13.4-bo'lim : Yuqori tartibli xususiy hosilalar
1. Quyidagi funksiya uchun Klerot teoremasini tasdiqlang.
\[f\left( {x,y} \right) = {x^3}{y^2} - \frac{{4{y^6}}}{{{x^3}}}\]Yechim
Birinchidan, bizga ikkita 1 kerak bo'lishini bilamizst xususiy hosilalarni buyurtma qiling. Mana ular,
\[{f_x} = 3{x^2}{y^2} + 12{x^{ - 4}}{y^6}\hspace{0.75in}{f_y} = 2{x^3}y - 24{x^{ - 3}}{y^5}\]2-qadam
Endi aralash ikkinchi tartibli xususiy hosilalarning har birini hisoblaylik.
\[{f_{x\,y}} = {\left( {{f_x}} \right)_y} = 6{x^2}y + 72{x^{ - 4}}{y^5}\hspace{0.5in}{f_{y\,x}} = {\left( {{f_y}} \right)_x} = 6{x^2}y + 72{x^{ - 4}}{y^5}\]OK, biz \({f_{x\,y}} = {f_{y\,x}}\) va shunga o'xshash Clairaut teoremasi ushbu funktsiya uchun tasdiqlanganligini ko'rishimiz mumkin.