TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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Linearity, product, reciprocal, and quotient rules for complex derivatives

Statement

Let $f, g : U \to C$ be complex differentiable at $a \in U$ , and let $α, β \in C$ . Then

$(α f + β g)^{'} (a) = α f^{'} (a) + β g^{'} (a),$

$(f g)^{'} (a) = f^{'} (a) g (a) + f (a) g^{'} (a) .$

If $g (a) \neq 0$ , then $g$ is nonzero on some neighbourhood of $a$ , the reciprocal $1 / g$ is complex differentiable at $a$ , and

$(1 / g)^{'} (a) = - \frac{g^{'} (a)}{g (a)^{2}}, (f / g)^{'} (a) = \frac{f^{'} (a) g (a) - f (a) g^{'} (a)}{g (a)^{2}} .$

Every constant function has derivative $0$ , and the identity function has derivative $1$ .

Facts & Assumptions

Given: An open set $U \subseteq C$ , a point $a \in U$ , functions $f, g : U \to C$ complex differentiable at $a$ , and scalars $α, β \in C$ .

[F1]

Complex differentiability at $a$ is existence of the difference-quotient limit at $a$ (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

[L1]

A complex-differentiable function is continuous at the point of differentiability (Complex differentiability at a point implies continuity there).

[L2]

Complex modulus is definite, multiplicative, and subadditive (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct

1.1

The difference quotients of a constant function and of the identity are respectively $0$ and $1$ , so their derivatives have those values.

F1algebra

1.2

Taking the finite linear combination of the two difference quotients gives $(α f + β g)^{'} (a) = α f^{'} (a) + β g^{'} (a)$ .

givenF1algebra

1.3

For nonzero allowed $h$ , $\frac{f (a + h) g (a + h) - f (a) g (a)}{h} = f (a + h) \frac{g (a + h) - g (a)}{h} + g (a) \frac{f (a + h) - f (a)}{h} .$

algebra

1.4

Suppose $g (a) \neq 0$ . Continuity [L1] supplies a neighbourhood of $a$ on which $∣ g (z) - g (a) ∣ < ∣ g (a) ∣$ , and on this neighbourhood $g (z) \neq 0$ by [L2].

givenL1L2

2.1

By [L1], $f (a + h) \to f (a)$ , while the two quotients in step 1.3 tend to $g^{'} (a)$ and $f^{'} (a)$ ; hence the product formula follows.

step 1.3givenF1L1algebra

2.2

For nonzero allowed $h$ in that neighbourhood, $\frac{g (a + h)^{- 1} - g (a)^{- 1}}{h} = - \frac{1}{g (a + h) g (a)} \frac{g (a + h) - g (a)}{h} .$

step 1.4algebra

3.1

The reciprocal factor in step 2.2 tends to $1 / g (a)^{2}$ , so the reciprocal derivative is $- g^{'} (a) / g (a)^{2}$ . Applying the product rule to $f (1 / g)$ and simplifying gives the quotient formula.

step 2.1step 2.2givenF1L2algebra∎

Depends on

Used by

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Sources

J. Lebl, Guide to Cultivating Complex Analysis, Proposition 2.2.4 (standard reference, not scraped)
J. Orloff, MIT 18.04 Topic 2, §2.6.1 (standard reference, not scraped)