Product Rule for Complex Derivatives

Theorem

Let $\map f z, \map j z, \map k z$ be single-valued continuous complex functions in a domain $D \subseteq \C$, where $D$ is open.

Let $f$, $j$, and $k$ be complex-differentiable at all points in $D$.

Let $\map f z = \map j z \, \map k z$.

Then:

$\forall z \in D: \map {f'} z = \map j z \, \map {k'} z + \map {j'} z \, \map k z$

Proof

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

\(\displaystyle \map {f'} {z_0}\)

\(=\)

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

\(\displaystyle \)

\(=\)

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

\(\displaystyle \)

\(=\)

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

\(\displaystyle \)

\(=\)

\(\displaystyle \lim_{h \mathop \to 0} \paren {\map j {z_0 + h} \frac {\map k {z_0 + h} - \map k {z_0} } h + \frac {\map j {z_0 + h} - \map j {z_0} } h \, \map k {z_0} }\)

\(\displaystyle \)

\(=\)

\(\displaystyle \map j {z_0} \, \map {k'} {z_0} + \map {j'} {z_0} \, \map k {z_0}\)

\(\displaystyle \leadsto \ \ \)

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

\(\displaystyle \map {f'} z\)

\(=\)

\(\displaystyle \map j z \, \map {k'} z + \map {j'} z \, \map k z\)

Definition of Derivative of Complex Function

$\blacksquare$