Derivative using Fundamental Theorem of Calculus when integrand has product of two functions? I want to find the derivative of the following:
$$exp \left( -\int_{t-\tau(t)}^t \frac{\mu(x)U(x)}{S} \,dx \right)$$
I tried to use the Fundamental theorem of calculus of the form:
$$\frac{d}{dx}\int_0^x t^3 \,dx = f(x)\frac{dx}{dx} - f(0)\frac{d0}{dx} = x^3$$
(from Wikipedia) and I got something, but I'm not confident I followed all the rules correctly. I know to start with the chain rule to deal with the $exp()$, but then finding the derivative of the integral is mixing me up. My question for this part is, is it correct to start like this:
$$-\left(\frac{\mu(t)U(t)}{S}\frac{dt}{dt} - \frac{\mu(t-\tau(t))U(t-\tau(t))}{S}\frac{d(t-\tau(t))}{dt}\right)$$
or do I need to deal with the product of functions first before applying the FTC (e.g., via substitution or integrating by parts).
Sorry if this is a silly question. It's been a long time since I took calculus and I couldn't find a similar example online.
Thanks!
 A: As you say, the $\exp$ part is straightforward, so let's look at the derivative of
$$
Q = \int_{t-\tau(t)}^t \frac{\mu(x)U(x)}{S} \,dx.
$$
A standard thing to do here is to write this as a sum of two integrals, splitting at some arbitrary (but fixed) point $b$:
\begin{align}
Q 
&= \int_{t-\tau(t)}^t \frac{\mu(x)U(x)}{S} \,dx\\
&= \int_{t-\tau(t)}^b \frac{\mu(x)U(x)}{S} \,dx
+ \int_b^t \frac{\mu(x)U(x)}{S} \,dx\\
&= -\int_b^{t-\tau(t)} \frac{\mu(x)U(x)}{S} \,dx
+ \int_b^t \frac{\mu(x)U(x)}{S} \,dx\\
\end{align}
Now you've got two terms, each of which is amenable to taking derivatives. For the first, you get the derivative
$$
-\frac{\mu(t-v(t))U(t-v(t))}{S}\left(1 - v'(t) \right).
$$
where that final factor comes from applying the chain rule. For the second, you get just
$$
\frac{\mu(t)U(t)}{S}.
$$
And that's the end of the story.
To answer the question you asked, though, suppose we define
$$
H(x) = \frac{\mu(x)U(x)}{S}
$$
Then you have
$$
Q = \int_{t-v(t)}^t H(x) dx
$$
to which you can apply the FTC, and the derivative involves a difference of expressions involving $H$, which you can then expand back out --- no need to integrate a product at all.
