Differentiation of multivariable function proof \frac{d}{dx} \int_{v(x)}^{u(x)} f(t,x)dt=u'(x)f(u(x),x)-v'(x)f(v(x),x)+\int_{v(x)}^{u(x)}\frac{\partial }{\partial x} f(t,x)dt

Walter Clyburn 2022-01-05 Answered
Differentiation of multivariable function proof
\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\int_{{{v}{\left({x}\right)}}}^{{{u}{\left({x}\right)}}}}{f{{\left({t},{x}\right)}}}{\left.{d}{t}\right.}={u}'{\left({x}\right)}{f{{\left({u}{\left({x}\right)},{x}\right)}}}-{v}'{\left({x}\right)}{f{{\left({v}{\left({x}\right)},{x}\right)}}}+{\int_{{{v}{\left({x}\right)}}}^{{{u}{\left({x}\right)}}}}{\frac{{\partial}}{{\partial{x}}}}{f{{\left({t},{x}\right)}}}{\left.{d}{t}\right.}\)

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Expert Answer

Melinda McCombs
Answered 2022-01-06 Author has 2906 answers
Start with
\(\displaystyle{I}{\left({x}\right)}={\int_{{v}}^{{u}}}{f{{\left({t},{x}\right)}}}{\left.{d}{t}\right.}={F}{\left({u},{x}\right)}-{F}{\left({v},{x}\right)}\)
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.
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