TITLE: If $f(x)$ has a vertical asymptote, does $f'(x)$ have one too?
QUESTION [26 upvotes]: So here is what I understand:

If $f(x)$ is increasing/decreasing, then its derivative $f'(x)$ is positive/negative

and...

If $f(x)$ is increasing/decreasing, then the derivative of $f'(x)$ (which is $f''(x)$) is concave up/concave down

So my question is: if a graph has a vertical asymptote, the derivative must also have a vertical asymptote, too, right? Does it also work vice versa? I feel like there is a trick to it, but I'm not sure.
I have a graph from GeoGebra here. The dotted line is the derivative.

REPLY [50 votes]: if a graph has vertical asymptote, the derivative must also have a vertical asymptote too, right? 

No. A counterexample: $$f(x)=\frac{1}{x}+\sin\left(\frac{1}{x}\right)$$ This function is monotone and has a vertical asymptote at $x=0$. But its derivative has no limit.