R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
15. nodarb%2łba
ln x
1. piemrs. Noteikt atvasinjumu funkcijai y = (x2 + 4) .
ln x
ln y = ln(x2 + 4) ,
ln y = ln x " ln(x2 + 4),
2
y 1 1
= ln(x2 + 4)+ ln x " " 2x ,
y x x2 + 4
# ś# ln x
ln(x2 + 4)+ 2x ln x
2 ś# ź#
y = "(x2 + 4) .
ś# ź#
x x2 + 4
# #
x3 -1
2. piemrs. Noteikt atvasinjumu funkcijai y = (arctg3x) .
x3 -1
ln y = ln(arctg3x) ,
ln y = x3 -1 " ln(arctg3x),
1
2
y 1 - 1 1
2
= (x3 -1) " 3x2 ln(arctg3x)+ x3 -1 " " " 3 ,
2
y 2 arctg3x
1+ (3x)
# ś#
3x2 3 x3 -1
x3 -1
ś# ź#
2
y = ln(arctg3x)+ "(arctg3x) .
ś# ź#
(1+ 9x2)arctg3x
2 x3 -1
# #
2
3. piemrs. Noteikt atvasinjumu funkcijai y = (7x - 9)"10x sin 5x " arccos3 x .
2
ln y = ln((7x - 9)"10x sin 5x " arccos3 x),
ln y = ln(7x - 9)+ ln10x + ln(sin2 5x)+ ln(arccos3 x),
ln y = ln(7x - 9)+ x ln10 + 2ln(sin 5x)+ 3ln(arccos x),
# ś#
2
y 1 1 1
ś#- 1 ź#
= " 7 +1" ln10 + 2 " " cos5x " 5 + 3" " ,
ś#
y 7x - 9 sin 5x arccos x
1- x2 ź#
# #
15. nodarb%2łba. 1. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
# ś#
7 3
2
ś# ź#
2
y = + ln10 +10ctg 5x - "(7x - 9)"10x sin 5x " arccos3 x
ś#
7x - 9
arccos x " 1- x2 ź#
# #
4
(6x3 + 5x) tg2 x
4. piemrs. Noteikt atvasinjumu funkcijai y = .
3
e-x + 8
4
(6x3 + 5x) tg2 x
ln y = ln ,
3
e-x + 8
1
4
3
ln y = ln(6x3 + 5x) + ln tg2 x - ln(e-x + 8) ,
1
ln y = 4ln(6x3 + 5x)+ 2ln tg x - ln(e-x + 8),
3
2
y 1 1 1 1 1
= 4 " "(18x2 + 5)+ 2 " " - " " e-x "(-1),
y 6x3 + 5x tg x cos2 x 3 e-x + 8
4
# ś#
4(18x2 + 5)+ 2 e-x (6x3 + 5x) tg2 x
2 ś#
y = +
ś# ź#
3
6x3 + 5x tg x " cos2 x 3(e-x + 8)ź# " e-x + 8 .
# #
5. piemrs. Funkcija uzdota aizklt veid ar viendojumu x2 + 2xy - y2 = 4 . Atrast
2
funkcijas atvasinjumu y .
Atvasinsim viendojumu x2 + 2xy - y2 = 4 pc x, Femot vr, ka y = y(x):
2 2
2x + 2y + 2xy - 2yy = 0,
2 2
2yy - 2xy = 2(x + y),
2
y (y - x) = x + y ,
x + y
2
y = .
y - x
y
6. piemrs. Funkcija uzdota aizklt veid ar viendojumu x2 - y2 + arcsin = 5.
x
2
Atrast funkcijas atvasinjumu y .
15. nodarb%2łba. 2. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
1
2
1 - 1 y x - y
2
2
(x2 - y2) (2x - 2yy )+ " = 0 ,
2
2 x2
y
# ś#
1- ś# ź#
x
# #
2 2
x - yy x y x - y
+ " = 0 ,
x2 - y2 x2 - y2 x2
2 2
x - yy xy - y
+ = 0 ,
x2 - y2 x x2 - y2
2 2
x2 - xyy + xy - y = 0 ,
2
y (x - xy) = y - x2 ,
y - x2
2
y = .
x - xy
7. piemrs. Funkcija uzdota aizklt veid ar viendojumu e2 x sin3 y -1 = e-3 y cos2 x .
2
Atrast funkcijas atvasinjumu y .
2
2 2
e2 x " 2sin3 y + e2 x " 3sin y " cos y " y - 0 = e-3 y "(- 3y )cos2 x + e-3 y " 2cos x "(- sin x),
2
2 2
3e2 x sin y cos y " y + 3e-3 y cos2 x " y = -2e-3 y sin x cos x - 2e2 x sin3 y ,
2
2
3y (e2 x sin y cos y + e-3 y cos2 x)= -2(e-3 y sin x cos x + e2 x sin3 y),
2(e-3 y sin x cos x + e2 x sin3 y)
2
y = -
2
3(e2 x sin y cos y + e-3 y cos2 x).
2
ż#
x = 3t - 2t
8. piemrs. Noteikt atvasinjumu parametriski dotai funkcijai .
# 2
5
(t3
#y = + 6)
2
xt = 6t - 2 ,
2
2 3 2
# ś# 2 - 6t
2
5 5
2
yt = ś#(t3 + 6) ź# = (t3 + 6) " 3t = ,
ś# ź#
3
5
# #
55 (t3 + 6)
15. nodarb%2łba. 3. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
2 2
2
yt 6t 3t
2
yx = = : (6t - 2) = .
