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

Question

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

Aicharya · Accepted Answer

Given that, $ f(x, y)=2(x-y)^2-x^4-y^4, 	ext { we get } $ On differentiating partially w.r.t. $x$, $ f_x=4(x-y)-4 x^3 $ Again differentiating partially, we get $ \begin{aligned} & & f_{x x} & =4-12 x^2 \ \Rightarrow & & \left(f_{x x}ight)_{(0,0)} & =4-0=4 \ 	ext { Similarly } & & f_{y y} & =4-12 y^2 \ \Rightarrow & & \left(f_{y y}ight)_{(0,0)} & =4-0=4 \ 	ext { and } & & f_{x y} & =-4+0 \ \Rightarrow & & \left(f_{x y}ight)_{(0,0)} & =-4 \ & 	herefore & \left(f_{x x} f_{y y}-f_{x y}^2ight)_{(0,0)} & =4(4)-(-4)^2=0 \end{aligned} $

$\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…

Solution