I'm working on a proof of the product rule for derivatives using Taylor Series, and the following questions have come up in the process. I'm suspecting that the answer to each of these questions is yes, but I have not idea how to prove such a thing.
Setup:
Let $I\subseteq \Bbb R$ be some interval. Let $f:\Bbb R \mapsto \Bbb R$ be continuous on $I$.
Question 1:
Does the continuity of $f(x)$ on $I$ imply the continuity of $f^{(n)}(x)$ on $I$, where $n\in N=\Bbb N\cup \{0\}$?
Question 2:
If $g:\Bbb R \mapsto \Bbb R$ is also continuous on $I$, is $u(x)=f(x)g(x)$ also continuous on $I$?
Could you give proofs?
Edit: I've gotten the second one figured out from the limit of a product is the product of the limits. Still need help with the proof of the first one.
Edit 2.0: The first one is FALSE! (dun dun dun) See the comment thread for a good counterexample by Randall.
In order to answer the first Question i'll elaborate Randalls (see comments) chosen function
$$ f(x)=\begin{cases} x^2\sin(\frac{1}{x}), & x\not= 0 \\ 0, & x=0 \end{cases}$$
$f$ is differentiable at every point. For its derivative $f'$ we get
$$ f'(x)=\begin{cases} 2x\sin(\frac{1}{x})-\cos(\frac{1}{x}), & x\not= 0 \\ 0, & x=0 \end{cases}$$
which is not continous at $x = 0$.
Proof:
In order to prove $f'$ is not continous for $x=0$, we show the following
$$\exists \varepsilon > 0 \forall\delta > 0\ \exists x \in \mathbb{R} : \vert x-0\vert < \delta \wedge \vert f'(x)-f'(0)\vert > \varepsilon$$
If we choose $x = \frac{1}{n\pi}$ we have
$$ \vert \frac{1}{n\pi}\vert < \delta \Leftrightarrow \frac{1}{\delta\pi} < n $$
which leads us to
$$ \vert f'(x) - f'(0) \vert = \left\vert f'\left(\frac{1}{n\pi}\right)\right\vert = \left\vert 2\frac{1}{n\pi}\sin(n\pi)-\cos(n\pi) \right\vert = \vert 0 \pm 1 \vert = 1 > \frac{1}{2}=:\varepsilon$$
Therefore $f'$ is not continous at $x=0$