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Question Number 36178 by prof Abdo imad last updated on 30/May/18

let f(x,y)=ln((√(x^2  +y^2 )))   calculate (∂^2 f/∂x^2 )(x,y)+(∂^2 f/∂y^2 )

$${let}\:{f}\left({x},{y}\right)={ln}\left(\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }\right)\: \\ $$$${calculate}\:\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }\left({x},{y}\right)+\frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} } \\ $$

Commented by abdo imad last updated on 31/May/18

we have f(x,y) =(1/2)ln(x^2  +y^2 ) ⇒  (∂f/∂x)(x,y) = (x/(x^2  +y^2 )) ⇒ (∂/∂x)((∂f/∂x)(x,y))=((x^2  +y^2  −2x^2 )/((x^2  +y^2 )^2 ))  ⇒ (∂^2 f/∂x^2 )(x,y) = ((−x^2  +y^2 )/((x^2  +y^2 )^2 ))  f is symetric so  (∂^2 f/∂x^2 )(x,y) =((−y^2  +x^2 )/((x^2  +y^2 ))) ⇒(∂^2 f/∂x^2 )(x,y) +(∂^2 f/∂y^2 )(x,y) =0

$${we}\:{have}\:{f}\left({x},{y}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\:=\:\frac{{x}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }\:\Rightarrow\:\frac{\partial}{\partial{x}}\left(\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\right)=\frac{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:−\mathrm{2}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }\left({x},{y}\right)\:=\:\frac{−{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{f}\:{is}\:{symetric}\:{so} \\ $$$$\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }\left({x},{y}\right)\:=\frac{−{y}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)}\:\Rightarrow\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }\left({x},{y}\right)\:+\frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} }\left({x},{y}\right)\:=\mathrm{0} \\ $$

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