Proof:

We have,

${\displaystyle {\cfrac {dF}{dt}}=\lim _{\Delta t\rightarrow 0}{\cfrac {F(t+\Delta t)-F(t)}{\Delta t}}~.}$

Now,

${\displaystyle {\begin{aligned}{\cfrac {F(t+\Delta t)-F(t)}{\Delta t}}&={\cfrac {1}{\Delta t}}\left[\int _{a(t+\Delta t)}^{b(t+\Delta t)}f(x,t+\Delta t)~{\text{dx}}-\int _{a(t)}^{b(t)}f(x,t)~{\text{dx}}\right]\\&\equiv {\cfrac {1}{\Delta t}}\left[\int _{a+\Delta a}^{b+\Delta b}f(x,t+\Delta t)~{\text{dx}}-\int _{a}^{b}f(x,t)~{\text{dx}}\right]\\&={\cfrac {1}{\Delta t}}\left[-\int _{a}^{a+\Delta a}f(x,t+\Delta t)~{\text{dx}}+\int _{a}^{b+\Delta b}f(x,t+\Delta t)~{\text{dx}}-\int _{a}^{b}f(x,t)~{\text{dx}}\right]\\&={\cfrac {1}{\Delta t}}\left[-\int _{a}^{a+\Delta a}f(x,t+\Delta t)~{\text{dx}}+\int _{a}^{b}f(x,t+\Delta t)~{\text{dx}}+\int _{b}^{b+\Delta b}f(x,t+\Delta t)~{\text{dx}}-\int _{a}^{b}f(x,t)~{\text{dx}}\right]\\&=\int _{a}^{b}{\cfrac {f(x,t+\Delta t)-f(x,t)}{\Delta t}}~{\text{dx}}+{\cfrac {1}{\Delta t}}\int _{b}^{b+\Delta b}f(x,t+\Delta t)~{\text{dx}}-{\cfrac {1}{\Delta t}}\int _{a}^{a+\Delta a}f(x,t+\Delta t)~{\text{dx}}~.\end{aligned}}}$

Since ${\displaystyle f(x,t)}$ is essentially constant over the infinitesimal intervals ${\displaystyle a<x<a+\Delta a}$ and ${\displaystyle b<x<b+\Delta b}$, we may write

${\displaystyle {\cfrac {F(t+\Delta t)-F(t)}{\Delta t}}\approx \int _{a}^{b}{\cfrac {f(x,t+\Delta t)-f(x,t)}{\Delta t}}~{\text{dx}}+f(b,t+\Delta t)~{\cfrac {\Delta b}{\Delta t}}-f(a,t+\Delta t)~{\cfrac {\Delta a}{\Delta t}}~.}$

Taking the limit as ${\displaystyle \Delta t\rightarrow 0}$, we get

${\displaystyle \lim _{\Delta t\rightarrow 0}\left[{\cfrac {F(t+\Delta t)-F(t)}{\Delta t}}\right]=\lim _{\Delta t\rightarrow 0}\left[\int _{a}^{b}{\cfrac {f(x,t+\Delta t)-f(x,t)}{\Delta t}}~{\text{dx}}\right]+\lim _{\Delta t\rightarrow 0}\left[f(b,t+\Delta t)~{\cfrac {\Delta b}{\Delta t}}\right]-\lim _{\Delta t\rightarrow 0}\left[f(a,t+\Delta t)~{\cfrac {\Delta a}{\Delta t}}\right]}$

or,

${\displaystyle {{\cfrac {dF(t)}{dt}}=\int _{a(t)}^{b(t)}{\frac {\partial f(x,t)}{\partial t}}~{\text{dx}}+f[b(t),t]~{\frac {\partial b(t)}{\partial t}}-f[a(t),t]~{\frac {\partial a(t)}{\partial t}}~.}\qquad \qquad \qquad \square }$