About the continuity of a partial derivative Suppose that $f\in C(\Gamma\times U,\Bbb C)$ and $f(w,\cdot)\in C^1(U,\Bbb C)$ for each $w\in\Gamma$ where $\Gamma,U\subset\Bbb C$ and $\Gamma$ is compact.
We can ensure that the partial derivative $\partial_2 f:\Gamma\times U\to\Bbb C$ is continuous? That is, we can says that $\partial_2 f$ is continuous respect to it first argument? I guess that no without more assumptions but anyway Im not completely sure, and I dont found an easy counterexample yet.
Can someone put some light to this question?

EDIT:
@charMD give this answer, that he deleted because he didnt noticed that this question involves $\partial_2 f(w,\cdot)$ being holomorphic for each $w\in\Gamma$

Define $$ f: (x,y)\in \mathbb{R}^2 \mapsto \begin{cases}
x \sin\big(\frac{y}{x}\big) &\mbox{ if $x\neq 0$}\\
0 &\mbox{ if $x=0$.}
\end{cases}$$
Then $f$ is continuous and for all $x$, $f(x,\cdot)$ is $\mathcal{C}^1$.
However, we have $\partial_2 f :(x,y) \mapsto \begin{cases}
\cos\big(\frac{y}{x}\big) &\mbox{ if $x\neq 0$}\\
0 &\mbox{ if $x=0$.}
\end{cases}$ and thus $\partial_2 f$ is not continuous at $(0,0)$.

However probably this example can be applied to my question. That is: we can see that $\partial_2 f(x,\cdot)$ is analytic for each $x\in\Bbb R$, so it have a complex analytic version of it for each $x\in\Bbb R$. 
Thus it remains to see if there is a complex version of $f$ that it would be continuous in $\Bbb C^2$ and if $\partial_2 f(x,\cdot)$ could be satisfactorily extended for $x\in\Bbb C$, what at first glance it seems.

UPDATE:
The function proposed my @charMD cannot be extended continuously to $\Bbb C^2$ because
$$|\lim_{w\to 0} w\sin(i/w)|=\lim_{w\to 0}|w||\sinh (1/w)|=\infty$$
 A: As far as I can see this answer is correct.
I think I found a satisfactory answer, at least for each interior point of $U$. Using the derivative formula of Cauchy for analytic functions we have that
$$\partial_2 f(w,z)=\frac1{2\pi i}\int_{\partial\Bbb D(z_0,\epsilon)}\frac{f(w,\zeta)}{(\zeta-z)^2}d\zeta\tag1$$
Here the circular path $\partial\Bbb D(z_0,\epsilon)$ is homotopic to some $\partial\Bbb D(z,s)\subset\partial\Bbb D(z_0,\epsilon)$ for $\overline{\Bbb D}(z_0,\epsilon)\subset U$.
Then from $(1)$ we can show that if $f\in C(\Gamma\times U^\circ,\Bbb C)$ then $\partial_2 f$ is continuous on $\Gamma \times\overline{\Bbb D}(z_1,\epsilon_1)$, for some $\overline{\Bbb D}(z_1,\epsilon_1)\subset\overline{\Bbb D}(z_0,\epsilon)$, that is
$$|\partial_2 f(w,z)-\partial_2 f(v,r)|=\frac1{2\pi}\left|\int_{\partial\Bbb D(z_0,\epsilon)}\left(\frac{f(w,\zeta)}{(\zeta-z)^2}-\frac{f(v,\zeta)}{(\zeta-r)^2}\right)d\zeta\right|\le\epsilon K\tag2$$
where $K$ is a chosen bound because the function
$$g:\Gamma\times\partial\Bbb D(z_0,\epsilon)\times\overline{\Bbb D}(z_1,\epsilon_1)\to\Bbb C,\quad (w,z,\zeta)\mapsto \frac{f(w,\zeta)}{(\zeta-z)^2}\tag3$$
is uniformly continuous. Hence $\partial_2 f$ is continuous on $\Gamma\times U^\circ$.
