Logarithmic Derivative of Product of Analytic Functions
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Theorem
Let $D \subset \C$ be open.
Let $f, g: D \to \C$ be analytic.
Let $z \in D$ with $f \left({z}\right) \ne 0 \ne g \left({z}\right)$.
Then:
- $\dfrac{\left({f g}\right)' \left({z}\right)} {\left({f g}\right) \left({z}\right)} = \dfrac{f' \left({z}\right)} {f \left({z}\right)} + \dfrac {g' \left({z}\right)} {g \left({z}\right)}$
Proof
Follows directly from Complex Derivative of Product.
$\blacksquare$