Here's an example that came up in practice.
Theorem: (Cauchy-Pompeiu) Let $U \subset \mathbb{C}$ be a bounded open set with piecewise- $C^1$ boundary $\partial U$ oriented positively (see appendix B ), and let $f: \bar{U} \rightarrow \mathbb{C}$ be continuous with bounded continuous partial derivatives in $U$. Then for $z \in U$ : $$ f(z)=\frac{1}{2 \pi i} \int_{\partial U} \frac{f(\zeta)}{\zeta-z} d \zeta+\frac{1}{2 \pi i} \int_U \frac{\frac{\partial f}{\partial \bar{\zeta}}(\zeta)}{\zeta-z} \, \text{Leb}_{\mathbb C}(d \zeta) $$
In the linked source (a libretexts book written by Jiří Lebl), there is the following question:
Why can we not differentiate in $\bar z$ under the integral in the second term of the Cauchy–Pompeiu formula? Notice that it would lead to an impossible result.
[Namely, applying $\partial_{\bar z}$ to both sides, if it was possible to move $\partial_{\bar z}$ from the outside to the inside of the integral in the 2nd term, then we would get that 2nd term $=0$ and so we would conclude $\partial_{\bar z} f=0$, i.e. every $C^1$ function $f$ on $\bar U$ is holomorphic. Which is of course totally false.]
The standard differentiation under the integral result is
Let $X$ be an open subset of $\mathbb R$, and $\Omega$ be a measure space. Suppose ${\displaystyle g\colon X\times \Omega \to \mathbb R}$ satisfies the following conditions:
${\displaystyle g(x,\omega )}$ is a Lebesgue-integrable function of ${\displaystyle \omega }$ for each ${\displaystyle x\in X}$.
For almost all ${\displaystyle \omega \in \Omega }$, the partial derivative ${\displaystyle g_{x}}$ exists for all ${\displaystyle x\in X}$.
There is an integrable function ${\displaystyle \theta \colon \Omega \to \mathbb {R} }$ such that ${\displaystyle |g_{x}(x,\omega)|\leq \theta (\omega )}$ for all ${\displaystyle x\in X}$ and almost every ${\displaystyle \omega \in \Omega }$.
Then, for all ${\displaystyle x\in X}$, $${\frac d{dx} \int _{\Omega }g(x,\omega )\,d\omega =\int _{\Omega}g_{x}(x,\omega )\,d\omega .}$$
The same proof (Dominated Convergence Theorem) shows the same thing if we replace $\mathbb R$ by $\mathbb C$ and $\partial_x$ by $\partial_{\bar z}$.
So in our Cauchy-Pompeiu setting, we have both $X,\Omega:= U$; and $g(z,\zeta):= \frac{\partial_{\bar \zeta} f(\zeta)}{\zeta-z} : U \times U \to \mathbb C$.
(1.) is satisfied, (2.) is almost satisfied: for all fixed $\zeta \in U$, the derivative $\partial_{\bar z} g$ exists (and equals $0$, and is in fact (3.) dominated by the integrable function $\theta \equiv 0$), for all $z\in U$ except at just one point $z=\zeta$!
So the Leibniz rule fails to apply for our Cauchy-Pompeiu integral, because a "for all" condition fails at just one point! I find this to be quite subtle: there are "for all" and "for almost all" conditions in the theorem, and it is of critical importance to remember which variable requires the "for all" condition!
(Just remember: the variable being differentiated needs "for all", and the variable being integrated just needs "for almost all". Which makes total sense.)