Chain rule doubt












1












$begingroup$


I have a doubt of appling the chain rule.
I have this $L$ function:
$$
L = ycdot log(frac{e^{a x+b}}{e^{ax+b} + exp^{cx+d}})
$$
I can rewrite it as:
$$
L = ycdot log(p)
$$
where
$$
p = frac{e^{v_{0}}}{e^{v_{0}}+e^{v_{1}}}
$$
$$
v_{0} = ax+b
$$
$$
v_{1} = cx+d
$$



If I apply the chain rule I have this:



$$
frac{partial L}{partial x} = frac{partial L}{partial p} cdot frac{partial p}{partial v_{0}} cdot frac{partial v_{0}}{partial x}
$$



But I know that I am missing somewhere the value of $frac{partial v_{1}}{partial x}$



What I am doing wrong?






derivatives chain-rule







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I have a doubt of appling the chain rule.
    I have this $L$ function:
    $$
    L = ycdot log(frac{e^{a x+b}}{e^{ax+b} + exp^{cx+d}})
    $$
    I can rewrite it as:
    $$
    L = ycdot log(p)
    $$
    where
    $$
    p = frac{e^{v_{0}}}{e^{v_{0}}+e^{v_{1}}}
    $$
    $$
    v_{0} = ax+b
    $$
    $$
    v_{1} = cx+d
    $$



    If I apply the chain rule I have this:



    $$
    frac{partial L}{partial x} = frac{partial L}{partial p} cdot frac{partial p}{partial v_{0}} cdot frac{partial v_{0}}{partial x}
    $$



    But I know that I am missing somewhere the value of $frac{partial v_{1}}{partial x}$



    What I am doing wrong?






    derivatives chain-rule







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I have a doubt of appling the chain rule.
      I have this $L$ function:
      $$
      L = ycdot log(frac{e^{a x+b}}{e^{ax+b} + exp^{cx+d}})
      $$
      I can rewrite it as:
      $$
      L = ycdot log(p)
      $$
      where
      $$
      p = frac{e^{v_{0}}}{e^{v_{0}}+e^{v_{1}}}
      $$
      $$
      v_{0} = ax+b
      $$
      $$
      v_{1} = cx+d
      $$



      If I apply the chain rule I have this:



      $$
      frac{partial L}{partial x} = frac{partial L}{partial p} cdot frac{partial p}{partial v_{0}} cdot frac{partial v_{0}}{partial x}
      $$



      But I know that I am missing somewhere the value of $frac{partial v_{1}}{partial x}$



      What I am doing wrong?






      derivatives chain-rule







      share|cite|improve this question











      $endgroup$




      I have a doubt of appling the chain rule.
      I have this $L$ function:
      $$
      L = ycdot log(frac{e^{a x+b}}{e^{ax+b} + exp^{cx+d}})
      $$
      I can rewrite it as:
      $$
      L = ycdot log(p)
      $$
      where
      $$
      p = frac{e^{v_{0}}}{e^{v_{0}}+e^{v_{1}}}
      $$
      $$
      v_{0} = ax+b
      $$
      $$
      v_{1} = cx+d
      $$



      If I apply the chain rule I have this:



      $$
      frac{partial L}{partial x} = frac{partial L}{partial p} cdot frac{partial p}{partial v_{0}} cdot frac{partial v_{0}}{partial x}
      $$



      But I know that I am missing somewhere the value of $frac{partial v_{1}}{partial x}$



      What I am doing wrong?






      derivatives chain-rule




      derivatives chain-rule






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 12:48









      idea

      2,16841125




      2,16841125










      asked Dec 13 '18 at 12:35









      MaikMaik

      212




      212






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Note that $$p=p(v_0,v_1)\
          frac{dp}{dx}=frac{partial p}{partial v_0}frac{dv_0}{dx}+frac{partial p}{partial v_1}frac{dv_1}{dx}$$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Your result is wrong because $p$ is a function of $x$ that contains the two functions $nu_0$ and $nu_1$ but not in the nested form $p(nu_0(nu_1))$. So we must write its derivative using also the quotient rule as:



            $$
            frac{partial p}{partial x}=frac{e^{nu_0}frac{partial nu_0}{partial x}(e^{v_{0}}+e^{v_{1}})-e^{nu_0}(e^{v_{0}}frac{partial nu_0}{partial x}+e^{v_{1}}frac{partial nu_1}{partial x})}{(e^{v_{0}}+e^{v_{1}})^2}
            $$



            and the correct final result is
            $$
            frac{partial L}{partial x} = frac{partial L}{partial p} cdot frac{e^{nu_0}frac{partial nu_0}{partial x}(e^{v_{0}}+e^{v_{1}})-e^{nu_0}(e^{v_{0}}frac{partial nu_0}{partial x}+e^{v_{1}}frac{partial nu_1}{partial x})}{(e^{v_{0}}+e^{v_{1}})^2}
            $$






