# 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 $g: \operatorname{Im} \left({f}\right) \to \C$ be a complex-differentiable function, where $\operatorname{Im} \left({f}\right)$ 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:

$\forall z \in D: h' \left({z}\right) = f' \left({g \left({z}\right)}\right) g' \left({z}\right)$

## Proof

Put $y = g \left({z}\right)$.

Let $\delta z \in \C \setminus \left\{ {0} \right\}$.

Put $\delta y = g \left({z + \delta z}\right) - y$, so:

$g \left({z + \delta z}\right) = y + \delta y$

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

$(1): \quad \delta y \to 0$ by definition of continuity, as $g$ is continuous.
$(2): \quad \dfrac {\delta y} {\delta z} \to g'\left({z}\right)$ by definition of complex-differentiability.

There are two cases to consider:

### Case 1

Suppose $g' \left({z}\right) \ne 0$.

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

Then:

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

hence the result.

$\Box$

### Case 2

Now suppose $g' \left({z}\right) = 0$.

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

### Case 2a

If $\delta y = 0$, then:

$\dfrac {h \left({z + \delta z}\right) - h \left({z}\right)} {\delta z} = 0 = f' \left({y}\right) g' \left({z}\right)$

Hence the result.

$\Box$

### Case 2b

If $\delta y \ne 0$, then:

$\dfrac {h \left({z + \delta z}\right) - h \left({z}\right)} {\delta z} = \dfrac {f \left({y + \delta y}\right) - f \left({y}\right)} {\delta y} \dfrac {\delta y} {\delta z}$

As $\delta y \to 0$:

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

Thus:

$\displaystyle \lim_{\delta z \mathop \to 0} \frac {h \left({z + \delta z}\right) - h \left({z}\right)} {\delta z} \to 0 = f' \left({y}\right) g' \left({z}\right)$

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).