What ins the derivative of int_0^tvarphi(t,u)du












-1














Let $f(t)=int_0^tvarphi(t,u)du$ where $varphi$ is a continuos function. Do yo know how to compute the derivative of f?
It is not possible to directly apply the fundamental theorem of calculus.






multivariable-calculus







share|cite|improve this question




















  • 1




    Do you mean the derivative of $f$?
    – Olivier Moschetta
    Nov 30 at 14:17
















-1














Let $f(t)=int_0^tvarphi(t,u)du$ where $varphi$ is a continuos function. Do yo know how to compute the derivative of f?
It is not possible to directly apply the fundamental theorem of calculus.






multivariable-calculus







share|cite|improve this question




















  • 1




    Do you mean the derivative of $f$?
    – Olivier Moschetta
    Nov 30 at 14:17














-1












-1








-1







Let $f(t)=int_0^tvarphi(t,u)du$ where $varphi$ is a continuos function. Do yo know how to compute the derivative of f?
It is not possible to directly apply the fundamental theorem of calculus.






multivariable-calculus







share|cite|improve this question















Let $f(t)=int_0^tvarphi(t,u)du$ where $varphi$ is a continuos function. Do yo know how to compute the derivative of f?
It is not possible to directly apply the fundamental theorem of calculus.






multivariable-calculus




multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 at 18:11

























asked Nov 30 at 14:15









Adrián Hinojosa Calleja

937




937








  • 1




    Do you mean the derivative of $f$?
    – Olivier Moschetta
    Nov 30 at 14:17














  • 1




    Do you mean the derivative of $f$?
    – Olivier Moschetta
    Nov 30 at 14:17








1




1




Do you mean the derivative of $f$?
– Olivier Moschetta
Nov 30 at 14:17




Do you mean the derivative of $f$?
– Olivier Moschetta
Nov 30 at 14:17










1 Answer
1






active

oldest

votes


















0














I'm assuming you meant the derivative of $f$ in your post.



In a simple setting we need some assumptions on $varphi$ namely that $dfrac{partialvarphi}{partial t}$ exists and is continuous (to simplify things). Then the derivative of $f$ is obtained by
$$f'(t)=int_0^t dfrac{partialvarphi}{partial t}(t,u),du+varphi(t,t)$$
We get two bits, in the first one we differentiate under the integral sign while the second looks like an application of the fundamental theorem of calculus.



Note that the integral above is well-defined since $dfrac{partialvarphi}{partial t}$ is continuous.






share|cite|improve this answer



















  • 1




    You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
    – Dog_69
    Nov 30 at 14:24












  • Thanks! I'll correct.
    – Olivier Moschetta
    Nov 30 at 14:25






  • 1




    I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
    – Dog_69
    Nov 30 at 19:26










  • You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
    – Olivier Moschetta
    Dec 1 at 15:42










  • Okey. Perfect. :)
    – Dog_69
    Dec 1 at 21:02











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%2f3020139%2fwhat-ins-the-derivative-of-int-0t-varphit-udu%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









0














I'm assuming you meant the derivative of $f$ in your post.



In a simple setting we need some assumptions on $varphi$ namely that $dfrac{partialvarphi}{partial t}$ exists and is continuous (to simplify things). Then the derivative of $f$ is obtained by
$$f'(t)=int_0^t dfrac{partialvarphi}{partial t}(t,u),du+varphi(t,t)$$
We get two bits, in the first one we differentiate under the integral sign while the second looks like an application of the fundamental theorem of calculus.



Note that the integral above is well-defined since $dfrac{partialvarphi}{partial t}$ is continuous.






share|cite|improve this answer



















  • 1




    You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
    – Dog_69
    Nov 30 at 14:24












  • Thanks! I'll correct.
    – Olivier Moschetta
    Nov 30 at 14:25






  • 1




    I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
    – Dog_69
    Nov 30 at 19:26










  • You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
    – Olivier Moschetta
    Dec 1 at 15:42










  • Okey. Perfect. :)
    – Dog_69
    Dec 1 at 21:02
















0














I'm assuming you meant the derivative of $f$ in your post.



