Consider the function $$f(x)=\sum _{k=0}^{\infty }\frac{\sin \left(k^2\pi x\right)}{k^2}.$$ It converges uniformly on $\mathbb{R}$ according to the Weierstrass M-test and is continuous on $\mathbb{R}$ due to the uniform limit theorem.
According to my lecture notes, the termwise differentiated series $\sum_{k=1}^\infty \pi \cos (k^2\pi x)$ diverges. It doesn't state for which values of $x$ it diverges, however, using a similar argument as in this answer, I believe it diverges for all $x\in \mathbb{R}$ (the argument goes like this; whenever $\cos (k^2 \pi x)$ is small, then $k^2 x$ will be close to an odd multiple of $1/2$, and therefor $(2k)^2 x$ will be close to an even multiple of $1/2$ and correspondingly larger).
Now, my lecture notes go on to claim that $f$ in fact is differentiable at some rational points with derivative $-1/2$. I was wondering if this is true and if yes, if it's easy to show (on an undergraduate level). Also, I'd be grateful if you could explain why the linked argument fails to show divergence on all of $\mathbb{R}$, if my lecture notes really are correct.
If sequence of functions converges, and sequence of derivatives converges uniformly, then limit of sequence is differentiable (with derivative equal to limit of derivatives). Converse is not true: it's possible that sequence of derivatives doesn't converge at all, but sum of sequence is still differentiable.
Example from "Counterexamples in analysis": consider $f_n(x) = \frac{\sin nx}{\sqrt n}$. Then $f_n$ converges uniformly to $0$ (which is differentiable) on $[0, 1]$, but $f'_n(x) = \sqrt{n}\cos nx$ which doesn't converge for any $x$.
So, it's possible that series of derivatives diverges, while the original series is still differentiable.