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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 )
letf(x,y)=ln(x2+y2)calculate2fx2(x,y)+2fy2
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
wehavef(x,y)=12ln(x2+y2)fx(x,y)=xx2+y2x(fx(x,y))=x2+y22x2(x2+y2)22fx2(x,y)=x2+y2(x2+y2)2fissymetricso2fx2(x,y)=y2+x2(x2+y2)2fx2(x,y)+2fy2(x,y)=0

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