20- mavzu: Funksiyaning differensiali

Quyidagi misollarni keltiramiz:

• Quyidagi misollarni keltiramiz:
• 1-misol. y=a0xn+a1xn-1+…+an-1x+an bo‘lsa,
• y=na0xn-1+(n-1)anxn-2+…+an-1 ,
• - - - - - - - - - - - - - - - - - - -
• y(n)=n.(n–1).….2*1* a0=a0 n! ,
• y(n+1)=y(n+2)=…=0 .
• Demak, n – darajali ko‘phadning n – tartibli hosilasi o‘zgarmas son bo‘lib, (n+1)- tartibli hosilasidan boshlab yuqori tartibli hosilalarining barchasi nolga teng bo‘lar ekan.

2-misol. f(x)=ekx , k – o‘zgarmas (k0).

• 2-misol. f(x)=ekx , k – o‘zgarmas (k0).
• f(x)=ekx(kx) =kekx;
• f (x)=(f(x)) =(kekx) =k(ekx)
• k*kekx=k2ekx
• va hokazo,
• f(n)(x)=knekx
• ni olamiz. Demak,
• (ekx)(n)= knekx, nN

3-misol. f(x)=sinx.

• 3-misol. f(x)=sinx.
• f(x)=cosx=sin(x+ ),
• f(x)=(f(x)) =(sin(x+ ))
• =cos(x+ )*1=sin(x+),
• - - - - - - - - - -- - - - - - - - - - - - -
• f(n)(x)=sin(x+n* ),
• ya’ni
• (sinx)(n)=sin(x+n* ), nN

4-misol. f(x)=cosx.

• 4-misol. f(x)=cosx.
• Yuqoridagiga o‘xshash,
• (cos x)(n)=cos(x+n* ), nN
• ni olish mumkin.
• 5-misol. f(x)=U*V, bu yerda U va V lar ixtiyoriy tartibli hosilalari mavjud funksiyalardir.
• (U*V) =UV+UV
• (UV) =(UV+UV) =UV+UV+UV+UV=UV+ 2UV+UV
• va hokazo.
• ni olish mumkin. Bu Leybnis formulasi deb yuritiladi. Bu yerda nolinchi tartibli hosila funksiyaning o‘zi ekanligini eslash lozim.

Endi, yuqori tartibli differensial tushunchasini kiritamiz. Buning uchun funksiya differensialini uning birinchi tartibli differensiali argument orttirmasini o‘zgarmas deb qabul qilgan holda (n–1) – tartibli differensialning differensialini n-tartibli differensial deb ataymiz va uning uchun dny , dnf(x) kabi belgilashlarni qo‘llaymiz.

• Endi, yuqori tartibli differensial tushunchasini kiritamiz. Buning uchun funksiya differensialini uning birinchi tartibli differensiali argument orttirmasini o‘zgarmas deb qabul qilgan holda (n–1) – tartibli differensialning differensialini n-tartibli differensial deb ataymiz va uning uchun dny , dnf(x) kabi belgilashlarni qo‘llaymiz.
• Demak, ta’rif bo‘yicha dny=d(dn-1y) ekan. Oxirgi formula asosida
• d2y=d(dy)=d[f (x)dx]=(f (x)dx)dx=f (x)dx2
• va hokazo,
• dny=f(n)(x)dxn
• formulani olamiz.