Tugas blog-matematika

KELOMPOK = Adi Pambudi
Dio Visiasa Silaen
Syamsul Bahri
Widia Afridani
Exercise 5.2
For problem 1 – 10, use the rule for sums and differences to find the derivative of given
function.
Question!
1. f(x) = 𝑥7
+ 2𝑥10
2. h(x) = 30 – 5𝑥2
3. g(x) = 𝑥100
– 40𝑥5
4. C(x) = 1.000 + 200x – 40𝑥2
5. y =
− 15
𝑥
+ 25
6. s(t) = 16𝑡2
–
2𝑡
3
+ 10
7. g(x) =
𝑥100
25
– 20 𝑥
8. y = 12𝑥0.2
+ 0.45x
9. q(v) = 𝑣
2
5 + 7 – 15𝑣
3
5
10. f(x) =
5
2𝑥2
+
5
2𝑥−2
–
5
2
For problem 11-15, find the indicated numerical derivative.
11. h’(1
2
) when h(x) = 30 – 5𝑥2
12. C’(300) when C(x) = 1.000 + 200x – 40𝑥2
13. s’(0) when s(t) = 16𝑡2
–
2𝑡
3
+ 10
14. q’(32) when q(v) = 𝑣
2
5 + 7 – 15𝑣
3
5
15. f’(6) when f(x) =
5
2𝑥2
+
5
2𝑥−2
–
5
2

Answer!
1. f(x) = 𝑥7
+ 2𝑥10
f’(x) = 7𝑥6
+ 20𝑥9
2. h(x) = 30 – 5𝑥2
h’(x) = –10x
3. g(x) = 𝑥100
– 40𝑥5
g’(x) = 100𝑥99
– 200𝑥4
4. C(x) = 1.000 + 200x – 40𝑥2
C’(x) = 200 – 80x
5. y =
− 15
𝑥
+ 25
= – 15𝑥−1
+ 25
y’ = – 15𝑥−2
6. s(t) = 16𝑡2
–
2𝑡
3
+ 10
s’(t) = 32t –
2
3
7. g(x) =
𝑥100
25
– 20 𝑥
=
𝑥100
25
– 20𝑥
1
2
g’(x) = 4𝑥99
– 10𝑥−
1
2
8. y = 12𝑥0.2
+ 0.45x
y’ = 2.4𝑥−0.8
+ 0.45
9. q(v) = 𝑣
2
5 + 7 – 15𝑣
3
5
q’(v) =
2
5
𝑣−
3
5 – 9𝑣−
2
5
10. f(x) =
5
2𝑥2
+
5
2𝑥−2
–
5
2
=
5
2
𝑥−2
+
5
2
𝑥2
–
5
2
f’(x) = – 5𝑥−3
+ 5x
11. h(x) = 30 – 5x
h’(x) = –10x
h’(1
2
) = –10(1
2
)
h’(1
2
) = –5

12. C(x) = 1.000 + 200x – 40𝑥2
C’(x) = 200 – 80x
C’(300) = 200 – (80 . 300)
= 200 – 24.000
= –23.800
13. s(t) = 16𝑡2
–
2
3
𝑡 + 10
s’(t) = 32t –
2
3
s’(0) = –
2
3
14. q(v) = 𝑣
2
5 + 7 – 15𝑣
3
5
q’(v) =
2
5
𝑣−
3
5 – 9𝑣−
2
5
q’(32) =
2
5
.
1
8
– 9
1
4
q’(32) =
2
40
–
9
4
q’(32) =
1
20
–
9
4
15. f(x) =
5
2𝑥2
+
5
2𝑥−2
–
5
2
=
5
2
𝑥−2
+
5
2
𝑥2
–
5
2
f’(x) = – 5𝑥−3
+ 5x
f’(6) = – 5(6)−3
+ 5(6)
f’(6) = –
5
216
+ 30

