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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 =

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(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 $(0,0)$: $(f_{xx}f_{yy} - f_{xy}^2)_{(0,0)} = (4)(4) - (-4)^2 = 16 - 16 = 0$.

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

Similar Questions

If $u = e^{-x^2 - y^2}$,then

If $z = \sec^{-1}\left(\frac{x^4+y^4-8x^2y^2}{x^2+y^2}\right)$, then $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}$ is equal to

If ${x^x}{y^y}{z^z} = c$,then ${{\partial z} \over {\partial x}} = $

If $u = x^2 \tan^{-1}(\frac{y}{x}) - y^2 \tan^{-1}(\frac{x}{y})$,then $\frac{\partial^2 u}{\partial x \partial y} = $

If $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$ then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module