Logarithmic Derivative of Product of Analytic Functions

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

Let $f,g:D\to\C$ be analytic.

Let $z\in D$ with $f(z)\neq0\neq g(z)$.

Then $\displaystyle\frac{(fg)'(z)}{(fg)(z)} = \frac{f'(z)}{f(z)} + \frac{g'(z)}{g(z)}$.

Proof
Follows directly from Complex Derivative of Product.

Also see

 * Product of Analytic Functions is Analytic
 * Complex Derivative of Product
 * Logarithmic Derivative of Infinite Product of Analytic Functions