derivative of a summation with variable upper limit Is the following statement correct and if yes does it need to satisfy specific requirement to be correct:
$${ d \over dt} \sum_{j=1}^{N(t)} f(t,j) = \sum_{j=1}^{N(t)} {df(t,j) \over dt} + f(t,N(t)) {dN(t) \over dt}$$
 A: There are two approaches for this problem: one is to define:
$$f_j(t)=\left\{\begin{array}{ll}f(t,j)&j\le N(t)\\0&j>N(t)\end{array}\right.$$
Then
$$\sum_{j=1}^{N(t)}f(t,j)=\sum_{j=1}^{\infty}f_j(t)$$
and continue as you already know.
$$\frac d{dt}\sum_{j=1}^{N(t)}f(t,j)=\sum_{j=1}^{\infty}f_j'(t)$$
Beware where each $f_j$ is non-continues.
The other approach is to have all points $\{t_i\}_{i\in I}$ in which $N(t)$ take a jump, meaning $N(t_i^-)\ne N(t_i^+)$.  If $\{t_i\}_{i\in I}$ is an well ordered set, you can have: $0<t_1<t_2<\cdots<t_k<\cdots<T$, ($t_0=0$) and we define $N_i=N(t)$ for $t_{i-1}<t<t_i$.  Note that $N_i$ should be well defined as there are no jumps in $N$ between $t_{i-1}$ and $t_i$.
So now you divide in this function defined by parts:
$$\left.\frac d{dt}\sum_{j=1}^{N(t)}f(t,j)\right|_{t_{k-q}<t<t_k}=\sum_{j=1}^{N_k}\frac{df(t,j)}{dt}$$
The derivative at each $t_k$ does not exist unless both limits exists and are equal.
A: Since $N(t)$ is an integer, $N'(t)$ only makes sense at points $t_0$ where $N(t)=N(t_0)$ for all $t$ in some neighbourhood of $t_0$; in such a case, $N'(t_0)=0$. On any point where $N(t)$ changes value, its derivative is not defined. 
