tugas kalkulus diferentiation

1

TUGAS KALKULUS ( DIFFERENTIATION )

( Halaman 23-31 )MATEMATIKA 2

Disusun Oleh :

Nama : 1. SIRILUS OKI SELPHADINATA2. SITI FATIMAH3. ODI BARKAH4. ANDEKI

Prodi : Teknik ElektronikaKelas : 1E ASemester : 2 (Genap)

POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNGKawasan Industri Air Kantung Sungailiat, Bangka 33211

Telp. (0717) 93586, Fax. (0717) 93585Email : [email protected] : www.polman-babel.ac.id

TAHUN AJARAN 2014/2015

POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG

http://www.polman-babel.ac.id/

mailto:[email protected]

2

Latihan 5.1

1. f ( x )=2x3

f ' ( x )=2 ddx

(x3 )

f ' ( x )=6 x2

2. g ( x )= x100

25

g ' ( x )= ddx ( x

100

25 ) g' (x )=100x99

3. f ( x )=20x12

f ' ( x )=20 ddx

(x¿¿12)¿

f ' ( x )=10 x−12

4. y=−16√x

y=−16x12

y '=−16 ddx

(x¿¿12)¿

y '=−8 x−12

5. f (t )=2 t3

f ' (t )= ddx ( 2t3 )

f ' (t )=2 t

6. f ( x )= x π

2π

f ( x )=2xπ π

f ' ( x )=2 ddx

(xπ¿¿ π)¿

f ' ( x )=2xπ 2π

7. f ( x )=10x5

f ( x )=10 x−5 POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG

3

f ' ( x )=10 ddx

(x¿¿−5)¿

f ' ( x )=−50x−6

8. s ( t )=100 t0,6

s ' ( t )=100 ddx

(t ¿¿0,6)¿

s ' (t )=60 t−0,4

9. h ( s)=−25 s12

h ' ( s)=−25 ddx

(s¿¿12)¿

h ' ( s)=−12,5 s−12

10. f ( x )= 1

43√x2

f ' ( x )=4 x23

f ' ( x )=4 ddx

(x¿¿23)¿

f ' ( x )=83x

−13

11. f ' (3 )when f ( x )=2 x3

f ' ( x )=2 ddx

(x3 )

f ' ( x )=6 x2 f ' (3 )=6(3)2 f ' (3 )=54

12. g' (x )when g (x )= x100

25

g ' ( x )= ddx ( x

100

25 ) g' (x )=100x99 g' (1 )=100(1)99 g' (1 )=100

13. f ' (81 )when f (x )=20x12


4

f ' ( x )=20 ddx

(x¿¿12)¿

f ' ( x )=10 x−12

f ' (81 )=10(81)−12

f ' (81 )=109

14. dydx

|25 y=−16 √x

y=−16x12

y '=−16 ddx

(x¿¿12)¿

y '=−8 x−12

y '=−8(25)−12

y '=−85

15. f ' (200 )when f ( t )=2t3

f ' (t )= ddx ( 2t3 )

f ' (t )=2 t f ' (200 )=2(200) f ' (200 )=400

Latihan 5.2


5

1. f ( x )=x7+2 x10

f ' ( x )= ddx

(x7 )+ ddx

(2 x10)

f ' ( x )=7 x6+20 x9

2. h ( x )=30−5 x2

h' ( x )= ddx

(30 )− ddx

(5 x2 )

h' ( x )=−10 x

3. g ( x )=x100−40 x5

g' (x )= ddx

(x100 )+ ddx

(40x5 )

g' (x )=100x99−200 x4

4. c ( x )=1000+200 x−40 x2

c ' ( x )= ddx

(1000 )+ ddx

(200x )+ ddx

(40 x2)

c ' ( x )=200−80 x

5. y=−15x

+25

y '= ddx (−15x )+ ddx (25 )

y '=−15 x−2

6. s ( t )=16 t2−2 t3

+10

s' (t )= ddx

(16 t2 )− ddx

( 2t3

)+ ddx

(10 )

s' (t )=32t−6 t

7. g ( x )= x100

25−20√x

g ' ( x )= ddx

( x100

25)− ddx

(20√ x)

g' (x )=100x99−2012

8. y=12 x0,2+0,45 x

y '= ddx

(12 x0,2 )+ ddx

(0,45 x )


