R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
21. nodarb%2łba
3
1. piemrs. Noteikt funkcijas z = ex y2 visus otrs krtas parcilos atvasinjumus.
Vispirms atrad%2łsim pirms krtas parcilos atvasinjumus:
3 3
"z "z
= 3x2 y2ex y2 , = 2x3 yex y2 .
"x "y
Otrs krtas parcilie atvasinjumi ir
3 3 3 3
"2 z
= 3" 2xy2ex y2 + 3x2 y2ex y2 " 3x2 y2 = 6xy2ex y2 + 9x4 y4ex y2 ;
"x2
3 3 3 3
"2 z
= 3x2 " 2yex y2 + 3x2 y2ex y2 " 2x3 y = 6x2 yex y2 + 6x5 y3ex y2 ;
"x"y
3 3 3 3
"2 z
= 2 " 3x2 yex y2 + 2x3 yex y2 " 3x2 y2 = 6x2 yex y2 + 6x5 y3ex y2 ;
"y"x
3 3 3 3
"2 z
= 2x3ex y2 + 2x3 yex y2 " 2x3 y = 2x3ex y2 + 4x6 y2ex y2 .
"y2
2 2 2 2
K redzams, dotaj piemr ir spk viend%2łba zxy = zyx .
2. piemrs. Noteikt funkcijas z = xsin y visus otrs krtas parcilos atvasinjumus.
Pirms krtas parcilie atvasinjumi:
2 2
zx = sin y " xsin y-1, zy = xsin y ln x " cos y .
Otrs krtas parcilie atvasinjumi ir
2
2 2 2
zxx = (zx ) = sin y(sin y -1)xsin y-2 ,
x
2
2 2 2 2 2
zxy = zyx = (zx ) = cos y " x + sin y " xsin y-1 ln x " cos y ,
y
2
2 2 2
zyy = (zy ) = xsin y ln x " cos y " ln x " cos y + xsin y ln x "(- sin y) = xsin y(ln2 x " cos2 y - ln x " sin y)
y
21. nodarb%2łba. 1. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
"3z
3. piemrs. Funkcijai z = sin(xy3) noteikt parcilo atvasinjumu .
"x2"y
T k dotajai funkcijai atvasinaanas krt%2łba nav svar%2łga, tad vispirms atvasinsim pc t
main%2łg, pc kura atvasinjums ir vienkraks, aoreiz tas ir pc x:
"z
= cos(xy3)" y3 .
"x
Iegkto funkciju vlreiz atvasinsim pc x:
"2 z
= -sin(xy3)" y3 " y3 = -sin(xy3)" y6 .
"x2
Tagad rezulttu atvasinsim pc y:
"3z
= -cos(xy3)"3xy2 " y6 - sin(xy3)"6y5 = -3xy8 cos(xy3)- 6y5 sin(xy3).
"x2"y
4. piemrs. Funkcijai z = arcsin3(x - x2 + 4)+ arctg(xy) noteikt parcilo atvasinjumu
"3z
.
"x"y2
Doto funkciju vispirms atvasinsim pc y, jo ais atvasinjums ir daudz vienkraks:
"z 1 x
= 0 + " x = .
2
"y 1+ x2 y2
1+ (xy)
Ar%2ł aoreiz varam izvlties, pc kura main%2łg atvasint. T k atvasinjums pc y atkal ir
vienkraks, atvasinsim pc t:
"2 z 0 - x " 2x2 y 2x3 y
= = - .
2
"y2 x2 y2)2
(1+ (1+ x2 y2)
Vairs izvles nav, a%2ł funkcija jatvasina pc x:
2
"3z 6x2 y(1+ x2 y2) - 2x3 y " 2(1+ x2 y2)" 2xy2 (1+ x2 y2)(6x2 y + 6x4 y3 - 8x4 y3)
= - = - =
4 4
"y2"x
(1+ x2 y2) (1+ x2 y2)
2x4 y3 - 6x2 y
= .
3
(1+ x2 y2)
21. nodarb%2łba. 2. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
x
y
5. piemrs. Uzrakst%2łt otrs krtas piln diferenci<a izteiksmi funkcijai z = xy + e .
