tugas 2(1)
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POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
TUGAS 2MATEMATIKA 2
DISUSUN
Oleh :
Nama : Aulia RamadhaniNPM : 003 14 03Prodi : TeknikElektronikaKelas : 1E ASemester : 2 (Dua)
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNGKawasanIndustri Air KantungSungailiat, Bangka 33211
Telp. (0717) 93586, Fax. (0717) 93585Email :polman@polman-babel.ac.idWebsite :www.polman-babel.ac.id
TAHUN AJARAN 2014/2015
Tugas 2 MTK2 Page 1
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Tentukanlahnilaidydx
darifungsiberikutini !
1. y=√x5+6 x2+3
2. y= 3√x4+6 x+1
3. y= 5√x2−5 x
4. y= 1
√ x4+2 x
5. y= 13√ x2−6 x
6. y= 15√ x2−5 x+2
7. y=sin√x2+6 x
8. y=cos3√ x3+2
9. y=sin1
√ x2+2
10. y=cos1
3√x2+6
Jawaban :
1. y=√x5+6 x2+3
Misalu¿ x5+6 x2+3 , maka dudx
=5 x4+12 x
y=√u=u12 , maka dy
du=1
2u
−12 =1
2(x5+6 x2+3)
−12
Makadydx
=dydu.dudx
=12(x5+6 x2+3)
−12 .(5 x4+12 x)
dydx
=
12(5 x4+12x )
(x5+6 x2+3)12
=
12(5x 4+12x )
√x5+6 x2+3
2. y= 3√x4+6 x+1
Misalu¿ x4+6 x+1 , maka dudx
=4 x3+6
y= 3√u=u13 , maka dy
du=1
3u
−23 =1
3(x4+6 x+1)
−23
Makadydx
=dydu.dudx
=13(x 4+6 x+1)
−23 .(4 x3+6)
dydx
=
13(4 x3+6)
(x4+6 x+1)23
=
13(4 x3+6)
3√(x¿¿4+6 x+1)2 ¿
3. y= 5√x2−5 x
Tugas 2 MTK2 Page 2
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Misalu¿ x2−5 x , maka dudx
=2 x−5
y= 5√u=u15 , maka dy
du=1
5u
−45 =1
5(x2−5x )
−45
Makadydx
=dydu.dudx
=15(x2−5 x )
−45 .(2 x−5)
dydx
=
15(2 x−5)
(x2−5 x)45
=
15(2 x−5)
5√(x2−5 x )4
4. y= 1
√ x4+2 x= 1
(x 4+2x )12
=(x4+2 x)−12
Misalu¿ x4+2 x , maka dudx
=4 x3+2
y=u−12 , maka dy
du=−1
2u
−32 =−1
2(x4+2 x)
−32
Makadydx
=dydu.dudx
=−12
(x4+2x )−32 .(4 x3+2)
dydx
=
−12
(4 x3+2)
(x4+2 x)32
=−2x3−1
√(x4+2x )3
5. y= 13√ x2−6 x
= 1
(x2−6 x)13
=(x2−6 x )−13
Misalu¿ x2−6 x , maka dudx
=2 x−6
y=u−13 , maka dy
du=−1
3u
−43 =−1
3(x2−6 x)
−43
Makadydx
=dydu.dudx
=−13
(x2−6 x )−4
3 .(2 x−6)
Tugas 2 MTK2 Page 3
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
dydx
=
−13.(2x−6)
(x2−6 x )−43
=
−13
(2 x−6)
3√(x2−6 x )4
6. y= 15√ x2−5 x+2
= 1
(x2−5 x+2)15
=(x2−5 x+2)−15
Misalu¿ x2−5 x+2 , maka dudx
=2 x−5
y=u−15 , maka dy
du=−1
5u
−65 =−1
5(x2−5 x+2)
−65
Makadydx
=dydu.dudx
=−15
(x2−5 x+2)−65 .(2x−5)
dydx
=
−15.(2 x−5)
(x2−5 x+2)65
=
−15
(2 x−5)
5√(x2−5 x+2)6
7. y=sin√x2+6 x
Misalu¿ x2+6 x , maka dudx
=2 x+6
v=√u=u12 , makadv
du=1
2u
−12 =1
2(x2+6x )
−12
y=sin v , maka dydv
=cos v=cos √u=cos√ x2+6 x
Makadydx
=dydv.dvdu.dudx
=cos√ x2+6 x .12(x2+6 x)
−12 .(2x+6)
dydx
=
12. (2 x+6 ) .cos √x2+6 x
(x2+6 x )12
=( x+3 ) .cos√ x2+6 x
√x2+6 x
8. y=cos3√ x3+2
Misalu¿ x3+2 , maka dudx
=3 x2
Tugas 2 MTK2 Page 4
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
v=3√u=u13 , maka dv
du=1
3u
−23 =1
3(x3+2)
−23
y=cosv , maka dydv
=−sin v=−sin 3√u=−sin3√ x3+2
Makadydx
=dydv.dvdu.dudx
=−sin3√x3+2 .
13(x3+2)
−23 .3 x2
dydx
=
13.3 x2 .−sin
3√x3+2
(x3+2)23
=x2 .−sin
3√x3+23√(x3+2)2
9. y=sin1
√ x2+2=sin
1
(x2+2)12
=sin(x2+2)−12
Misalu¿ x2+2 , maka dudx
=2 x
v=u−1
2 , maka dvdu
=−12u
−32 =−1
2(x2+2)
−32
y=sin v , maka dydv
=cos v=cosu−12 =cos (x2+2)
−12
Makadydx
=dydv.dvdu.dudx
=cos( x2+2)−12 .−1
2(x2+2)
−32 .2 x
dydx
=
−12.2 x .cos (x2+2)
−12
(x2+2)32
=−x .cos( x2+2)
−12
√(x2+2)3
10. y=cos1
3√x2+6=cos
1
(x2+6)13
=cos (x2+6)−13
Misalu¿ x2+6 , maka dudx
=2 x
v=u−1
3 , maka dvdu
=−13u
−43 =−1
3(x2+6)
−43
y=cosv , maka dydv
=−sin v=−sinu−1
3 =−sin(x2+6)−13
Tugas 2 MTK2 Page 5
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Makadydx
=dydv.dvdu.dudx
=−sin(x2+6)−13 .−1
3(x2+6)
−43 .2 x
dydx
=
−13.2 x .−sin(x2+6)
−13
(x2+6)43
=
13.2 x . sin(x2+6)
−13
3√(x2+6)4
Tugas 2 MTK2 Page 6
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