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Tugas matematika2
- 1. Nama Mahasiswa :Muhammad PachroniSuryana Kelas : 1 Elektronika A TUGAS MATEMATIKA 2 ( Turunan menggunakan Dalil Rantai ) Tentukan ππ¦ ππ₯ dari: 1. y = β π₯5 + 6 π₯2 + 3 6. y = 1 β π₯2β5π₯+2 5 2. y = β π₯4 + 6π₯ + 13 7. y = sin β π₯2 + 6π₯ 3. y= β π₯2 β 5π₯ 5 8. y = cos β π₯3 + 23 4. y= 1 β π₯4+2π₯ 9. y = sin 1 β π₯2+2 5. y= 1 β π₯2β6π₯ 3 10. y = cos 1 β π₯2+6 3 Jawab : 1. y = β π₯5 + 6 π₯2 + 3 Misal u = x5 +6x2 +3 maka ππ’ ππ₯ = 5x4 +12x y = β π’ = u1/2 maka ππ¦ ππ’ = 1 2 u-1/2 Jadi, ππ¦ ππ₯ = ππ’ ππ₯ ππ¦ ππ’ = (5x4 +12x)( 1 2 u-1/2 ) = (5x4 +12x)( 1 2 (x5 +6x2 +3)1/2 ) = (5 π₯4 +12x) 2β5+6 π₯2+3 2. y = β π₯4 + 6x + 13 Misal u = x4 + 6x+ 1 maka ππ’ ππ₯ = 4x3 + 6 y = β π’3 = u1/3 maka ππ¦ ππ’ = 1 3 u-2/3 Jadi, ππ¦ ππ₯ = ππ’ ππ₯ ππ¦ ππ’ = (4 π₯3 + 6)( 1 3 u-2/3 ) = (4 π₯3 + 6)( 1 3 ( π₯4 + 6x+ 1)-2/3 )
- 2. = (4 π₯3 + 6) 3β( π₯4 + 6x+ 1)3 2 3. y= β π₯2 β 5π₯ 5 Misal u = x2 - 5x maka ππ’ ππ₯ = 2x-5 y = β π’5 = u1/5 maka ππ¦ ππ’ = 1 5 u-4/5 Jadi, ππ¦ ππ₯ = ππ’ ππ₯ ππ¦ ππ’ = (2x-5)( 1 5 u-4/5 ) = (2x-5)( 1 5 (x2 - 5x)-4/5 ) = (2xβ5) 5 β( π₯2 β 5x)5 4 4. y= 1 β π₯4+2π₯ Misal u = x4 + 2x maka ππ’ ππ₯ = 4x3 +2 y= 1 βπ’ = 1 π’ -1/2 = u-1/2 maka ππ¦ ππ’ =- 1 2 u-3/2 Jadi, ππ¦ ππ₯ = ππ’ ππ₯ ππ¦ ππ’ = (4x3 +2)( - 1 2 u-3/2 ) = (4x3 +2)(- 1 2 (x4 + 2x)-3/2 ) = - (4 π₯3 +2) 2β( π₯4 + 2x) 3 5. y= 1 β π₯2β6π₯ 3 Misal u = x2 -6x maka ππ’ ππ₯ =2x-6 y= 1 βπ’ 3 = 1 π’ 1/3 = u-1/3 maka ππ¦ ππ’ =- 1 3 u-4/3 Jadi, ππ¦ ππ₯ = ππ’ ππ₯ ππ¦ ππ’ = (2x-6)( - 1 3 u-4/3 ) = (2x-6)( - 1 3 (x2 -6x)-4/3 )
- 3. = - (2xβ6) 3 β(π₯2 β6x)3 4 6. y = 1 β π₯2β5π₯+2 5 Misal u = x2 -5x+2 maka ππ’ ππ₯ =2x-5 y= 1 βπ’ 5 = 1 π’ 1/5 = u-1/5 maka ππ¦ ππ’ =- 1 5 u-6/5 Jadi, ππ¦ ππ₯ = ππ’ ππ₯ ππ¦ ππ’ = (2x-5)( - 1 5 u-6/5 ) = (2x-5)( - 1 5 (x2 -5x+2)-6/5 ) = - (2xβ5) 5 β( π₯2β5x+2)5 6 7. y = sin β π₯2 + 6π₯ Misal u = x2 +6x maka ππ’ ππ₯ = 2x+6 v = β π’ = u1/2 maka ππ£ ππ’ = 1 2 u-1/2 = 1 2 (x2 +6x)-1/2 y = sin v maka ππ¦ ππ£ = cos v = cos β π’ = cos β π₯2 + 6x Jadi, ππ¦ ππ₯ = ππ’ ππ₯ ππ£ ππ’ ππ¦ ππ£ = (2x+6)( 1 2 (x2 +6x)-1/2 )( cos β π₯2 + 6x) = (2x+6)(cosβ π₯2+6x) 2β π₯2+6x 8. y = cos β π₯3 + 23 Misal u = x3 +2 maka ππ’ ππ₯ = 3x2 v= β π’3 = u1/3 maka ππ£ ππ’ = 1 3 u-2/3 = 1 3 (x3 +2)-2/3 y = cos v maka ππ¦ ππ£ = - sin v = - sin β π’3 = - sin β π₯3 + 23 Jadi, ππ¦ ππ₯ = ππ’ ππ₯ ππ£ ππ’ ππ¦ ππ£ = (3x2 )( 1 3 (x3 +2)-2/3 )( - sin β π₯3 + 23 )
- 4. =- (3 π₯2 )( sinβ π₯3+2 3 ) 3 β( π₯3+2)3 2 9. y = sin 1 β π₯2+2 Misal u = x2 +2 maka ππ’ ππ₯ = 2x v= 1 βπ’ = 1 π’ 1/2 = u-1/2 maka ππ£ ππ’ =- 1 2 u-3/2 = - 1 2 (x2 +2)-3/2 y = sin v maka ππ¦ ππ£ = cos v = cos 1 βπ’ = cos 1 β π₯2+2 Jadi, ππ¦ ππ₯ = ππ’ ππ₯ ππ£ ππ’ ππ¦ ππ£ = (2x) (- 1 2 (x2 +2)-3/2 ) (cos 1 β π₯2+2 ) =- (2x)( cos 1 β π₯3+2 ) 2β( π₯2+2) 3 10. y = cos 1 β π₯2+6 3 Misal u = x2 +6 maka ππ’ ππ₯ = 2x v= 1 βπ’ 3 = 1 π’ 1/3 = u-1/3 maka ππ¦ ππ’ =- 1 3 u-4/3 = - 1 3 (x2 +6)-4/3 y = cos v maka ππ¦ ππ£ = - sin v = - sin 1 βπ’ 3 = - sin 1 β π₯2+6 3 Jadi, ππ¦ ππ₯ = ππ’ ππ₯ ππ£ ππ’ ππ¦ ππ£ = (2x)( - 1 3 (x2 +6)-4/3 )( - sin 1 β π₯2+6 3 ) =- (2x)( sin 1 β π₯2+6 3 ) 3 β( π₯2+6)3 4