Chain rule - Find $frac{partial z}{partial x}$ and $frac{partial z}{partial y} $ if $z=pq+qw$












0












$begingroup$



Find $frac{partial z}{partial x}$ and $frac{partial z}{partial y} $ if



$z=pq+qw$



$p=2x-y$, $q=x-2y$ and $w=-2x-2y$




Is $frac{partial z}{partial x}$ equals to:



$frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
frac{partial z}{partial q} frac{partial q}{partial x}+ frac{partial z}{partial w} frac{partial w}{partial x}$



Or
$frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
frac{partial z}{partial w} frac{partial w}{partial x}$



I can complete the solution, I need only the correct rule. Thanks






calculus chain-rule







share|cite|improve this question









$endgroup$

















    0












    $begingroup$



    Find $frac{partial z}{partial x}$ and $frac{partial z}{partial y} $ if



    $z=pq+qw$



    $p=2x-y$, $q=x-2y$ and $w=-2x-2y$




    Is $frac{partial z}{partial x}$ equals to:



    $frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
    frac{partial z}{partial q} frac{partial q}{partial x}+ frac{partial z}{partial w} frac{partial w}{partial x}$



    Or
    $frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
    frac{partial z}{partial w} frac{partial w}{partial x}$



    I can complete the solution, I need only the correct rule. Thanks






    calculus chain-rule







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      Find $frac{partial z}{partial x}$ and $frac{partial z}{partial y} $ if



      $z=pq+qw$



      $p=2x-y$, $q=x-2y$ and $w=-2x-2y$




      Is $frac{partial z}{partial x}$ equals to:



      $frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
      frac{partial z}{partial q} frac{partial q}{partial x}+ frac{partial z}{partial w} frac{partial w}{partial x}$



      Or
      $frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
      frac{partial z}{partial w} frac{partial w}{partial x}$



      I can complete the solution, I need only the correct rule. Thanks






      calculus chain-rule







      share|cite|improve this question









      $endgroup$





      Find $frac{partial z}{partial x}$ and $frac{partial z}{partial y} $ if



      $z=pq+qw$



      $p=2x-y$, $q=x-2y$ and $w=-2x-2y$




      Is $frac{partial z}{partial x}$ equals to:



      $frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
      frac{partial z}{partial q} frac{partial q}{partial x}+ frac{partial z}{partial w} frac{partial w}{partial x}$



      Or
      $frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
      frac{partial z}{partial w} frac{partial w}{partial x}$



      I can complete the solution, I need only the correct rule. Thanks






      calculus chain-rule




      calculus chain-rule






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 3 '18 at 17:32









      DimaDima

      615416




      615416






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          $$frac{partial z}{partial x}=frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
          frac{partial z}{partial w} frac{partial w}{partial x}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much
            $endgroup$
            – Dima
            Dec 3 '18 at 20:00



















          1












          $begingroup$

          $frac{partial z}{partial x}=frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
          frac{partial z}{partial w} frac{partial w}{partial x}=$          , then



          $frac{partial z}{partial x}=p+w$ and



          $frac{partial z}{partial y}=-3q-2p-2w$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much, for $frac{partial z}{partial y}$ I think it is $-q-2(p+w)$, what do you think?
            $endgroup$
            – Dima
            Dec 3 '18 at 19:59










          • $begingroup$
            Thankful. no $frac{partial z}{partial w}frac{partial w}{partial y}=-2q$
            $endgroup$
            – yavar
            Dec 4 '18 at 13:58











          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%2f3024392%2fchain-rule-find-frac-partial-z-partial-x-and-frac-partial-z-partia%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












          $begingroup$

          $$frac{partial z}{partial x}=frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
          frac{partial z}{partial w} frac{partial w}{partial x}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much
            $endgroup$
            – Dima
            Dec 3 '18 at 20:00
















          1












          $begingroup$

          $$frac{partial z}{partial x}=frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
          frac{partial z}{partial w} frac{partial w}{partial x}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much
            $endgroup$
            – Dima
            Dec 3 '18 at 20:00














          1












          1








          1





          $begingroup$

          $$frac{partial z}{partial x}=frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
          frac{partial z}{partial w} frac{partial w}{partial x}$$






          share|cite|improve this answer









          $endgroup$



          $$frac{partial z}{partial x}=frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
          frac{partial z}{partial w} frac{partial w}{partial x}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 17:37









