Tugas 2 MTK2

1. Nama : Siti Fatimah
NPM : 0031427
Kelas : 1 EA
Mata Kuliah : Matematika2
TUGAS 2
Tentukandx/dydari:
1 . y= √𝑥5 + 6𝑥2 + 3
2 . y=√𝑥4 + 6𝑥 + 1
3
3 . y=√𝑥2 − 5𝑥
5
4 . y=
1
√𝑥4+2𝑥
5 . y=
1
√𝑥2−6𝑥
3
6 . y=
1
√𝑥2−5𝑥+2
5
7 . y=𝑠𝑖𝑛√𝑥2 + 6𝑥
8 . y=𝑐𝑜𝑠√𝑥3 + 2
3
9 . y=𝑠𝑖𝑛
1
√𝑥2+2
10 . y=𝑐𝑜𝑠
1
√𝑥2+6
3
Penyelesaian:
1 . y= √𝑥5 + 6𝑥2 + 3
misal u= 𝑥5 + 6𝑥2 + 3 , du/dx = 5𝑥4 + 12𝑥
y= √ 𝑢 = 𝑢
1
2 , dy/du=
1
2
𝑢−
1
2 =
1
2
(𝑥5 + 6𝑥2 + 3)−
1
2
dy/dx = du/dx .dy/du
= (5𝑥4 + 12𝑥).
1
2
(𝑥5 + 6𝑥2 + 3)−
1
2
=
1
2
.(5𝑥4 + 12𝑥). (𝑥5 + 6𝑥2 + 3)−
1
2

2. 2 . y=√𝑥4 + 6𝑥 + 1
3
misal u= 𝑥4 + 6𝑥 + 1 , du/dx = 4𝑥3 + 6
y= √ 𝑢3
= 𝑢
1
3 , dy/du=
1
3
𝑢−
2
3 =
1
3
(𝑥4 + 6𝑥 + 1)−
2
3
dy/dx = du/dx .dy/du
= (4𝑥3 + 6).
1
3
(𝑥4 + 6𝑥 + 1)−
2
3
=
1
3
.( 4𝑥3 + 6).(𝑥4 + 6𝑥 + 1)−
2
3
3 . y=√𝑥2 − 5𝑥
5
misal u= 𝑥2 − 5𝑥 , du/dx = 2𝑥 − 5
y= √ 𝑢5
= 𝑢
1
5 , dy/du=
1
5
𝑢−
4
5 =
1
5
( 𝑥2 − 5𝑥)−
4
5
dy/dx = du/dx .dy/du
= (2𝑥 − 5).
1
5
( 𝑥2 − 5𝑥)−
4
5
=
1
5
.(2𝑥 − 5). ( 𝑥2 − 5𝑥)−
4
5
4 . y=
1
√𝑥4+2𝑥
misal u= 𝑥4 + 2𝑥 , du/dx = 4𝑥3 + 2
misal v= √ 𝑢= 𝑢
1
2 , dv/du=
1
2
𝑢−
1
2 =
1
2
( 𝑥4 + 2𝑥) −
1
2
y=
1
𝑣
= 𝑣−1 , dy/dv= −1𝑣−2 = −1(√𝑥4 + 2𝑥 )−2
dy/dx = du/dx .dv/du. dy/dv
= (4𝑥3 + 2).
1
2
( 𝑥4 + 2𝑥)−
1
2 . −1(√𝑥4 + 2𝑥 )−2
=−
1
2
.(4𝑥3 + 2).( 𝑥4 + 2𝑥)−
1
2.(√𝑥4 + 2𝑥 )−2

