This document contains mathematics differential equations problems and solutions from Group 3 of Class 1 Electronics A at the Polytechnic Manufacturing State College of Bangka Belitung. It includes 10 problems each on functions, derivatives, and finding derivatives of various functions at given values. The problems cover topics like linear, quadratic, cubic, square root, and inverse functions. The document provides the step-by-step workings to find the derivatives of each function.

1. TUGAS MATEMATIKA
KELOMPOK 3
DISUSUN OLEH :
1. TIA LESTARI DEWI YANTI
2. MIZA PISARI
3. NIKMAH UTAMI
4. RIA AYU WAN
KELAS :
1 ELEKTRONIKA A
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
TAHUN AJARAN 2014/2015
Industri air kantung Sungailiat, 33211
Bangka induk propinsi kepulauan Bangka Belitung
Telpon : (0717) 431335 ext. 2281, 2126
FAX : (0717) 93585 email “polman-babel@yahoo.co.id
http//www.polman-babel.ac.id

2. SOAL MATEMATIKA DIFERENSIAL
Soal 4.I
1. f(x) = 4 6. f(x) = 5x2+x-3
2. f(x) = 7x+2 7. f(x) = x3+13x
3. f(x) = -3x-9 8. f(x) = 2x3+15
4. f(x) = 10-3x 9. f(x) = -1/x
5. f(x) = -3/4x 10. f(x) =1/√𝑥
Soal 4.2
1.f(x) = 7 6. g(x)= 25
2. y = 5 7. s(t) = 100
3. f(x) = 0 8. z(x) = 23
4. f(t) = -3 9. y = - 1/2
5. f(x) = 𝜋 10. f(x) = √41
Soal 4.3
1. f(x) = 9x 6.f(x) = 𝜋𝑥 − 25
2. g(x) =-75x 7. f(x) = -3/4x
3. f(x) = x+1 8. s(t) = 100t-45
4. y = 50x+30 9. z(x) = 0.08x+400
5. f(t) = 2t+5 10. f(x) = √41𝑥+1
Soal 4.4
1.f(x) = x3 6. F(x) = 5𝑥 𝜋
2. f(x) = x100 7.f(x) = 1/x5
3.f(x) = x1/4 8.s(t) = 𝑡0,6
4. y = √ 𝑥 9.h (s) = 𝑠
4
5
5.f(t) = t’ 10.f(x) = 1/∛𝑥2
Soal 4.5
1. if f(x) = x3 find f’ (5) 6. if F(x)=xπ find f’ (10)
2. if g(x) = -100 find g’ (25) 7. if f(x)=
1
𝑥5 find f’ (2)
3. if f(x) = x1/4 find f’ (81) 8. if s(t) = 𝑡0.6
find s’ (32)
4. if y = √𝑥 find dy/dx 9. if h(s) = 𝑠
4
5 fin d h’ (32)
5. if f(t)=t, find f’(19) 10.if y =
1
√𝑥23 find
𝑑𝑦
𝑑𝑥
(64)

3. PEMBAHASAN
Pembahasan 4.1
1.f(x) = 4 =
𝑑𝑢
𝑑𝑥
= 0
2. f(x) = 7x +2 =
𝑑𝑢
𝑑𝑥
= 7
3. f(x) = -3x-9 =
𝑑𝑢
𝑑𝑥
= -3
4. f(x) = 10 – 3x =
𝑑𝑢
𝑑𝑥
= -3
5. f(x) =
−3
4
x =
𝑑𝑢
𝑑𝑥
=
−3
4
6. f(x) = 5𝑥2
+ x-3 =
𝑑𝑢
𝑑𝑥
= 10x + 1
7. f (x) = 𝑥3
+ 13x =
𝑑𝑢
𝑑𝑥
= 3𝑥2
+ 13
8. f (x) = 2𝑥3
+ 13x =
𝑑𝑢
𝑑𝑥
= 6𝑥2
9. f(x) =
−1
𝑥
=
𝑑𝑢
𝑑𝑥
= −𝑥−1
= x-2
10. f (x) =
1
√ 𝑥
=
𝑑𝑢
𝑑𝑥
=𝑥
−1
2 = -
1
2
𝑥
−3
2
Pembahasan 4.2
1. f(x) = 7 =
𝒅𝒖
𝒅𝒙
= 0
2. y = 𝟓 =
𝒅𝒖
𝒅𝒚
= 0
3. f (x) = 0 =
𝒅𝒖
𝒅𝒙
= 0
4. f (t) = –3 =
𝒅𝒖
𝒅𝒕
= 0
5. f (x) = π =
𝒅𝒖
𝒅𝒙
= 0
6. g (x) = 25 =
𝒅𝒖
𝒅𝒙
= 0
7. s (t) = 100 =
𝒅𝒖
𝒅𝒙
= 0
8. z (x) = 𝟐 𝟑
=
𝒅𝒖
𝒅𝒙
= 0
9. y = -
𝟏
𝟐
=
𝒅𝒖
𝒅𝒚
= 0
10. f (x) = √ 𝟒𝟏 =
𝒅𝒖
𝒅𝒙
= 0
Pembahasan 4.3
1. f (x) = 9x =
𝑑𝑢
𝑑𝑥
= 9
2. g (x) = -75x =
𝑑𝑢
𝑑𝑥
= -75
3. f (x) = x + 1 =
𝑑𝑢
𝑑𝑥
= 1
4. y = 50x + 30 =
𝑑𝑢
𝑑𝑥
= 50
5. f (t) = 2t + 5 =
𝑑𝑢
𝑑𝑡
= 2
6. f (x) = πx – 25 =
𝑑𝑢
𝑑𝑥
= π
7. f (x) =
−3
4
x =
𝑑𝑢
𝑑𝑥
= -
3
4