In a simple setting we need some assumptions on $varphi$ namely that $dfrac{partialvarphi}{partial t}$ exists and is continuous (to simplify things). Then the derivative of $f$ is obtained by
$$f'(t)=int_0^t dfrac{partialvarphi}{partial t}(t,u),du+varphi(t,t)$$
We get two bits, in the first one we differentiate under the integral sign while the second looks like an application of the fundamental theorem of calculus.



Note that the integral above is well-defined since $dfrac{partialvarphi}{partial t}$ is continuous.






share|cite|improve this answer



















  • 1




    You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
    – Dog_69
    Nov 30 at 14:24












  • Thanks! I'll correct.
    – Olivier Moschetta
    Nov 30 at 14:25






  • 1




    I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
    – Dog_69
    Nov 30 at 19:26










  • You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
    – Olivier Moschetta
    Dec 1 at 15:42










  • Okey. Perfect. :)
    – Dog_69
    Dec 1 at 21:02














0












0








0






I'm assuming you meant the derivative of $f$ in your post.



In a simple setting we need some assumptions on $varphi$ namely that $dfrac{partialvarphi}{partial t}$ exists and is continuous (to simplify things). Then the derivative of $f$ is obtained by
$$f'(t)=int_0^t dfrac{partialvarphi}{partial t}(t,u),du+varphi(t,t)$$
We get two bits, in the first one we differentiate under the integral sign while the second looks like an application of the fundamental theorem of calculus.



Note that the integral above is well-defined since $dfrac{partialvarphi}{partial t}$ is continuous.






share|cite|improve this answer














I'm assuming you meant the derivative of $f$ in your post.



In a simple setting we need some assumptions on $varphi$ namely that $dfrac{partialvarphi}{partial t}$ exists and is continuous (to simplify things). Then the derivative of $f$ is obtained by
$$f'(t)=int_0^t dfrac{partialvarphi}{partial t}(t,u),du+varphi(t,t)$$
We get two bits, in the first one we differentiate under the integral sign while the second looks like an application of the fundamental theorem of calculus.



Note that the integral above is well-defined since $dfrac{partialvarphi}{partial t}$ is continuous.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 1 at 15:42

























answered Nov 30 at 14:21









Olivier Moschetta

2,7761411




2,7761411








  • 1




    You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
    – Dog_69
    Nov 30 at 14:24












  • Thanks! I'll correct.
    – Olivier Moschetta
    Nov 30 at 14:25






  • 1




    I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
    – Dog_69
    Nov 30 at 19:26










  • You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
    – Olivier Moschetta
    Dec 1 at 15:42










  • Okey. Perfect. :)
    – Dog_69
    Dec 1 at 21:02














  • 1




    You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
    – Dog_69
    Nov 30 at 14:24












  • Thanks! I'll correct.
    – Olivier Moschetta
    Nov 30 at 14:25






  • 1




    I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
    – Dog_69
    Nov 30 at 19:26










  • You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
    – Olivier Moschetta
    Dec 1 at 15:42










  • Okey. Perfect. :)
    – Dog_69
    Dec 1 at 21:02








1




1




You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
– Dog_69
Nov 30 at 14:24






You mean $varphi(t,t)$ and $varphi(t,0)$. But the last one should not appear in my opinion.
– Dog_69
Nov 30 at 14:24














Thanks! I'll correct.
– Olivier Moschetta
Nov 30 at 14:25




Thanks! I'll correct.
– Olivier Moschetta
Nov 30 at 14:25




1




1




I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
– Dog_69
Nov 30 at 19:26




I don't understand why you write $varphi(t,0)$. In a single variable the theorem States that $d/dt int_0^t f(u)du= f(t)$. The value of $f$ at the initial point doesn't matter.
– Dog_69
Nov 30 at 19:26












You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
– Olivier Moschetta
Dec 1 at 15:42




You're right, it shouldn't be there. I wrongly applied a more general result where the bounds of the integral are two functions $g(t)$ and $h(t)$, then you get one term for each bound.
– Olivier Moschetta
Dec 1 at 15:42












Okey. Perfect. :)
– Dog_69
Dec 1 at 21:02




Okey. Perfect. :)
– Dog_69
Dec 1 at 21:02


















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3020139%2fwhat-ins-the-derivative-of-int-0t-varphit-udu%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

To store a contact into the json file from server.js file using a class in NodeJS

Image
0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
Read more

Redirect URL with Chrome Remote Debugging Android Devices

Image
1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
Read more

Dieringhausen

Image
Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
Read more