Question

Найти дифференциал функции указанного порядка $d^{3}u=? u=frac{x}{y}$

Answer 1

$u=frac yx\ du=frac{partial^3u}{partial x^3}+3frac{partial^3u}{partial x^2partial y}+3frac{partial^3u}{partial xpartial y^2}+frac{partial^3u}{partial y^3}\ frac{partial u}{partial x}=-frac y{x^2}\ frac{partial u}{partial y}=frac1x\ frac{partial^2u}{partial x^2}=frac{2y}{x^3}\ frac{partial^2u}{partial y^2}=0\ frac{partial^2u}{partial xpartial y}=-frac1{x^2}\ frac{partial^3u}{partial x^3}=-frac{6y}{x^4}\ frac{partial^3u}{partial x^2partial y}=frac2{x^3}$

$frac{partial^3u}{partial xpartial y^2}=0\ frac{partial^3u}{partial y^2}=0\d^3u=-frac{6y}{x^4}+3cdotfrac2{x^3}+3cdot0+0=frac6{x^3}left(1-frac yxright)$