# Two varibale function. Continuity, derivability and differentiability

1. Aug 22, 2014

### Dansuer

1. The problem statement, all variables and given/known data
Discuss the continuity, derivability and differentiability of the function

$f(x,y) = \frac{x^3}{x^2+y^2}$ if (x,y)≠(0,0) and 0 otherwise

2. Relevant equations
if f is differentiable then $∇f.v=\frac{∂f}{∂v}$
if f has both continuous partial derivative in a neighbourhood of $x_0$ then it's differentiable in $x_0$

3. The attempt at a solution
I have no problem with continuity.
For derivability i consider the definition of directional derivative in an arbitray direction (α,β)
$\frac{f(αt,βt)-f(0,0)}{t}=\frac{α^3t^3-0}{t^3}=α^3$
all the directional derivatives exist.
this equation $∇f.v=\frac{∂f}{∂v}$ tells me that the directional derivative should depend linearly on α and β which is not the case, f is not differentialble in $x_0$

on the other hand it's easy to calculate the partial derivatives in a neighborood of (0,0) and see that they are continous.
$f(x,0)=x \ \ \ f(0,y)=0$
$\frac{∂}{∂x}f(x,0)=1 \ \ \ \frac{∂}{∂y}f(0,y)=0$

i get two different results with two different approaches and i can't figure out what's wrong with one of them, or both :tongue2:

any help is appreciated

2. Aug 22, 2014

The partial derivatives are not continuous. To see this, compute either of them at $p \neq (0,0)$ and take the limit as $p \rightarrow 0$.

If I am not mistaken, what you showed is that the functions $f_x(x,0), f_y(0,y)$ of one variable (the other being fixed to zero) are continuous.

3. Aug 22, 2014

### Dansuer

Oh i see what's the problem. Thanks a lot for the help