Combination Theorem for Complex Derivatives/Product Rule/Proof 1

Theorem
Let $D$ be an open subset of the set of complex numbers.

Let $f, g: D \to \C$ be complex-differentiable functions on $D$

Let $f g$ denote the pointwise product of the functions $f$ and $g$.

Then $f g$ is complex-differentiable in $D$, and its derivative $\paren {f g}'$ is defined by:
 * $\map {\paren {f g}'} z = \map {f'} z \map g z + \map f z \map {g'} z$

for all $z \in D$.

Proof
Define $k: D \to \C$ as the pointwise product of $f$ and $g$, so $k = fg$.

Let $z_0 \in D$ be a point in $D$.