Derivative of Infinite Product of Analytic Functions

Theorem
Let $D\subset\C$ be open.

Let $(f_n)$ be a sequence of analytic functions $f_n:D\to\C$.

Let the product $\displaystyle \prod_{n \mathop = 1}^\infty f_n$ converge locally uniformly to $f$.

Then $\displaystyle f' = \sum_{n \mathop = 1}^\infty f_n'\cdot \prod_{\substack{k\mathop =1\\ k\mathop \neq n}}^\infty f_k$

and the series converges locally uniformly in $D$.

Outline of Proof
Using Logarithmic Derivative of Infinite Product of Analytic Functions we write $\displaystyle \frac{f'}f = \sum_{n \mathop = 1}^\infty \frac{f_n'}{f_n}$, and we multiply this by $f$.

Proof
We may suppose none of the $f_n$ is identically zero on any open subset of $D$.

Let $E = D\setminus\{z\in D : f(z) = 0\}$.

By Logarithmic Derivative of Infinite Product of Analytic Functions, $\displaystyle \frac{f'}f = \sum_{n \mathop = 1}^\infty \frac{f_n'}{f_n}$ locally uniformly in $E$.

By Linear Combination of Convergent Series, $\displaystyle f' = \sum_{n \mathop = 1}^\infty f_n'\cdot \prod_{\substack{k\mathop =1\\ k\mathop \neq n}}^\infty f_k$ on $E$.

By Uniformly Convergent Sequence Multiplied with Function, the series converges locally uniformly in $E$.

By Uniformly Convergent Series on Dense Subset, the series converges locally uniformly in $D$.

Let $g$ denote its limit.

By Uniform Limit of Analytic Functions is Analytic, $g$ is analytic in $D$.

By Uniqueness of Analytic Continuation, $f=g$.

Also see

 * Logarithmic Derivative of Infinite Product of Analytic Functions