Lai uzrakst%2łtu otrs krtas piln diferenci<a izteiksmi, nepiecieaami visi otrs krtas
parcilie atvasinjumi:
x x
1 # ś#
x
y y
2 2
zx = y + e " , zy = x + e " ś#- ź# ;
ś# ź#
y y2
# #
x 2
# ś#
1
y
2 2 2
zxx = (zx )2 = e " ś# ź# ,
x
ś# ź#
y
# #
x x x
# ś# # ś# # ś#
x 1 1 1 x
y y y
2 2 2 2
zxy = (zx )2 = (zy )2 = 1+ e " ś#- ź# " + e " ś#- ź# = = 1- e ś# +1ź# ,
y x
ś# ź# ś# ź# ź#
y2 y y2 y2 ś# y
# # # # # #
2
x x x
# ś# # - 2x x x
x ś# # ś#
y y y
2 2 2 ś#
zyy = (zy )2 = e " - ź# - e " ś# ź# = e ś# + 2ź# .
y
ś# ź# ś# ź# ź#
y2 y3 y3 ś# y
# # # # # #
Divu argumentu funkcijas otrs krtas piln diferenci<a apr7inaanas formula ir
2
2 2 2 2 2 2
d z = zxxdx2 + 2zxydxdy + zyydy2 , ttad dots funkcijas otrs krtas pilnais diferencilis
ir
x x x
#
# ś#ś# # ś#
1 1 x x x
2 y y y
d z = e dx2 + 2ś#1- e ś# +1ź#ź#dxdy + e ś# + 2ź#dy2 .
ź#ź# ś# ź#
ś#
y2 y2 ś# y y
# # # y3 # #
#
x + 5y
6. piemrs. Uzrakst%2łt otrs krtas piln diferenci<a izteiksmi funkcijai z = .
y - 5x
Dots funkcijas pirms un otrs krtas parcilie atvasinjumi ir
1"(y - 5x)- (x + 5y)"(- 5) y - 5x + 5x + 25y 26y
2
zx = = = ,
2 2 2
(y - 5x) (y - 5x) (y - 5x)
5 "(y - 5x)- (x + 5y)"1 5y - 25x - x - 5y - 26x
2
zy = = = ;
2 2 2
(y - 5x) (y - 5x) (y - 5x)
260y
-2 2 -3
2 2
zxx = (26y(y - 5x) ) = 26y "(- 2)(y - 5x) "(- 5) = ,
x
3
(y - 5x)
21. nodarb%2łba. 3. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
2
26(y - 5x) - 26y " 2(y - 5x) (y - 5x)(26y -130x - 52y) 26y +130x
2 2
zxy = = = - ,
4 4 3
(y - 5x) (y - 5x) (y - 5x)
52x
-2 2 -3
2 2
zyy = (- 26x(y - 5x) ) = -26x "(- 2)(y - 5x) = .
y
3
(y - 5x)
Dots funkcijas otrs krtas pilnais diferencilis ir
260y 2(26y +130x)dxdy + 52x
2
d z = dx2 - dy2 .
3 3 3
(y - 5x) (y - 5x) (y - 5x)
x2 - y2
7. piemrs. Pierd%2łt, ka funkcija u = ln apmierina viendojumu
xy
#
"3u "3u "3u "3u 1 1 ś#
+ - - = 2ś# - ź# .
"x3 "x2"y "x"y2 "y3 ś# y3 x3 ź#
# #
a
Vispirms prveidosim doto funkciju, izmantojot logaritmu %2łpaa%2łbu ln = ln a - ln b , tad
b
u = ln(x2 - y2)- ln x - ln y .
Noteiksim visus atvasinjumus, kuri ieiet dotaj viendojum, tas ir visus treas krtas
ata7ir%2łgus parcilos atvasinjumus:
"u 2x 1
= - ,
"x x2 - y2 x
"2u 2(x2 - y2)- 2x " 2x - 2x2 - 2y2 1
= + x-2 = + ,
2 2
"x2 x2
(x2 - y2) (x2 - y2)
2
"3u - 4x(x2 - y2) - (- 2x2 - 2y2)" 2(x2 - y2)" 2x
= - 2x-3 =
4
"x3
(x2 - y2)
(x2 - y2)(- 4x(x2 - y2)+ (2x2 + 2y2)" 4x) 2 4x3 +12xy2 2
= - = - ;
4 3
x3 x3
(x2 - y2) (x2 - y2)
"u - 2y 1
= - ,
"y x2 - y2 y
21. nodarb%2łba. 4. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
"2u - 2(x2 - y2)+ 2y "(- 2y) - 2x2 - 2y2 1
= + y-2 = + ,
2 2
"y2 y2
(x2 - y2) (x2 - y2)
2
"3u - 4y(x2 - y2) -(- 2x2 - 2y2)" 2(x2 - y2)"(- 2y)
= - 2y-3 =
4
"y3
(x2 - y2)
(x2 - y2)(- 4y(x2 - y2)-(2x2 + 2y2)" 4y) 2 - 4y3 -12x2 y 2
= - = - ;
4 3
y3 y3
(x2 - y2) (x2 - y2)
2
# ś#
"3u " "2u - 4y(x2 - y2) -(- 2x2 - 2y2)" 2(x2 - y2)"(- 2y)
= ś# ź# = =
ś# 4
"x2"y "y "x2 ź#
# # (x2 - y2)
(x2 - y2)(- 4y(x2 - y2)-(2x2 + 2y2)" 4y) - 4y3 -12x2 y
= = ;
4 3
(x2 - y2) (x2 - y2)
2
# ś#
"3u " "2u - 4x(x2 - y2) -(- 2x2 - 2y2)" 2(x2 - y2)" 2x
= ś# ź# = =
ś# 4
"x"y2 "x "y2 ź#
# # (x2 - y2)
(x2 - y2)(- 4x(x2 - y2)+ (2x2 + 2y2)" 4x) 4x3 +12xy2
= = .
