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) = 𝑣25 + 7 – 15𝑣

35

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) = 𝑣25 + 7 – 15𝑣

35

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) = 𝑣25 + 7 – 15𝑣

35

q’(v) = 2

5𝑣−

3

5 – 9𝑣−2

5

10. f(x) = 5

2𝑥2 +

5

2𝑥−2 –

5

2

= 5

2𝑥−2 +

5

2𝑥2 –

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

3𝑡 + 10

s’(t) = 32t – 2

3

s’(0) = – 2

3

14. q(v) = 𝑣2

5 + 7 – 15𝑣3

5

q’(v) = 2

5𝑣−

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

2𝑥−2 +

5

2𝑥2 –

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 𝑥3

)

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

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)(𝑥1

2 + 5)

y’ = (15

2𝑥−

3

2)( 𝑥 + 5) + (– 15𝑥−1

2 + 25)( 1

2𝑥−

1

2)

= 15

2𝑥−1 +

25

2𝑥−

3

2 – 15

2𝑥−1 +

25

2𝑥−

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 𝑥3

)

g’(x) = (6𝑥2 +4x)( 2𝑥1

3) + (2𝑥3 + 2𝑥2)(2

3𝑥−

2

3)

= 12𝑥21

3 + 8𝑥11

3 + 4

3𝑥2

1

3 + 4

3𝑥1

1

3

= 40

3𝑥2

1

3 + 28

3𝑥1

1

3

8. f(x) = 10

𝑥5 . 𝑥3 + 1

5

= 10𝑥−5 . 𝑥

5

3 +

1

5

f’(x) = −50𝑥−6 . (𝑥5

3 +

1

5) + 10𝑥−5 .

3

5𝑥2

= −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 - 𝑥2

3) −2

3𝑥−

1

3(2𝑥3 + 3)

= 18𝑥2 – 𝑥2

3 – 4

3𝑥2

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)(𝑥1

2 + 5)

y’ = (15

2𝑥−

3

2)( 𝑥 + 5) + (–15𝑥−1

2 + 25)( 1

2𝑥−

1

2)

= 15

2𝑥−1 +

25

2𝑥−

3

2 – 15

2𝑥−1 +

25

2𝑥−

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 . (𝑥5

3 +

1

5) + 10𝑥−5 .

3

5𝑥2

= −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

32 − 1

2𝑥12 + 6

5. y = −15

𝑥

6. s(t) = 2𝑡

32 − 3

4𝑡12 + 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

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

2𝑥−

3

2

4. f(x) = 3𝑥

32 − 1

2𝑥12 + 6

f’(x) =

9

2𝑥

12 2𝑥

12+6 − 𝑥

−12 3𝑥

32−1

(2𝑥12+6)2

= 9𝑥 + 27𝑥

12 − 3𝑥 + 𝑥

−12

4𝑥 + 24𝑥12 + 36

= 6𝑥 + 27 𝑥 + 𝑥

−12

4𝑥 + 24 𝑥 + 36

5. y = −15

𝑥

=−15𝑥−1

y’ = 15𝑥−2

6. s(t) = 2𝑡

32 − 3

4𝑡12 + 6

s’(t) = 3𝑡

12 4𝑡

12+6 − 2𝑡

−12 2𝑡

32−3

(4𝑡12+6)2

= 12𝑡 + 18𝑡

12 − 4𝑡 + 6𝑡

−12

16𝑡 + 48𝑡12 + 36

= 8𝑡 + 18𝑡

12 − 6𝑦

−12

16 + 48𝑡12 + 36

7. g(x) = 𝑥100

𝑥−5 + 10

g’(x) = 100𝑥99 𝑥−5+10 + 5𝑥−6(𝑥100

(𝑥−5 + 10)2

= 100𝑥94 + 1000𝑥99 + 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

4𝑥2 +

1

8)

