Analytic function over an open disk This confusion arises while I am studying complex analysis. 

Let $f$ be analytic on an open disk, would $f'$ also be analytic on the same disk? 

My guess is that this is the case. Since analytic implies infinitely differentiable, then if $f$ is infinitely differentiable over every point in the disk, so is $f'$.  Could someone let me know whether this is a valid argument?
 A: 
Let $f$ be analytic on an open disk, would $f'$ also be analytic on the same disk? 

That is correct, but you should justify the last step in your argument

$$
 \text{$f$ analytic} \implies  \text{$f$ infinitely differentiable} \\
\implies \text{$f'$ infinitely differentiable} \implies \text{$f'$ analytic} \, .
$$

“Analytic” means that the function is locally given by a convergent power series. “Analytic” implies “infinitely differentiable” but a-priori not the other way around. 
It is however correct that if a function is holomorphic in an open set $D$, i.e. complex differentiable in every point of $D$ then the same is true for the derivative, and the function is equal to its Taylor series at every point $D$, i.e. it is analytic in $D$. This is a non-trivial fact about holomorphic functions, and a consequence of Cauchy's integral formula.
With this knowledge one can argue that
$$
 \text{$f$ analytic in $D$} \implies  \text{$f$ holomorphic in $D$} \\
\implies \text{$f'$ holomorphic in $D$} \implies \text{$f'$ analytic in $D$}
$$
Note that this is a stronger statement: If $f$ is complex differentiable in every point of $D$ then $f$ and all its derivatives are analytic in $D$.
An alternative is to show that a convergent power series can be differentiated term-by-term in its disk of convergence, Then
$$
\text{$f$ analytic} \implies \text{$f'$ analytic}
$$
is an immediate consequence.  
