13.4-bo'lim : Yuqori tartibli xususiy hosilalar
7. \(f\left( {x,y,z} \right) = {x^4}{y^3}{z^6}\) berilgan \(\displaystyle \frac{{{\partial ^6}f}}{{\partial y\partial {z^2}\partial y\partial {x^2}}}\) toping.
Yechim
Kleraut teoremasining tabiiy kengayishi orqali biz bu xususiy hosilalarni istalgan tartibda bajarishimiz mumkinligini bilamiz. Biroq, bu holatda muammo bayonotida ko'rsatilgan tartibdan boshqa hech qanday sabab yo'qdek tuyuladi.
Bu erda biz olishimiz kerak bo'lgan birinchi hosila.
\[\frac{{\partial f}}{{\partial x}} = 4{x^3}{y^3}{z^6}\]2-qadam
Ikkinchi hosila,
\[\frac{{{\partial ^2}f}}{{\partial {x^2}}} = 12{x^2}{y^3}{z^6}\]3-qadam
Uchinchi hosila,
\[\frac{{{\partial ^3}f}}{{\partial y\partial {x^2}}} = 36{x^2}{y^2}{z^6}\]4-qadam
To'rtinchi hosila,
\[\frac{{{\partial ^4}f}}{{\partial z\partial y\partial {x^2}}} = 216{x^2}{y^2}{z^5}\]5-qadam
Beshinchi hosila,
\[\frac{{{\partial ^5}f}}{{\partial {z^2}\partial y\partial {x^2}}} = 1080{x^2}{y^2}{z^4}\]6-qadam
Ushbu muammo uchun bizga kerak bo'lgan oltinchi va oxirgi hosila:
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{{\partial ^6}f}}{{\partial y\partial {z^2}\partial y\partial {x^2}}} = 2160{x^2}y{z^4}}}\]