Exercise 5.3
For problem 1 – 10, use the productruleto find the derivaative of given function.
Question!
1. f(x) = (2𝑥2
+ 3)(2x – 3)
2. h(x) = (4𝑥3
+ 1)( – 𝑥2
+ 2x + 5)
3. g(x) = (𝑥2
– 5)(3
𝑋
)
4. C(x) = (50 + 20x)(100 – 2x)
5. y = (−15
𝑥
+ 25)( 𝑥 + 5)
6. s(t) = (4𝑡 −
1
2
)(5t +
3
4
)
7. g(x) = (2𝑥3
+ 2𝑥2
)(2 𝑥
3
)
8. f(x) =
10
𝑥5
.
𝑥3 + 1
5
9. q(v) = (𝑣2
+ 7)(−5𝑣−2
+ 2)
10. f(x) = (2𝑥3
+ 3)(3 – 𝑥23
)
11. f’(1.5) when f(x) = (2𝑥2
+ 3)(2x – 3)
12. g’(10) when g(x) = (𝑥2
– 5)(3
𝑋
)
13. C’(150) when C(x) = (50 + 20x)(100 – 2x)
14.
𝑑𝑦
𝑑𝑥
|x=25 when y = (−15
𝑥
+ 25)( 𝑥 + 5)
15. f’(2) when f(x) =
10
𝑥5
.
𝑥3 + 1
5

Answer!
1. f(x) = (2𝑥2
+ 3)(2x – 3)
f’(x) = (4x)(2x-3) + (2𝑥2
+ 3)(2)
= 8𝑥2
– 12x + 4𝑥2
+ 6
= 12𝑥2
– 12x + 6
2. h(x) = (4𝑥3
+ 1)( – 𝑥2
+ 2x + 5)
h’(x) = (12𝑥2
)( – 𝑥2
+ 2x + 5) + (4𝑥3
+ 1)( –2x +2)
= –12𝑥4
+ 24𝑥3
+ 60𝑥2
+ (– 8𝑥4
) + 8𝑥3
+ (–2x) + 2
= –20𝑥4
+ 32𝑥3
+ 60𝑥2
– 2x + 2
3. g(x) = (𝑥2
– 5)(3
𝑋
)
= (𝑥2
– 5) ( 3𝑥−1
)
g’(x) = (2x)(3𝑥−1
)+ (𝑥2
– 5)(−3𝑥−2
)
= 6 – 3 + 15𝑥−2
4. C(x) = (50 + 20x)(100 – 2x)
C’(x) = (20)(100 – 2x) + (50 + 20x)( –2)
= 2000 – 40x – 100 – 40x
= –80x + 1900
5. y = (−15
𝑥
+ 25)( 𝑥 + 5)
= (– 15𝑥−
1
2 + 25)(𝑥
1
2 + 5)
y’ = (15
2
𝑥−
3
2)( 𝑥 + 5) + (– 15𝑥−
1
2 + 25)(1
2
𝑥−
1
2)
=
15
2
𝑥−1
+
25
2
𝑥−
3
2 –
15
2
𝑥−1
+
25
2
𝑥−
1
2
=
75
2 . 𝑥 𝑥
+
25
2 𝑥
6. s(t) = (4𝑡 −
1
2
)(5t +
3
4
)
s’(t) = (4)(5t +
3
4
) + (4𝑡 −
1
2
)(5)
= 20t + 3 + 20t –
5
2
= 40t +
6
2
–
5
2
= 40t +
1
2

7. g(x) = (2𝑥3
+ 2𝑥2
)(2 𝑥
3
)
g’(x) = (6𝑥2
+4x)( 2𝑥
1
3) + (2𝑥3
+ 2𝑥2
)(2
3
𝑥−
2
3)
= 12𝑥2
1
3 + 8𝑥1
1
3 +
4
3
𝑥2
1
3 +
4
3
𝑥1
1
3
=
40
3
𝑥2
1
3 +
28
3
𝑥1
1
3
8. f(x) =
10
𝑥5
.
𝑥3 + 1
5
= 10𝑥−5
.
𝑥
5
3
+
1
5
f’(x) = −50𝑥−6
. (𝑥
5
3
+
1
5
) + 10𝑥−5
.
3
5
𝑥2
= −10𝑥−3
– 10𝑥−6
+ 6𝑥−3
= −4𝑥−3
– 10𝑥−6
9. q(v) = (𝑣2
+ 7)(−5𝑣−2
+ 2)
q’(v) = 2x(−5𝑥−2
+ 2) + (𝑣2
+ 7)(10𝑥−3
)
= −10𝑣−1
+ 4𝑣 + 10𝑥−1
+ 70𝑥−3
= 4v + 70𝑣−3
10. f(x) = (2𝑥3
+ 3)(3 – 𝑥23
)
f’(x) = 6𝑥2
(3 - 𝑥
2
3) −
2
3
𝑥−
1
3(2𝑥3
+ 3)
= 18𝑥2
– 𝑥
2
3 –
4
3
𝑥2
2
3 – 2𝑥−
1
3
11. f(x) = (2𝑥2
+ 3)(2x – 3)
f’(x) = (4x)(2x-3) + (2𝑥2
+ 3)(2)
= 8𝑥2
– 12x + 4𝑥2
+ 6
= 12𝑥2
– 12x + 6
f’(1.5) = 12(1.5)2
– 12(1.5) + 6
= 27 – 18 + 6
= 15
12. g(x) = (𝑥2
– 5)(3
𝑋
)
g’(x) = (2x)(3𝑥−1
)+ (𝑥2
– 5)(−3𝑥−2
)
= 6 – 3 + 15𝑥−2
g’(10) = 6 – 3 +
15
100
= 3 +
15
100

