Why is the derevative of meromorphic function is meromorphic?

Question

I know that if $f$ is meromorphic then $\exists A\subset \Omega$ s.t $f$ is holomorphic on $\Omega \setminus A$ and $A$ is discrete, and $A$ are the poles of $f$. I want to show that $f'$ is meromorphic, but this seems trivial for $\forall x\in \Omega\setminus A$ we know that $f$ is analytic on a region of $x$, so $f$ is analytic there and thus $f'$ is holomorphic in $x$. this implies $f'$ holomorphic on $\Omega\setminus A$,hence meromorphic. Am I missing something? This seems too obvious.

Answer 1

Your argument is incomplete. You have to show that $f'$ doesn't have an essential singularity at points of $A$. Let $c \in A$. We can write $(z-c)^{n}f(z)=g(z)$ for some non-negative integer $n$ and some holomorphic function $g$ in a neighborhood of $c$. We then have $(z-c)^{n}f'(z)+n(z-c)^{n-1}f(z)=g'(z)$ from which it is easy to see that $c$ is indeed a pole of $f'$.

Answer 2

Hint
Expand the function as Laurent series at the poles. You know the rest part is holomorphic, and expanding it as Laurent series gives the meromorphicity of the poles in the original function.