Derivative of double integral of the same function I am still learning about integration so please excuse the stupid questions. How should I approach the following problem? 
$\frac{d}{dt} \big\{ \int_a^t ~ [\int_a^t f(\tilde{t}) d\tilde{t}] ~ f(\tilde{t})d\tilde{t}\big\} $
Should I treat them as derivative of a product of 2 exactly the same integrals? Or is there another way to solve for it? I would like to characterize $~f(t)~$ eventually. 
Thanks a lot! 
 A: The question contains $\frac{d}{dt} \big\{ \int_a^t ~ [\int_a^t f(\tilde{t}) d\tilde{t}] ~ f(\tilde{t})d\tilde{t}\big\} $. I will assume that this is intended to mean something equal to $\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\displaystyle{\int_{a}^{t}}\left(\displaystyle{\int_{a}^{t}}f\left(T\right)\,\mathrm{d}T\right)f\left(\widetilde{t}\right)\,\mathrm{d}\widetilde{t}\right)$, since there's no context that suggests something like $\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\displaystyle{\int_{a}^{t}}\left(\displaystyle{\int_{a}^{t}}f\left(\widetilde{t}\right)\,\mathrm{d}T\right)f\left(\widetilde{t}\right)\,\mathrm{d}\widetilde{t}\right)$ was intended.

For convenience, set $I\left(t\right)=\displaystyle{\int_{a}^{t}}f\left(T\right)\,\mathrm{d}T$. Note that $I\left(t\right)=\displaystyle{\int_{a}^{t}}f\left(\widetilde{t}\right)\,\mathrm{d}\widetilde{t}$ since this integral of $f$ doesn't depend on what we call the variable in the integral.
Since $I\left(t\right)=\displaystyle{\int_{a}^{t}}f\left(T\right)\,\mathrm{d}T$ doesn't depend on $\widetilde{t}$ in any way, it's a constant as far as the $\int\mathrm{d}\widetilde{t}$ integral is concerned. Then we have 
$$\begin{align} &\phantom{=}\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\displaystyle{\int_{a}^{t}}\left(\displaystyle{\int_{a}^{t}}f\left(T\right)\,\mathrm{d}T\right)f\left(\widetilde{t}\right)\,\mathrm{d}\widetilde{t}\right)\\
&=\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\displaystyle{\int_{a}^{t}}I\left(t\right)f\left(\widetilde{t}\right)\,\mathrm{d}\widetilde{t}\right)\text{ by def. of }I(t)\\
&=\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\left(I\left(t\right)\right)\left(\displaystyle{\int_{a}^{t}}f\left(\widetilde{t}\right)\,\mathrm{d}\widetilde{t}\right)\right)\text{ as }I(t)\text{ is like a constant}\\
&=\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\left(I\left(t\right)\right)\left(I\left(t\right)\right)\right)\text{by def. of }I(t)\\
&=\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\left(I\left(t\right)\right)^{2}\right)\text{ by algebra}
\end{align}$$
If $I\left(t\right)$ is differentiable, we can go one step further to get $2I\left(t\right)I'\left(t\right)$. And if we're in a nice situation where $f$ is continuous on an interval like $[a,t+\varepsilon)$, then by the Fundamental Theorem of Calculus, $I'\left(t\right)=\dfrac{\mathrm{d}}{\mathrm{d}t}\displaystyle{\int_{a}^{t}}f\left(T\right)\,\mathrm{d}T=f\left(t\right)$, so the final result is $\boxed{2f\left(t\right)\displaystyle{\int_{a}^{t}}f\left(\widetilde{t}\right)\,\mathrm{d}\widetilde{t}}$. 
Note the factor of $2$ that comes from properly taking into account the dependence on t of the inner integral.