13. C(x) = (50 + 20x)(100 – 2x)
C’(x) = (20)(100 – 2x) + (50 + 20x)( –2)
= 2000 – 40x – 100 – 40x
= –80x + 1900
C’(150) = –80(150) + 1900
= –1200 +1900
= 700
14. y = (−15
𝑥
+ 25)( 𝑥 + 5)
= (– 15𝑥−
1
2 + 25)(𝑥
1
2 + 5)
y’ = (15
2
𝑥−
3
2)( 𝑥 + 5) + (– 15𝑥−
1
2 + 25)(1
2
𝑥−
1
2)
=
15
2
𝑥−1
+
25
2
𝑥−
3
2 –
15
2
𝑥−1
+
25
2
𝑥−
1
2
=
75
2 . 𝑥 𝑥
+
25
2 𝑥
𝑑𝑦
𝑑𝑥
|x=25 =
75
2 . 25 25
+
25
2 25
=
3
10
+
5
2
=
3+25
10
=
28
10
15. f(x) =
10
𝑥5 .
𝑥3 + 1
5
= 10𝑥−5
.
𝑥
5
3
+
1
5
f’(x) = −50𝑥−6
. (𝑥
5
3
+
1
5
) + 10𝑥−5
.
3
5
𝑥2
= −10𝑥−3
– 10𝑥−6
+ 6𝑥−3
= −4𝑥−3
– 10𝑥−6
f’(2) = –
4
23 –
10
26
= –
4
8
–
10
64
= – (4
8
+
10
64
)
= – (32 + 10
64
)
= –
42
64

Exercise 5.4
For problem 1 – 10, use the quotient rule to find the derivative of given function.
Question!
1. f(x) =
5𝑥 + 2
3𝑥 − 1
2. h(x) =
4 − 5𝑥2
8𝑥
3. g(x) =
5
𝑥
4. f(x) =
3𝑥
3
2 − 1
2𝑥
1
2 + 6
5. y =
−15
𝑥
6. s(t) =
2𝑡
3
2 − 3
4𝑡
1
2 + 6
7. g(x) =
𝑥100
𝑥−5 + 10
8. y(x) =
4 − 5𝑥3
8𝑥2 − 7
9. q(v) =
𝑣3+ 2
𝑣2 −
1
𝑣3
10. f(x) =
−4𝑥2
4
𝑥2 + 8
11. f’(25) when f(x) =
5𝑥 + 2
3𝑥 − 1
12. h’(0.2) when h(x) =
4 − 5𝑥2
8𝑥
13. g’(0.25) when g(x) =
5
𝑥
14.
𝑑𝑦
𝑑𝑥
|10 when y =
−15
𝑥
15. g’(10) when g(x) =
𝑥100
𝑥−5 + 10

