1. The document provides 10 problems to calculate the derivative (dy/dx) of various functions involving radicals, trigonometric functions, and composition of functions. 2. The solutions show setting the radical or function within a radical equal to a variable u, then calculating du/dx and applying the chain rule and power rule to determine dy/dx. 3. The answers are provided in fractional form with radicals, involving variables and parameters from the original functions.

1. Nama : Monica roselina
Prodi : Teknik Elektronika
Kelas : 1 EA
Semester : 2 (Dua)
TUGAS MATEMATIKA 2
Tentukanlah nilai
𝒅𝒚
𝒅𝒙
dari fungsi berikut ini !
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

2. JAWABAN
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
=
𝑑𝑦
𝑑𝑢
.
𝑑𝑢
𝑑𝑥
=
1
2
(𝑥5
+ 6𝑥2
+ 3)−
1
2 . (5𝑥4
+ 12𝑥)
𝑑𝑦
𝑑𝑥
=
1
2
(5𝑥4
+ 12𝑥)
(𝑥5 + 6𝑥2 + 3)
1
2
=
1
2
(5𝑥4
+ 12𝑥)
√𝑥5 + 6𝑥2 + 3
2. 𝑦 = √𝑥4 + 6𝑥 + 1
3
Misal:
u= 𝑥4
+ 6𝑥 + 1 →
𝑑𝑢
𝑑𝑥
= 4𝑥3
+ 6
𝑦 = √ 𝑢
3
= 𝑢
1
3 →
𝑑𝑦
𝑑𝑢
=
1
3
𝑢−
2
3 =
1
3
(𝑥4
+ 6𝑥 + 1)−
2
3
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑢
.
𝑑𝑢
𝑑𝑥
=
1
3
(𝑥4
+ 6𝑥 + 1)−
2
3 . (4𝑥3
+ 6)
𝑑𝑦
𝑑𝑥
=
1
3
(4𝑥3
+ 6)
(𝑥4 + 6𝑥 + 1)
2
3
=
1
3
(4𝑥3
+ 6)
√(𝑥4 + 6𝑥 + 1)23
3. 𝑦 = √𝑥2 − 5𝑥
5
Misal :
u= 𝑥2
− 5𝑥 →
𝑑𝑢
𝑑𝑥
= 2𝑥 − 5
𝑦 = √ 𝑢
5
= 𝑢
1
5 →
𝑑𝑦
𝑑𝑢
=
1
5
𝑢−
4
5 =
1
5
(𝑥2
− 5𝑥)−
4
5
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑢
.
𝑑𝑢
𝑑𝑥
=
1
5
(𝑥2
− 5𝑥)−
4
5 . (2𝑥 − 5)

3. 𝑑𝑦
𝑑𝑥
=
1
5
(2𝑥 − 5)
(𝑥2 − 5𝑥)
4
5
=
1
5
(2𝑥 − 5)
√(𝑥2 − 5𝑥)45
4. 𝑦 =
1
√𝑥4+2𝑥
=
1
(𝑥4+2𝑥)
1
2
= (𝑥4
+ 2𝑥)−
1
2
Misal :
u= 𝑥4
+ 2𝑥 →
𝑑𝑢
𝑑𝑥
= 4𝑥3
+ 2
𝑦 = 𝑢−
1
2 →
𝑑𝑦
𝑑𝑢
= −
1
2
𝑢−
3
2 = −
1
2
(𝑥4
+ 2𝑥)−
3
2
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑢
.
𝑑𝑢
𝑑𝑥
= −
1
2
(𝑥4
+ 2𝑥)−
3
2 . (4𝑥3
+ 2)
𝑑𝑦
𝑑𝑥
=
−
1
2
(4𝑥3
+ 2)
(𝑥4 + 2𝑥)
3
2
=
−2𝑥3
− 1
√(𝑥4 + 2𝑥)3
5. 𝑦 =
1
√𝑥2−6𝑥
3 =
1
(𝑥2−6𝑥)
1
3
= (𝑥2
− 6𝑥)−
1
3
Misal:
u= 𝑥2
− 6𝑥 →
𝑑𝑢
𝑑𝑥
= 2𝑥 − 6
𝑦 = 𝑢−
1
3 →
𝑑𝑦
𝑑𝑢
= −
1
3
𝑢−
4
3 = −
1
3
(𝑥2
− 6𝑥)−
4
3
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑢
.
𝑑𝑢
𝑑𝑥
= −
1
3
(𝑥2
− 6𝑥)−
4
3 . (2𝑥 − 6)
𝑑𝑦
𝑑𝑥
=
−
1
3
. (2𝑥 − 6)
(𝑥2 − 6𝑥)−
4
3
=
−
1
3
(2𝑥 − 6)
√(𝑥2 − 6𝑥)43

