Walter Clyburn

2022-01-05

Differentiation of multivariable function proof

$\frac{d}{dx}{\int}_{v\left(x\right)}^{u\left(x\right)}f(t,x)dt={u}^{\prime}\left(x\right)f(u\left(x\right),x)-{v}^{\prime}\left(x\right)f(v\left(x\right),x)+{\int}_{v\left(x\right)}^{u\left(x\right)}\frac{\partial}{\partial x}f(t,x)dt$

Melinda McCombs

Beginner2022-01-06Added 38 answers

Start with

$I\left(x\right)={\int}_{v}^{u}f(t,x)dt=F(u,x)-F(v,x)$

Where F is the antiderivative of f

$\frac{d}{dx}I=\frac{d}{dx}F(u,x)-\frac{d}{dx}F(v,x)$

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

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

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

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

$\frac{d}{dx}F(u,x)=\frac{\partial u}{\partial x}f(u,x)+\frac{\partial x}{\partial x}f(u,x)$

Therefore

${I}^{\prime}=f(v,x)+\frac{\partial v}{\partial x}f(v,x)-f(u,x)-\frac{\partial u}{\partial x}f(u,x)$

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

${I}^{\prime}={v}^{\prime}f(v,x)-{u}^{\prime}f(u,x)+f(v,x)-f(u,x)$

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

$f(v,x)-f(u,x)={\int}_{u}^{v}{f}^{\prime}(t,x)dt$

Which then can all come together to give

$I}^{\prime}={}^{$

Where F is the antiderivative of f

Now for a function

In terms of F we have

because

Therefore

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

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

Which then can all come together to give

0