How to differentiate $int_{B(t)} f(x,t) dx$ with respect to $t$?












-2












$begingroup$


$int_{B(t)} f(x,t) dx$ is given and assume $x$ is an $n$ dimensional vector variable, $t$ is positive real variable and $f(x,t)$ is a sufficiently smooth function of both $x$ and $t$. Also $B(t)$ is a time varying region in the $d$ dimensional Euclidean space with sufficient smoothness conditions. Then what is the formula for differentiating the integral $int_{0}^{t} f(x,t) dx$ with respect to $t$? Could anyone please explain?






calculus multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Leibniz_integral_rule
    $endgroup$
    – Cameron Williams
    Dec 6 '18 at 16:15








  • 1




    $begingroup$
    Oh wait: how can $x$ be an $n$ dimensional vector if its extreme of integrations are $0$ and $t$?
    $endgroup$
    – Federico
    Dec 6 '18 at 16:21










  • $begingroup$
    Oops... I made a mistake. I will correct it now.
    $endgroup$
    – Keith
    Dec 6 '18 at 16:50
















-2












$begingroup$


$int_{B(t)} f(x,t) dx$ is given and assume $x$ is an $n$ dimensional vector variable, $t$ is positive real variable and $f(x,t)$ is a sufficiently smooth function of both $x$ and $t$. Also $B(t)$ is a time varying region in the $d$ dimensional Euclidean space with sufficient smoothness conditions. Then what is the formula for differentiating the integral $int_{0}^{t} f(x,t) dx$ with respect to $t$? Could anyone please explain?






calculus multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Leibniz_integral_rule
    $endgroup$
    – Cameron Williams
    Dec 6 '18 at 16:15








  • 1




    $begingroup$
    Oh wait: how can $x$ be an $n$ dimensional vector if its extreme of integrations are $0$ and $t$?
    $endgroup$
    – Federico
    Dec 6 '18 at 16:21










  • $begingroup$
    Oops... I made a mistake. I will correct it now.
    $endgroup$
    – Keith
    Dec 6 '18 at 16:50














-2












-2








-2





$begingroup$


$int_{B(t)} f(x,t) dx$ is given and assume $x$ is an $n$ dimensional vector variable, $t$ is positive real variable and $f(x,t)$ is a sufficiently smooth function of both $x$ and $t$. Also $B(t)$ is a time varying region in the $d$ dimensional Euclidean space with sufficient smoothness conditions. Then what is the formula for differentiating the integral $int_{0}^{t} f(x,t) dx$ with respect to $t$? Could anyone please explain?






calculus multivariable-calculus







share|cite|improve this question











$endgroup$




$int_{B(t)} f(x,t) dx$ is given and assume $x$ is an $n$ dimensional vector variable, $t$ is positive real variable and $f(x,t)$ is a sufficiently smooth function of both $x$ and $t$. Also $B(t)$ is a time varying region in the $d$ dimensional Euclidean space with sufficient smoothness conditions. Then what is the formula for differentiating the integral $int_{0}^{t} f(x,t) dx$ with respect to $t$? Could anyone please explain?






calculus multivariable-calculus




calculus multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 16:53







Keith

















asked Dec 6 '18 at 16:11









KeithKeith

1,418922




1,418922








  • 1




    $begingroup$
    en.wikipedia.org/wiki/Leibniz_integral_rule
    $endgroup$
    – Cameron Williams
    Dec 6 '18 at 16:15








  • 1




    $begingroup$
    Oh wait: how can $x$ be an $n$ dimensional vector if its extreme of integrations are $0$ and $t$?
    $endgroup$
    – Federico
    Dec 6 '18 at 16:21










  • $begingroup$
    Oops... I made a mistake. I will correct it now.
    $endgroup$
    – Keith
    Dec 6 '18 at 16:50














  • 1




    $begingroup$
    en.wikipedia.org/wiki/Leibniz_integral_rule
    $endgroup$
    – Cameron Williams
    Dec 6 '18 at 16:15








  • 1




    $begingroup$
    Oh wait: how can $x$ be an $n$ dimensional vector if its extreme of integrations are $0$ and $t$?
    $endgroup$
    – Federico
    Dec 6 '18 at 16:21










  • $begingroup$
    Oops... I made a mistake. I will correct it now.
    $endgroup$
    – Keith
    Dec 6 '18 at 16:50








1




1




$begingroup$
en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– Cameron Williams
Dec 6 '18 at 16:15






$begingroup$
en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– Cameron Williams
Dec 6 '18 at 16:15






1




1




$begingroup$
Oh wait: how can $x$ be an $n$ dimensional vector if its extreme of integrations are $0$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:21




$begingroup$
Oh wait: how can $x$ be an $n$ dimensional vector if its extreme of integrations are $0$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:21












$begingroup$
Oops... I made a mistake. I will correct it now.
$endgroup$
– Keith
Dec 6 '18 at 16:50




$begingroup$
Oops... I made a mistake. I will correct it now.
$endgroup$
– Keith
Dec 6 '18 at 16:50










1 Answer
1






active

oldest

votes


















1












$begingroup$

Some clever guy has already thought about it: look here.



ADDENDUM



In case you are interested in differentiating with respect to $t$ the function
$$
tmapstoint_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n,
$$
just notice that
$$
int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n
= int_0^t g(x_n,t) ,dx_n
$$
where
$$
g(x_n,t) = int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_{n-1},
$$
so Leibniz rule can be applied iteratively.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 16:18












  • $begingroup$
    What do you mean? Swap the role of $x$ and $t$?
    $endgroup$
    – Federico
    Dec 6 '18 at 16:20










  • $begingroup$
    That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 16:25










  • $begingroup$
    He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
    $endgroup$
    – A. Pongrácz
    Dec 6 '18 at 16:26












  • $begingroup$
    Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
    $endgroup$
    – Federico
    Dec 6 '18 at 16:29













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3028688%2fhow-to-differentiate-int-bt-fx-t-dx-with-respect-to-t%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Some clever guy has already thought about it: look here.



