If $f(x)$ has a vertical asymptote, does $f'(x)$ have one too?

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.

Not quite. Suppose $y = f(x)$ has a vertical asymptote at $x=a$ in the sense that $\lim_{x \to a\pm} f(x) = +\infty$ or $-\infty$. Then $\lim_{x \to a\pm} f'(x)$ can't be a finite number. It could be $+\infty$ or $-\infty$, but it also might not exist at all. In fact, $f'$ might not exist at all.

In the other direction, $f'$ can have a vertical asymptote without $f$ having one. For example, consider $f(x) = x^{2/3}$, $f'(x) = \dfrac{2}{3} x^{-1/3}$, with $a=0$.