Chain rule when applying L'Hopital's rule












2












$begingroup$


I have a very basic question regarding derivation function:



$$f(omega(t)) = frac{2 +x(t)cdot frac{domega(t)}{dt}}{omega(t)} $$
when I check for
$$= lim_{omega(t)to 0}frac{2 +x(t)cdotfrac{domega(t)}{dt}}{omega(t)} = frac{2}{0} $$
now if we apply L'Hopital's rule
we get



$$= lim_{omega(t)to 0}(frac{frac{d(2)}{domega(t)} +[frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt} + frac{d(frac{omega(t)}{dt})}{domega(t)}cdot frac{dx(t)}{dt}]}{frac{domega(t)}{domega(t)}} )$$
So here is my question this
$$ frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}$$
should become
$$ frac{dx(t)}{dt}$$
according to the chain rule , or am I mathematically wrong.

Please also let me know if I have done something mathematically wrong during the derivation.






derivatives partial-derivative limits-without-lhopital chain-rule







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I don't think L'Hospitals is applicable because 2/0 is not indeterminate form.
    $endgroup$
    – Zach
    Dec 6 '18 at 12:14










  • $begingroup$
    @Zach Oh thanks for the correction!
    $endgroup$
    – Robert
    Dec 6 '18 at 13:51
















2












$begingroup$


I have a very basic question regarding derivation function:



$$f(omega(t)) = frac{2 +x(t)cdot frac{domega(t)}{dt}}{omega(t)} $$
when I check for
$$= lim_{omega(t)to 0}frac{2 +x(t)cdotfrac{domega(t)}{dt}}{omega(t)} = frac{2}{0} $$
now if we apply L'Hopital's rule
we get



$$= lim_{omega(t)to 0}(frac{frac{d(2)}{domega(t)} +[frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt} + frac{d(frac{omega(t)}{dt})}{domega(t)}cdot frac{dx(t)}{dt}]}{frac{domega(t)}{domega(t)}} )$$
So here is my question this
$$ frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}$$
should become
$$ frac{dx(t)}{dt}$$
according to the chain rule , or am I mathematically wrong.

Please also let me know if I have done something mathematically wrong during the derivation.






derivatives partial-derivative limits-without-lhopital chain-rule







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I don't think L'Hospitals is applicable because 2/0 is not indeterminate form.
    $endgroup$
    – Zach
    Dec 6 '18 at 12:14










  • $begingroup$
    @Zach Oh thanks for the correction!
    $endgroup$
    – Robert
    Dec 6 '18 at 13:51














2












2








2





$begingroup$


I have a very basic question regarding derivation function:



$$f(omega(t)) = frac{2 +x(t)cdot frac{domega(t)}{dt}}{omega(t)} $$
when I check for
$$= lim_{omega(t)to 0}frac{2 +x(t)cdotfrac{domega(t)}{dt}}{omega(t)} = frac{2}{0} $$
now if we apply L'Hopital's rule
we get



$$= lim_{omega(t)to 0}(frac{frac{d(2)}{domega(t)} +[frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt} + frac{d(frac{omega(t)}{dt})}{domega(t)}cdot frac{dx(t)}{dt}]}{frac{domega(t)}{domega(t)}} )$$
So here is my question this
$$ frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}$$
should become
$$ frac{dx(t)}{dt}$$
according to the chain rule , or am I mathematically wrong.

Please also let me know if I have done something mathematically wrong during the derivation.






derivatives partial-derivative limits-without-lhopital chain-rule







share|cite|improve this question











$endgroup$




I have a very basic question regarding derivation function:



$$f(omega(t)) = frac{2 +x(t)cdot frac{domega(t)}{dt}}{omega(t)} $$
when I check for
$$= lim_{omega(t)to 0}frac{2 +x(t)cdotfrac{domega(t)}{dt}}{omega(t)} = frac{2}{0} $$
now if we apply L'Hopital's rule
we get



$$= lim_{omega(t)to 0}(frac{frac{d(2)}{domega(t)} +[frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt} + frac{d(frac{omega(t)}{dt})}{domega(t)}cdot frac{dx(t)}{dt}]}{frac{domega(t)}{domega(t)}} )$$
So here is my question this
$$ frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}$$
should become
$$ frac{dx(t)}{dt}$$
according to the chain rule , or am I mathematically wrong.

