View
50
Download
1
Category
Preview:
Citation preview
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 : polman@polman-babel.ac.idWebsite : www.polman-babel.ac.id
TAHUN AJARAN 2014/2015
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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 )
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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)
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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 ¿¿¿
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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)
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
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
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Recommended