Download - math 2nd task
Nama : M. Habiburrakhman
Kelas : 1 EA
Tugas Matematika 2
Tentukanlah nilai dydx
dari fungsi berikut ini !
1. y=βx5+6 x2+3
2. y= 3βx4+6 x+1
3. y= 5βx2β5 x
4. y= 1
β x4+2 x
5. y= 13β x2β6 x
6. y= 15β x2β5 x+2
7. y=sinβx2+6 x
8. y=cos3β x3+2
9. y=sin1
β x2+2
10. y=cos1
3βx2+6
[Type text]
Jawaban :
1. y=βx5+6 x2+3
Misal uΒΏ x5+6 x2+3 , maka dudx
=5 x4+12 x
y=βu=u12 , maka dy
du=1
2u
β12 =1
2(x5+6 x2+3)
β12
Makadydx
=dydu
.dudx
=12(x5+6 x2+3)
β12 .(5 x4+12 x)
dydx
=
12(5 x4+12 x )
(x5+6 x2+3)12
=
12(5x 4+12x )
βx5+6 x2+3
2. y= 3βx4+6 x+1
Misal u ΒΏ x4+6 x+1 , maka dudx
=4 x3+6
y= 3βu=u13 , maka dy
du=1
3u
β23 =1
3(x4+6 x+1)
β23
Maka dydx
=dydu
.dudx
=13(x 4+6 x+1)
β23 .(4 x3+6)
dydx
=
13(4 x3+6)
(x4+6 x+1)23
=
13(4 x3+6)
3β(xΒΏΒΏ4+6 x+1)2 ΒΏ
3. y= 5βx2β5 x
Misal u ΒΏ x2β5 x , maka dudx
=2 xβ5
y= 5βu=u15 , maka dy
du=1
5u
β45 =1
5(x2β5x )
β45
Maka dydx
=dydu
.dudx
=15(x2β5 x )
β45 .(2 xβ5)
[Type text]
dydx
=
15(2 xβ5)
(x2β5 x)45
=
15(2 xβ5)
5β(x2β5 x )4
4. y= 1
β x4+2 x= 1
(x 4+2x )12
=(x4+2 x)β12
Misal u ΒΏ x4+2 x , maka dudx
=4 x3+2
y=uβ12 , maka dy
du=β1
2u
β32 =β1
2(x4+2 x)
β32
Maka dydx
=dydu
.dudx
=β12
(x4+2 x )β32 .(4 x3+2)
dydx
=
β12
(4 x3+2)
(x4+2 x)32
=β2x3β1
β(x4+2x )3
5. y= 13β x2β6 x
= 1
(x2β6 x)13
=(x2β6 x )β13
Misal u ΒΏ x2β6 x , maka dudx
=2 xβ6
y=uβ13 , maka dy
du=β1
3u
β43 =β1
3(x2β6 x)
β43
Maka dydx
=dydu
.dudx
=β13
(x2β6 x )β4
3 .(2 xβ6)
dydx
=
β13
.(2xβ6)
(x2β6 x )β43
=
β13
(2 xβ6)
3β(x2β6 x )4
[Type text]
6. y= 15β x2β5 x+2
= 1
(x2β5 x+2)15
=(x2β5 x+2)β15
Misal u ΒΏ x2β5 x+2 , maka dudx
=2 xβ5
y=uβ15 , maka dy
du=β1
5u
β65 =β1
5(x2β5 x+2)
β65
Maka dydx
=dydu
.dudx
=β15
(x2β5 x+2)β65 .(2 xβ5)
dydx
=
β15
.(2 xβ5)
(x2β5 x+2)65
=
β15
(2 xβ5)
5β(x2β5 x+2)6
7. y=sinβx2+6 x
Misal u ΒΏ x2+6 x , maka dudx
=2 x+6
v=βu=u12 , makadv
du=1
2u
β12 =1
2(x2+6 x )
β12
y=sin v , maka dydv
=cos v=cos βu=cosβ x2+6 x
Maka dydx
=dydv
.dvdu
.dudx
=cosβ x2+6 x .12(x2+6 x)
β12 .(2x+6)
dydx
=
12
. (2 x+6 ) .cos βx2+6 x
(x2+6 x )12
=( x+3 ) .cosβ x2+6 x
βx2+6 x
8. y=cos3β x3+2
Misal u ΒΏ x3+2 , maka dudx
=3 x2
[Type text]
v=3βu=u13 , maka dv
du=1
3u
β23 =1
3(x3+2)
β23
y=cosv , maka dydv
=βsin v=βsin 3βu=βsin3β x3+2
Maka dydx
=dydv
.dvdu
.dudx
=βsin3βx3+2 .
13(x3+2)
β23 . 3 x2
dydx
=
13
.3 x2 .βsin3βx3+2
(x3+2)23
=x2 .βsin
3βx3+23β(x3+2)2
9. y=sin1
β x2+2=sin
1
(x2+2)12
=sin(x2+2)β12
Misal u ΒΏ x2+2 , maka dudx
=2 x
v=uβ1
2 , maka dvdu
=β12
uβ32 =β1
2(x2+2)
β32
y=sin v , maka dydv
=cos v=cosuβ12 =cos (x2+2)
β12
Maka dydx
=dydv
.dvdu
.dudx
=cos( x2+2)β12 .β1
2(x2+2)
β32 . 2 x
dydx
=
β12
. 2 x . cos (x2+2)β1
2
(x2+2)32
=βx .cos( x2+2)
β12
β(x2+2)3
10. y=cos1
3βx2+6=cos
1
(x2+6)13
=cos (x2+6)β13
Misal u ΒΏ x2+6 , maka dudx
=2 x
[Type text]
v=uβ1
3 , maka dvdu
=β13
uβ43 =β1
3(x2+6)
β43
y=cosv , maka dydv
=βsin v=βsin uβ1
3 =βsin(x2+6)β13
Maka dydx
=dydv
.dvdu
.dudx
=βsin(x2+6)β13 .β1
3(x2+6)
β43 . 2 x
dydx
=
β13
. 2 x .βsin(x2+6)β13
(x2+6)43
=
13
.2 x . sin(x2+6)β13
3β(x2+6)4
[Type text]