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Question-103790




Question Number 103790 by ugwu Kingsley last updated on 17/Jul/20
Commented by ugwu Kingsley last updated on 17/Jul/20
i need help with this asap
$${i}\:{need}\:{help}\:{with}\:{this}\:{asap} \\ $$
Answered by bemath last updated on 17/Jul/20
(a) (∂f/∂x) = e^y −e^x  ; (∂f/∂y) = xe^y   critical point  → { (((∂f/∂x) = 0 ⇒e^y  = e^x )),(((∂f/∂y) = 0 ⇒xe^(y ) =0; x=0 ∧y=0)) :}
$$\left({a}\right)\:\frac{\partial{f}}{\partial{x}}\:=\:{e}^{{y}} −{e}^{{x}} \:;\:\frac{\partial{f}}{\partial{y}}\:=\:{xe}^{{y}} \\ $$$${critical}\:{point} \\ $$$$\rightarrow\begin{cases}{\frac{\partial{f}}{\partial{x}}\:=\:\mathrm{0}\:\Rightarrow{e}^{{y}} \:=\:{e}^{{x}} }\\{\frac{\partial{f}}{\partial{y}}\:=\:\mathrm{0}\:\Rightarrow{xe}^{{y}\:} =\mathrm{0};\:{x}=\mathrm{0}\:\wedge{y}=\mathrm{0}}\end{cases} \\ $$$$ \\ $$
Commented by bemath last updated on 17/Jul/20
(c)(∂^2 f/∂x^2 ) = −e^x  ; (∂^2 f/∂y^2 ) = xe^y      (∂^2 f/(∂x∂y)) = e^y
$$\left({c}\right)\frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }\:=\:−{e}^{{x}} \:;\:\frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} }\:=\:{xe}^{{y}} \\ $$$$\:\:\:\frac{\partial^{\mathrm{2}} {f}}{\partial{x}\partial{y}}\:=\:{e}^{{y}} \: \\ $$

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