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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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 45 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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)