6

y '=2,4 x−0,8+0,45

9. q ( v )=v12+7−15v

12

q ' ( v )= ddx

(v12 )+ ddx

(7 )− ddx

(15 v12)

q ' (v )=12v32−7,5v

32

10. f ( x )= 5

2 x2+ 5

2x−2−52

f ( x )= ddx

(10 x−2)+ ddx

(10x2)− ddx

( 52)

f ' ( x )=20 x−3+20 x

11. h' ( 12 )whenh ( x )=30−5 x2

h' ( x )= ddx

(30 )− ddx

(5 x2)

h' ( x )=−10 x

h' ( x )=−10( 12 )=−5

12. c ' (300 )whenc ( x )=1000+200 x−40x2

c ' ( x )= ddx

(1000 )+ ddx

(200 x)− ddx

(40x2)

c ' ( x )=200−80 x c ' (300 )=200−80 (300 )=−23800

13. s' (0 )when s ( t )=16 t 2−2t3

+10

s' (t )= ddx

(16 t2 )− ddx

( 2t3

)+ ddx

(10 )

s' (t )=32t−6 t s' (0 )=32 (0 )−6 (0 )=0

14. q ' (32 )whenq (v )=v12+7−15v

12

q ' ( v )= ddx

(v12 )+ ddx

(7 )− ddx

(15 v12)

q ' (v )=12v32−7,5v

32


7

q ' (32 )=12(32)

32−7,5 (32 )

32=−1267,14

15. f ' (6 )when f ( x )= 5

2 x2+ 5

2x−2−52

f ( x )= ddx

(10 x−2)+ ddx

(10x2)− ddx

( 52)

f ' ( x )=20 x−3+20 x f ' ( x )=20(6)−3+20 (6 )=121,85

Latihan 5.3


8

1. f ( x )=(2 x¿¿2+3)+(2x−3)¿

f ' ( x )=(2x¿¿2+3) ddx

(2x−3 )+ (2x−3 ) ddx

(2x¿¿2+3)¿¿

f ' ( x )=(2x¿¿2+3) (2 )+ (2x−3 )(4 x )¿ f ' ( x )=4 x2+6+8 x2−12x f ' ( x )=12 x2−12x+6

2. h ( x )=(4 x¿¿2+1)+(−x2+2 x+5)¿

h ' ( x )=(4 x¿¿2+1) ddx

(−x2+2x+5)+(−x2+2 x+5) ddx

(4 x¿¿2+1)¿¿

h' ( x )=(4 x¿¿2+1) (−2 x+2 )+(−x2+2x+5)(8 x )¿ h' ( x )=(−8x¿¿3+8x2−2 x+2)+(−8 x¿¿3+16 x2+40 x)¿¿ h' ( x )=−16 x3+24 x2+38 x+2

3. g ( x )=(x¿¿2−5)+( 3x)¿

g' (x )=(x¿¿2−5)(3 x¿¿−1)¿¿

g' (x )=(x¿¿2−5) ddx

(3 x¿¿−1)+(3 x¿¿−1) ddx

(x¿¿2−5)¿¿¿¿

g' (x )=(x¿¿2−5)(−3 x¿¿−2)+(3 x¿¿−1)(2x )¿¿¿ g' (x )=−3x+15 x−2+6 x g' (x )=3x−15x−2

4. c ( x )=(50+20x )(100−2x )

c ' ( x )=(50+20 x ) ddx

(100−2x )+(100−2x ) ddx

(50+20 x )

c ' ( x )= (50+20 x ) (−2 )+(100−2x )(20) c ' ( x )=−100x−40 x+2000−40x c ' ( x )=−80 x+1900

