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Question Number 36179 by prof Abdo imad last updated on 30/May/18

let f(x,y) = ((xy)/(x+y)) 1) find D_f 2)calcule x(∂f/∂x)(x,y) +y (∂f/∂y)(x,y) interms of f(x,y)

$${let}\:{f}\left({x},{y}\right)\:=\:\frac{{xy}}{{x}+{y}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right){calcule}\:{x}\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\:+{y}\:\frac{\partial{f}}{\partial{y}}\left({x},{y}\right)\:{interms}\:{of}\:{f}\left({x},{y}\right) \\ $$

Commented by maxmathsup by imad last updated on 19/Aug/18

1) D_f ={(x,y)∈R^2 / x+y ≠0} 2) we have (∂f/∂x)(x,y) =((y(x+y)−xy(1))/((x+y)^2 )) =(y^2 /((x+y)^2 )) and (∂f/∂y)(x,y) =((x(x+y) −xy(1))/((x+y)^2 )) =(x^2 /((x+y)^2 )) ⇒ x (∂f/∂x)(x,y) +y(∂f/∂y)(x,y) = ((xy^2 )/((x+y)^2 )) +((yx^2 )/((x+y)^2 )) =((xy(x+y))/((x+y)^2 )) =((xy)/(x+y)) .

$$\left.\mathrm{1}\right)\:{D}_{{f}} =\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:/\:{x}+{y}\:\neq\mathrm{0}\right\} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\:=\frac{{y}\left({x}+{y}\right)−{xy}\left(\mathrm{1}\right)}{\left({x}+{y}\right)^{\mathrm{2}} }\:=\frac{{y}^{\mathrm{2}} }{\left({x}+{y}\right)^{\mathrm{2}} }\:\:{and} \\ $$$$\frac{\partial{f}}{\partial{y}}\left({x},{y}\right)\:=\frac{{x}\left({x}+{y}\right)\:−{xy}\left(\mathrm{1}\right)}{\left({x}+{y}\right)^{\mathrm{2}} }\:=\frac{{x}^{\mathrm{2}} }{\left({x}+{y}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${x}\:\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\:+{y}\frac{\partial{f}}{\partial{y}}\left({x},{y}\right)\:=\:\frac{{xy}^{\mathrm{2}} }{\left({x}+{y}\right)^{\mathrm{2}} }\:+\frac{{yx}^{\mathrm{2}} }{\left({x}+{y}\right)^{\mathrm{2}} }\:=\frac{{xy}\left({x}+{y}\right)}{\left({x}+{y}\right)^{\mathrm{2}} }\:=\frac{{xy}}{{x}+{y}}\:. \\ $$

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