= −𝑥4 − 1

2𝑥2

f’(x) = −4𝑥3 – 𝑥

11. f(x) = 5𝑥 + 2

3𝑥 − 1 = 5x + 2 (

1

3𝑥−1 − 1) =

5

3− 5𝑥 +

2

3𝑥−1 − 2

f’(x) = −5 − 2

3𝑥−2

f’(25) = −5 − 2

3.(25)2

−2

= −5 − 2

625 . 3

12. h(x) = 4 − 5𝑥

8𝑥 = 4 − 5𝑥(

1

8𝑥−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

2𝑥−

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

2𝑥−

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 12

= 50 15𝑡 + 120 − 1

2

C’(t) = −25 15𝑡 + 120 − 3

2 . 15

= −375 15𝑡 + 120 − 3

2

= −375

15𝑡+120 32

= −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

2𝑥−

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. 𝑥2𝑦 = 1

2. 𝑥𝑦3 = 3𝑥2𝑦 + 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 𝑥2𝑦 = 1

7. 𝑑𝑦

𝑑𝑥|(5,2) when 𝑥𝑦3 = 3𝑥2𝑦 + 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. 𝑥2𝑦 = 1

2𝑥𝑑𝑥

𝑑𝑥 𝑦 + 𝑥2 𝑑𝑦

𝑑𝑥= 0

2𝑥𝑦 + 𝑥2 𝑑𝑦

𝑑𝑥= 0

2𝑥𝑦 = −𝑥2 𝑑𝑦

𝑑𝑥

𝑑𝑦

𝑑𝑥= −

2𝑥𝑦

𝑥2

2. 𝑥𝑦3 = 3𝑥2𝑦 + 5𝑦

𝑥𝑦3 − 3𝑥2𝑦 − 5𝑦 = 0

𝑦3 𝑑𝑥

𝑑𝑥+ 3𝑦2𝑥

𝑑𝑦

𝑑𝑥− 6𝑥𝑦

𝑑𝑥

𝑑𝑥− 3𝑥2 𝑑𝑦

𝑑𝑥− 5𝑦

𝑑𝑦

𝑑𝑥= 0

3𝑦2𝑥 − 3𝑥2 − 5 𝑑𝑦

𝑑𝑥= 6𝑥𝑦 − 𝑦3

𝑑𝑦

𝑑𝑥=

6𝑥𝑦−𝑦3

3𝑦2𝑥−3𝑥2−5

3. 𝑥 + 𝑦 = 25

1

2𝑥−

1

2𝑑𝑥

𝑑𝑥+

1

2𝑦−

1

2𝑑𝑦

𝑑𝑥= 0

1

2𝑦−

1

2𝑑𝑦

𝑑𝑥= −

1

2𝑥−

1

2𝑑𝑥

𝑑𝑥

𝑑𝑦

𝑑𝑥=

−1

2𝑥−

12

−1

2𝑦−

12

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

2𝑥𝑑𝑥

𝑑𝑥+ 2𝑦

𝑑𝑦

𝑑𝑥= 0

2𝑦𝑑𝑦

𝑑𝑥= −2𝑥

𝑑𝑦

𝑑𝑥=

−2𝑥

2𝑦

6. 𝑥2𝑦 = 1

2𝑥𝑑𝑥

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 = 3𝑥2𝑦 + 5𝑦

𝑥𝑦3 − 3𝑥2𝑦 − 5𝑦 = 0

𝑦3 𝑑𝑥

𝑑𝑥+ 3𝑦2𝑥

𝑑𝑦

𝑑𝑥− 6𝑥𝑦

𝑑𝑥

𝑑𝑥− 3𝑥2 𝑑𝑦

𝑑𝑥− 5𝑦

𝑑𝑦

𝑑𝑥= 0

3𝑦2𝑥 − 3𝑥2 − 5 𝑑𝑦

𝑑𝑥= 6𝑥𝑦 − 𝑦3

𝑑𝑦

𝑑𝑥=

6𝑥𝑦−𝑦3

3𝑦2𝑥−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

2𝑥−

1

2𝑑𝑥

𝑑𝑥+

1

2𝑦−

1

2𝑑𝑦

𝑑𝑥= 0

1

2𝑦−

1

2𝑑𝑦

𝑑𝑥= −

1

2𝑥−

1

2𝑑𝑥

𝑑𝑥

𝑑𝑦

𝑑𝑥=

−1

2𝑥−

12

−1

2𝑦−

12

𝑑𝑦

𝑑𝑥|(4,9) =

4−

12

9−

12

=1

21

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

2𝑥𝑑𝑥

𝑑𝑥+ 2𝑦

𝑑𝑦

𝑑𝑥= 0

2𝑦𝑑𝑦

𝑑𝑥= −2𝑥

𝑑𝑦

𝑑𝑥=

−2𝑥

2𝑦

𝑑𝑦

𝑑𝑥|(2,1) =

−2(2)

2(1)

=−4

2

= −2