 = cos(๐ฅ2 + 2)โ 1 2 ๐๐ฆ ๐๐ฅ = ๐๐ฆ ๐๐ฃ . ๐๐ฃ ๐๐ข . ๐๐ข ๐๐ฅ = cos(๐ฅ2 + 2)โ 1 2 . โ 1 2 (๐ฅ2 + 2)โ 3 2 .2๐ฅ
5.
๐๐ฆ ๐๐ฅ = โ 1 2 . 2๐ฅ .
cos(๐ฅ2 + 2)โ 1 2 (๐ฅ2 + 2) 3 2 = โ๐ฅ . cos(๐ฅ2 + 2)โ 1 2 โ(๐ฅ2 + 2)3 10. ๐ฆ = cos 1 โ๐ฅ2+6 3 = cos 1 (๐ฅ2+6) 1 3 = cos(๐ฅ2 + 6)โ 1 3 u = ๐ฅ2 + 6 , maka ๐๐ข ๐๐ฅ = 2๐ฅ ๐ฃ = ๐ขโ 1 3 , maka ๐๐ฃ ๐๐ข = โ 1 3 ๐ขโ 4 3 = โ 1 3 (๐ฅ2 + 6)โ 4 3 ๐ฆ = cos ๐ฃ , maka ๐๐ฆ ๐๐ฃ = โsin ๐ฃ = โsin ๐ขโ 1 3 = โsin(๐ฅ2 + 6)โ 1 3 ๐๐ฆ ๐๐ฅ = ๐๐ฆ ๐๐ฃ . ๐๐ฃ ๐๐ข . ๐๐ข ๐๐ฅ = โsin(๐ฅ2 + 6)โ 1 3 . โ 1 3 (๐ฅ2 + 6)โ 4 3 .2๐ฅ ๐๐ฆ ๐๐ฅ = โ 1 3 . 2๐ฅ . โsin(๐ฅ2 + 6)โ 1 3 (๐ฅ2 + 6) 4 3 = 1 3 . 2๐ฅ . sin(๐ฅ2 + 6)โ 1 3 โ(๐ฅ2 + 6)43
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