Fundamental theorem of Calculus Part 1 If I'm taking the derivative with respect to $t$ of the integral: 
$$\int_{-\infty}^t (t-\tau)u(\tau)d\tau$$
Does the Fundamental theorem of Calculus result in the answer of $0$?
I am trying to write the system in implicit form which is why I'm asking. I believe the implicit form would then be: $\dfrac{dy}{dt} = 0$.
Edit: The reason I don't feel comfortable using the Fundamental Theorem of Calculus is since the lower bound is not constant. Instead it is negative infinity.
Thank you for all the responses.
 A: Don't forget that you have $t$ also inside the integrand. You can deal with this in several different ways.
The simplest way here is to write $$\int_{-\infty}^t(t-\tau)u(\tau)d\tau  = t\int_{-\infty}^tu(\tau)d\tau - \int_{-\infty}^t\tau u(\tau)d\tau$$
Now by taking the derivative using the product rule you get
$$\frac{d}{dt}\left(\int_{-\infty}^t(t-\tau)u(\tau)d\tau\right)  = \int_{-\infty}^tu(\tau)d\tau  + t \frac{d}{dt}\left(\int_{-\infty}^tu(\tau)d\tau\right) - \frac{d}{dt}\left(\int_{-\infty}^t\tau u(\tau)d\tau\right)$$
and use the fundamental theorem of calculus (see also Andre's answer here).
The other way to do it is to use Leibniz rule which says that
$$\frac{d}{dt} \int_a^t f(t,\tau) d\tau = f(t,t)  + \int_a^t \frac{\partial }{\partial t}f(t,\tau) d\tau$$
where for your case $f(t,\tau) = (t-\tau)u(\tau) \to \frac{\partial}{\partial t} f(t,\tau) = u(\tau)$. In either case, if you do it correctly, you should end up with $\int_{-\infty}^tu(\tau)d\tau$ as the final answer.
${\bf Edit:}$
As a previous answer that is now removed tried to say: it does not matter that its $-\infty$ in the lower limit of the integral, you can still use the fundamental theorem of calculus. To see this write (for any $a$) $$\int_{-\infty}^t g(x) dx = \int_{-\infty}^a g(x) dx + \int_{a}^t g(x) dx$$
and now take the derivative to get
$$\frac{d}{dt}\left(\int_{-\infty}^t g(x) dx\right) = g(t)$$
since the derivative of the first term (which is a constant) vanishes.
