Combination Theorem for Complex Derivatives/Product Rule/Proof 2

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
Denote the open ball of $0$ by radius $r \in \R_{>0}$ as $\map {B_r} 0$.

Let $z \in D$.

By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:


 * $\map f {z + h} = \map f z + \map h {\map {f'} z + \map {\epsilon_f} h}$


 * $\map g {z + h} = \map g z + \map h {\map {g'} z + \map {\epsilon_g} h}$

where $\epsilon_f, \epsilon_g: \map {B_r} 0 \setminus \set 0 \to \C$ are complex functions that converge to $0$ as $h$ tends to $0$.

Then:

Define $\epsilon: \map {B_r} 0 \setminus \set 0 \to \C$ by:
 * $\map \epsilon h = h \paren {\map {f'} z + \map {\epsilon_f} h} \paren {\map {g'} z + \map {\epsilon_g} h}$

From Product Rule for Limits of Complex Functions and Combined Sum Rule for Limits of Complex Functions:


 * $\ds \lim_{h \mathop \to 0} \map \epsilon h = \paren {\lim_{h \mathop \to 0} h} \paren {\lim_{h \mathop \to 0} \paren {\map {f'} z + \map {\epsilon_f} h} } \paren {\lim_{h \mathop \to 0} \paren {\map {g'} z + \map {\epsilon_g} h} } = 0$

Then the Epsilon-Function Complex Differentiability Condition shows that:


 * $\map {\paren {f g}'} z = \map {f'} z \map g z + \map f z \map {g'} z$