1. The document contains 10 problems involving calculating derivatives of various functions containing radicals and trigonometric functions. 2. The functions include square roots, cube roots and inverse square roots of polynomials, and trigonometric functions like sine and cosine of square roots. 3. For each problem, the function is defined, its derivative is set up using chain rule and partial derivatives, and the final derivative is written out.

1. Tugas 2 MTK2
Nama : Andeki
Kelas : 1 EA
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 . 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

2. =
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
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

3. = (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
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

4. 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
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