Calc III - is this differentiable in all points?

[Telemachus]

Homework Statement

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

So, I thought trying with the partial derivatives.

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}

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}

And here is the deal. As you see, I've calculated by definition that for all points of the form (x_0,-x_0) 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.

 

[lanedance]

how did you get that the derivative at y=-x is zero?

 

[HallsofIvy]

The partial derivative, with respect to y, at point (x_0, -x_0), is given by
\lim_{h\to 0}\frac{f(x_0, -x_0+h) - f(x_0, -x0)}{h}= \lim_{h\to 0}\frac{(x_0-x_0+h)^2sin(\frac{\pi}{x_0- x_0+y})- 0}{h}
= \lim_{h\to 0}\frac{h^2 sin(\frac{\pi}{h})}{h}= \lim_{h\to 0} hsin(\frac{\pi}{h})= 0

And similarly for the partial derivative with respect to x.

Telemachus, what you have shown is that the partial derivatives exist which is NOT the same as saying the function is differentiable. There is a theorem that says a function is differentiable at a point if and only if its partial derivatives are continuous in some neighborhood of that point.

 

[Telemachus]

Yes, you're right. Thanks HallsOfIvy. I've misunderstood this theorem and thought that the inverse of it was truth too, which it isn't. The condition suffices for the differentiability, but it isn't a necessary condition.