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Question Number 36173 by prof Abdo imad last updated on 30/May/18
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 ))
$${calculate}\:\frac{\partial{f}}{\partial{x}}\:{and}\:\frac{\partial{f}}{\partial{y}}\:{in}\:{this}\:{cases} \\ $$$$\left.\mathrm{1}\right)\:{f}\left({x},{y}\right)=\:{e}^{−{x}} \:{sin}\left(\mathrm{2}{y}\:+\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right){f}\left({x},{y}\right)\:=\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right){e}^{−{xy}} \\ $$$$\left.\mathrm{3}\right){f}\left({x},{y}\right)\:=\:\frac{{x}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 13/Apr/19
1) we have f(x,y) =e^(−x) sin(2y+1) ⇒(∂f/∂x)(x,y) =−e^(−x) sin(2y+1) and (∂f/∂y)(x,y) =2 e^(−x) cos(2y+1) 2)we have f(x,y) =(x^2 +y^2 )e^(−xy) ⇒(∂f/∂x)(x,y) =2x e^(−xy) −y(x^2 +y^2 )e^(−xy) =(2x−yx^2 −y^3 )e^(−xy) and (∂f/∂y)(x,y) = 2y e^(−xy) −x (x^2 +y^2 )e^(−xy) =(2y−x^3 −xy^2 )e^(−xy) 3) we have f(x,y) =(x/(x^2 +y^2 )) ⇒(∂f/∂x)(x,y) =((x^2 +y^2 −x(2x))/((x^2 +y^2 )^2 )) =((y^2 −x^2 )/((x^2 +y^2 )^2 )) and (∂f/∂y)(x,y) =x((−2y)/((x^2 +y^2 )^2 )) =((−2xy)/((x^2 +y^2 )^2 )) .
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x},{y}\right)\:={e}^{−{x}} {sin}\left(\mathrm{2}{y}+\mathrm{1}\right)\:\Rightarrow\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\:=−{e}^{−{x}} {sin}\left(\mathrm{2}{y}+\mathrm{1}\right)\:{and} \\ $$$$\frac{\partial{f}}{\partial{y}}\left({x},{y}\right)\:=\mathrm{2}\:{e}^{−{x}} {cos}\left(\mathrm{2}{y}+\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:{f}\left({x},{y}\right)\:=\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right){e}^{−{xy}} \:\Rightarrow\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\:=\mathrm{2}{x}\:{e}^{−{xy}} \:−{y}\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right){e}^{−{xy}} \\ $$$$=\left(\mathrm{2}{x}−{yx}^{\mathrm{2}} \:−{y}^{\mathrm{3}} \right){e}^{−{xy}} \:\:\:{and} \\ $$$$\frac{\partial{f}}{\partial{y}}\left({x},{y}\right)\:=\:\mathrm{2}{y}\:{e}^{−{xy}} \:−{x}\:\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right){e}^{−{xy}} \:=\left(\mathrm{2}{y}−{x}^{\mathrm{3}} \:−{xy}^{\mathrm{2}} \right){e}^{−{xy}} \\ $$$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left({x},{y}\right)\:=\frac{{x}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }\:\Rightarrow\frac{\partial{f}}{\partial{x}}\left({x},{y}\right)\:=\frac{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} −{x}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${and}\:\frac{\partial{f}}{\partial{y}}\left({x},{y}\right)\:={x}\frac{−\mathrm{2}{y}}{\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{−\mathrm{2}{xy}}{\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right)^{\mathrm{2}} }\:. \\ $$

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