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If-V-log-x-2-y-2-then-V-xx-V-yy-

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Question Number 295 by defg last updated on 25/Jan/15
If V=log (x^2 +y^2 ) then V_(xx) +V_(yy) =?
$$\mathrm{If}\:{V}=\mathrm{log}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:\mathrm{then}\:{V}_{{xx}} +{V}_{{yy}} =? \\ $$
Answered by 123456 last updated on 19/Dec/14
V_x =((2x)/(x^2 +y^2 )) V_(xx) =((2(x^2 +y^2 )−2x∙2x)/((x^2 +y^2 )^2 ))=((−2x^2 +2y^2 )/(x^4 +2x^2 y^2 +y^4 )) V_y =((2y)/(x^2 +y^2 )) V_(yy) =((2(x^2 +y^2 )−2y∙2y)/((x^2 +y^2 )^2 ))=((2x^2 −2y^2 )/(x^4 +2x^2 y^2 +y^4 )) V_(xx) +V_(yy) =((−2x^2 +2y^2 )/(x^4 +2x^2 y^2 +y^4 ))+((2x^2 −2y^2 )/(x^4 +2x^2 y^2 +y^4 ))=0
$${V}_{{x}} =\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$${V}_{{xx}} =\frac{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\mathrm{2}{x}\centerdot\mathrm{2}{x}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} } \\ $$$${V}_{{y}} =\frac{\mathrm{2}{y}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$${V}_{{yy}} =\frac{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\mathrm{2}{y}\centerdot\mathrm{2}{y}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} } \\ $$$${V}_{{xx}} +{V}_{{yy}} =\frac{−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} }+\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{4}} }=\mathrm{0} \\ $$

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