5. y=(−15√x

+25)(√x+5)

y=(−15 x−12 +25)(x

12+5)

y '=(−15 x−12 +25) d

dx( x

12+5)+( x

12+5) d

dx(−15 x

−12 +25)

y '=(−15 x−12 +25)( 12 x

−12 )+( x 12+5)(7,5 x−3

2 )

y '=7,5 x−1+22,5x−12 +7,5 x−1+37,5x

−32


9

y '=22,5 x−12 +15x−1+37,5 x

−32

6. s ( t )=(4 t−12 )(5 t+ 34 )s' (t)=(4 t−12 ) ddx (5 t+ 34 )+(5 t+ 34 ) ddx (4 t−12 ) s' (t )=(4 t−12 )(5 )+(5 t+ 34 )(4) s' (t )=(20 t−52 )+(20 t+3 )

s' (t )=40 t−112

7. g ( x )=(2 x3+2 x2 ) (2 3√x )

g ( x )=(2 x3+2 x2 )(2x13 )

g' (x )=(2x3+2 x2) ddx

(2 x13)+(2x

13 ) ddx

(2x3+2 x2)

g' (x )=(2x3+2 x2) ( 23 x−23 )+(2 x 13) (6 x2+4 x )

g' (x )=( 43 x73+ 43x43)+(12 x 73+8 x 43)

g' (x )=403x73+283x43

8. f ( x )=( 10x5 )( x3+15 )

f ( x )=(10x−5 ) (3 x2)

f '( x)=(10 x−5 ) ddx

(3 x2 )+ (3 x2 ) ddx

(10 x−5 )

f ' ( x )=(10 x−5 ) (6 x )+(3 x2) (−50 x−6 ) f ' ( x )=60 x−5−150 x−4

9. q ( v )=(v¿¿2+7)(−5 v−2+2)¿

q ' (v )=(v¿¿2+7) ddx

(−5 v−2+2)+(−5v−2+2) ddx

(v¿¿2+7)¿¿

q ' (v )=(v¿¿2+7)(10 v−3)+(−5v−2+2)(2v )¿ q '(v )=(10v¿¿−1+70v−3)+(−10 v¿¿−2+4 v )¿¿ q ' ( v )=4 v+10v−1−10v−2+70 v−3


10

10. f ( x )=(2 x¿¿3+3) (3−3√ x2)¿

f ( x )=(2 x¿¿3+3) (3−x23 )¿

f ' ( x )=(2 x¿¿3+3) ddx

(3−x23 )+(3−x

23) ddx

(2x¿¿3+3)¿¿

f ' ( x )=(2 x¿¿3+3)(23x

−13 )+(3−x

23 )(6x2)¿

f ' ( x )=( 43 x83+2x

−13 )+(18 x2−683 )

f ' ( x )=223x83+18x2+2 x

−13

11. f ' (15 )when f ( x )=(2x¿¿2+3)+(2x−3)¿

f ' ( x )=(2x¿¿2+3) ddx

(2x−3 )+ (2x−3 ) ddx

(2x¿¿2+3)¿¿

f ' ( x )=(2x¿¿2+3) (2 )+ (2x−3 )(4 x )¿ f ' ( x )=4 x2+6+8 x2−12x f ' ( x )=12 x2−12x+6 f ' (15 )=12(15)2−12 (15 )+6=2526

12. g' (10)wheng ( x )=(x¿¿2−5)+( 3x)¿

g' (x )=(x¿¿2−5)(3 x¿¿−1)¿¿

g' (x )=(x¿¿2−5) ddx

(3 x¿¿−1)+(3 x¿¿−1) ddx

(x¿¿2−5)¿¿¿¿

g' (x )=(x¿¿2−5)(−3 x¿¿−2)+(3 x¿¿−1)(2x )¿¿¿ g' (x )=−3x+15 x−2+6 x g' (x )=3x−15x−2 g' (10 )=3 (10 )−15 (10 )−2=30,15