Answer!
1. f(x) =
5𝑥 + 2
3𝑥 − 1
f’(x) =
5 3𝑥−1 – 3(5𝑥+3)
(3𝑥−1)2
=
15𝑥 – 5 − 15𝑥 − 6
9𝑥 − 6𝑥 + 1
=
−11
9𝑥 − 6𝑥 + 1
2. h(x) =
4 − 5𝑥2
8𝑥
h’(x) =
−10𝑥 8𝑥 – 8(4−5𝑥2)
64𝑥2
=
−80𝑥2 + 40𝑥2 − 32
64𝑥2
=
−40𝑥2 − 32
64𝑥2
=
−5𝑥2 − 4
8𝑥2
3. g(x) =
5
𝑥
= 5𝑥−
1
2
g’(x) = −
5
2
𝑥−
3
2
4. f(x) =
3𝑥
3
2 − 1
2𝑥
1
2 + 6
f’(x) =
9
2
𝑥
1
2 2𝑥
1
2+6 − 𝑥
−
1
2 3𝑥
3
2−1
(2𝑥
1
2+6)2
=
9𝑥 + 27𝑥
1
2 − 3𝑥 + 𝑥
−
1
2
4𝑥 + 24𝑥
1
2 + 36
=
6𝑥 + 27 𝑥 + 𝑥
−
1
2
4𝑥 + 24 𝑥 + 36
5. y =
−15
𝑥
=−15𝑥−1
y’ = 15𝑥−2

6. s(t) =
2𝑡
3
2 − 3
4𝑡
1
2 + 6
s’(t) =
3𝑡
1
2 4𝑡
1
2+6 − 2𝑡
−
1
2 2𝑡
3
2−3
(4𝑡
1
2+6)2
=
12𝑡 + 18𝑡
1
2 − 4𝑡 + 6𝑡
−
1
2
16𝑡 + 48𝑡
1
2 + 36
=
8𝑡 + 18𝑡
1
2 − 6𝑦
−
1
2
16 + 48𝑡
1
2 + 36
7. g(x) =
𝑥100
𝑥−5 + 10
g’(x) =
100𝑥99 𝑥−5+10 + 5𝑥−6(𝑥100
(𝑥−5 + 10)2
=
100𝑥94 + 1000 𝑥99 + 5𝑥94
𝑥−10 + 20𝑥−5 + 100
8. y(x) =
4 − 5𝑥3
8𝑥2 − 7
y’(x) =
−15𝑥2 8𝑥2−7 − 16𝑥(4−5𝑥3)
(8𝑥2 – 7)2
𝑑𝑦
𝑑𝑥
=
−120𝑥4 − 105𝑥2 − 64𝑥 + 80𝑥4
64𝑥4 − 112𝑥2 + 39
=
−40𝑥4 − 105𝑥2 − 64𝑥
64𝑥4 − 112𝑥2 + 39
9. q(v) =
𝑣3+ 2
𝑣2 −
1
𝑣3
=
𝑣3+ 2
𝑣2 − 𝑣−3
q’(v) =
3𝑣2 𝑣2−𝑣−3 − 2𝑣 − 3𝑣−4(𝑣3+2)
(𝑣2 − 𝑣−3)2
=
3𝑣4 − 3𝑣−1 − 2𝑣4 − 2𝑣 − 3𝑣−1 −6𝑣
𝑣4− 2𝑣−1 + 𝑣−6
=
𝑣4 − 6𝑣−1 − 8𝑣
𝑣4− 2𝑣−1 + 𝑣−6
10. f(x) =
−4𝑥2
4
𝑥2 + 8
=
−4𝑥2
4𝑥−2 + 8
= (−4𝑥2
)(
1
4
𝑥2
+
1
8
)
= −𝑥4
−
1
2
𝑥2
f’(x) = −4𝑥3
– 𝑥

11. f(x) =
5𝑥 + 2
3𝑥 − 1
= 5x + 2 (
1
3
𝑥−1
− 1) =
5
3
− 5𝑥 +
2
3
𝑥−1
− 2
f’(x) = −5 −
2
3
𝑥−2
f’(25) = −5 −
2
3.(25)2
−2
=
−5 − 2
625 . 3
12. h(x) =
4 − 5𝑥
8𝑥
= 4 − 5𝑥(
1
8
𝑥−1
) =
𝑥
2
−1
−
5
8
h’(x) = −
𝑥
2
−2
h’(0.2) =
−(0.2)
2
−2
=
0.04
2
= 0.02
13. g(x) =
5
𝑥
= 5𝑥−
1
2
g’(x) = −
5
2
𝑥−
3
2
g’(0.25) =
−5
2 0.25 . 0.25
=
−5
2 0.25 . (−0.5)
= 20
14. y =
−15
𝑥
=−15𝑥−1
𝑑𝑦
𝑑𝑥
= 15𝑥−2
𝑑𝑦
𝑑𝑥
|10 =
15
(10)2
= 0,15
15. g(x) =
𝑥100
𝑥−5 + 10
=𝑥100
(𝑥5
+
1
10
) = 𝑥105
+
𝑥
10
100
g’(x) = 105𝑥104
+ 10𝑥99
g’(1) = 105 + 10
= 115

