Nama : M. Habiburrakhman

Kelas : 1 EA

Tugas Matematika 2

Tentukanlah nilai dydx

dari fungsi berikut ini !

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

[Type text]

Jawaban :

1. y=√x5+6 x2+3

Misal u¿ 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+12 x )

(x5+6 x2+3)12

=

12(5x 4+12x )

√x5+6 x2+3

2. y= 3√x4+6 x+1

Misal u ¿ 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

Maka 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

Misal u ¿ x2−5 x , maka dudx

=2 x−5

y= 5√u=u15 , maka dy

du=1

5u

−45 =1

5(x2−5x )

−45

Maka dydx

=dydu

.dudx

=15(x2−5 x )

−45 .(2 x−5)

[Type text]

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

Misal u ¿ x4+2 x , maka dudx

=4 x3+2

y=u−12 , maka dy

du=−1

2u

−32 =−1

2(x4+2 x)

−32

Maka dydx

=dydu

.dudx

=−12

(x4+2 x )−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

Misal u ¿ x2−6 x , maka dudx

=2 x−6

y=u−13 , maka dy

du=−1

3u

−43 =−1

3(x2−6 x)

−43

Maka dydx

=dydu

.dudx

=−13

(x2−6 x )−4

3 .(2 x−6)

dydx

=

−13

.(2x−6)

(x2−6 x )−43

=

−13

(2 x−6)

3√(x2−6 x )4

[Type text]

6. y= 15√ x2−5 x+2

= 1

(x2−5 x+2)15

=(x2−5 x+2)−15

Misal u ¿ 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

Maka dydx

=dydu

.dudx

=−15

(x2−5 x+2)−65 .(2 x−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

Misal u ¿ x2+6 x , maka dudx

=2 x+6

v=√u=u12 , makadv

du=1

2u

−12 =1

2(x2+6 x )

−12

y=sin v , maka dydv

=cos v=cos √u=cos√ x2+6 x

Maka 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=cos3√ x3+2

Misal u ¿ x3+2 , maka dudx

=3 x2

[Type text]

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

Maka dydx

=dydv

.dvdu

.dudx

=−sin3√x3+2 .

13(x3+2)

−23 . 3 x2

dydx

=

13

.3 x2 .−sin3√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

Misal u ¿ x2+2 , maka dudx

=2 x

v=u−1

2 , maka dvdu

=−12

u−32 =−1

2(x2+2)

−32

y=sin v , maka dydv

=cos v=cosu−12 =cos (x2+2)

−12

Maka dydx

=dydv

.dvdu

.dudx

=cos( x2+2)−12 .−1

2(x2+2)

−32 . 2 x

dydx

=

−12

. 2 x . cos (x2+2)−1

2

(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

Misal u ¿ x2+6 , maka dudx

=2 x

[Type text]

v=u−1

3 , maka dvdu

=−13

u−43 =−1

3(x2+6)

−43

y=cosv , maka dydv

=−sin v=−sin u−1

3 =−sin(x2+6)−13

Maka 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

[Type text]