13.6-bo'lim : Zanjir qoidasi
10. Quyidagi tenglama uchun \(\displaystyle \frac{{\partial z}}{{\partial x}}\) va \(\displaystyle \frac{{\partial z}}{{\partial y}}\) ni hisoblang.
\[{{\bf{e}}^{z\,y}} + x{z^2} = 6x{y^4}{z^3}\]Yechim
Avval tenglamani tezda qayta yozing.
\[{{\bf{e}}^{z\,y}} + x{z^2} - 6x{y^4}{z^3} = 0\]2-qadam
Oldingi bosqichdagi qayta yozishdan shuni ko'rishimiz mumkinki,
\[F\left( {x,y} \right) = {{\bf{e}}^{z\,y}} + x{z^2} - 6x{y^4}{z^3}\]Endi biz hosilalarni olish uchun eslatmalarda olingan formulalardan oddiygina foydalanishimiz mumkin.
\[\frac{{\partial z}}{{\partial x}} = - \frac{{{F_x}}}{{{F_z}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{{z^2} - 6{y^4}{z^3}}}{{y{{\bf{e}}^{z\,y}} + 2xz - 18x{y^4}{z^2}}} = \frac{{6{y^4}{z^3} - {z^2}}}{{y{{\bf{e}}^{z\,y}} + 2xz - 18x{y^4}{z^2}}}}}\] \[\frac{{\partial z}}{{\partial y}} = - \frac{{{F_y}}}{{{F_z}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{z{{\bf{e}}^{z\,y}} - 24x{y^3}{z^3}}}{{y{{\bf{e}}^{z\,y}} + 2xz - 18x{y^4}{z^2}}} = \frac{{24x{y^3}{z^3} - z{{\bf{e}}^{z\,y}}}}{{y{{\bf{e}}^{z\,y}} + 2xz - 18x{y^4}{z^2}}}}}\]E'tibor bering, har bir javobning ikkinchi shaklida biz kasr oldidagi "-" belgisini suratga ko'chirdik va uni ko'paytirdik. Agar xohlasak, biz buni maxraj yordamida osonlikcha bajarishimiz mumkin edi.