Consider a complex-valued function $f(\sigma+it)$ which satisfies $$f(\sigma+it)\sim g(\sigma+it),$$ as $|t|\to\infty.$ Must it be the case that $f'(\sigma+it)\sim g'(\sigma+it)$?
Consider, for example, $$f(\sigma+it)\sim\left(\frac{t}{2\pi}\right)^{1-c-it}e^{it+i\frac{\pi}{4}};$$ this is actually a result due to Stirling's approximation for the $\chi$ factor of the Riemann zeta functional equation! But if one wants to find an asymptotic for $f'(\sigma+it)$, are they permitted to just differentiate this above asymptotic for $f(\sigma+it)$?
Of course if this were an equality things would be much more straightforward and we could differentiate if we were so inclined, but since we have an error term with an asymptotic it's not obvious to me that this is legitimate.
On the other hand, if we have two functions being asymptotic to each other, they have the same growth rate asymptotically so it is natural to expect their derivatives to be asymptotically equivalent?