I'm trying to prove the chain rule for two complex-differentiable functions, from first principles, meaning a $\delta z$ proof.
We haven't covered Wirtinger derivatives, which keep popping up in various proofs.
I've made a start by making use of the small difference formula:
$$
\lim_{\delta z\to 0} g(z) + \delta z g'(z) = \lim_{\delta z\to0} g(z + \delta z)
$$
And I know that I could use this to complete the proof, just repeating the process on f, but I'd like it to be more rigorous as this formula is not something that's been presented in the course, I'm just using it as tool in this proof and so I want to be able to fully explain. So this is what I have:
$$
(f \circ g)(z) = \lim_{\delta z\to0} \frac{f(g(z + \delta z)) – f(g(z))}{\delta z}\\
=\lim_{\delta z\to0} \frac{f(g(z)+\delta z g'(z)) – f(g(z))}{\delta z}\\
=\lim_{\delta z\to0} \frac{f(g(z)) + \delta z g'(z) f'(g(z)) – f(g(z))}{\delta z}\\
=\lim_{\delta z\to0} g'(z) f'(g(z))\\
= g'(z) f'(g(z))
$$
So the trouble I'm having is expanding on what happens between line 2 and 3. I can do this for line 1 to 2, using the above formula, but for the next line I'm really stuggling to prove the formula holds.
Am I overthinking this?
Best Answer
The proof goes the same way as for the real chain rule : $$ \frac{\mathrm{d}}{\mathrm{d}z}f(g(z)) := \lim_{z'\rightarrow z}\frac{f(g(z'))-f(g(z))}{z'-z} = \lim_{z'\rightarrow z}\frac{f(g(z'))-f(g(z))}{g(z')-g(z)}\frac{g(z')-g(z)}{z'-z} = f'(g(z))g'(z) $$ by definition of $f'$ and $g'$ (which are supposed to exist).