Partial derivative problem on absolute value function Find the first and second order partial derivatives of 
$f(x,y)=|2x^2-y|$.
I start with limit definition but not able to solve.
please help.
 A: When $2x^2-y>0$, $f(x,y)=2x^2-y$, and$$f'_x=4x, f'_y=-1, f''_{xx}=4,f''_{xy}=f''_{yx}=0,f''_{yy}=0.$$
When $2x^2-y<0$, $f(x,y)=-2x^2+y$, and$$f'_x=-4x, f'_y=1, f''_{xx}=-4,f''_{xy}=f''_{yx}=0,f''_{yy}=0.$$
When $2x^2-y=0$,


*

*$f'_x$ does not exist except at $x=0$ where $f'_x=0$,

*$f'_y$ does not exist,

*$f''_{xx}$ does not exist, 

*$f'_{xy}=0$, 

*$f''_{yx}$ does not exist,

*$f''_{yy}$ does not exist.

A: $$\frac{\delta f}{dx} = 4x, -\sqrt{2}x \lt y\lt \sqrt{2}x$$
$$\frac{\delta f}{dy} = -1, x \lt -\sqrt{\frac{y}{2}}, x\gt \sqrt{\frac{y}{2}}$$
$$\frac{\delta f}{dx}= -4x, y\lt -\sqrt{2}x, y \gt \sqrt{2}x$$
$$\frac{\delta f}{dy}= 1, -\sqrt{\frac{y}{2}} \lt x\lt \sqrt{\frac{y}{2}}$$
$$\frac{\delta^2 f}{dx^2} = 4, -\sqrt{2}x \lt y\lt \sqrt{2}x$$
$$\frac{\delta^2 f}{dx^2} = -4, y\lt -\sqrt{2}x, y \gt \sqrt{2}x$$
$$\frac{\delta^2 f}{dy^2} = 0 -\infty \lt x\lt \infty$$
$$\frac{\delta^2 f}{dxdy} = \frac{\delta^2 f}{dydx} = 0 -\infty \lt x\lt \infty$$
A: We define $g$ as the function inside the absolute value. (i.e: $f(x,y) = |g(x,y)|$)
$$\frac{\partial f(x,y)}{\partial x} = \frac{g(x,y)\frac{\partial g(x,y)}{\partial x}}{|g(x,y)|}$$
Therefore, applying this to our function (i.e: $g(x,y) = 2x^2-y$) we get:
$$\frac{\partial |2x^2-y|}{\partial x} = \frac{(2x^2-y)(4x)}{|2x^2-y|},$$
$$\frac{\partial |2x^2-y|}{\partial y} = \frac{-2x^2+y}{|2x^2-y|},$$
for the first partial derivatives and
$$\frac{\partial\frac{(2x^2-y)(4x)}{|2x^2-y|}}{\partial x} = \frac{|2x^2-y|(24x^2-4y) - \frac{8x^3-4xy}{|2x^2-y|}}{|2x^2-y|^2},$$
$$\frac{\partial \frac{-2x^2+y}{|2x^2-y|}}{\partial y} = \frac{|2x^2-y| - \frac{(-2x^2+y)^2}{|2x^2-y|}}{|2x^2-y|^2},$$
for the second partial derivatives.
Finally, if we apply the definition of absolute value function to our results we get exactly what Statish Ramanathan said.
