TUGAS 2MATEMATIKA 2

Nama : Cinjy Saylendra MonicaProdi : Teknik ElektronikaKelas : 1E ASemester : 2 (Dua)Tahun Ajaran : 2014/2015

Soal :

1. y=√x5+6 x2+32. y= 3√x4+6 x+13. 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 x8. y=cos 3√ x3+2

9. y=sin 1

√ x2+2

10. y=cos 13√x2+6

Jawaban :

1. y=√x5+6 x2+3

u ¿ x5+6 x2+3 , maka dudx

=5 x4+12 x

y=√u=u12 , maka dy

du=12u

−12 =12

(x5+6 x2+3 )−12

dydx

=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

u ¿ x4+6 x+1 , maka dudx

=4 x3+6

y= 3√u=u13 , maka dy

du=13u

−23 =13(x4+6 x+1)

−23

dydx

=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

u ¿ x2−5 x , maka dudx

=2 x−5

y= 5√u=u15 , maka dy

du=15u

−45 =1

5(x2−5x )

−45

dydx

=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

u ¿ x4+2 x , maka dudx

=4 x3+2

y=u−12 , maka dy

du=−12u

−32 =−1

2( x4+2x )

−32

dydx

=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

u ¿ x2−6 x , maka dudx

=2 x−6

y=u−13 , maka dy

du=−13u

−43 =−1

3(x2−6 x)

−43

dydx

=dydu.dudx

=−13

(x2−6 x )−43 .(2 x−6)

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

u ¿ x2−5 x+2 , maka dudx

=2 x−5

y=u−15 , maka dy

du=−15u

−65 =−1

5(x2−5 x+2)

−65

dydx

=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

u ¿ x2+6 x , maka dudx

=2 x+6

v=√u=u12, maka dv

du=12u

−12 =12(x2+6x )

−12

y=sin v , maka dydv

=cos v=cos √u=cos√ x2+6 x

dydx

=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=cos 3√ x3+2

u ¿ x3+2 , maka dudx

=3 x2

v=3√u=u13 , maka dv

du=13u

−23 =13(x3+2)

−23

y=cosv , maka dydv

=−sin v=−sin 3√u=−sin 3√ x3+2

dydx

=dydv.dvdu.dudx

=−sin3√x3+2 . 1

3(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=sin 1

√ x2+2=sin 1

(x2+2)12

=sin(x2+2)−12

u ¿ x2+2 , maka dudx

=2 x

v=u−12 , maka dv

du=−12u

−32 =−1

2(x2+2)

−32

y=sin v , maka dydv

=cos v=cosu−12 =cos (x2+2)

−12

dydx

=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=cos 13√x2+6

=cos 1

(x2+6)13

=cos (x2+6)−13

u ¿ x2+6 , maka dudx

=2 x

v=u−13 , maka dv

du=−13u

−43 =−1

3(x2+6)

−43

y=cosv , maka dydv

=−sin v=−sinu−13 =−sin(x2+6)

−13

dydx

=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