Derivative of Complex Composite Function

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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


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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:

\(\ds \lim_{\delta z \mathop \to 0} \frac {\map h {z + \delta z} - \map h z} {\delta z}\) \(=\) \(\ds \lim_{\delta z \mathop \to 0} \frac {\map f {\map g {z + \delta z} } - \map f {\map g z} } {\map g {z + \delta z} - \map g z} \frac {\map g {z + \delta z} - \map g z} {\delta z}\)
\(\ds \) \(=\) \(\ds \lim_{\delta z \mathop \to 0} \frac {\map f {y + \delta y} - \map f y} {\delta y} \frac {\delta y} {\delta z}\)
\(\ds \) \(=\) \(\ds \map {f'} y \map {g'} z\) As $\delta y \to 0$, then $\dfrac {\map f {y + \delta y} - \map f y} {\delta y} \to \map {f'} y$

hence the result.

$\Box$


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.

$\Box$


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.

$\Box$


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

$\blacksquare$


Also known as

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


Also see


Sources

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