Compute integral of this form using Leibniz Rule? I was just told about Leibniz rule for computing something like $\frac{d}{du}\int_{t1}^{t0+u}f(u,t)\, dt $.
\begin{align}
{d\over du}\int_{t1}^{t0+u} f(u,t)\,dt&=\int_{t1}^{t0+u} {\partial f\over \partial u}(u,t)\,dt+f(u,t0+u)-f(u,t1){d\over du}[t1]\\
&=\int_{t1}^{t0+u} {\partial f\over \partial u}(u,t)\,dt+f(u,t0+u).
\end{align} 
Can use this in conjunction with other rules to compute the following?
$\frac{d}{du}\mid_{u=0}$  $\int_{t1}^{t0+u}F(f(u,t),g(u,t))\, dt$.

Denote $F_1(u, \; t)=F(f(u,t), \; g(u,t)).$ Then by Leibniz rule, 
$$\dfrac{d}{du}\int\limits_{t1}^{t_0+u}F(f(u,t), \; g(u,t))\, dt =\dfrac{d}{du}\int\limits_{t1}^{t_0+u}F_{1}(u,t)\, dt = \\
=  \int\limits_{t1}^{t_0+u} \dfrac{\partial{}}{\partial{u}}\left(F_{1}(u,t) \right)\, dt +F_{1}(u, \; t_{0}+u)\; {\dfrac{d(t_{0}+u )}{du}} - F_{1}(u, \; t_{1}) \;{\dfrac{dt_{1}}{du}} = \\
= \int\limits_{t1}^{t_0+u} \dfrac{\partial{}}{\partial{u}}\left(F_{1}(u,t) \right)\, dt +F_{1}(u, \; t_{0}+u)= \\ 
= \int\limits_{t1}^{t_0+u} \dfrac{\partial{}}{\partial{u}}\left(F_{1}(u,t) \right)\, dt+F(f(u, t_{0}+u), \; g(u, t_{0}+u)),$$ 
since  ${\dfrac{d(t_{0}+u )}{du}}=1, \quad {\dfrac{dt_{1}}{du}}=0. $ 

Applying the chain rule $$ \dfrac{\partial}{\partial{u}}\left(F_{1}(u,t) \right)= \dfrac{\partial}{\partial{u}}\left( F(f(u,t), \; g(u,t)) \right) = \\
 = \left.\dfrac{\partial{F(s, \; r)}}{\partial{s}}\right|_{\matrix {s=f(u,t)\\ r=g(u,t)}} \cdot \dfrac{\partial{f(u,t)}}{\partial{u}} + \left.\dfrac{\partial{F(s, \; r)}}{\partial{r}} \right|_{\matrix {s=f(u,t)\\ r=g(u,t)}} \cdot \dfrac{\partial{g(u,t)}}{\partial{u}} .$$
