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).
Also see
Sources
- 2001: Christian Berg: Kompleks funktionsteori: $\S 1.1$