13. c ' (150 )whwnc ( x )=(50+20 x)(100−2 x)

c ' ( x )=(50+20 x ) ddx

(100−2x )+(100−2x ) ddx

(50+20 x )

c ' ( x )= (50+20 x ) (−2 )+(100−2x )(20) c ' ( x )=−100x−40 x+2000−40x c ' ( x )=−80 x+1900 c ' (150 )=−80 (150 )+1900=−10100

14.dydx

|x=25| y=(−15√ x

+25)(√ x+5)


11

y=(−15 x−12 +25)(x

12+5)

y '=(−15 x−12 +25) d

dx( x

12+5)+( x

12+5) d

dx(−15 x

−12 +25)

y '=(−15 x−12 +25)( 12 x

−12 )+( x 12+5)(7,5 x−3

2 )

y '=7,5 x−1+22,5x−12 +7,5 x−1+37,5x

−32

y '=22,5 x−12 +15x−1+37,5 x

−32

15. f ' (2 )when f ( x )=(10x5 )( x3+15 )

f ( x )=(10x−5 ) (3 x2)

f '( x)=(10 x−5 ) ddx

(3 x2 )+ (3 x2 ) ddx

(10 x−5 )

f ' ( x )=(10 x−5 ) (6 x )+(3 x2) (−50 x−6 ) f ' ( x )=60 x−5−150 x− 4

f ' (2 )=60(2)−5−150 (2 )−4=−14716

Latihan 5.4


12

1. f ( x )= 5 x+23 x−1

f ' ( x )=(3 x−1 ) d

dx(5x+2 )−(5x+2 ) d

dx(3 x−1 )

(3x−1)2

f ' ( x )= (3 x−1 ) (5 )−(5 x+2 ) (3 )(3 x−1)2

f '( x)=5 x−5−15 x+69 x2−6 x+1

f '( x)= −10x+19 x2−6 x+1

2. h ( x )=4−5 x2

8 x

h ' ( x )=(8x ) d

dx(4−5 x2)−(4−5 x2) d

dx(8x )

(8 x)2

h ' ( x )=(8x )(−10 x)−(4−5 x2)(8)64 x2

h ' ( x )=−80 x2−32−40x2

64 x2

h ' ( x )=−120 x2−3264 x2

3. g ( x )= 5

√x

g ' ( x )=(√x ) d

dx(5 )−(5) d

dx(√x )

(√ x)2

g ' ( x )=(√x ) (0 )−(5)(x

12)

(√x )2

g' (x)=− (5 )(x

12 )

x

g' (x )= 5

2 x32

4. f ( x )=3 x32−1

2 x12+6

f ' ( x )=(2 x12+6)

ddx

(3 x¿¿32−1)−(3 x¿¿

32−1)

ddx

(2x¿¿12+6)

(2 x 12+6)2 ¿¿¿


13

f ' ( x )=(2 x12+6)¿¿

5. y=−15x

y '=(−15 ) d

dx(−x )−(−x) d

dx(−15)

(−x)2

y '=(−15 ) (−1 )−(1)(0)

x2

y '=15x2

6. s(t )=2t32−3

4 t12+6

s '(t )=(4 t12+6) d

dx(2t ¿¿

32−3)−

(2 t¿¿32−3) d

dx(4 t

12+6)

(4 t12+6)2

¿¿

s '(t )=(4 t12+6)(3 t¿¿

12)−

(2 t ¿¿32−3)(2 t

−12 )

(4 t12+6)2

¿¿

s' (t )=12 t+18 t

12−4 t−6 t

−12

16 t+48 t12+36

7. g ( x )= x100

x−5+10

g ' ( x )=(x−5+10)ddx

(x¿¿100)−(x¿¿100) d

dx(x−5+10)

(x−5+10)2¿¿

g ' ( x )=(x−5+10)(100 x¿¿99)−(x¿¿100)(−5 x−6)