4. 8. s (t) = 100𝑡 - 45 =
𝑑𝑢
𝑑𝑡
= 100
9. z (x) = 0,08x + 400 =
𝑑𝑢
𝑑𝑥
= 0,08
10. f(x) = √41𝑥 + 1 =
𝑑𝑢
𝑑𝑥
= 41𝑥
1
2 =
41
2
𝑥
−1
2
Pembahasan 4.4
1.f(x) = 𝑥3
=
𝑑𝑢
𝑑𝑥
= 3𝑥2
2.g(x) =𝑥100
=
𝑑𝑢
𝑑𝑥
=100𝑥99
3.f(x) = 𝑥1/4
=
𝑑𝑢
𝑑𝑥
=
𝑖
4
𝑥
−
3
4
4.y =f(x) = √ 𝑥 = 𝑥
1
2 =
𝑑𝑢
𝑑𝑥
=
1
2
𝑥
−
1
2
5.f(t) = 𝑡1
=
𝑑𝑢
𝑑𝑡
=1
6.f(x) = 𝑥 𝜋
=
𝑑𝑢
𝑑𝑥
= π𝑥 𝜋−1
7.f(x) =
1
𝑥5 = 𝑥−5
=
𝑑𝑢
𝑑𝑥
= −5𝑥−6
8.s(t) = 𝑡0,6
= 𝑡
3
5 =
𝑑𝑢
𝑑𝑥
=
3
5
𝑡
−2
5
9.h(s) = 𝑠
4
5 =
𝑑𝑢
𝑑𝑠
=
4
5
𝑠
−1
5
10.f(x) =
1
√𝑥23 =
𝑑𝑢
𝑑𝑥
=
−2
3
𝑥
−5
3
Pembahasan 4.5
1. if f(x) = 𝑥3
f’(5)
F’(x) = 3𝑥2
F’(5) = 3 (5)2
F’(5) = 75
2. if g (x) = -100 find g’ (25)
= 0
3.if f (x) = 𝑥
1
4 find f’ (81)
f(x) = 𝑥
1
4
F’ (x) =
1
4
𝑥
−3
4
F’ (81) =
1
4
𝑥
−3
4
=
1
4
√𝑥34
=
1
4
√(81)34
=
1
4
√531441
4
=
1
4
27

5. = 1
4
×
1
27
=
1
108
= 0,009
4.if y=√𝑥 find
𝑑𝑦
𝑑𝑥
Y = 𝑥
1
2
Y’ =
1
2
𝑥
−1
2
5. if f(t)=t, find f’(19)
=0
6. if F(x)=xπ find f’ (10)
f’ = 𝜋𝑥 𝜋−1
𝜋 =
22
7
f’ =
22
7
𝑥
22
7
−1
f’ =
22
7
𝑥
15
7
f’ =
22
7
√𝑥157
f’ =
22
7
√(10)157
f’ =
22
7
× 13894954,9
f’ = 43669858,3
7. if f(x)=
1
𝑥5 find f’ (2)
F(x) = 1.x-5
F’(x) = -5x-6
F’(2) =
−5
𝑥6
=
−5
26
=
−5
64
8. if s(t) = 𝑡0.6
find s’ (32)
t’ = 𝑡0,6
t’ =
3
5
𝑡
−2
5
t’ (32) =
3
5
(32)
−2
5
=
3
5
1
√3225
=
3
5
1
√1024
5
=
3
5
.
1
4
=
3
20
= 0,5

6. 9. if h(s) = 𝑠
4
5 find h’ (32)
h’ =
4
5
𝑥
−1
5
h’(32) =
4
5
32
−1
5
=
4
5
.
1
√325
=
4
5
.
1
2
=
4
10
10.if y =
1
√𝑥23 find
𝑑𝑦
𝑑𝑥
(64)
Y = 𝑥
−2
3
y’ =
−2
3
𝑥
−5
3
y’ =
−2
3
𝑥
−5
3
y’ =
−2
3
.
1
√(64)35
y’ =
−2
3
.
1
√2621445
y’ =
−2
3
.
1
12,13
y’ =
−2
36,39
y’ = 0,055