Exercise 5.5
For problem 1 – 10, use the chain rule to find the derivative of given function.
Question!
1. f(x) = (3𝑥2
− 10)3
2. g(x) = 40(3𝑥2
− 10)3
3. h(x) = 10(3𝑥2
− 10)−3
4. h(x) = ( 𝑥 + 3)2
5. f(u) = (
1
𝑢2
− 𝑢)3
6. y =
1
(𝑥2−8)3
7. y = 2𝑥3 + 5𝑥 + 1
8. s(t) = (2𝑡3
+ 5𝑡)
1
3
9. f(x) =
10
(2𝑥−6)5
10. C(t) =
50
15𝑡+120
For problem 11 – 15, find the indicated numerical derivative.
11. f’(10) when f(x) = (3𝑥2
− 10)3
12. h’(3) when h(x) = 10(3𝑥2
− 10)−3
13. f’(144) when h(x) = ( 𝑥 + 3)2
14. f’(2) when f(u) = (
1
𝑢2
− 𝑢)3
15.
𝑑𝑦
𝑑𝑥
|4 when y =
1
(𝑥2−8)3

Answer!
1. f(x) = (3𝑥2
− 10)3
f’(x) = 3(3𝑥2
− 10)2
. (6𝑥)
= 18𝑥(3𝑥2
− 10)2
2. g(x) = 40(3𝑥2
− 10)3
g’(x) = 120(3𝑥2
− 10)2
. (6𝑥)
= 720𝑥(3𝑥2
− 10)2
3. h(x) = 10(3𝑥2
− 10)−3
h’(x) = −30(3𝑥2
− 10)−4
. (6𝑥)
= −180(3𝑥2
− 10)−4
4. h(x) = ( 𝑥 + 3)2
h’(x) = 2 𝑥 + 3 . (
1
2
𝑥−
1
2)
= 𝑥−
1
2( 𝑥 + 3)
5. f(u) = (
1
𝑢2
− 𝑢)3
f’(u) = 3(
1
𝑢2
− 𝑢)2
. (−2𝑢−3
− 1)
= 3 −2𝑢−3
− 1 . (
1
𝑢2
− 𝑢)2
= 3
2
𝑢3
− 1 .
1
𝑢2
− 𝑢 2
6. y =
1
(𝑥2−8)3
= (𝑥2
− 8)−3
𝑑𝑦
𝑑𝑥
= −3(𝑥2
− 8)−4
. (2𝑥)
=
−6𝑥
(𝑥2−8)4
7. y = 2𝑥3 + 5𝑥 + 1
= 2𝑥3
+ 5𝑥 + 1
1
2
𝑑𝑦
𝑑𝑥
=
1
2
2𝑥3
+ 5𝑥 + 1 −
1
2 . (6𝑥 + 5)
=
1
2
6𝑥 + 5 . 2𝑥3
+ 5𝑥 + 1 −
1
2
=
1
2
.
6𝑥+5
2𝑥3+5𝑥+1

8. s(t) = (2𝑡3
+ 5𝑡)
1
3
s’(t) =
1
3
2𝑡3
+ 5𝑡 −
2
3 . (4𝑡 + 5)
=
1
3
2𝑡3
+ 5𝑡 −
2
3 . (4𝑡 + 5)
=
1
3
4𝑡 + 5 . 2𝑡3
+ 5𝑥 + 1 −
2
3
9. f(x) =
10
(2𝑥−6)5
= 10(2𝑥 − 6)−5
f’(x) = −50(2𝑥 − 6)−6
. (2)
=
−100
(2𝑥−6)6
10. C(t) =
50
15𝑡+120
=
50
15𝑡+120
1
2
= 50 15𝑡 + 120 −
1
2
C’(t) = −25 15𝑡 + 120 −
3
2 . 15
= −375 15𝑡 + 120 −
3
2
=
−375
15𝑡+120
3
2
=
−375
15𝑡+120 . 15𝑡+120
11. f(x) = (3𝑥2
− 10)3
f’(x) = 3(3𝑥2
− 10)2
. (6𝑥)
= 18𝑥(3𝑥2
− 10)2
f’(10) = 18 10 (3 10 2
− 10)2
= 180(3 100 − 10)2
= 180(300 − 10)2
= 180(290)2
= 15.138.000
12. h(x) = 10(3𝑥2
− 10)−3
h’(x) = −30(3𝑥2
− 10)−4
. (6𝑥)
= −180(3𝑥2
− 10)−4
=
−180𝑥
(3𝑥2−10)4
h’(3) =
−180(3)
(3(3)2−10)4
=
−540
(27−10)4
=
−540
(17)4
=
−540
4913

