\(\displaystyle{I}{\left({x}\right)}={\int_{{v}}^{{u}}}{f{{\left({t},{x}\right)}}}{\left.{d}{t}\right.}={F}{\left({u},{x}\right)}-{F}{\left({v},{x}\right)}\)

Where F is the antiderivative of f

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{I}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{F}{\left({u},{x}\right)}-{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{F}{\left({v},{x}\right)}\)

Now for a function \(\displaystyle{w}{\left({a}{\left({x}\right)},{b}{\left({x}\right)}\right)}\) its derivative with respect to x can be written as

\(\displaystyle{\frac{{\partial{a}}}{{\partial{x}}}}{\left({x}\right)}{\frac{{\partial{w}}}{{\partial{x}}}}{\left({a},{\left({x}\right)},{b}{\left({x}\right)}\right)}+{\frac{{\partial{b}}}{{\partial{x}}}}{\left({x}\right)}{\frac{{\partial{w}}}{{\partial{x}}}}{\left({a},{\left({x}\right)},{b}{\left({x}\right)}\right)}\)

In terms of F we have \(\displaystyle{F}{\left({u},{x}\right)}\) and \(\displaystyle{F}{\left({v},{x}\right)}\) which remember u and v are functions of x. Therefore for each of them we can write this by plugging into the above formula. \(\displaystyle{w}{\left({x}\right)}={F}{\left({x}\right)},{a}{\left({x}\right)}={v}{\left({x}\right)}\) or \(\displaystyle{u}{\left({x}\right)}\) and \(\displaystyle{b}{\left({x}\right)}={x}\).

because \(\displaystyle{\frac{{\partial{x}}}{{\partial{x}}}}={1}\) Now the above also holds for \(\displaystyle{F}{\left({u},{x}\right)}\) so

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{F}{\left({u},{x}\right)}={\frac{{\partial{u}}}{{\partial{x}}}}{f{{\left({u},{x}\right)}}}+{\frac{{\partial{x}}}{{\partial{x}}}}{f{{\left({u},{x}\right)}}}\)

Therefore

\(\displaystyle{I}'={f{{\left({v},{x}\right)}}}+{\frac{{\partial{v}}}{{\partial{x}}}}{f{{\left({v},{x}\right)}}}-{f{{\left({u},{x}\right)}}}-{\frac{{\partial{u}}}{{\partial{x}}}}{f{{\left({u},{x}\right)}}}\)

Now rearranging you can see that it starts to take on your form.

\(\displaystyle{I}'={v}'{f{{\left({v},{x}\right)}}}-{u}'{f{{\left({u},{x}\right)}}}+{f{{\left({v},{x}\right)}}}-{f{{\left({u},{x}\right)}}}\)

Now the last two terms can be written in terms of an integral.

\(\displaystyle{f{{\left({v},{x}\right)}}}-{f{{\left({u},{x}\right)}}}={\int_{{u}}^{{v}}}{f}'{\left({t},{x}\right)}{\left.{d}{t}\right.}\)

Which then can all come together to give

\(\displaystyle{I}'={v}'{f{{\left({v},{x}\right)}}}-{u}'{f{{\left({u},{x}\right)}}}+{\int_{{u}}^{{v}}}{\frac{{\partial}}{{\partial{x}}}}{f{{\left({t},{x}\right)}}}{\left.{d}{t}\right.}\)

\(\displaystyle\Rightarrow{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\int_{{v}}^{{u}}}{f{{\left({t},{x}\right)}}}{\left.{d}{t}\right.}={v}'{f{{\left({v},{x}\right)}}}-{u}'{f{{\left({u},{x}\right)}}}+{\int_{{v}}^{{u}}}{\frac{{\partial}}{{\partial{x}}}}{f{{\left({t},{x}\right)}}}{\left.{d}{t}\right.}\)

Which is the formula you want to prove.