(x−5+10)2¿¿

g' (x )=100 x94+1000 x99+5 x94

x−25+20 x−5+100

g' (x )= 105x94+1000 x99

x−25+20 x−5+100

8. y= 4−5 x3

8x2−7

y '=(8 x2−7 ) d

dx(4−5 x3 )−(4−5 x3) d

dx(8 x2−7)

(8 x2−7)2


14

y '=(8 x2−7 )(−15 x2)−(4−5 x3)(16x )

(8 x2−7)2

y '=−120 x4−80 x3+105 x2−64 x64 x4−112 x2+49

9.q (v )= v3+2

v2− 1v2

q (v )= v3+2v2−v−2

q ' (v )=(v2−v−2) d

dx(v3+2 )−(v3+2) d

dx(v2−v−2)

(v2−v−2)2

q ' (v )=(v2−v−2) (3v2)−(v3+2)(2v+2v−3)

v4−2v+v−4

q ' (v )=3v4−3v−2v4+2v+4 v+4 v−3

v4−2 v+v−4

q ' (v )= v4+3v+4v−3

v 4−2v+v−4

10.f (x)=−4 x2

4x2

+8

f (x)= −4 x2

4 x−2+8

f ' (x)=(4 x−2+8)ddx

(−4 x¿¿2)−(−4 x¿¿2) ddx

(4 x−2+8)

(4 x−2+8)2¿¿

f ' (x)=(4 x−2+8)(−8x )−(−4 x¿¿2)(−8 x−3)

(4 x−2+8)2¿

f ' (x)=−36 x−1−64 x−36 x−1

16x−4+64 x−2+64

f ' (x)= −64 x−1−64 x16 x−4+64 x−2+64

11. f ' (25 )when f ( x )= 5 x+23 x−1

f ' ( x )=(3 x−1 ) d

dx(5x+2 )−(5x+2 ) d

dx(3 x−1 )

(3x−1)2

f ' ( x )= (3 x−1 ) (5 )−(5 x+2 ) (3 )(3 x−1)2

f '( x)=5 x−5−15 x+69 x2−6 x+1


15

f '( x)= −10x+19 x2−6 x+1

f ' (25 )= −10 (25 )+19 (25 )2−6 (25 )+1

=−2495476

12. h' (0,2 )whenh (x )= 4−5x2

8 x

h ' ( x )=(8x ) d

dx(4−5 x2)−(4−5 x2) d

dx(8x )

(8 x)2

h ' ( x )=(8x )(−10 x)−(4−5 x2)(8)64 x2

h ' ( x )=−80 x2−32−40x2

64 x2

h ' ( x )=−120 x2−3264 x2

h' (0,2 )=−120 (0,2 )2−3264 (0,2 )2

=−36,82,56

13. g' (0,25 )wheng ( x )= 5

√ x

g ' ( x )=(√x ) d

dx(5 )−(5) d

dx(√x )

(√ x)2

g ' ( x )=(√x ) (0 )−(5)(x

12)

(√x )2

g' (x)=− (5 )(x

12 )

x

g' (x )= 5

2 x32

g' (0,25 )= 5

2(0,25)32

=20

14. y ' (10 )when y=−15x

y '=(−15 ) d

dx(−x )−(−x) d

dx(−15)

(−x)2

y '=(−15 ) (−1 )−(1)(0)

x2

y '=15x2


16

y '= 15

(10 )2= 15100

15. g' (1 )when g ( x )= x100

x−5+10

g ' ( x )=(x−5+10)ddx

(x¿¿100)−(x¿¿100) d

dx(x−5+10)

(x−5+10)2¿¿

g ' ( x )=(x−5+10)(100 x¿¿99)−(x¿¿100)(−5 x−6)