            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%2f3037950%2fchain-rule-doubt%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Note that $$p=p(v_0,v_1)\
              frac{dp}{dx}=frac{partial p}{partial v_0}frac{dv_0}{dx}+frac{partial p}{partial v_1}frac{dv_1}{dx}$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Note that $$p=p(v_0,v_1)\
                frac{dp}{dx}=frac{partial p}{partial v_0}frac{dv_0}{dx}+frac{partial p}{partial v_1}frac{dv_1}{dx}$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Note that $$p=p(v_0,v_1)\
                  frac{dp}{dx}=frac{partial p}{partial v_0}frac{dv_0}{dx}+frac{partial p}{partial v_1}frac{dv_1}{dx}$$






                  share|cite|improve this answer









                  $endgroup$



                  Note that $$p=p(v_0,v_1)\
                  frac{dp}{dx}=frac{partial p}{partial v_0}frac{dv_0}{dx}+frac{partial p}{partial v_1}frac{dv_1}{dx}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 13:05









                  Empy2Empy2

                  33.6k12362




                  33.6k12362























                      0












                      $begingroup$

                      Your result is wrong because $p$ is a function of $x$ that contains the two functions $nu_0$ and $nu_1$ but not in the nested form $p(nu_0(nu_1))$. So we must write its derivative using also the quotient rule as:



                      $$
                      frac{partial p}{partial x}=frac{e^{nu_0}frac{partial nu_0}{partial x}(e^{v_{0}}+e^{v_{1}})-e^{nu_0}(e^{v_{0}}frac{partial nu_0}{partial x}+e^{v_{1}}frac{partial nu_1}{partial x})}{(e^{v_{0}}+e^{v_{1}})^2}
                      $$



                      and the correct final result is
                      $$
                      frac{partial L}{partial x} = frac{partial L}{partial p} cdot frac{e^{nu_0}frac{partial nu_0}{partial x}(e^{v_{0}}+e^{v_{1}})-e^{nu_0}(e^{v_{0}}frac{partial nu_0}{partial x}+e^{v_{1}}frac{partial nu_1}{partial x})}{(e^{v_{0}}+e^{v_{1}})^2}
                      $$






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        Your result is wrong because $p$ is a function of $x$ that contains the two functions $nu_0$ and $nu_1$ but not in the nested form $p(nu_0(nu_1))$. So we must write its derivative using also the quotient rule as:



                        $$
                        frac{partial p}{partial x}=frac{e^{nu_0}frac{partial nu_0}{partial x}(e^{v_{0}}+e^{v_{1}})-e^{nu_0}(e^{v_{0}}frac{partial nu_0}{partial x}+e^{v_{1}}frac{partial nu_1}{partial x})}{(e^{v_{0}}+e^{v_{1}})^2}
                        $$



                        and the correct final result is
                        $$
                        frac{partial L}{partial x} = frac{partial L}{partial p} cdot frac{e^{nu_0}frac{partial nu_0}{partial x}(e^{v_{0}}+e^{v_{1}})-e^{nu_0}(e^{v_{0}}frac{partial nu_0}{partial x}+e^{v_{1}}frac{partial nu_1}{partial x})}{(e^{v_{0}}+e^{v_{1}})^2}
                        $$






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Your result is wrong because $p$ is a function of $x$ that contains the two functions $nu_0$ and $nu_1$ but not in the nested form $p(nu_0(nu_1))$. So we must write its derivative using also the quotient rule as:



                          $$
                          frac{partial p}{partial x}=frac{e^{nu_0}frac{partial nu_0}{partial x}(e^{v_{0}}+e^{v_{1}})-e^{nu_0}(e^{v_{0}}frac{partial nu_0}{partial x}+e^{v_{1}}frac{partial nu_1}{partial x})}{(e^{v_{0}}+e^{v_{1}})^2}
                          $$



                          and the correct final result is
                          $$
                          frac{partial L}{partial x} = frac{partial L}{partial p} cdot frac{e^{nu_0}frac{partial nu_0}{partial x}(e^{v_{0}}+e^{v_{1}})-e^{nu_0}(e^{v_{0}}frac{partial nu_0}{partial x}+e^{v_{1}}frac{partial nu_1}{partial x})}{(e^{v_{0}}+e^{v_{1}})^2}
                          $$






                          share|cite|improve this answer











                          $endgroup$



                          Your result is wrong because $p$ is a function of $x$ that contains the two functions $nu_0$ and $nu_1$ but not in the nested form $p(nu_0(nu_1))$. So we must write its derivative using also the quotient rule as:



                          $$
                          frac{partial p}{partial x}=frac{e^{nu_0}frac{partial nu_0}{partial x}(e^{v_{0}}+e^{v_{1}})-e^{nu_0}(e^{v_{0}}frac{partial nu_0}{partial x}+e^{v_{1}}frac{partial nu_1}{partial x})}{(e^{v_{0}}+e^{v_{1}})^2}
                          $$



                          and the correct final result is
                          $$
                          frac{partial L}{partial x} = frac{partial L}{partial p} cdot frac{e^{nu_0}frac{partial nu_0}{partial x}(e^{v_{0}}+e^{v_{1}})-e^{nu_0}(e^{v_{0}}frac{partial nu_0}{partial x}+e^{v_{1}}frac{partial nu_1}{partial x})}{(e^{v_{0}}+e^{v_{1}})^2}
                          $$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 13 '18 at 13:10

























                          answered Dec 13 '18 at 12:59









                          Emilio NovatiEmilio Novati

                          52.1k43474




                          52.1k43474






























                              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%2f3037950%2fchain-rule-doubt%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

                              Plaza Victoria

                              Image
                              Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                              Read more

                              Puebla de Zaragoza

                              Image
                              Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                              Read more

                              Musa

                              Image
                              Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                              Read more