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

1,413920




1,413920








  • 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













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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

4,889514




4,889514












  • $begingroup$
    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




















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Fiat S.p.A.

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Cet article court présente un sujet plus développé dans : Fiat et Fiat Chrysler Automobiles. Fiat S.p.A. Fiat S.p.A. ( Fabbrica Italiana Automobili Torino , acronyme FIAT) est un groupe, notamment connu comme société-mère du constructeur automobile italien, Fiat. Portail des entreprises Portail de Turin Portail de l’automobile This page is only for reference, If you need detailed information, please check here
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