(x−5+10)2¿¿

g' (x )=100 x94+1000 x99+5 x94

x−25+20 x−5+100

g' (x )= 105x94+1000 x99

x−25+20 x−5+100

g' (1 )= 105(1)94+1000(1)99

(1)−25+20(1)−5+100=1105121


17

Latihan 5.51. f ( x ) ¿(3 x2−10)3

u=3x2−10→du /dx=6 x y=u3→dy /du=3u2=3(3 x2−10)2 dydx

=dudx.dydu

¿6 x .3(3x2−10)2 ¿18 x (3 x2−10)2

2. g ( x )¿ 40(3 x2−10)3

u=3x2−10→du /dx=6 x y=40u3→dy /du=120u2=120 (3 x2−10)2 dydx

=dudx.dydu

¿6 x .120(3x2−10)2 ¿720 x (3 x2−10)2

3. h ( x )¿10 (3 x2−10)−3

u=3x2−10→du /dx=6 x y=10u−3→dy /du=−30u− 4=−30(3x2−10)−4 dydx

=dudx.dydu

¿6 x .(−30)(3 x2−10)−4 ¿−180 x(3x2−10)− 4

4. h ( x )¿ (√ x+3)2

u=√x+3→x12+3→du/dx=1

2x

−12

y=u2→dy /du=2u=2(√ x+3) dydx

=dudx.dydu

¿ 12x

−12 .2(√x+3)


18

¿ x−12 .2(√x+3)

5. f ( v ) ¿( 1v2−v )2

u= 1

v2−v→v−2−v→du/dx=−2v−3

y=u2→dy /du=2u=2( 1v2

−v)

dydx

=dudx.dydu

¿−2v−3 .2( 1v2

−v)

¿− 2v3.2( 1v2

−v )

6. y=1

( x2−8)3→y=(x2−8)−3

u=x2−8→du /dx=2 x y=u−2→dy /du=−3u−4=−3 (x2−8)−4 dydx

=dudx.dydu

¿2 x .−3(x2−8)−4 ¿−6 x (x2−8)−4

¿− 6x

(x2−8)−4

7. y=√2 x3+5x+1u=2x3+5 x+1→du/dx=6 x2+5

y=√u→u12→

12u

−12 →dy /du= 1

2√u= 1

2√2 x3+5x+1

dydx

=dudx.dydu

¿6 x2+5. 1

2√2 x3+5x+1

8. s ( t )=(2 t 3+5t )12

u=2t 3+5 t→du/dx=6 t2+5

y=u12→dy /du=1

2u

−12 =1

2(2 t 3+5t )

−12

dydx

=dudx.dydu


19

¿6 t 2+5. 12(2t 3+5 t)

−12

9. f (x)=10

(2x−6)5

f (x)=10(2 x−6)−5 u=2x−6→du /dx=2

y=10u−5→dy /du=−50u−6=−50(2 x−6)−6 dydx

=dudx.dydu

¿2 .−50 (2x−6)−6 ¿−100(2 x−6)−6

¿− 100

(2 x−6)−6

10. c (t )= 50

√15 t+120

c (t )=50(15 t+120)−12

u=15t+120→du/dx=15

y=50u−12 →dy /du=−25u

−32 =−25 (15 t+120)

−32

dydx

=dudx.dydu

¿15 .−25(15 t+120)−32

¿−375(15 t+120)−32

11. f ' (10 )when f ( x ) ¿(3 x2−10)3

u=3x2−10→du /dx=6 x y=u3→dy /du=3u2=3(3 x2−10)2 dydx

=dudx.dydu

¿6 x .3(3x2−10)2 ¿18 x (3 x2−10)2 ¿18(10)(3 (10)2−10)2 ¿15138000

12. h' (3 )whenh (x ) ¿10(3 x2−10)−3

u=3x2−10→du /dx=6 x


20

y=10u−3→dy /du=−30u− 4=−30(3x2−10)−4 dydx

=dudx.dydu

¿6 x .(−30)(3 x2−10)−4 ¿−180 x(3x2−10)− 4 ¿−180(3)(3(3)2−10)−4

¿− 54083521

13. f ' (144 )when f (x ) ¿(√x+3)2

u=√x+3→x12+3→du/dx=1

2x

−12

y=u2→dy /du=2u=2(√ x+3) dydx

=dudx.dydu

¿ 12x

−12 .2(√x+3)

¿ x−12 .2(√x+3)