13. h(x) = ( 𝑥 + 3)2
h’(x) = 2 𝑥 + 3 . (
1
2
𝑥−
1
2)
= 𝑥−
1
2( 𝑥 + 3)
=
( 𝑥+3)
𝑥
h’(144) =
( 144+3)
144
=
12+3
12
=
15
12
14. f(u) =
1
𝑢2
− 𝑢
3
f’(u) = 3
1
𝑢2
− 𝑢
2
. −2𝑢−3
− 1
= 3 −2𝑢−3
− 1 .
1
𝑢2
− 𝑢
2
= 3
2
𝑢3
− 1 .
1
𝑢2
− 𝑢 2
f’(2) = 3
2
23
− 1 .
1
22
− 2 2
= 3
2
8
−
8
8
.
1
4
−
8
4
2
= 3 −
6
8
. −
7
4
2
=−
18
8
49
16
= −
882
128
= −
441
64
15. y =
1
(𝑥2−8)3
= (𝑥2
− 8)−3
𝑑𝑦
𝑑𝑥
= −3(𝑥2
− 8)−4
. (2𝑥)
=
−6𝑥
(𝑥2−8)4
𝑑𝑦
𝑑𝑥
|4 =
−6(4)
(42−8)4
=
−24
(16−8)4
=
−24
(8)4
= −
24
4096

Exercise 5.6
For problem 1 – 5, use implicit differentiation to find
𝑑𝑦
𝑑𝑥
.
Question!
1. 𝑥2
𝑦 = 1
2. 𝑥𝑦3
= 3𝑥2
𝑦 + 5𝑦
3. 𝑥 + 𝑦 = 25
4.
1
𝑥
+
1
𝑦
= 9
5. 𝑥2
+ 𝑦2
= 16
For problem 6 – 10, find the indicated numerical derivative.
6.
𝑑𝑦
𝑑𝑥
|(3,1) when 𝑥2
𝑦 = 1
7.
𝑑𝑦
𝑑𝑥
|(5,2) when 𝑥𝑦3
= 3𝑥2
𝑦 + 5𝑦
8.
𝑑𝑦
𝑑𝑥
|(4,9) when 𝑥 + 𝑦 = 25
9.
𝑑𝑦
𝑑𝑥
|(5,10) when
1
𝑥
+
1
𝑦
= 9
10.
𝑑𝑦
𝑑𝑥
|(2,1) when 𝑥2
+ 𝑦2
= 16