Please also let me know if I have done something mathematically wrong during the derivation.






derivatives partial-derivative limits-without-lhopital chain-rule




derivatives partial-derivative limits-without-lhopital chain-rule






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 11:42









idea

2,15841125




2,15841125










asked Dec 6 '18 at 11:37









RobertRobert

1285




1285








  • 2




    $begingroup$
    I don't think L'Hospitals is applicable because 2/0 is not indeterminate form.
    $endgroup$
    – Zach
    Dec 6 '18 at 12:14










  • $begingroup$
    @Zach Oh thanks for the correction!
    $endgroup$
    – Robert
    Dec 6 '18 at 13:51














  • 2




    $begingroup$
    I don't think L'Hospitals is applicable because 2/0 is not indeterminate form.
    $endgroup$
    – Zach
    Dec 6 '18 at 12:14










  • $begingroup$
    @Zach Oh thanks for the correction!
    $endgroup$
    – Robert
    Dec 6 '18 at 13:51








2




2




$begingroup$
I don't think L'Hospitals is applicable because 2/0 is not indeterminate form.
$endgroup$
– Zach
Dec 6 '18 at 12:14




$begingroup$
I don't think L'Hospitals is applicable because 2/0 is not indeterminate form.
$endgroup$
– Zach
Dec 6 '18 at 12:14












$begingroup$
@Zach Oh thanks for the correction!
$endgroup$
– Robert
Dec 6 '18 at 13:51




$begingroup$
@Zach Oh thanks for the correction!
$endgroup$
– Robert
Dec 6 '18 at 13:51










1 Answer
1






active

oldest

votes


















1












$begingroup$

Recall we can apply l'Hopital for expressions $frac{f(x)}{g(x)}$ in the form $frac{infty}{infty}$ or $frac{0}{0}$ or also, as an extension, for the case $f(x)^{g(x)}=e^{g(x)log(f(x))}$ when $g(x)log(f(x))$ is in the indeterminate form $frac{infty}{infty}$ or $frac{0}{0}$.



Note also that for the $frac{infty}{infty}$ case it is not strictly necessary that the numerator approaches $infty$ to apply l'Hopital, indeed the numerator need not even have a limit provided that the other conditions hold and the denominator approaches $infty$. Refer to this wiki article for a discussion on that.



With reference to your main doubt we have



$$frac{d(x(t))}{domega(t)}=frac{d(x(t))}{dt}frac{dt}{domega(t)}=frac{frac{d(x(t))}{dt}}{frac{domega(t)}{dt}}$$



and therefore



$$frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}=frac{d(x(t))}{dt}$$



Refer also to the related




  • Derivative of a function with respect to another function.






share|cite|improve this answer











$endgroup$













    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%2f3028384%2fchain-rule-when-applying-lhopitals-rule%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












    $begingroup$

    Recall we can apply l'Hopital for expressions $frac{f(x)}{g(x)}$ in the form $frac{infty}{infty}$ or $frac{0}{0}$ or also, as an extension, for the case $f(x)^{g(x)}=e^{g(x)log(f(x))}$ when $g(x)log(f(x))$ is in the indeterminate form $frac{infty}{infty}$ or $frac{0}{0}$.



    Note also that for the $frac{infty}{infty}$ case it is not strictly necessary that the numerator approaches $infty$ to apply l'Hopital, indeed the numerator need not even have a limit provided that the other conditions hold and the denominator approaches $infty$. Refer to this wiki article for a discussion on that.



