Question Number 191867 by Spillover last updated on 02/May/23
$${Prove}\:{that}\:{if}\:\:\:{u}={f}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right),{where}\:{f}\:\:{is}\:{arbitry} \\ $$$${function}\:{then}\:\:\:\:{x}^{\mathrm{2}} \:\frac{\partial{u}}{\partial{y}}\:=\:{y}^{\mathrm{2}} \frac{\partial{u}}{\partial{x}} \\ $$
Answered by qaz last updated on 02/May/23
$$\partial{u}={f}\:'\left(\mathrm{3}{x}^{\mathrm{2}} {dx}+\mathrm{3}{y}^{\mathrm{2}} {dy}\right) \\ $$$$\Rightarrow\frac{\partial{u}}{\partial{x}}=\mathrm{3}{x}^{\mathrm{2}} {f}\:'\:\:\:\:\:\:\:\:\:\frac{\partial{u}}{\partial{y}}=\mathrm{3}{y}^{\mathrm{2}} {f}\:' \\ $$$$\Rightarrow{x}^{\mathrm{2}} \frac{\partial{u}}{\partial{y}}={y}^{\mathrm{2}} \frac{\partial{u}}{\partial{x}} \\ $$