Define $\text{D}$ to be the set of density matrices on some finite dimensional Hilbert space and suppose that I have a density matrix $\rho$ parametrized by some scalar $\theta$. I have an integral that looks like this $\int_{\text{D}} \frac{d\rho(\theta)}{d\theta} \, d\rho(\theta)$ where $d\rho$ is the uniform distribution on the set of density matrices. Is it true that \begin{equation} \int_{\text{D}} \frac{d\rho(\theta)}{d\theta} \, d\rho(\theta) = \frac{d}{d\theta} \int_{\text{D}}\rho(\theta) \, d\rho(\theta) . \end{equation}
The integral doesn't precisely fit into the requirements for the Leibniz rule because $\theta$ also appears in the differential. If the statement isn't true, do the two integrals differ by some small error?
Let's take a simple case where $D$ is an interval and $\rho$ is differentiable. Then I expect $$ \int_D \frac{d\rho(\theta)}{d\theta}\;d\rho(\theta) \quad\text{means}\quad \int_D \frac{d\rho(\theta)}{d\theta}\;\frac{d\rho(\theta)}{d\theta}\;d\theta \tag1$$ and $$ \int_D\rho(\theta)\;d\rho(\theta) \quad\text{means}\quad \int_D\rho(\theta)\;\frac{d\rho(\theta)}{d\theta}\;d\theta \tag2$$ Then the Leibniz rule would say (from the product rule) $$ \frac{d}{d\theta}\int_D\rho(\theta)\;d\rho(\theta) = \int_D\frac{d}{d\theta}\left(\rho(\theta)\;\frac{d\rho(\theta)}{d\theta}\right)\;d\theta = \int_D \left(\rho(\theta)\;\frac{d^2\rho(\theta)}{d\theta^2} +\frac{d\rho (\theta)}{d\theta}\;\frac{d\rho(\theta)}{d\theta} \right)\;d\theta $$ This is not $(1)$.
If $\rho(\theta)$ has the form $a\theta+b$, then the second derivative is $0$, so in that case you will get what you propose.