4 3
(x2 - y2) (x2 - y2)
Atrastos atvasinjumus ievietosim dotaj viendojum:
ś# ?
4x3 +12xy2 2 - 4y3 -12x2 y 4x3 +12xy2 # - 4y3 -12x2 y 2
ś# ź#
- + - - - =
3 3 3 3
ś#
x3 y3 ź#
(x2 - y2) (x2 - y2) (x2 - y2) (x2 - y2)
# #
?
# ś#
1 1
ś#
= 2ś# - ź# ,
y3 x3 ź#
# #
#
2 2 1 1 ś#
Iegksim - + = 2ś# - ź# , kas bija jpierda.
x3 y3 ś# y3 x3 ź#
# #
8. piemrs. Dota funkcija z = u3 sin2 v , kur u = x2 + sin y , v = ln(3x + 4y). Atrast
2 2
parcilos atvasinjumus zx un zy .
T k
"z "z
= 3u2 sin2 v , = 2u3 sin vcosv = u3 sin 2v ,
"u "v
21. nodarb%2łba. 5. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
"u "v 3 "u "v 4
= 2x , = , = cos y , = ,
"x "x 3x + 4y "y "y 3x + 4y
"z "z "u "z "v "z "z "u "z "v
tad saskaF ar formulm = + un = + iegksim
"x "u "x "v "x "y "u "y "v "y
"z 3 "z 4
= 3u2 sin2 v " 2x + u3 sin 2v " , = 3u2 sin2 v " cos y + u3 sin 2v " ,
"x 3x + 4y "y 3x + 4y
vai, ievietojot atvasinjumos u = x2 + sin y , v = ln(3x + 4y):
"z 2 3 3
= 3(x2 + sin y) sin2(ln(3x + 4y))" 2x + (x2 + sin y) sin(2ln(3x + 4y))" ,
"x 3x + 4y
"z 2 3 4
= 3(x2 + sin y) sin2(ln(3x + 4y))"cos y + (x2 + sin y) sin(2ln(3x + 4y))" .
"y 3x + 4y
9. piemrs. Dota funkcija z = ln3(x + xy), kur x = tg2 3t , y = 2 - ctg 6t . Noteikt
dz
atvasinjumu .
dt
Noteiksim funkcijas z parcilos atvasinjumus pc x un y:
1
# ś#
3ln2(x + xy) y
"z 1 1 -
ź#
2
= 3ln2(x + xy)" "#1+ (xy) " yś# = "ś#1+ ,
ś# ź#
ś# ź#
"x 2
x + xy # # x + xy 2 x
# #
1
3ln2(x + xy)
"z 1 1 -
x
2
= 3ln2(x + xy)" " (xy) " x = " .
"y 2
x + xy x + xy 2 y
Noteiksim funkciju x un y atvasinjumus:
dx 1 6sin 3t dy 1 6
#
= 2tg 3t " " 3 = , = 0 - ś#- "6ś# = .
ź#
dt cos2 3t cos3 3t dt sin2 6t sin2 6t
# #
dz "z dx "z dy
Pc formulas = + iegksim
dt "x dt "y dt
# ś#
3ln2(x + xy) y
dz
ź#" 6sin 3t + 3ln2(x + xy) x " 6
= "ś#1+ " =
ś# ź#
dt cos3 3t sin2 6t
x + xy 2 x x + xy 2 y
# #
21. nodarb%2łba. 6. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
ś#
ś#
18ln2(x + xy)#ś#1+ y
ś## ź#
ź#" sin 3t + x
=
ś# ź#
ś## 2 x # cos3 3t 2 y sin2 6t ź#
x + xy
# #
vai, ievietojot aaj atvasinjum x = tg2 3t , y = 2 - ctg 6t :
ś#
ś#
18ln2(tg2 3t + tg 3t 2 - ctg6t)#ś#1+ 2 - ctg 6t
dz sin 3t tg 3t
ś## ź#.
