Differentiation of multivariable function proofddx∫v(x)u(x)f(t,x)dt=u′(x)f(u(x),x)−v′(x)f(v(x),x)+∫v(x)u(x)∂∂xf(t,x)dt

Start with
${\displaystyle I\left(x\right)={\int }_v^{u}f(t,x)dt=F(u,x)-F(v,x)}$
Where F is the antiderivative of f
${\displaystyle \frac{d}{dx}I=\frac{d}{dx}F(u,x)-\frac{d}{dx}F(v,x)}$
Now for a function ${\displaystyle w(a\left(x\right),b\left(x\right))}$ its derivative with respect to x can be written as
${\displaystyle \frac{\partial a}{\partial x}\left(x\right)\frac{\partial w}{\partial x}(a,\left(x\right),b\left(x\right))+\frac{\partial b}{\partial x}\left(x\right)\frac{\partial w}{\partial x}(a,\left(x\right),b\left(x\right))}$
In terms of F we have ${\displaystyle F(u,x)}$ and ${\displaystyle F(v,x)}$ 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 ${\displaystyle F(u,x)}$ so
${\displaystyle \frac{d}{dx}F(u,x)=\frac{\partial u}{\partial x}f(u,x)+\frac{\partial x}{\partial x}f(u,x)}$
Therefore
${\displaystyle I^{\prime }=f(v,x)+\frac{\partial v}{\partial x}f(v,x)-f(u,x)-\frac{\partial u}{\partial x}f(u,x)}$
Now rearranging you can see that it starts to take on your form.
${\displaystyle I^{\prime }=v^{\prime }f(v,x)-u^{\prime }f(u,x)+f(v,x)-f(u,x)}$
Now the last two terms can be written in terms of an integral.
${\displaystyle f(v,x)-f(u,x)={\int }_u^v{f}^{\prime }(t,x)dt}$
Which then can all come together to give

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