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)


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

$l e t f (x, y) = \frac{x y}{x + y}$ $1) f i n d D_{f}$ $2) c a l c u l e x \frac{\partial f}{\partial x} (x, y) + y \frac{\partial f}{\partial y} (x, y) i n t e r m s o f f (x, y)$
Commented by maxmathsup by imad last updated on 19/Aug/18
$1) D_f ={(x,y)∈R^2 / x+y ≠0} 2) we have (∂f/∂x)(x,y) =((y(x+y)−xy(1))/((x+y)^2 )) =(y^2 /((x+y)^2 )) and (∂f/∂y)(x,y) =((x(x+y) −xy(1))/((x+y)^2 )) =(x^2 /((x+y)^2 )) ⇒ x (∂f/∂x)(x,y) +y(∂f/∂y)(x,y) = ((xy^2 )/((x+y)^2 )) +((yx^2 )/((x+y)^2 )) =((xy(x+y))/((x+y)^2 )) =((xy)/(x+y)) .$
$1) D_{f} = {(x, y) \in R^{2} / x + y \neq 0}$ $2) w e h a v e \frac{\partial f}{\partial x} (x, y) = \frac{y (x + y) - x y (1)}{{(x + y)}^{2}} = \frac{y^{2}}{{(x + y)}^{2}} a n d$ $\frac{\partial f}{\partial y} (x, y) = \frac{x (x + y) - x y (1)}{{(x + y)}^{2}} = \frac{x^{2}}{{(x + y)}^{2}} \Rightarrow$ $x \frac{\partial f}{\partial x} (x, y) + y \frac{\partial f}{\partial y} (x, y) = \frac{{x y}^{2}}{{(x + y)}^{2}} + \frac{{y x}^{2}}{{(x + y)}^{2}} = \frac{x y (x + y)}{{(x + y)}^{2}} = \frac{x y}{x + y} .$
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