Question Number 41028 by Choudharyvishal155@gmail.com last updated on 31/Jul/18
$${if}\:\mathrm{1}/{u}\:=\:\sqrt{\left({x}\mathrm{2}\:+\:{y}\mathrm{2}\:+{z}\mathrm{2}\right)} \\ $$$${then}\:{x}\partial{u}/\partial{x}\:+\:{y}\partial{u}/\partial{y}\:+\:{z}\partial{u}/\partial{z}\:=\:? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 31/Jul/18
$${u}=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\:\:{so}\:\:\frac{\partial{u}}{\partial{x}}=\frac{−\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\frac{−\mathrm{3}}{\mathrm{2}}} ×\mathrm{2}{x} \\ $$$${x}\frac{\partial{u}}{\partial{x}}+{y}\frac{\partial{u}}{\partial{y}}+{z}\frac{\partial{u}}{\partial{z}}=−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\frac{−\mathrm{3}}{\mathrm{2}}} \\ $$$$\:\:=−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}} =−\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}\:=−{u} \\ $$