calculate (∂f/∂x) and (∂f/∂y) in this cases 1) f(x,y)= e^(−x) sin(2y +1) 2)f(x,y) =(x^2 +y^2 )e^(−xy) 3)f(x,y) = (x/(x^2 +y^2 ))


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Question Number 36173 by prof Abdo imad last updated on 30/May/18

$c a l c u l a t e \frac{\partial f}{\partial x} a n d \frac{\partial f}{\partial y} i n t h i s c a s e s$ $1) f (x, y) = e^{- x} s i n (2 y + 1)$ $2) f (x, y) = (x^{2} + y^{2}) e^{- x y}$ $3) f (x, y) = \frac{x}{x^{2} + y^{2}}$
Commented by maxmathsup by imad last updated on 13/Apr/19

$1) w e h a v e f (x, y) = e^{- x} s i n (2 y + 1) \Rightarrow \frac{\partial f}{\partial x} (x, y) = - e^{- x} s i n (2 y + 1) a n d$ $\frac{\partial f}{\partial y} (x, y) = 2 e^{- x} c o s (2 y + 1)$ $2) w e h a v e f (x, y) = (x^{2} + y^{2}) e^{- x y} \Rightarrow \frac{\partial f}{\partial x} (x, y) = 2 x e^{- x y} - y (x^{2} + y^{2}) e^{- x y}$ $= (2 x - {y x}^{2} - y^{3}) e^{- x y} a n d$ $\frac{\partial f}{\partial y} (x, y) = 2 y e^{- x y} - x (x^{2} + y^{2}) e^{- x y} = (2 y - x^{3} - {x y}^{2}) e^{- x y}$ $3) w e h a v e f (x, y) = \frac{x}{x^{2} + y^{2}} \Rightarrow \frac{\partial f}{\partial x} (x, y) = \frac{x^{2} + y^{2} - x (2 x)}{{(x^{2} + y^{2})}^{2}} = \frac{y^{2} - x^{2}}{{(x^{2} + y^{2})}^{2}}$ $a n d \frac{\partial f}{\partial y} (x, y) = x \frac{- 2 y}{{(x^{2} + y^{2})}^{2}} = \frac{- 2 x y}{{(x^{2} + y^{2})}^{2}} .$
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