    With reference to your main doubt we have



    $$frac{d(x(t))}{domega(t)}=frac{d(x(t))}{dt}frac{dt}{domega(t)}=frac{frac{d(x(t))}{dt}}{frac{domega(t)}{dt}}$$



    and therefore



    $$frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}=frac{d(x(t))}{dt}$$



    Refer also to the related




    • Derivative of a function with respect to another function.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Recall we can apply l'Hopital for expressions $frac{f(x)}{g(x)}$ in the form $frac{infty}{infty}$ or $frac{0}{0}$ or also, as an extension, for the case $f(x)^{g(x)}=e^{g(x)log(f(x))}$ when $g(x)log(f(x))$ is in the indeterminate form $frac{infty}{infty}$ or $frac{0}{0}$.



      Note also that for the $frac{infty}{infty}$ case it is not strictly necessary that the numerator approaches $infty$ to apply l'Hopital, indeed the numerator need not even have a limit provided that the other conditions hold and the denominator approaches $infty$. Refer to this wiki article for a discussion on that.



      With reference to your main doubt we have



      $$frac{d(x(t))}{domega(t)}=frac{d(x(t))}{dt}frac{dt}{domega(t)}=frac{frac{d(x(t))}{dt}}{frac{domega(t)}{dt}}$$



      and therefore



      $$frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}=frac{d(x(t))}{dt}$$



      Refer also to the related




      • Derivative of a function with respect to another function.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Recall we can apply l'Hopital for expressions $frac{f(x)}{g(x)}$ in the form $frac{infty}{infty}$ or $frac{0}{0}$ or also, as an extension, for the case $f(x)^{g(x)}=e^{g(x)log(f(x))}$ when $g(x)log(f(x))$ is in the indeterminate form $frac{infty}{infty}$ or $frac{0}{0}$.



        Note also that for the $frac{infty}{infty}$ case it is not strictly necessary that the numerator approaches $infty$ to apply l'Hopital, indeed the numerator need not even have a limit provided that the other conditions hold and the denominator approaches $infty$. Refer to this wiki article for a discussion on that.



        With reference to your main doubt we have



        $$frac{d(x(t))}{domega(t)}=frac{d(x(t))}{dt}frac{dt}{domega(t)}=frac{frac{d(x(t))}{dt}}{frac{domega(t)}{dt}}$$



        and therefore



        $$frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}=frac{d(x(t))}{dt}$$



        Refer also to the related




        • Derivative of a function with respect to another function.






        share|cite|improve this answer











        $endgroup$



        Recall we can apply l'Hopital for expressions $frac{f(x)}{g(x)}$ in the form $frac{infty}{infty}$ or $frac{0}{0}$ or also, as an extension, for the case $f(x)^{g(x)}=e^{g(x)log(f(x))}$ when $g(x)log(f(x))$ is in the indeterminate form $frac{infty}{infty}$ or $frac{0}{0}$.



        Note also that for the $frac{infty}{infty}$ case it is not strictly necessary that the numerator approaches $infty$ to apply l'Hopital, indeed the numerator need not even have a limit provided that the other conditions hold and the denominator approaches $infty$. Refer to this wiki article for a discussion on that.



        With reference to your main doubt we have



        $$frac{d(x(t))}{domega(t)}=frac{d(x(t))}{dt}frac{dt}{domega(t)}=frac{frac{d(x(t))}{dt}}{frac{domega(t)}{dt}}$$



        and therefore



        $$frac{d(x(t))}{domega(t)}cdot frac{domega(t)}{dt}=frac{d(x(t))}{dt}$$



        Refer also to the related




        • Derivative of a function with respect to another function.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 6 '18 at 13:46

























        answered Dec 6 '18 at 12:41









        gimusigimusi

        92.8k84494




        92.8k84494






























            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%2f3028384%2fchain-rule-when-applying-lhopitals-rule%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

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
            Read more