          Anubhab GhosalAnubhab Ghosal

          81618




          81618












          • $begingroup$
            Thank you so much
            $endgroup$
            – Dima
            Dec 3 '18 at 20:00


















          • $begingroup$
            Thank you so much
            $endgroup$
            – Dima
            Dec 3 '18 at 20:00
















          $begingroup$
          Thank you so much
          $endgroup$
          – Dima
          Dec 3 '18 at 20:00




          $begingroup$
          Thank you so much
          $endgroup$
          – Dima
          Dec 3 '18 at 20:00











          1












          $begingroup$

          $frac{partial z}{partial x}=frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
          frac{partial z}{partial w} frac{partial w}{partial x}=$          , then



          $frac{partial z}{partial x}=p+w$ and



          $frac{partial z}{partial y}=-3q-2p-2w$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much, for $frac{partial z}{partial y}$ I think it is $-q-2(p+w)$, what do you think?
            $endgroup$
            – Dima
            Dec 3 '18 at 19:59










          • $begingroup$
            Thankful. no $frac{partial z}{partial w}frac{partial w}{partial y}=-2q$
            $endgroup$
            – yavar
            Dec 4 '18 at 13:58
















          1












          $begingroup$

          $frac{partial z}{partial x}=frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
          frac{partial z}{partial w} frac{partial w}{partial x}=$          , then



          $frac{partial z}{partial x}=p+w$ and



          $frac{partial z}{partial y}=-3q-2p-2w$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much, for $frac{partial z}{partial y}$ I think it is $-q-2(p+w)$, what do you think?
            $endgroup$
            – Dima
            Dec 3 '18 at 19:59










          • $begingroup$
            Thankful. no $frac{partial z}{partial w}frac{partial w}{partial y}=-2q$
            $endgroup$
            – yavar
            Dec 4 '18 at 13:58














          1












          1








          1





          $begingroup$

          $frac{partial z}{partial x}=frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
          frac{partial z}{partial w} frac{partial w}{partial x}=$          , then



          $frac{partial z}{partial x}=p+w$ and



          $frac{partial z}{partial y}=-3q-2p-2w$






          share|cite|improve this answer









          $endgroup$



          $frac{partial z}{partial x}=frac{partial z}{partial p} frac{partial p}{partial x}+ frac{partial z}{partial q} frac{partial q}{partial x}+
          frac{partial z}{partial w} frac{partial w}{partial x}=$          , then



          $frac{partial z}{partial x}=p+w$ and



          $frac{partial z}{partial y}=-3q-2p-2w$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 19:57









          yavaryavar

          843




          843












          • $begingroup$
            Thank you so much, for $frac{partial z}{partial y}$ I think it is $-q-2(p+w)$, what do you think?
            $endgroup$
            – Dima
            Dec 3 '18 at 19:59










          • $begingroup$
            Thankful. no $frac{partial z}{partial w}frac{partial w}{partial y}=-2q$
            $endgroup$
            – yavar
            Dec 4 '18 at 13:58


















          • $begingroup$
            Thank you so much, for $frac{partial z}{partial y}$ I think it is $-q-2(p+w)$, what do you think?
            $endgroup$
            – Dima
            Dec 3 '18 at 19:59










          • $begingroup$
            Thankful. no $frac{partial z}{partial w}frac{partial w}{partial y}=-2q$
            $endgroup$
            – yavar
            Dec 4 '18 at 13:58
















          $begingroup$
          Thank you so much, for $frac{partial z}{partial y}$ I think it is $-q-2(p+w)$, what do you think?
          $endgroup$
          – Dima
          Dec 3 '18 at 19:59




          $begingroup$
          Thank you so much, for $frac{partial z}{partial y}$ I think it is $-q-2(p+w)$, what do you think?
          $endgroup$
          – Dima
          Dec 3 '18 at 19:59












          $begingroup$
          Thankful. no $frac{partial z}{partial w}frac{partial w}{partial y}=-2q$
          $endgroup$
          – yavar
          Dec 4 '18 at 13:58




          $begingroup$
          Thankful. no $frac{partial z}{partial w}frac{partial w}{partial y}=-2q$
          $endgroup$
          – yavar
          Dec 4 '18 at 13:58


















          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%2f3024392%2fchain-rule-find-frac-partial-z-partial-x-and-frac-partial-z-partia%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

          How do I know what Microsoft account the skydrive app is syncing to?

          Image
          up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
          Read more

          Grease: Live!

          Image
          A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
          Read more

          When does type information flow backwards in C++?

          Image
          up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
          Read more