3
2
xt 55 + 6)3 5
(t3 5(3t -1) (t3 + 6)
ż#
x = e2t cos3t
9. piemrs. Noteikt atvasinjumu parametriski dotai funkcijai .
#
#y = e-2t sin 3t
2
xt = e2t " 2cos3t + e2t "(- sin 3t)" 3 = e2t (2cos3t - 3sin 3t),
2
yt = e-2t "(- 2)sin 3t + e-2t cos3t " 3 = e-2t (3cos3t - 2sin 3t),
2
yt e2t (2cos3t - 3sin 3t) e4t (2cos3t - 3sin 3t).
2
yx = = =
2
xt e-2t (3cos3t - 2sin 3t) 3cos3t - 2sin 3t
2
ż#
x = t log2 t
10. piemrs. Noteikt atvasinjumu parametriski dotai funkcijai .
#
y = 25t + 5
#
1 t
2
2
xt = 2t log2 t + t " = 2t log2 t + ,
t ln 2 ln 2
2
yt = 25t ln 2 " 5,
2
yt 5 " 25t ln 2 ln t 5 " 25t ln 2 5 " 25t ln2 2
Ą#log ń#
2
yx = = = t = = = .
2
xt 2t log2 t + t ó# 2 ln 2Ą# 2t ln t + t 2t ln t + t
Ł# Ś#
ln 2 ln 2 ln 2
11. piemrs. Atrast funkcijas y = 5x + x2 pieaugumu "y un diferencili dy punkt
x = 2 , ja "x = 0,01.
Funkcijas y = 5x + x2 pieaugums ir:
"y = y(x + "x) y(x) = 5(x + "x) + (x + "x)2 5x x2 = 5x + 5"x + x2 + + 2x"x + "x2
5x x2 = 5"x + 2x"x + "x2.
Vai ar%2ł "y = (5 + 2x)"x + "x2.
2
Ja x = 2 un "x = 0,01, tad "y = = (5 + 2 " 2)" 0,01+ (0,01) =
x = 2
"x = 0,01
15. nodarb%2łba. 4. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
= (5 + 4)"0,01 + 0,0001 = 0,0901.
Funkcijas diferencili var iegkt divos veidos:
" T k funkcijas diferencilis pc defin%2łcijas ir funkcijas pieauguma galven
liner da<a, tad to var iegkt no funkcijas pieauguma izteiksmes
2
"y = (5 + 2x)"x + ("x) . Galven liner da<a izteiksm ir (5 + 2x)"x , ttad
funkcijas diferencilis ir dy = (5 + 2x)dx . Atgdinsim, ka "x = dx.
2
" Varam izmantot formulu dy = f (x)dx , tad
dy = (5x + x2)2 dx =(5 + 2x)dx .
Ja x = 2 un "x = dx = 0,01, tad dy = (5 + 2"2)"0,01 = 0,09.
x = 2
"x = 0,01
3
12. piemrs. Noteikt funkcijas y = tg 2x +1 diferencili.
Vispirms noteiksim dots funkcijas atvasinjumu:
2
1 1 3tg 2x +1
2
2
y = 3tg 2x +1 " " " 2 = .
cos2 2x +1 2 2x +1 2x +1cos2 2x +1
Mekltais diferencilis ir
2
3tg 2x +1
dy = dx .
2x +1cos2 2x +1
2
(2,027) - 3
13. piemrs. Aptuveni apr7int .
2
(2,027) + 5
x2 - 3
Apskat%2łsim funkciju f (x) = . Apz%2łmsim x0 = 2, "x = 0,027. Td gad%2łjum
x2 + 5
mekltais lielums
2
(2,027) - 3
= f(x0+"x) = f(2 + 0,027) = f(2,027).
2
(2,027) + 5
K zinms, f(x0 + "x) H" f(x0) + df(x0) jeb
2
f (x0 + "x) H" f (x0 )+ f (x0 )"x .
15. nodarb%2łba. 5. lpp. Augstk matemtika.
I. Volodko
R%2łgas Tehnisk universitte. In~eniermatemtikas katedra. Uzdevumu
risinjumu paraugi.
x2 - 3
Atrad%2łsim funkcijas f (x) = atvasinjumu:
x2 + 5
1 2x(x2 + 5)- 2x(x2 - 3)=
2
f (x) = "
2
x2 - 3 (x2 + 5)
2 "
x2 + 5
1 x2 + 5 2x(x2 + 5 - x2 + 3) x2 + 5 8x
= " " = " .
2 2
2 x2 - 3 x2 - 3
(x2 + 5) (x2 + 5)
Ievietojot aaj izteiksm x = x0 = 2, iegksim
22 + 5 8 " 2 9 16 16
2
f (2) = " = " = .
2
22 - 3
(22 + 5)1 81 27
L%2łdz ar to
2
(2,027) - 3 1 16 1 1
= f(2,027) H" + "0,027 = +16 " 0,001 = + 0,016 =
2
9 27 3 3
(2,027) + 5
= 0,333 + 0,016 = 0,349.
2
(2,027) - 3 1,108729
Prec%2łz vrt%2łba ir = = 0,34889.
2
9,108729
(2,027) + 5
15. nodarb%2łba. 6. lpp. Augstk matemtika.
I. Volodko
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