3. 5 . y=
1
√𝑥2−6𝑥
3
misal u= 𝑥2 − 6𝑥 , du/dx = 2𝑥 − 6
misal v= √ 𝑢3
= 𝑢
1
3 , dv/du=
1
3
𝑢−
2
3 =
1
3
(𝑥2 − 6𝑥 ) −
2
3
y=
1
𝑣
= 𝑣−1 , dy/dv= −1𝑣−2 = −1(√𝑥2 − 6𝑥
3
)−2
dy/dx = du/dx .dv/du. dy/dv
= (2𝑥 − 6).
1
3
(𝑥2 − 6𝑥 ) −
2
3 . −1(√𝑥2 − 6𝑥
3
)−2
=−
1
3
.(2𝑥 − 6).(𝑥2 − 6𝑥 )−
2
3.(√𝑥2 − 6𝑥
3
)−2
6 . y=
1
√𝑥2−5𝑥+2
5
misal u=𝑥2 − 5𝑥 + 2 , du/dx = 2𝑥 − 5
misal v = √ 𝑢5
= 𝑢
1
5 , dv/du=
1
5
𝑢−
4
5 =
1
5
(𝑥2 − 5𝑥 + 2) −
4
5
y=
1
𝑣
= 𝑣−1 , dy/dv= −1𝑣−2 = −1(√𝑥2 − 5𝑥 + 2
5
)−2
dy/dx = du/dx .dv/du. dy/dv
= (2𝑥 − 5).
1
5
(𝑥2 − 5𝑥 + 2) −
4
5 . −1(√𝑥2 − 5𝑥 + 2
5
)−2
=−
1
5
.(2𝑥 − 5).(𝑥2 − 5𝑥 + 2 )−
4
5.(√𝑥2 − 5𝑥 + 2
5
)−2
7 . y=𝑠𝑖𝑛√𝑥2 + 6𝑥
misal u=𝑥2 + 6𝑥 , du/dx =2x+6
misal v=√ 𝑢 = 𝑢
1
2 , dv/du=
1
2
𝑢−
1
2 =
1
2
(𝑥2 + 6𝑥) −
1
2
y= sin 𝑣 , dy/dv= cos 𝑣 = cos√ 𝑢 = cos√𝑥2 + 6𝑥
dy/dx = du/dx . dv/du. dy/dv
=(2x+6).
1
2
(𝑥2 + 6𝑥) −
1
2 . cos√𝑥2 + 6𝑥
=
1
2
(2x+6). (𝑥2 + 6𝑥) −
1
2 .cos√𝑥2 + 6𝑥

4. 8 . y=𝑐𝑜𝑠√𝑥3 + 2
3
misal u=𝑥3 + 2 , du/dx =3𝑥2
misal v=√ 𝑢3
= 𝑢
1
3 , dv/du=
1
3
𝑢−
2
3 =
1
3
(𝑥3 + 2) −
2
3
y= cos 𝑣 , dy/dv=−𝑠𝑖𝑛 𝑣= −𝑠𝑖𝑛 √ 𝑢3
= −sin √𝑥3 + 2
3
dy/dx = du/dx . dv/du. dy/dv
=(3𝑥2).
1
3
(𝑥3 + 2) −
2
3 .−sin √𝑥3 + 2
3
=−
1
3
(3𝑥2).(𝑥3 +2) −
2
3 . sin √𝑥3 + 2
3
9 . y=𝑠𝑖𝑛
1
√𝑥2+2
y=sin(√𝑥2 + 2 )−1
misal u = 𝑥2 + 2 , du/dx = 2x
misal v =√ 𝑢 = 𝑢
1
2 , dv/du=
1
2
𝑢−
1
2 = =
1
2
(𝑥2 + 2 )−
1
2
y=sin 𝑣−1 , dy/dv= cos 𝑣−1 = cos(√ 𝑢 )−1 = cos(√𝑥2 + 2 )−1
dy/dx = du/dx . dv/du. dy/dv
=(2x).
1
2
(𝑥2 + 2 )−
1
2 . cos (√𝑥2 + 2 )−1
= 𝑥 (𝑥2 + 2 )−
1
2 . cos (√𝑥2 + 2 )−1
= 𝑥 (𝑥2 + 2 )−
1
2 . cos
1
√𝑥2+2
10 . y=𝑐𝑜𝑠
1
√𝑥2+6
3
y=cos(√𝑥2 + 6
3
)−1
misal u = 𝑥2 + 6 , du/dx = 2x
misal v =√ 𝑢3
= 𝑢
1
3 , dv/du=
1
3
𝑢−
2
3 =
1
3
(𝑥2 + 6 )−
2
3
y=cos 𝑣−1 , dy/dv= −sin 𝑣−1 = −sin (√ 𝑢3
)−1 = −sin (√𝑥2 + 6
3
)−1

5. dy/dx = du/dx . dv/du. dy/dv
=(2x).
1
3
(𝑥2 + 6 )−
2
3 . −sin (√𝑥2 + 6
3
)−1
= −
1
3
(2x). (𝑥2 + 6 )−
2
3 .sin (√𝑥2 + 6
3
)−1
= −
1
3
(2x). (𝑥2 + 6 )−
2
3 .sin
1
√𝑥2+6
3