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 $D$.

\(\ds \map {k'} {z_0}\)

\(=\)

\(\ds \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h\)

\(\ds \)

\(=\)

\(\ds \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} \, \map g {z_0 + h} - \map f {z_0} \, \map g {z_0} } h\)

\(\ds \)

\(=\)

\(\ds \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} \, \map g {z_0 + h} - \map f {z_0 + h} \, \map g {z_0} + \map f {z_0 + h} \, \map g {z_0} - \map f {z_0} \, \map g {z_0} } h\)

\(\ds \)

\(=\)

\(\ds \lim_{h \mathop \to 0} \paren {\map f {z_0 + h} \frac {\map g {z_0 + h} - \map g {z_0} } h + \frac {\map f {z_0 + h} - \map f {z_0} } h \, \map g {z_0} }\)

\(\ds \)

\(=\)

\(\ds \map f {z_0} \, \map {g'} {z_0} + \map {f'} {z_0} \, \map g {z_0}\)

\(\ds \leadsto \ \ \)

\(\ds \forall z \in D: \, \)

\(\ds \map {k'} z\)

\(=\)

\(\ds \map f z \, \map {g'} z + \map {f'} z \, \map g z\)

Definition of Derivative of Complex Function

$\blacksquare$