# Determine whether a multivariable function is infinitely differentiable

I am having trouble trying to determine whether the following function is infinitely differentiable at the origin $$f(x,y)=\begin{cases}{xy(x^4-y^4)\over x^4+y^4}\quad &\text{ if } (x,y)\ne (0,0)\\ 0 &\text{ at }(0,0)\end{cases}$$ Clearly all partial deravatives exist, so I think (correct me if I am wrong) to prove that $f$ is infinitely differentiable at the origin, I just need to show that all the partial dervatives (of all orders) are continuous at the origin, so in this case they need to tend to 0 as we move towards the origin. And conversely to show that $f$ is not infinitely differentiable, it's enough to show that there exist a partial dervative that is not continuous at the origin. This is what I got so far, I find it very difficult to show either of these is true. I tried finding the first few dervative manually but the algrebra quickly becomes unmanageable!

• What does "infinitely differentiable" means for a function with two or more variables? – DonAntonio May 21 '17 at 21:31

The function is not infinitely differentiable. The numerator has degree six and the numerator four. So unfortunately you need to take two partial derivatives to get to the same degree (quotients of polynomials of same degree which both vanish at the origin usually do not have a limit). Consider $$\frac{\partial^{2}}{\partial x\partial y}\left( \frac{xy(x^{4}-y^{4})} {x^{4}+y^{4}}\right) =\frac{x^{12}+33x^{8}y^{4}-33x^{4}y^{8}-y^{12}}{\left( x^{4}+y^{4}\right) ^{3}}$$ For $x=0$ you get$$\frac{-y^{12}}{\left( y^{4}\right) ^{3}}=-1\rightarrow-1$$ while for $y=0$ you get$$\frac{x^{12}}{\left( x^{4}\right) ^{3}}=1\rightarrow1.$$ This shows that $\frac{\partial^{2}f}{\partial x\partial y}$ does not have a limit as $(x,y)\to (0,0)$ and so it cannot be continuous.