Solving this odd mix of integrals/derivatives? I've got a homework assignment that looks something like this: (numbers changed)
$$F(x) =\int_1^x f(t) \, \mathrm{d} t$$
$$f(t) =\int_x^{t^2} \frac{\sqrt{7+u^4}}{u} \, \mathrm{d} u$$
Find $ F\;''(1) $
I changed the numbers because I don't want the answer.  This seems simple but I can't wrap my head around it.  Any ideas?
Edit: I entered this in completely wrong for some reason. Will the same answer still apply?
 A: By the fundamental theorem of calculus, if $\displaystyle F(x)=\int_a^x f(t)\ dt$, then $F'(x)=f(x)$. So here
$$
F'(x)=\frac{\sqrt{7+x^4}}{x}.
$$
You can take the second derivative to find $F''(x)$, and then plug in $x=1$.
A: With respect to your modified question, things are a little stickier. One thing that might help you in getting the answer is to consider the following representation of $f(t)$:
$$f(t)=\int_x^{t^2} \frac{\sqrt{7+u^4}}{u} \, \mathrm du=\int_p^{t^2} \frac{\sqrt{7+u^4}}{u} \, \mathrm du-\int_p^{x} \frac{\sqrt{7+u^4}}{u} \, \mathrm du$$
where $p$ is some constant. (Why this is justified is something you'll have to explain.) We then have
$$\begin{align*}F(x)=\int_1^x f(t) \, \mathrm dt&=\int_1^x \left(\int_p^{t^2} \frac{\sqrt{7+u^4}}{u} \, \mathrm du-\int_p^{x} \frac{\sqrt{7+u^4}}{u} \, \mathrm du\right) \, \mathrm dt\\&=\int_1^x \int_p^{t^2} \frac{\sqrt{7+u^4}}{u} \, \mathrm du\,\mathrm dt-\left(\int_p^{x} \frac{\sqrt{7+u^4}}{u} \, \mathrm du\right)\left(\int_1^x\mathrm dt\right)\end{align*}$$
(You'll also have to explain how I got that last bit.) Differentiating the second term requires that you use the product rule in addition to the Fundamental Theorem; differentiating the first term will require the careful use of the chain rule...
