 Walter Clyburn

2022-01-05

Differentiation of multivariable function proof
$\frac{d}{dx}{\int }_{v\left(x\right)}^{u\left(x\right)}f\left(t,x\right)dt={u}^{\prime }\left(x\right)f\left(u\left(x\right),x\right)-{v}^{\prime }\left(x\right)f\left(v\left(x\right),x\right)+{\int }_{v\left(x\right)}^{u\left(x\right)}\frac{\partial }{\partial x}f\left(t,x\right)dt$ Melinda McCombs

$I\left(x\right)={\int }_{v}^{u}f\left(t,x\right)dt=F\left(u,x\right)-F\left(v,x\right)$
Where F is the antiderivative of f
$\frac{d}{dx}I=\frac{d}{dx}F\left(u,x\right)-\frac{d}{dx}F\left(v,x\right)$
Now for a function $w\left(a\left(x\right),b\left(x\right)\right)$ its derivative with respect to x can be written as
$\frac{\partial a}{\partial x}\left(x\right)\frac{\partial w}{\partial x}\left(a,\left(x\right),b\left(x\right)\right)+\frac{\partial b}{\partial x}\left(x\right)\frac{\partial w}{\partial x}\left(a,\left(x\right),b\left(x\right)\right)$
In terms of F we have $F\left(u,x\right)$ and $F\left(v,x\right)$ which remember u and v are functions of x. Therefore for each of them we can write this by plugging into the above formula. $w\left(x\right)=F\left(x\right),a\left(x\right)=v\left(x\right)$ or $u\left(x\right)$ and $b\left(x\right)=x$.
because $\frac{\partial x}{\partial x}=1$ Now the above also holds for $F\left(u,x\right)$ so
$\frac{d}{dx}F\left(u,x\right)=\frac{\partial u}{\partial x}f\left(u,x\right)+\frac{\partial x}{\partial x}f\left(u,x\right)$
Therefore
${I}^{\prime }=f\left(v,x\right)+\frac{\partial v}{\partial x}f\left(v,x\right)-f\left(u,x\right)-\frac{\partial u}{\partial x}f\left(u,x\right)$
Now rearranging you can see that it starts to take on your form.
${I}^{\prime }={v}^{\prime }f\left(v,x\right)-{u}^{\prime }f\left(u,x\right)+f\left(v,x\right)-f\left(u,x\right)$
Now the last two terms can be written in terms of an integral.
$f\left(v,x\right)-f\left(u,x\right)={\int }_{u}^{v}{f}^{\prime }\left(t,x\right)dt$
Which then can all come together to give

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