I am trying to figure out why my function is differentiable and therefore continuous in the point $(0,0)$ which is also a critical point and a saddle point.
Considering my function: $f(x, y) = x^3 – 3xy^2$
The partial derivative of $x$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial x}(x,y) = 3x^2-3y^2 $$
The partial derivative of $y$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial y}(x,y) = -6xy $$
Therefore I get: $$\nabla f(x,y) = \begin{pmatrix} \frac{\partial f }{\partial x}\\ \frac{\partial f }{\partial y} \end{pmatrix} = \begin{pmatrix} 3x^2-3y^2 \\ -6xy \end{pmatrix} $$
How can I show that my function is point-wise differentiable in $(0,0)$?
Best Answer
Since the partial derivatives exist and are continuous at every point, then the function is differentiable at every point.