Bringing a derivative inside an integral?

[AxiomOfChoice]

I'm trying to show that a function $f(z)$ is analytic by showing $f'(z)$ exists. But $f(z)$ is defined in terms of a contour integral:
$$f(z) = \oint_{|\zeta - z_0| = r} g(z,\zeta) d\zeta.$$
Since the integral is being carried out with respect to $\zeta$ and not $z$, am I allowed to bring the $d/dz$ operator inside the integral? Or is it more complicated than that? Are there certain conditions that $g(z,\zeta)$ must satisfy? If so, what are they?

THANKS!

 

[Gib Z]

I'm not sure if it will work the same for Contour Integrals, but the conditions for bringing a derivative inside are given nicely here: http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

 

[Phyisab****]

I think you might want to use leibnitz rule.