Apparently smooth function, discontinuous derivative This question is about the existence of discontinuous derivatives, but it doesn't provide much examples except this one and $y = |x|$. 
The function
$$f(x) = \left\{ \begin{array}{lr}
\cos(ax) & 0 \leq x \leq c\\
\cos(ac) e^{-b (x - c)} & x > c
\end{array}
\right.$$
has $[0, +\infty)$ as its domain. $a, b, c \in \mathbb (0,+\infty)$ are real constants. Its first derivative is
$$f'(x) = \left\{ \begin{array}{lr}
-a \sin(ax) & 0 \leq x \leq c\\
- b \cos(ac) e^{-b (x - c)} & x > c
\end{array}
\right.$$
$f'(x)$ is continuous only when $a/b = \cot(ac)$ and not in general.
$f(x)$ is a continuous function, without a vertical tangent, infinitely differentiable in its domain, and $a,b$ can be chosen such that the function doesn't have corners in $x = c$; despite this, it has a discontinuous derivative. 
1) How can (even graphically) the derivative be not continuous where the function is so?
2) Are there any other similar examples that can be done?
 A: The first thing to know is that a derivative still has the intermediate value property. This means that it cannot have what some analysts call discontinuities "of the first kind." That is, a derivative cannot have a jump discontinuity or removable discontinuity. One type of discontinuity that they can have is called an essential discontinuity. Using $\sin 1/x$, you can create a function which is differentiable, but whose derivative oscillates too much to be continuous. I wrote the following last year:

Consider the function $g\colon[-1,1]\to \Bbb R$ where
  $$g(x)=\begin{cases}
 x^k\sin\frac{1}{x},& \text{ if }x\neq 0,\\
 0,&\text{ if }x=0,
\end{cases}$$
  for $1<k<2$. Then, away from the origin, we can differentiate $g$ using elementary methods. At the origin, we have
  \begin{align*}
 \lim_{h\to 0}\left(\frac{h^k\sin \frac{1}{h}}{h}\right)&=\lim_{h\to 0}\left(h^{k-1}\sin\frac{1}{h}\right)\\
 &=0.
\end{align*}
  Thus
  $$g'(x)=\begin{cases}
 kx^{k-1}\sin\frac{1}{x}-x^{k-2}\cos\frac{1}{x},& \text{ if }x\neq 0,\\
 0,&\text{ if }x=0.
\end{cases}$$
  On the domain $[-1,1]$, $g'$ is defined. However, while the term $kx^{k-1}\sin\frac{1}{x}$ is bounded, $x^{k-2}\cos\frac{1}{x}$ is not. Thus the derivative of $g$ is not Riemann integrable, despite the fact that $g$ is differentiable everywhere.

Volterra famously constructed a highly discontinuous derivative. Pompeiu has an interesting example too. The details are ugly, but here is a sketch:

First, note that the function $x^3$ has derivative $0$ at the origin even though it is strictly increasing. We want to take advantage of this "kink" and make a differentiable function which has one of these kinks at every point on a dense set. We can enumerate the rationals with $q_i$ and write a function like $$\sum_{i=1}^\infty 2^{-i}(x-q_i)^3.$$
  This is almost the function we want, but there is a problem because every time we add a "copy" of $x^3$, we "ruin" the derivative vanishing for another term. What we actually want to do is the opposite:
  $$f(x)=\sum_{i=1}^\infty 2^{-i}(x-q_i)^{1/3}.$$
  We can show that this converges to a differentiable function, and the derivative is $\infty$ on a dense set. Now let $h=f^{-1}$. We can show that $h$ is differentiable, strictly increasing (since $f$ is strictly increasing), and the derivative is $0$ wherever the derivative of $f$ was $\infty$. Now we have an interesting contradiction: $h$ is strictly increasing, so its derivative cannot be zero, but $h'$ attains $0$ on a dense set. This means that $h'$ is discontinuous whenever $h'(x)>0$, which must happen at infinitely many points.


TL;DR: there are several examples, but none of them are easy to graph since they necessarily violate the intermediate value property.

As an aside, for this reason, some people reserve the word "smooth" for infinitely differentiable functions. This captures most of the "normal" functions we encounter like exponential, polynomial, and trig functions, but excludes some of the nastier pathological functions.
A: Disclaimer: This is not exactly an answer to the two questions that you asked, but rather an explanation of what may have caused your confusion.
There two incorrect statements in your question, which may be part of the reason you're confused. First of all, the correct derivative of the function in your example is
$$f'(x) = \left\{\begin{array}{lr} -a\sin(ax) & 0\le x\color{red}{<}c \\ -b\cos(ac)e^{-b(x-c)} & x>c \end{array}\right.$$
Placing "$\le$" instead of the red "$<$" was wrong precisely because in general the derivative of this function at $c$ doesn't exist (except in the special case when $a/b=\cot(ac)$, as you observed).
Because of that, your statement that 

$f(x)$ is a continuous function, without a vertical tangent, $\color{red}{\text{infinitely differentiable}}$ in its domain

is false, since in fact this function isn't even differentiable once at $c$ (again, in general).
There's a bit of ambiguity in these discussions of "discontinuous" derivatives. The example that you linked to and $y=|x|$ are two very different things!


*

*The derivative of $y=|x|$ doesn't exist at $x=0$; and your example is similar to it, since the derivative of your function doesn't exist at $x=c$ either.

*But the linked example, as well as the great answer from @ElliotG, address a much more interesting, but a different question: functions for which the derivative exists everywhere but is discontinuous.
