One of the first results of complex analysis is that differentiability is a very strong condition: If $U\subseteq \Bbb C$ is open, then every complex differentiable function $f:U\to\Bbb C$ is automatically analytic.
If $X,Y$ are complex Banach spaces, $U\subseteq X$ open and $f:U\to Y$ differentiable, is $f$ analytic?
It makes sense to define what analytic means in this context. We say that a map $f:U\to Y$ is analytic if each point $x_0\in U$ admits a neighbourhood $U'$ so that $$f\lvert_{U'} (x)= \sum_{n=0}^\infty a_n(x-x_0)^n,$$ where $a_n : \times^n X\to Y$ is a continuous $n$-multilinear map (in particular $a_0$ is a constant element of $Y$) and the sum converges uniformly on $U'$. $a_n(x-x_0)^n$ is then shorthand for $a_n(x-x_0,...,x-x_0)$.