Derivative of Complex Composite Function

Theorem
Let $f: D \to \C$ be a complex-differentiable function, where $D \subseteq \C$ is an open set.

Let $g: \Img f \to \C$ be a complex-differentiable function, where $\Img f$ denotes the image of $f$.

Define $h = f \circ g: D \to C$ as the composite of $f$ and $g$.

Then $h$ is complex-differentiable, and its derivative is defined as:


 * $\forall z \in D: \map {h'} z = \map {f'} {\map g z} \map {g'} z$

Proof
Put $y = \map g z$.

Let $\delta z \in \C \setminus \set 0$.

Put $\delta y = \map g {z + \delta z} - y$, so:


 * $\map g {z + \delta z} = y + \delta y$

As $\delta z \to 0$, we have:


 * $(1): \quad \delta y \to 0$ by definition of continuity, as $g$ is continuous.


 * $(2): \quad \dfrac {\delta y} {\delta z} \to \map {g'} z$ by definition of complex-differentiability.

There are two cases to consider:

Case 1
Suppose $\map {g'} z \ne 0$.

When $\delta z \to 0$, we have $\delta y \ne 0$ from $(2)$, if $\cmod {\delta z}$ is sufficiently small.

Then:

hence the result.

Case 2
Now suppose $\map {g'} z = 0$.

When $\delta z \to 0$, there are two possibilities:

Case 2a
If $\delta y = 0$, then:
 * $\dfrac {\map h {z + \delta z} - \map h z} {\delta z} = 0 = \map {f'} y \map {g'} z$

Hence the result.

Case 2b
If $\delta y \ne 0$, then:
 * $\dfrac {\map h {z + \delta z} - \map h z} {\delta z} = \dfrac {\map f {y + \delta y} - \map f y} {\delta y} \dfrac {\delta y} {\delta z}$

As $\delta y \to 0$:


 * $(1): \quad \dfrac {\map f {y + \delta y} - \map f y} {\delta y} \to \map {f'} y$
 * $(2): \quad \dfrac {\delta y} {\delta z} \to 0$

Thus:
 * $\ds \lim_{\delta z \mathop \to 0} \frac {\map h {z + \delta z} - \map h z} {\delta z} \to 0 = \map {f'} y \map {g'} z$

Again, hence the result.

All cases have been covered, so by Proof by Cases, the result is complete.

Also known as
This is often informally referred to as the chain rule (for differentiation).

Also see

 * Derivative of Composite Function.