4. 6. 𝑦 =
1
√𝑥2−5𝑥+2
5 =
1
(𝑥2−5𝑥+2)
1
5
= (𝑥2
− 5𝑥 + 2)−
1
5
Misal:
u= 𝑥2
− 5𝑥 + 2 →
𝑑𝑢
𝑑𝑥
= 2𝑥 − 5
𝑦 = 𝑢−
1
5 →
𝑑𝑦
𝑑𝑢
= −
1
5
𝑢−
6
5 = −
1
5
(𝑥2
− 5𝑥 + 2)−
6
5
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑢
.
𝑑𝑢
𝑑𝑥
= −
1
5
(𝑥2
− 5𝑥 + 2)−
6
5 . (2𝑥 − 5)
𝑑𝑦
𝑑𝑥
=
−
1
5
. (2𝑥 − 5)
(𝑥2 − 5𝑥 + 2)
6
5
=
−
1
5
(2𝑥 − 5)
√(𝑥2 − 5𝑥 + 2)65
7. 𝑦 = sin √𝑥2 + 6𝑥
Misal:
u= 𝑥2
+ 6𝑥 →
𝑑𝑢
𝑑𝑥
= 2𝑥 + 6
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
=
𝑑𝑦
𝑑𝑣
.
𝑑𝑣
𝑑𝑢
.
𝑑𝑢
𝑑𝑥
= cos √ 𝑥2 + 6𝑥 .
1
2
(𝑥2
+ 6𝑥)−
1
2 . (2𝑥 + 6)
𝑑𝑦
𝑑𝑥
=
1
2
. (2𝑥 + 6) . cos √𝑥2 + 6𝑥
(𝑥2 + 6𝑥)
1
2
=
(𝑥 + 3) . cos √𝑥2 + 6𝑥
√𝑥2 + 6𝑥
8. 𝑦 = cos √𝑥3 + 2
3
Misal:
u= 𝑥3
+ 2 →
𝑑𝑢
𝑑𝑥
= 3𝑥2
𝑣 = √ 𝑢
3
= 𝑢
1
3 →
𝑑𝑣
𝑑𝑢
=
1
3
𝑢−
2
3 =
1
3
(𝑥3
+ 2)−
2
3

5. 𝑦 = cos 𝑣 →
𝑑𝑦
𝑑𝑣
= −sin 𝑣 = −sin √ 𝑢
3
= −sin √𝑥3 + 2
3
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑣
.
𝑑𝑣
𝑑𝑢
.
𝑑𝑢
𝑑𝑥
= −sin √𝑥3 + 2
3
.
1
3
(𝑥3
+ 2)−
2
3 . 3𝑥2
𝑑𝑦
𝑑𝑥
=
1
3
. 3𝑥2
. −sin √𝑥3 + 2
3
(𝑥3 + 2)
2
3
=
𝑥2
. −sin √𝑥3 + 2
3
√(𝑥3 + 2)23
9. 𝑦 = sin
1
√𝑥2+2
= sin
1
(𝑥2+2)
1
2
= sin(𝑥2
+ 2)−
1
2
Misal:
u= 𝑥2
+ 2 →
𝑑𝑢
𝑑𝑥
= 2𝑥
𝑣 = 𝑢−
1
2 →
𝑑𝑣
𝑑𝑢
= −
1
2
𝑢−
3
2 = −
1
2
(𝑥2
+ 2)−
3
2
𝑦 = sin 𝑣 →
𝑑𝑦
𝑑𝑣
= cos 𝑣 = cos 𝑢−
1
2 = cos(𝑥2
+ 2)−
1
2
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑣
.
𝑑𝑣
𝑑𝑢
.
𝑑𝑢
𝑑𝑥
= cos(𝑥2
+ 2)−
1
2 . −
1
2
(𝑥2
+ 2)−
3
2 . 2𝑥
𝑑𝑦
𝑑𝑥
=
−
1
2
.2𝑥 .cos(𝑥2+2)
−
1
2
(𝑥2+2)
3
2
=
−𝑥 .cos(𝑥2+2)
−
1
2
√(𝑥2+2)3
10. 𝑦 = cos
1
√𝑥2+6
3 = cos
1
(𝑥2+6)
1
3
= cos(𝑥2
+ 6)−
1
3
Misal:
u= 𝑥2
+ 6 →
𝑑𝑢
𝑑𝑥
= 2𝑥
𝑣 = 𝑢−
1
3 →
𝑑𝑣
𝑑𝑢
= −
1
3
𝑢−
4
3 = −
1
3
(𝑥2
+ 6)−
4
3
𝑦 = cos 𝑣 →
𝑑𝑦
𝑑𝑣
= −sin 𝑣 = −sin 𝑢−
1
3 = −sin(𝑥2
+ 6)−
1
3
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑣
.
𝑑𝑣
𝑑𝑢
.
𝑑𝑢
𝑑𝑥
= −sin(𝑥2
+ 6)−
1
3 . −
1
3
(𝑥2
+ 6)−
4
3 . 2𝑥
𝑑𝑦
𝑑𝑥
=
−
1
3
. 2𝑥 . −sin(𝑥2
+ 6)−
1
3
(𝑥2 + 6)
4
3
=
1
3
. 2𝑥 . sin(𝑥2
+ 6)−
1
3
√(𝑥2 + 6)43