Suppose that $g$ is analytic in $B'(a;r)$ in the sense that $g$ is continuously differentiable on $B'(a;r)$. If $g$ is differentiable at $a$ then can we say that $g$ is analytic in $B(a;r)$ ( again in the sense that $g$ is continuously differentiable on $B(a;r)$), where $B'(a;r) = B(a;r) \setminus \{a \}$.
I think the fact is true but I can't actually figure out why? Please help me.
Thank you in advance.
If I understand you correctly, $g$ has an isolated singularity at $z=a$, but in fact $g'(a)$ also exists?
In that case $g$ is certainly analytic on $B(a;r)$. In fact, by Riemann's removable singularity theorem it is enough to assume that $g$ is bounded near $a$. (In that case $\lim_{z\to a} g(z) = L$ exists, and if we let $g(a) = L$, we get a function which is analytic on $B(a;r)$.)