# Derivative of Infinite Product of Analytic Functions

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## Theorem

Let $D \subset \C$ be open.

Let $\sequence {f_n}$ be a sequence of analytic functions $f_n: D \to \C$.

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

Then:

- $\ds f' = \sum_{n \mathop = 1}^\infty f_n' \cdot \prod_{\substack {k \mathop = 1 \\ k \mathop \ne n} }^\infty f_k$

and the series converges locally uniformly in $D$.

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## Outline of Proof

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

## Proof

By Infinite Product of Analytic Functions is Analytic, $f$ is analytic.

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

Let $E = D \setminus \set {z \in D: \map f z = 0}$.

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

By Linear Combination of Convergent Series:

- $\ds f' = \sum_{n \mathop = 1}^\infty f_n' \cdot \prod_{\substack {k \mathop = 1 \\ k \mathop \ne n}}^\infty f_k$ on $E$

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

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

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Let $g$ denote its limit.

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

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

$\blacksquare$