Ta`rif: Agar f’(X)-xosila funksiya -nuqtada hosilaga EGA bo`lsa, uni f(X) funksiyaning nuqtadagi ikkinchi tartibli hosilasi

22-AMALIY MASHG`ULOT MAVZU:FUNKSIYANING YUQORI TARTIBLI HOSILALARI VA DIFFERENSIALLARI
f(x) funksiya nuqtaning biror atrofida aniqlangan bo`lib, shu atrof nuqtalarida f’(x)-hosilaga ega bo`lsin. Ta`rif:Agar f’(x)-xosila funksiya -nuqtada hosilaga ega bo`lsa, uni f(x) funksiyaning nuqtadagi ikkinchi tartibli hosilasi deyiladi va
y yoki f’’(x) yoki
kabi belgilanadi.
1-izoh. 3,4,5,...-taritbli hosilalari o`xshash ta`riflanadi.
1-eslatma. f(x) funksiyaning (n-1) –tartibli (x)-hosilasi nuqtaning biror atrofida mavjud bo`lib, bu f^n-1(x)-hosila funksiya x₀ -nuqtada hosilaga ega bo`lsa, uni f(x) funksiyaning nuqtadagi n- tartibli hosilasi deyiladi va
( ]’
kabi belgilanadi.
1-misol. y=ln(sinx) funksiyani uchinchi tartibli hosilasi topilsin.
Yechilishi: y’=[ln(sinx)]’= (sinx)’= =ctgx y’’=(ctgx)’= , y’’’ kabi bo`ladi. 2-eslatma. 1) y= bo`lsa, y’= lna, y’’= lna)<sup>2</sup>, ... , y<sup>(n)</sup>= lna)<sup>n</sup> kabi bo`ladi.
2) y= bo`lsa, y’=( = <sup>x</sup>)’=e<sup>x</sup>, ... , y<sup>(n)</sup>=( = kabi bo`ladi
3) y=sinx bo`lsa, y’=cosx=sin(x+ ), y’=-sinx= sin(x+ ), ... , y<sup>(n)</sup>= sin(x+ ) kabi bo`ladi.
4) y=cosx bo`lsa, y’=-sinx=cos(x+ ), y’=-cosx=cos(x+2 , ... , y<sup>(n)</sup>= cos(x+ ) kabi bo`ladi.
5) y=lnx, y’= y’’= - y⁽ⁿ⁾= kabi bo`ladi.
6) y= bo`lsa , y’= - , y’’= , ... , y⁽ⁿ⁾= .
7) y=x^m bo`lsa , y<sup>(n)</sup>=m (m-1) ...[m-n+1] x<sup>m-n</sup> kabi bo`ladi
8) y=(1+x)^α bo`lsa , y<sup>(n)</sup>= α (α -1) ... [α -n+1] (1+x) <sup>α-n</sup> kabi bo`ladi.
2-misol. y=f(x)=sinαx funksiyaning y⁽ⁿ⁾-hosilasi topilsin. Yechilishi: y’=αcosαx, y”=α(cosαx)=α(-αsinαx)=-a²sinαx ,... ,y⁽ⁿ⁾=aⁿsin(αx+n kabi bo`ladi. 3-misol. y= funksiyaning y<sup>(n)</sup>-hosilasi topilsin. Yechish:Berilgan funksiyani y= kabi yoziladi. Demak, y<sup>(n)</sup>= ]<sup>(n)</sup> = ( <sup>(n)</sup> – <sup>(n)</sup> = [(x-2)<sup>-1</sup>]<sup>(n )</sup>- [(x-1)<sup>-1</sup>]<sup>(n)</sup>= =(-1) (-1-1) ... [-1-(n-1)] (x-2)<sup>-1-n</sup>- [(-1) (-1-1) ... {-1-(n-1)} (x-1)<sup>-1-n</sup>]= = (x-2)<sup>-n-1</sup> – - (x-1)<sup>-n-1</sup>]=(-1)<sup>n</sup> n! [(x-2)<sup>-n-1</sup>- (x-1)<sup>-n-1</sup>]= =(-1)<sup>n</sup> n! . Demak , y<sup>(n)</sup>= ]<sup>(n)</sup>= (-1)<sup>n</sup> n! kabi bo`ladi. 4 –misol. funksiyaning y’’’-hosilasi topilsin. Yechilishi: y’=( x³ + + y”= [ (e^2x )’( )+(e^2x ) ( )’=e^2x 2( + +e^2x (6x² +6x)=e^2x[4x³+6x²+6x²+6x]=e^2x(4x³+12x²+6x) , y’’’= (e^2x)’(4x³+12x²+6x)+ e^2x(4x³+12x²+6x)’= e^2x 2(4x³+12x²+6x)+ +e^2x(12x²+24x+6)=e^2x[8x³+24x²+12x+12x²+24x+6]=e^2x[8x³+36x²+36x+6] kabi bo`ladi.
Endi funksiyaning yuqori tartibli differensialini topishga misollar qaraymiz. 3-eslatma: (n-1)-tartibli differensialdan olingan differensialga n-tartibli differensial deyiladi, ya`ni