Is f(x,y) Differentiable at (0,0)?

[Barioth]

Hi,

I'm having trouble évaluation the differentiability at (0,0) of the function

\(\displaystyle f(x,y)=\frac{x^3}{x^2+y^2}\) for (x,y) not nul, and \(\displaystyle f(x,y)=0\) if (x,y)=0

I know f is differentiable if (x,y) isn't nul since the partial derivative are continuous, but I don't know how to evaluate it at (0,0)

Thanks for passing by

 

[I like Serena]

Hi,

I'm having trouble évaluation the differentiability at (0,0) of the function

\(\displaystyle f(x,y)=\frac{x^3}{x^2+y^2}\) for (x,y) not nul, and \(\displaystyle f(x,y)=0\) if (x,y)=0

I know f is differentiable if (x,y) isn't nul since the partial derivative are continuous, but I don't know how to evaluate it at (0,0)

Thanks for passing by

Let's take the partial derivative with respect to y:
$$\frac{\partial}{\partial y}f(x,y) = - \frac{2x^3 y}{(x^2+y^2)^2}$$

Is it the same in all directions when approaching (0,0)?
You may want to switch to polar coordinates.

 

[Barioth]

$$\frac{\partial}{\partial y}f(x,y) = - \frac{2x^3 y}{(x^2+y^2)^2}$$

tacking y=m*x we end up with

\(\displaystyle \frac{-2*m*x^4}{(m^2+1)^2*x^4}= \frac{-2*m}{(m²+1)²}\)

This telling me that the limite when (x,y)->(0,0) isn't contiunous.

There for the partial derivative doesn't exist at 0. I can conclude that f(x,y) has no derivative at (0,0).

Thank you for the help!