Combination Theorem for Complex Derivatives/Product Rule

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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 $\left({f g}\right)'$ is defined by:

$\left({f g}\right)' \left({z}\right) = f' \left({z}\right) g \left({z}\right) + f \left({z}\right) g' \left({z}\right)$

for all $z \in D$.


Proof

Denote the open ball of $0$ by radius $r \in \R_{>0}$ as $B_r \left({0}\right)$.

Let $z \in D$.

By the Alternative Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:

$f \left({z + h}\right) = f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right)$
$g \left({z + h}\right) = g \left({z}\right) + h \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$

where $\epsilon_f, \epsilon_g: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ are continuous functions that converge to $0$ as $h$ tends to $0$.

Then:

\(\displaystyle \left({fg}\right) \left({z + h}\right)\) \(=\) \(\displaystyle f \left({z}\right) g \left({z}\right) + h \left({f \left({z}\right) g' \left({z}\right) + f \left({z}\right) \epsilon_g \left({h}\right) }\right) + h \left({g \left({z}\right) f' \left({z}\right) + g \left({z}\right) \epsilon_f \left({h}\right) }\right) + h^2 \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({fg}\right) \left({z}\right) + h \left({f' \left({z}\right) g \left({z}\right) + f \left({z}\right) g' \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right) }\right)\)

Define $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C $ by $\epsilon \left({h}\right) = h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$.

From Product Rule for Continuous Functions and Combined Sum Rule for Continuous Functions, it follows that $\epsilon$ is continuous.

From Product Rule for Limits of Functions and Combined Sum Rule for Limits of Functions, it follows that:

$\displaystyle \lim_{h \to 0} \epsilon \left({h}\right) = \left({\lim_{h \to 0} h}\right) \left({\lim_{h \to 0} \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) }\right) \left({\lim_{h \to 0} \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right) }\right)= 0$

Then the Alternative Differentiability Condition shows that:

$\left({f g}\right)' \left({z}\right) = f' \left({z}\right) g \left({z}\right) + f \left({z}\right) g' \left({z}\right)$

$\blacksquare$


Sources