Leibniz integral rule can be applied like following if $N$ is a function of $t$
$$ \frac{dN}{dt} =\frac{d}{dt}\left(\int_{a(t)}^{b(t)}\rho(x,t)dx\right) $$
$$ \frac{dN}{dt} =\int_a^b \frac{\partial \rho}{\partial t}dx+ \rho(b,t)\frac{db}{dt}-\rho(a,t)\frac{da}{dt} $$
Now, if $N$ is a function of $x$ and $t$, what will the Leibniz integral rule look like?
I think it will be like this, am I right?
$$ \frac{\partial N}{\partial t} = \frac{\partial }{\partial t}\left(\int_{a(t)}^{b(t)}\rho(x,t)dx\right) $$
$$ \frac{\partial N}{\partial t} = \int_a^b \frac{\partial \rho}{\partial t}dx+ \rho(b,t)\frac{\partial b}{\partial t}-\rho(a,t)\frac{\partial a}{\partial t} $$
Let's expand the right-hand side: $$\frac{\partial N}{\partial t} = \frac{\partial}{\partial t} \left[R(b(t), t) - R(a(t), t)\right], ~~~~\text{ where } \frac{\partial R}{\partial x} = \rho(x, t)$$ Continuing to simplify the right-hand side, we have \begin{align*} \frac{\partial N}{\partial t} &= \frac{\partial b}{\partial t} \cdot \frac{\partial R}{\partial x}(b(t), t) + \frac{\partial R}{\partial t}(b(t), t) - \frac{\partial a}{\partial t} \cdot \frac{\partial R}{\partial x}(a(t), t) -\frac{\partial R}{\partial t}(a(t), t) \\ \\ &= [b'(t) \cdot \rho(b(t), t) - a'(t) \cdot \rho(a(t), t)] + \left[\frac{\partial R}{\partial t}(b(t), t) - \frac{\partial R}{\partial t}(a(t), t)\right] \\ \\ &= [b'(t) \cdot \rho(b(t), t) - a'(t) \cdot \rho(a(t), t)] + \int_{a(t)}^{b(t)} \frac{\partial^2 R}{\partial x \partial t}(x, t)~ dx \end{align*} And assuming the function $R$ is smooth enough to have mixed partials be equivalent (see Clairaut Theorem), the integral can be simplified and the resulting expression is $$[b'(t) \cdot \rho(b(t), t) - a'(t) \cdot \rho(a(t), t)] + \int_{a(t)}^{b(t)} \frac{\partial}{\partial t}\frac{\partial R}{\partial x}(x, t)~ dx \\ = [b'(t) \cdot \rho(b(t), t) - a'(t) \cdot \rho(a(t), t)] + \int_{a(t)}^{b(t)} \frac{\partial \rho}{\partial t}(x, t) ~ dx$$ So the answer to your question is yes! Now if you'd be a little keener than me, you'd have notice of course this must be the case, because the value $N$ is actually only a function of $t$, so of course $dN / dt = \partial N / \partial t$... (Recall $a$ and $b$ are also only functions of $t$, so their partial derivatives are also their total derivatives.)