ADDENDUM



In case you are interested in differentiating with respect to $t$ the function
$$
tmapstoint_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n,
$$
just notice that
$$
int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n
= int_0^t g(x_n,t) ,dx_n
$$
where
$$
g(x_n,t) = int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_{n-1},
$$
so Leibniz rule can be applied iteratively.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 16:18












  • $begingroup$
    What do you mean? Swap the role of $x$ and $t$?
    $endgroup$
    – Federico
    Dec 6 '18 at 16:20










  • $begingroup$
    That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 16:25










  • $begingroup$
    He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
    $endgroup$
    – A. Pongrácz
    Dec 6 '18 at 16:26












  • $begingroup$
    Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
    $endgroup$
    – Federico
    Dec 6 '18 at 16:29


















1












$begingroup$

Some clever guy has already thought about it: look here.



ADDENDUM



In case you are interested in differentiating with respect to $t$ the function
$$
tmapstoint_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n,
$$
just notice that
$$
int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n
= int_0^t g(x_n,t) ,dx_n
$$
where
$$
g(x_n,t) = int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_{n-1},
$$
so Leibniz rule can be applied iteratively.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 16:18












  • $begingroup$
    What do you mean? Swap the role of $x$ and $t$?
    $endgroup$
    – Federico
    Dec 6 '18 at 16:20










  • $begingroup$
    That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 16:25










  • $begingroup$
    He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
    $endgroup$
    – A. Pongrácz
    Dec 6 '18 at 16:26












  • $begingroup$
    Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
    $endgroup$
    – Federico
    Dec 6 '18 at 16:29
















1












1








1





$begingroup$

Some clever guy has already thought about it: look here.



ADDENDUM



In case you are interested in differentiating with respect to $t$ the function
$$
tmapstoint_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n,
$$
just notice that
$$
int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n
= int_0^t g(x_n,t) ,dx_n
$$
where
$$
g(x_n,t) = int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_{n-1},
$$
so Leibniz rule can be applied iteratively.






share|cite|improve this answer











$endgroup$



Some clever guy has already thought about it: look here.



ADDENDUM



In case you are interested in differentiating with respect to $t$ the function
$$
tmapstoint_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n,
$$
just notice that
$$
int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_n
= int_0^t g(x_n,t) ,dx_n
$$
where
$$
g(x_n,t) = int_0^t dotsi int_0^t f(x_1,dots,x_n,t) ,dx_1dots dx_{n-1},
$$
so Leibniz rule can be applied iteratively.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 6 '18 at 16:54

























answered Dec 6 '18 at 16:16









FedericoFederico

4,954514




4,954514












  • $begingroup$
    That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 16:18












  • $begingroup$
    What do you mean? Swap the role of $x$ and $t$?
    $endgroup$
    – Federico
    Dec 6 '18 at 16:20










  • $begingroup$
    That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 16:25










  • $begingroup$
    He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
    $endgroup$
    – A. Pongrácz
    Dec 6 '18 at 16:26












  • $begingroup$
    Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
    $endgroup$
    – Federico
    Dec 6 '18 at 16:29




















  • $begingroup$
    That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 16:18












  • $begingroup$
    What do you mean? Swap the role of $x$ and $t$?
    $endgroup$
    – Federico
    Dec 6 '18 at 16:20










  • $begingroup$
    That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
    $endgroup$
    – José Carlos Santos
    Dec 6 '18 at 16:25










  • $begingroup$
    He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
    $endgroup$
    – A. Pongrácz
    Dec 6 '18 at 16:26












  • $begingroup$
    Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
    $endgroup$
    – Federico
    Dec 6 '18 at 16:29


















$begingroup$
That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:18






$begingroup$
That's for functions of the type $xmapstoint_a^bf(x,t),mathrm dt$. How do you apply it in this context?
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:18














$begingroup$
What do you mean? Swap the role of $x$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:20




$begingroup$
What do you mean? Swap the role of $x$ and $t$?
$endgroup$
– Federico
Dec 6 '18 at 16:20












$begingroup$
That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:25




$begingroup$
That's not important. What's important is that in the OP's question one must differentiate with respect to $t$ in two spots, since we are dealing with$$tmapstoint_0^tf(x,y),mathrm dx.$$If it was, say$$tmapstoint_0^pi f(x,y),mathrm dx,$$then I would agree with you.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:25












$begingroup$
He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
$endgroup$
– A. Pongrácz
Dec 6 '18 at 16:26






$begingroup$
He means that $f$ is supposed to be $(n+1)$-ary, as $x$ is the collection of the first $n$ variables. Your link points to the solution if $f$ is binary. Of course there is some problem with the notation, maybe all the first $n$ variables should run from $0$ to $t$?
$endgroup$
– A. Pongrácz
Dec 6 '18 at 16:26














$begingroup$
Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
$endgroup$
– Federico
Dec 6 '18 at 16:29






$begingroup$
Guys, I don't understand. Leibniz allows to differentiate with respect to $t$ the function $tmapsto int_0^t f(x,t),dx$. $x$ of course has to be a scalar variable, since the extremes of integrations are $0$ and $t$
$endgroup$
– Federico
Dec 6 '18 at 16:29




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3028688%2fhow-to-differentiate-int-bt-fx-t-dx-with-respect-to-t%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more