ź#
= " +
ś# ź#
ś## 2tg 3t # cos3 3t 2sin2 6t 2 - ctg 6t ź#
dt
tg2 3t + tg 3t 2 - ctg 6t
# #
10. piemrs. Dota funkcija z = arctg x2 + 2y , kur y = xln5x . Noteikt funkcijas z pilno
atvasinjumu pc x.
Vispirms noteiksim funkcijas z parcilos atvasinjumus pc x un y:
1
"z 1 1 - x
2
= " (x2 + 2y) " 2x = ,
2
"x 2
(1+ x2 + 2y) x2 + 2y
1+( x2 + 2y)
1
"z 1 1 - 1
2
= " (x2 + 2y) " 2 = .
2
"y 2
(1+ x2 + 2y) x2 + 2y
1+( x2 + 2y)
Noteiksim funkcijas y = xln5x atvasinjumu:
dy 1
2
y = = ln5x + x " "5 = ln5x +1.
dx 5x
dz "z "z dy
Ievietojot aos atvasinjumus formul = + , iegksim dots funkcijas pilno
dx "x "y dx
atvasinjumu pc x:
dz x 1 x + ln 5x +1
= + "(ln 5x +1) =
dx
(1+ x2 + 2y) x2 + 2y (1+ x2 + 2y) x2 + 2y (1+ x2 + 2y) x2 + 2y
jeb, ievietojot y = xln5x :
dz x + ln5x +1
= .
dx
(1+ x2 + 2xln5x) x2 + 2xln5x
dy dy
11. piemrs. Noteikt , ja xex +1- ey - ey-1 = 0 . Apr7int .
dx dx
x=1
21. nodarb%2łba. 7. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
Apz%2łmsim F(x, y) = xex +1 - ey - ey-1 . Viegli prliecinties, ka punkt x = 1, y = 1 no
dots viend%2łbas iegkstam identitti: e +1- e -1 = 0 . Turklt funkcijai F(x, y) eksist
"F
neprtraukti parcilie atvasinjumi pc x un pc y un `" 0 punkt (1,1), tad punkta
"y
(1,1) apkrtn eksist viena vien%2łga funkcija y = y(x), pie tam y(1)=1. T k
y-1
2 2
Fx = ex + xex , Fy = -e - e ,
2
dy Fx(x, y) iegkstam
pc formulas = -
2
dx Fy(x, y)
dy ex + xex (x +1)ex
= - = .
dx - e - ey-1 e + ey-1
No kurienes seko
dy (x +1)ex 2e
= = , jo y(1)=1.
dx e + ey-1 x=1 e +1
x=1
2
"z "z "z
12. piemrs. Noteikt un , ja xyzex yz +1 - ey - yey-1 = 0 . Apr7int ,
x=1
"x "y "x
y=1
"z
.
x=1
"y
y=1
2
Mksu gad%2łjum F(x, y, z) = xyzex yz +1 - ey - yey-1 . Viegli prliecinties, ka punkt
x =1, y =1, z =1 no dots viend%2łbas iegkstam identitti: e +1- e -1 = 0 . Turklt
funkcija F(x, y, z) ir diferencjama visos punktos (x, y, z), jo funkcijai F(x, y, z) eksist
"F
neprtraukti parcilie atvasinjumi un F(1,1,1)= 0 , bet `" 0 punkt (1,1,1) . Ttad
"z
visi nosac%2łjumi ir spk un punkta (1,1,1) apkrtn eksist viena vien%2łga funkcija
z = z(x, y), pie tam z(1,1)=1. Noteiksim funkcijas F(x, y, z) parcilos atvasinjumus:
2 2 2
2
Fx = yzex yz + xyzex yz " 2xyz = yzex yz(1+ 2x2 yz),
2 2 2
y-1 y-1
2
Fy = xzex yz + xyzex yz " x2 z - e - e - yey-1 = xzex yz(1+ x2 yz)- e - e (1+ y),
2 2 2
2
Fz = xyex yz + xyzex yz " x2 y = xyex yz(1+ x2 yz).
21. nodarb%2łba. 8. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
2
Fy
2
"z Fx "z
Izmantojot formulas = - , = - , iegkstam:
2 2
"x Fz "y Fz
2
"z yzex yz(1+ 2x2 yz) z(1+ 2x2 yz),
= - = -
2
"x x(1+ x2 yz)
xyex yz(1+ x2 yz)
2
y-1
"z xzex yz(1+ x2 yz)- e - e (1+ y)
= -
2
"y
xyex yz(1+ x2 yz),
no kurienes seko
"z 1"(1+ 2) 3
= - = - , jo z(1,1)=1,
x=1
"x 1"(1+1) 2
y=1
"z e(1+1)- e -1"(1 +1) e - 2
= - = - .
x=1
"y e(1+1) 2e
y=1
21. nodarb%2łba. 9. lpp. Augstk matemtika.
I. Volodko
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