- #1

Telemachus

- 835

- 30

## Homework Statement

Hi there. I got next problem, and I must say if it is differentiable in all of its domain.

[tex]f(x,y)=\begin{Bmatrix} (x+y)^2\sin(\displaystyle\frac{\pi}{x+y}) & \mbox{ if }& y\neq{-x}\\0 & \mbox{if}& y=-x\end{matrix}[/tex]

So, I thought trying with the partial derivatives.

[tex]f_x=\begin{Bmatrix} 2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{x+y}) & \mbox{ if }& y\neq{-x}\\0 & \mbox{if}& y=-x\end{matrix}[/tex]

[tex]f_y=\begin{Bmatrix} 2(x+y)\sin(\displaystyle\frac{\pi}{x+y})-\pi\cos(\displaystyle\frac{\pi}{x+y}) & \mbox{ if }& y\neq{-x}\\0 & \mbox{if}& y=-x\end{matrix}[/tex]

And here is the deal. As you see, I've calculated by definition that for all points of the form [tex](x_0,-x_0)[/tex] the derivative is zero. But the limit of the derivative tending to that kind of points doesn't exists because of the cosine. Is there something wrong with this? I mean is there any kind of contradiction with this, or its alright and just the derivative isn't continuous at that kind of points?

Bye and thanks.

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