13.4-bo'lim : Yuqori tartibli xususiy hosilalar
2. Quyidagi funksiya uchun Klerot teoremasini tasdiqlang.
\[A\left( {x,y} \right) = \cos \left( {\frac{x}{y}} \right) - {x^7}{y^4} + {y^{10}}\]Yechim
Birinchidan, bizga ikkita 1 kerak bo'lishini bilamizst xususiy hosilalarni buyurtma qiling. Mana ular,
\[{A_x} = - \frac{1}{y}\sin \left( {\frac{x}{y}} \right) - 7{x^6}{y^4}\hspace{0.5in}{A_y} = \frac{x}{{{y^2}}}\sin \left( {\frac{x}{y}} \right) - 4{x^7}{y^3} + 10{y^9}\]2-qadam
Endi aralash ikkinchi tartibli xususiy hosilalarning har birini hisoblaylik.
\[\begin{align*}{A_{x\,y}} & = {\left( {{A_x}} \right)_y} = \frac{1}{{{y^2}}}\sin \left( {\frac{x}{y}} \right) + \frac{x}{{{y^3}}}\cos \left( {\frac{x}{y}} \right) - 28{x^6}{y^3}\\ {A_{y\,x}} & = {\left( {{A_y}} \right)_x} = \frac{1}{{{y^2}}}\sin \left( {\frac{x}{y}} \right) + \frac{x}{{{y^3}}}\cos \left( {\frac{x}{y}} \right) - 28{x^6}{y^3}\end{align*}\]OK, biz \({A_{x\,y}} = {A_{y\,x}}\) va shunga o'xshash Clairaut teoremasi ushbu funktsiya uchun tasdiqlanganligini ko'rishimiz mumkin.