Answer!
1. 𝑥2
𝑦 = 1
2𝑥
𝑑𝑥
𝑑𝑥
𝑦 + 𝑥2 𝑑𝑦
𝑑𝑥
= 0
2𝑥𝑦 + 𝑥2 𝑑𝑦
𝑑𝑥
= 0
2𝑥𝑦 = −𝑥2 𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
= −
2𝑥𝑦
𝑥2
2. 𝑥𝑦3
= 3𝑥2
𝑦 + 5𝑦
𝑥𝑦3
− 3𝑥2
𝑦 − 5𝑦 = 0
𝑦3 𝑑𝑥
𝑑𝑥
+ 3𝑦2
𝑥
𝑑𝑦
𝑑𝑥
− 6𝑥𝑦
𝑑𝑥
𝑑𝑥
− 3𝑥2 𝑑𝑦
𝑑𝑥
− 5𝑦
𝑑𝑦
𝑑𝑥
= 0
3𝑦2
𝑥 − 3𝑥2
− 5
𝑑𝑦
𝑑𝑥
= 6𝑥𝑦 − 𝑦3
𝑑𝑦
𝑑𝑥
=
6𝑥𝑦 −𝑦3
3𝑦2 𝑥−3𝑥2−5
3. 𝑥 + 𝑦 = 25
1
2
𝑥−
1
2
𝑑𝑥
𝑑𝑥
+
1
2
𝑦−
1
2
𝑑𝑦
𝑑𝑥
= 0
1
2
𝑦−
1
2
𝑑𝑦
𝑑𝑥
= −
1
2
𝑥−
1
2
𝑑𝑥
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
−
1
2
𝑥
−
1
2
−
1
2
𝑦
−
1
2
4.
1
𝑥
+
1
𝑦
= 9
𝑥−1
+ 𝑦−1
= 9
−𝑥−2 𝑑𝑥
𝑑𝑥
− 𝑦−2 𝑑𝑦
𝑑𝑥
= 0
−𝑦−2 𝑑𝑦
𝑑𝑥
= 𝑥−2
𝑑𝑦
𝑑𝑥
= −
𝑥−2
𝑦−2
5. 𝑥2
+ 𝑦2
= 16
2𝑥
𝑑𝑥
𝑑𝑥
+ 2𝑦
𝑑𝑦
𝑑𝑥
= 0
2𝑦
𝑑𝑦
𝑑𝑥
= −2𝑥
𝑑𝑦
𝑑𝑥
=
−2𝑥
2𝑦

6. 𝑥2
𝑦 = 1
2𝑥
𝑑𝑥
𝑑𝑥
𝑦 + 𝑥2 𝑑𝑦
𝑑𝑥
= 0
2𝑥𝑦 + 𝑥2 𝑑𝑦
𝑑𝑥
= 0
2𝑥𝑦 = −𝑥2 𝑑𝑦
𝑑𝑥
2𝑥𝑦
𝑥2 =
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
|(3,1) =
2 3 (1)
(3)2
=
6
9
=
2
3
7. 𝑥𝑦3
= 3𝑥2
𝑦 + 5𝑦
𝑥𝑦3
− 3𝑥2
𝑦 − 5𝑦 = 0
𝑦3 𝑑𝑥
𝑑𝑥
+ 3𝑦2
𝑥
𝑑𝑦
𝑑𝑥
− 6𝑥𝑦
𝑑𝑥
𝑑𝑥
− 3𝑥2 𝑑𝑦
𝑑𝑥
− 5𝑦
𝑑𝑦
𝑑𝑥
= 0
3𝑦2
𝑥 − 3𝑥2
− 5
𝑑𝑦
𝑑𝑥
= 6𝑥𝑦 − 𝑦3
𝑑𝑦
𝑑𝑥
=
6𝑥𝑦 −𝑦3
3𝑦2 𝑥−3𝑥2−5
𝑑𝑦
𝑑𝑥
|(5,2) =
6 5 (2)−(2)3
3 2 2(5)−3(5)2−5
=
60−8
60−75−5
=
52
−20
=
26
−10
8. 𝑥 + 𝑦 = 25
1
2
𝑥−
1
2
𝑑𝑥
𝑑𝑥
+
1
2
𝑦−
1
2
𝑑𝑦
𝑑𝑥
= 0
1
2
𝑦−
1
2
𝑑𝑦
𝑑𝑥
= −
1
2
𝑥−
1
2
𝑑𝑥
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
−
1
2
𝑥
−
1
2
−
1
2
𝑦
−
1
2
𝑑𝑦
𝑑𝑥
|(4,9) =
4
−
1
2
9
−
1
2
=
1
2
1
3
=
3
2
9.
1
𝑥
+
1
𝑦
= 9
𝑥−1
+ 𝑦−1
= 9
−𝑥−2
− 𝑦−2 𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
|(5,10) = −(5)−2
− (10)−2
= 0
= −
1
25
−
1
100
=
3
100

10. 𝑥2
+ 𝑦2
= 16
2𝑥
𝑑𝑥
𝑑𝑥
+ 2𝑦
𝑑𝑦
𝑑𝑥
= 0
2𝑦
𝑑𝑦
𝑑𝑥
= −2𝑥
𝑑𝑦
𝑑𝑥
=
−2𝑥
2𝑦
𝑑𝑦
𝑑𝑥
|(2,1) =
−2(2)
2(1)
=
−4
2
= −2

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