I'm studying analytical continuation using Conway's textbook but I'm very lost. I'm trying to work on some exercises but nothing is making sense so I've decided to ask here one of those exercises to help me with some guidance. The exercise is the following
Let $\gamma\colon [0,1] \to \mathbb{C}$ be a path and let $\{ (f_t, D_t):0 \leq t \leq 1 \}$ be an analytic continuation along $\gamma$. Show that $\{ (f^\prime_t, D_t):0 \leq t \leq 1 \}$ is also an analytic continuation along $\gamma$.
What I thought was the following: By hypothesis we have that for each $t \in [0,1]$ there is a function element $(f_t, D_t)$ such that $\gamma(t) \in D_t$ and there exists $\delta > 0$ such that if $s \in [0,1]$ implies $|s - t| < \delta$ then $\gamma(s) \in D_t$ and $[f_s]_{\gamma(s)} = [f_t]_{\gamma(s)}$, where $[f_t]_a$ is the germ of $f_t$ at the point $a$ (I basically transcribed the definition of analytic continuation that Conway gives). So picking $(g_t, D_t) \in [f_t]_{\gamma(s)}$ and $(g_s, D_s) \in [f_s]_{\gamma(s)}$ then we have that
$$ g_t(z) = g_s(z) \quad \forall z \in D_t \cap D_s. $$
But both $g_t$ and $g_s$ are analytic functions on $D_t \cap D_s$, so their derivative exists and must coincide on the same domain, that is,
$$ g^\prime_t(z) = g^\prime_s(z)\quad \forall z \in D_t \cap D_s. $$
Since $g_t$ and $g_s$ are arbitrary it follows that $[f^\prime_t]_{\gamma(s)} = [f^\prime_s]_{\gamma(s)}$ for every fixed $t \in [0,1]$ and every $s \in[0,1]$ that satisfies $|s - t| < \delta$.
It feels like this proof is wrong and I'm missing a lot of thing. How can I improve this? Thanks in advance!