begin{aligned} & f(x, y)=2(x-y)^2-x^4-y^4 & | vedclass

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(C) Given the function $f(x, y) = 2(x-y)^2 - x^4 - y^4$. First,find the partial derivatives with respect to $x$ and $y$: $f_x = 4(x-y) - 4x^3$ $f_y =

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(C) Given the function $f(x, y) = 2(x-y)^2 - x^4 - y^4$. First,find the partial derivatives with respect to $x$ and $y$: $f_x = 4(x-y) - 4x^3$ $f_y = -4(x-y) - 4y^3$ Now,find the second-order partial derivatives: $f_{xx} = \frac{\partial}{\partial x}(4x - 4y - 4x^3) = 4 - 12x^2$ $f_{yy} = \frac{\partial}{\partial y}(-4x + 4y - 4y^3) = 4 - 12y^2$ $f_{xy} = \frac{\partial}{\partial y}(4x - 4y - 4x^3) = -4$ Evaluating these at the point $(0,0)$: $(f_{xx})_{(0,0)} = 4 - 12(0)^2 = 4$ $(f_{yy})_{(0,0)} = 4 - 12(0)^2 = 4$ $(f_{xy})_{(0,0)} = -4$ Finally,calculate the value of the expression $(f_{xx}f_{yy} - f_{xy}^2)$ at $(0,0)$: $(f_{xx}f_{yy} - f_{xy}^2)_{(0,0)} = (4)(4) - (-4)^2 = 16 - 16 = 0$.

$\begin{aligned} & f(x, y)=2(x-y)^2-x^4-y^4 \\ & \left|\left(f_{x x} f_{y y}-f_{x y}^2\right)\right|_{(0,0)} \end{aligned}$

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