# Homework Help: Calc III - is this differentiable in all points?

1. Sep 21, 2010

### Telemachus

1. The problem statement, all variables and given/known data
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.

Last edited: Sep 21, 2010
2. Sep 22, 2010

### lanedance

Re: Differentiability

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

3. Sep 22, 2010

### HallsofIvy

Re: Differentiability

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 continous in some neighborhood of that point.

4. Sep 22, 2010

### Telemachus

Re: Differentiability

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.