¿(144)−12 .2(√144+3)

¿ 52

14. f ' (2 )when f (v )¿( 1v2−v )2

u= 1

v2−v→v−2−v→du/dx=−2v−3

y=u2→dy /du=2u=2( 1v2

−v)

dydx

=dudx.dydu

¿−2v−3 .2( 1v2

−v)

¿− 2v3.2( 1v2

−v )

¿− 2

(2 )3.2 ( 1

(2 )2−(2))

¿−178

15. y (4 )when y= 1

(x2−8)3→ y=(x2−8)−3

u=x2−8→du /dx=2 x POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG

21

y=u−2→dy /du=−3u−4=−3 (x2−8)−4 dydx

=dudx.dydu

¿2 x .−3(x2−8)−4 ¿−6 x (x2−8)− 4

¿− 6x

(x2−8)−4

¿−6(4)

((4)2−8)−4

¿−¿98304


22

Latihan 5.61. x2 y=1

ddx

(x2 y )= ddx

(1)

2 xdxdx. y+x2 . dy

dx=0

2 xy+ x2 dydx

=0

dydx

=2 xyx2

2. xy3=3 x2 y+5 y3 x3 y4−5 y=0 ddx

(3 x¿¿3 y4−5 y)= ddx

(0)¿

ddx

(3 x¿¿3 y4)−dydx

(5 y )= ddx

(0)¿

9 x2dxdx. y4+3 x3 .4 y3 dy

dx=0

9 x2 y 4+3 x34 y3 dydx

=0

dydx

= 9x2 y4

3 x34 y3

3. √ x+√ y=25

ddx

(x12+ y

12)= d

dx(25)

ddx

(x12)+ dydx

( y12)= d

dx(25)

12x

−12 + 12y

−12 dydx

=0

dydx

=−12x

−12

12y

−12

4. 1x +1y=9


23

ddx

(x−1+ y−1 )= ddx

(9 )

ddx

(x−1 )+ dydx

( y−1)= ddx

(9 )

−x−2− y−2 dydx

=0

dydx

= x−2

− y−2

5. x2+ y2=16ddx

(x2+ y2)= ddx

(16)

ddx

(x2 )+ dydx

( y2 )= ddx

(16)

2 x+2 y dydx

=0

dydx

=−2x2 y

6. dydx

|(3,1 )when x2

y=1

ddx

(x2 y )= ddx

(1)

2 xdxdx. y+x2 . dy

dx=0

2 xy+ x2 dydx

=0

dydx

=2 xyx2

dydx

=2 (3)(1)

(3)2=69

7. dydx

|(5,2 )when xy3

=3 x2 y+5 y

3 x3 y4−5 y=0 ddx

(3 x¿¿3 y4−5 y)= ddx

(0)¿

ddx

(3 x¿¿3 y4)−dydx

(5 y )= ddx

(0)¿

9 x2dxdx. y4+3 x3 .4 y3 dy

dx=0

9 x2 y 4+3 x34 y3 dydx

=0

dydx

= 9x2 y4

3 x34 y3


24

8. √ x+√ y=25ddx

(x12+ y

12)= d

dx(25)

ddx

(x12)+ dydx

( y12)= d

dx(25)

12x

−12 + 12y

−12 dydx

=0

dydx

=−12x

−12

12y

−12

dydx

=−12

(4 )−12

12

(9 )−12

=−1416

9. 1x +1y=9

ddx

(x−1+ y−1 )= ddx

(9 )

ddx

(x−1 )+ dydx

( y−1)= ddx

(9 )

−x−2− y−2 dydx

=0

dydx

= x−2

− y−2

dydx

=− (5 )−2

(10 )−2=

− 1251100

=−4

10. x2+ y2=16ddx

(x2+ y2)= ddx

(16)

ddx

(x2 )+ dydx

( y2 )= ddx

(16)

2 x+2 y dydx

=0

dydx

=−2x2 y

dydx

=−2 (2 )2 (1 )

=−2


tugas kalkulus diferentiation

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