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

dⁿy=d(d^n-1y) kabi yoziladi. 5-misol. y=e^4x funksiya uchun d³y-differensial topilsin. Yechish: d³y=(e^4x)’’’dx^3, , (e^4x)’=e^4x (4x)’=4 e^4x, (e^4x)”= (4 e^4x)’=4 e^4x=16e^4x, (e^4x)’’’=(16e^4x)’=16(e^4x)’=16 e^4x 4=16 ^4x=64 ^4x. Demak, d³y=(e^4x)’’’dx³=64e^4xdx³ kabi bo`ladi. 4-eslatma: Differensiallashning sodda qoidalari quyidagicha yoziladi.
a) d<sup>n</sup>(cf(x))=cd<sup>n</sup>f(x) ,
b) d<sup>n</sup>(f(x)+g(x))=d<sup>n</sup>f(x)±d<sup>n</sup>g(x) ,
d) d<sup>n</sup>(f(x) =d<sup>n</sup>f(x) +c<sub>n</sub>d<sup>n-1</sup>f(x) +c<sub>n</sub><sup>2</sup>d<sup>n-2</sup>f(x)d<sup>2</sup>g(x)+ + ... +c<sub>n</sub><sup>k</sup>d<sup>n-k</sup>f(x) d<sup>k</sup>g(x)+ ... +f(x) d<sup>n</sup>g(x). 6-misol: cos4x bo`lsa, d²y-topilsin. Yechilishi: dy=d( + cos4x)= d ) +d(cos4x)= )’dx+(cos4x)’dx= dx+ +(-sin4x) =(2x -4sin4x)dx , d²y=d(dy)=d[(2x -4sin4x)dx]=d [2x - 4sin4x]dx=[d(2x - d(4sin4x)]dx= =[(2x dx –(4sin4x)’dx]dx=[2(x ’dx-4(sin4x)’dx]dx= =[2(1 +x ) - 4 cos4x =(2 +4x² -16cos4x)dx²
7-misol: y=x²e^2x uchun d²y-topilsin. dy=d(x²e^2x))’=e^2xd(x²)+x²d(e^2x)=(x²)’dx e^2x+(e^2x)’dx x²=2xdx e^2x+2e^2x dx x²= =(2xe^2z+2x²e^2x)dx , d²y=d(dy)=d[2xe^2x+2x²e^2x]dx=[d(2xe^2x)+d(2x²e^2x)]dx=[2d(xe^2x)+2d(x²e^2x)]dx= =2[(x e^2x)’dx+(x²e^2x)’dx]dx=2[1 e^2x+x e^2x 2)dx+(2 xe^2x+x² e^2x 2)dx]dx= =2[e^2x+2x e^2x+2x e^2x+2x² e^2x]dx =2e^2x[1+4x+2x²]dx² (bu yerda dx-o`zgarmas ko`paytuvchi deb qaraladi).
Endi murakkab funksiyani yuqori tartibli hosilasiga misol qaraymiz. 8-misol.u=f(x)=e^2x , y=F(u)=u⁴ bo`lsa, y’, y”- lar topilsin. Yechilishi: 1) Ravshanki, y<sub>x</sub><sup>’</sup>=F’(u u<sub>x</sub><sup>’</sup> y<sub>x</sub>’=(u<sup>4</sup>)<sub>u</sub>’ (e<sup>2x</sup> )<sub>x</sub>’=4u<sup>3</sup> e<sup>2x</sup> 2=4(e<sup>2x</sup>)<sup>3</sup> e<sup>2x</sup> 2=8 e<sup>6x</sup> e<sup>2x</sup>=8e<sup>8x</sup>. 2)y<sub>xx</sub>”=F”(f(x)) f’<sup>2</sup>(x)+F’(f(x)) f”(x)=F<sub>uu</sub>”(u) (u<sub>x</sub>’)<sup>2</sup>+F<sub>u</sub>’(u) u<sub>xx</sub>”=12u<sup>2</sup> (2e<sup>2x</sup>)<sup>2</sup>+4u<sup>3</sup> 4e<sup>2x</sup>= =12 (e<sup>2x</sup>)<sup>2</sup> 4 e<sup>4x</sup>+4 (e<sup>2x</sup>)<sup>3</sup> 4 e<sup>2x</sup>=12e<sup>4x</sup> 4 e<sup>4x</sup>+4 e<sup>6x</sup> 4 e<sup>2x</sup>=48 e<sup>8x</sup>+16 e<sup>8x</sup>=64 e<sup>8x</sup> kabi bo`ladi.
Mustaqil yechish uchun uyga vazifa.
1) y=cos²x bo`lsa, y<sub>x</sub>’’’-topilsin. 2) y=x<sup>6</sup>-4x<sup>3</sup>+4 bo`lsa, y^Іv(1)-topilsin. 3)y=arctgx bo`lsa, y”(1)- topilsin. 4) y= bo`lsa, y^v(x)-topilsin. 5) y= bo`lsa, y<sup>v</sup>(x)-topilsin. 6) y=x<sup>3</sup> lnx bo`lsa, y^Іv(x)-topilsin. 7) y=(1+x²)arctgx bo`lsa, y”(x)- topilsin. 8) y=x<sup>x</sup> bo`lsa, y”(x)- topilsin. 9) y=x lnx bo`lsa, y<sub>(x)</sub><sup>(n)</sup>- topilsin. 10) y= bo`lsa, y⁽ⁿ⁾(x)-topilsin. 11) y= bo`lsa d<sup>2</sup>y topilsin. 12) bo`lsa, d²y topilsin. 13) y= sinx bo`lsa d<sup>2</sup>y topilsin. 14) u=f(x)=2<sup>x</sup> , y=F(u)=lnu bo`lsa , y’ , y” – lar topilsin