Using Leibniz’ rule for differentiating under the integral sign for line integrals Is there a reference which proves the validity of differentiating under the line integral to prove Cauchy’s integral formulae
$$f’(w)=\frac{1}{2\pi i}\int_C \frac d{dw}\frac{f(u)}{u-w}du$$
 A: You can use Theorem 2.27 From Folland's Real Analysis text. A simplified version of that theorem for complex numbers would say that if $C,D$ are compact, $h(z,w):C\times D\to \mathbb{C}$ is analytic for all $w$, $\partial h/\partial w (z,w)$ is continuous in both arguments, then for all $w\in D$ it follows that
$$\frac{\partial}{\partial w} \int_C h(z,w) dz=\int_C\frac{\partial}{\partial 
 w}h(z,w)dz$$
Essentially why this works is because
$$\frac{\int_C h(z,w)dz-\int_C h(z,w_0)}{w-w_0}=\int_C\frac{h(z,w)-h(z,w_0)}{w-w_0}dz$$
Folland uses Dominated Convergence Theorem to guarantee the above works. In our case as $C\times D$ is compact by Tychonoff's Theorem, and $\partial h/\partial w (z,w)$ is continuous on $C\times D$, then $|\partial h/\partial w (z,w)|$ is bounded above by a constant, say $M$. Since $C$ has finite measure (compact) it follows that $M\in L^1(C)$ so we are free to use Dominated Converges to justify differentiating under the integral sign.
In your case, $C$ is a circle, which is compact. Now for $f(u)/(u-w)$, you might say this isn't defined on a compact set, but if we limit the values of $w$ to a small closed disc and the values of $u$ to the circle, then our function is defined on a domain of the form $C\times D$ where $C,D$ are compact.
A: You can find a careful proof here
Here is another way: using simple facts about power series, we have, fixing an integer $n,$ and writing  $f(z) = \sum_{k=0}^\infty a_k(z-w)^k$ inside $C,$ we have
$f(z) = \sum_{k=0}^\infty a_k (z-w)^{k-n-1}(z-w)^{n+1}\Rightarrow \frac{f(z)}{(z-w)^{n+1}}=\sum_{k=0}^\infty a_k (z-w)^{k-n-1}.$
It follows that $\displaystyle \int_C\frac{f(z)}{(z-w)^{n+1}}dz=2\pi i a_k.\ $  But  $a_k=\frac{f^{(k)}(w)}{k!}.\ $ The result follows.
