Second derivative of f(x,y)

Question

I try do finde the Matrix $D^3f(a)$ where $f(x,y)=x^4-3x^3y^2$

What i tried is

$Df(x,y)=\left( \begin{array}{c} 4x^3-9x^2y^2 && -6x^3y\\ \end{array} \right)$

This is a $1\times 2 $ matrix. Then following this post I get:

$ D^2f(x,y)= \left( \begin{array}{c} 12x^2-18xy^2 && -18x^2y && -18x^2y && -6x^3\\ \end{array} \right) $

and

$ D^3f(x,y)= \left( \begin{array}{c} 24x-18y^2 && -36xy^2 && -36xy && -18x^2 && 36xy && -36x^2 && -18x^2 && -18x^3\\\\ \end{array} \right) $

which is a $8\times1$ matrix. I have troubles by understanding how to get $D^rf(x,y)$ for arbitrary functions

Answer 1

About your last comment: you can observe that $f(x,y) = x^3(x-3y^2)$. So in the half plane $x \geq 0$ you have that $f \geq 0$ if $x \geq 3y^2$ or $x=0$, otherwise $f < 0$. If $x < 0$, then $f$ is a sum of a non-negative and a positive term, so $f(x,y) > 0$. So $(0,0)$ is not a local extremum.

Answer 2

What you are looking for is called the Hessian matrix. You can read more on wikipedia here. Basically it allows you to have a matrix in which all of the variables of your multivariable function are partially derived. The Hessian for a 2 variable function with be: $$\pmatrix { \frac { \partial^2f }{\partial x^2 }& \frac {\partial^2f}{\partial x \partial y} \\ \frac {\partial^2f}{\partial y \partial x } &\frac { \partial^2f }{\partial y^2} }$$