I am trying to solve the following and would like some guidance on whether or not my attempt is satisfactory. I maybe missing an argument for why $g'(z)$ exists inside $C$ despite the derivation. Any help would be appreciated.
Let $\phi(s)$ be a continuous function on a simple closed curve $C$ and define $g(z)$ by $$ g(z) = \frac{1}{2\pi i} \int_C \frac{\phi(s)}{s-z} ds$$ Use the limit definition to show $g'(x)$ exist for all $z$ inside $C$ and find $g'(z)$.
Attempt: Using the limit definition of derivative,
\begin{align} g'(z) &= \lim_{h \to 0} \frac{g(z+h)-g(z)}{h} \\ &= \frac{1}{2\pi i} \lim_{h \to 0} \int_C \frac{\phi(s) \left[ \frac{1}{s-(z+h)} - \frac{1}{s-z} \right] ds }{h} \\ &= \frac{1}{2\pi i} \lim_{h \to 0} \int_C \frac{\phi(s) \left[ \frac{s-z-s+z+h}{(s-(z+h))(s-z)} \right] ds }{h} \\ &= \frac{1}{2\pi i} \lim_{h \to 0} \int_C \frac{\phi(s) \left[ \frac{h}{(s-(z+h))(s-z)} \right] ds }{h}\\ &= \frac{1}{2\pi i} \lim_{h \to 0} \int_C \frac{\phi(s)}{(s-(z+h))(s-z)} ds\\ &= \frac{1}{2\pi i} \int_C \frac{\phi(s)}{(s-z)^2} ds \end{align}
Therefore $g'(z)$ exists inside $C$.
Edit: apparently this is an analogue of Leibniz's rule and is exercise number 2 in chapter 4.2 of Conway's complex analysis.
Edit: I think the key part is recognizing $C$ is compact (closed and bounded) and any continuous function on a compact set is uniformly continuous, hence the limit can be interchanged with the integral.