tugas matematika 1 - semester 2 - polman babel
TRANSCRIPT
![Page 1: Tugas Matematika 1 - Semester 2 - Polman Babel](https://reader036.vdokumen.com/reader036/viewer/2022082515/55c8e929bb61ebdd4c8b4815/html5/thumbnails/1.jpg)
Nama :Miza Pisari
NPM : 0031417
Kelas : 1 Elektronika A
Tugas : MATEMATIKA
Carilah nilai turunan dari fungsi berikut ini.
1. y=x7+6 x5+12x3
2. y=xβ5+2 xβ3+3 xβ7
3. y=βx2+ 3β x7
4. y= 2
β x+ 6
βx3
5. y=x2 . sin2 x
6. y=( x2+1 ) .cos3 x7. y=sin5 x . tan 2x
8. y=cos3 x .cot 4 x
9. y= sin 5 xcos7 x
10. y= cos6 xtan 2 x
Jawab:
1. y '=7 x6+30 x4+36 x2
2. y '=β5 xβ6β6 xβ4β21 xβ8
3. y=x22+x
73=x+x
73
y '=1+ 73x73β33=1+ 7
3x43
y '=1+ 733βx4
4. y= 2
x12
+ 6
x32
=2 xβ12 +6 x
β32
y '=2.β12. x
β12
β22+6.β3
2. x
β32
β22=βx
β32 β9x
β52 =β1
x32
β 9
x52
y '= β1
β x3β 9
β x55. u=x2βu '=2x
v=sin 2xβv '=2. cos xy '=u' . v+u . v '=2 x . sin 2x+x2.2 .cos xy '=2x .sin 2 x+2x2 .cos x
6. u=x2+1βu '=2xv=cos3 xβ v '=β3.sin 3 xy '=u' . v+u . v '=2 x .cos3 x+(x2+1 ).β3. sin3 xy '=2x .cos 3x+(β3x2β3 ). sin 3 x
7. u=sin 5xβu '=5.cos 5xv=tan 2 xβv '=2. sec22 xy '=u' . v+u . v '=5. cos5 x . tan 2 x+sin 5x .2 . sec22 xy '=5. cos5 x . tan2 x+2. sin5 x . sec22x
8. u=cos3 xβu'=β3. sin 3 xv=cot 4 xβv '=β4.csc24 x
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y '=u' . v+u . v '=β3. sin 3x .cot 4 x+cos 3x .β4.csc24 xy '=β3. sin 3 x .cot 4 xβ4. cos3 x .csc 24 x
9. u=sin 5xβu '=5.cos 5xv=cos7 xβv '=β7. sin7 x
y '=u' . vβu . v '
v2=5. cos5 x .cos7 xβsin5 x .β7. sin 7 xΒΏΒΏΒΏ
y '=5.cos 5x .cos7 x+7.sin 5 x .sin 7 xcos27x
10. u=cos6 xβu'=β6. sin 6 xv=tan 2 xβv '=2. sec22 x
y '=u' . vβu . v '
v2=β6. sin 6 x . tan 2xβcos 6 x .2. sec22x
ΒΏΒΏΒΏ
y '=β6. sin 6 x . tan 2 xβ2. cos6 x . sec22 xtan22x