Derivative of Complex Composite Function

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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:

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


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.


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


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.


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 $\delta y \to 0$:

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


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

Again, hence the result.


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