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


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


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*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.


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.

