I am trying to study the function $f(x,y) = (x^2 + y^2)\sin(\frac{1}{x^2 + y^2})$
for $(x,y) \neq (0,0)$ and $f(x,y) = 0$ otherwise.
I've been asked to show that the function is differentiable in $R^2$ but the partial derivatives at $(0,0)$ do not exist.
What I know, what I tried:
For a function from $R^2 \rightarrow R$, if the partial derivatives exist in the neighbourhood of $(x_0,y_0)$, and are continuous at $(x_0,y_0)$, then the function is differentiable at $(x_0,y_0)$
However, the partial derivatives at $(0,0)$ do not even exist here, so what's going on?
What is the sufficient condition to ensure the differentiability of a multivariate function?
P.S. If partial derivatives at a point exist, and if they are also continuous at that point, can we say that the function is differentiable? Is the converse valid, that is, is it necessary for the partial derivatives to be continuous if a function is differentiable at a point? (From this example, clearly not so)
Best Answer
At $(0,0)$ both partial derivatives exist and are $0$. Also $f$ is differentiable at the origin. This follows from the fact $\frac {|f(x,y)|} {\sqrt {x^{2}+y^{2}}} \leq \sqrt {x^{2}+y^{2}}$. However the partial derivatives are not continuous at the origin. To see this take limit through $y=0$ of the partial derivative w.r.t. $x$.