List all points in $\mathbb{R}^2$ at which $f$ is differentiable as well as ALL points in $\mathbb{R}^2$ where $f$ is not differentiable (implied by the first list) when
\begin{equation} f(x) = \left\{ \begin{array}{lr} |x|^2(\sin(\pi|x|))^2 & \text{ if } x \in \mathbb{Q}^2\\ 0 & \text{otherwise} \end{array} \right. \end{equation}
Solution So I know this function is only continuous at those places where the function takes on the value $0$ for rational values of $x$, because for all the irrationals it's $0$. Then I think, where is the function $0$ for rational $x$? That's at $x=0$ and where $\sin(\pi |x|)^2=0$, i.e. for all integer multiples of $\pi$. Therefore where the euclidean length $|x|$ is an integer. This gives a bunch of concentric circles where the function is continuous.
Now I just need to write this set of points and prove it is differentiable at these points (I know it can't be differentiable anywhere else because it needs to be continuous in order to be differentiable). Can anyone help me prove that the function is differentiable at these circles where it is continuous? Thanks!
Edit: To prove $f$ is differentiable where it is continuous, that means showing either showing there exists a linear map $A$ such that $\displaystyle\lim_{h\to 0} \frac{|f(x+h)-f(x)-Ah|}{|h|}=0$, or else showing all of the partial derivatives exist and are continuous... My prof gave a hint that $0\leq \sin(t)\leq t$ for all $t\in[0,\pi/2]$.
Still haven't figured this one out... Thank you for your help!
I'd suggest changing coordinates to $x=r\cos \theta, y = r\sin \theta$. This will make it easier to find the partial derivatives which are tangent and perpendicular to the concentric circles. Thus:
\begin{equation} f(r) = \left\{ \begin{array}{lr} r^2(\sin(\pi r))^2 & \text{ if } r^2 \in \mathbb{Q}\\ 0 & \text{otherwise} \end{array} \right. \end{equation}
Which is a function of $r$ only. Since any change in $h$ in the $\theta$ has no effect on $f$, we can take $h$ to be a scalar: $$f'(n)=\lim_{h\to 0} \frac{f(n+h)-f(n)-Ah}{h}$$ And use the fact that: $$\sin(n\pi+h)-\sin(h)=\sin(h) (-1)^{n}$$