The document provides exercises on calculating derivatives using various rules including the sum and difference rule, product rule, and quotient rule. It includes 15 problems calculating derivatives of given functions and finding numerical derivatives at given values. The answers provide the step-by-step work to arrive at the derivatives using the appropriate rules.

1. 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

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

3. 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

4. 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
)
For problem 11-15, find the indicated numerical derivative.
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

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

6. 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

7. 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

8. 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
For problem 11-15, find the indicated numerical derivative.
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

9. 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

10. 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. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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