$frac{partial{z}}{partial{y}}$ if $z$ is given implicitly












0












$begingroup$


Suppose $z$ is given implicitly as:



$$e^z-x^2y-y^2z = 0$$



Find $$frac{partial{z}}{partial{y}}$$.



I let $F(x,y) = e^z-x^2y-y^2z$. Then,



$$frac{partial{F}}{partial{y}} = -x^2-2yz$$
$$frac{partial{F}}{partial{z}} = e^z-y^2$$



$$frac{frac{partial{F}}{partial{y}}}{frac{partial{F}}{partial{z}}} = frac{partial{F}}{partial{y}} cdot frac{partial{z}}{partial{F}} = frac{partial{z}}{partial{y}}$$



Therefore,



$$frac{partial{z}}{partial{y}} = frac{-x^2-2yz}{e^z-y^2}$$



However, the exercise cites the answer as



$$frac{x^2+2yz}{e^z-y^2}$$



For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.






multivariable-calculus chain-rule







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
    $endgroup$
    – Dair
    Dec 10 '18 at 5:17










  • $begingroup$
    I think so too. But, is there some reasoning as to why the negative just cancels out?
    $endgroup$
    – Art
    Dec 10 '18 at 5:21
















0












$begingroup$


Suppose $z$ is given implicitly as:



$$e^z-x^2y-y^2z = 0$$



Find $$frac{partial{z}}{partial{y}}$$.



I let $F(x,y) = e^z-x^2y-y^2z$. Then,



$$frac{partial{F}}{partial{y}} = -x^2-2yz$$
$$frac{partial{F}}{partial{z}} = e^z-y^2$$



$$frac{frac{partial{F}}{partial{y}}}{frac{partial{F}}{partial{z}}} = frac{partial{F}}{partial{y}} cdot frac{partial{z}}{partial{F}} = frac{partial{z}}{partial{y}}$$



Therefore,



$$frac{partial{z}}{partial{y}} = frac{-x^2-2yz}{e^z-y^2}$$



However, the exercise cites the answer as



$$frac{x^2+2yz}{e^z-y^2}$$



For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.






multivariable-calculus chain-rule







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
    $endgroup$
    – Dair
    Dec 10 '18 at 5:17










  • $begingroup$
    I think so too. But, is there some reasoning as to why the negative just cancels out?
    $endgroup$
    – Art
    Dec 10 '18 at 5:21














0












0








0





$begingroup$


Suppose $z$ is given implicitly as:



$$e^z-x^2y-y^2z = 0$$



Find $$frac{partial{z}}{partial{y}}$$.



I let $F(x,y) = e^z-x^2y-y^2z$. Then,



$$frac{partial{F}}{partial{y}} = -x^2-2yz$$
$$frac{partial{F}}{partial{z}} = e^z-y^2$$



$$frac{frac{partial{F}}{partial{y}}}{frac{partial{F}}{partial{z}}} = frac{partial{F}}{partial{y}} cdot frac{partial{z}}{partial{F}} = frac{partial{z}}{partial{y}}$$



Therefore,



$$frac{partial{z}}{partial{y}} = frac{-x^2-2yz}{e^z-y^2}$$



However, the exercise cites the answer as



$$frac{x^2+2yz}{e^z-y^2}$$



For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.






multivariable-calculus chain-rule







share|cite|improve this question









$endgroup$




Suppose $z$ is given implicitly as:



$$e^z-x^2y-y^2z = 0$$



Find $$frac{partial{z}}{partial{y}}$$.



I let $F(x,y) = e^z-x^2y-y^2z$. Then,



$$frac{partial{F}}{partial{y}} = -x^2-2yz$$
$$frac{partial{F}}{partial{z}} = e^z-y^2$$



$$frac{frac{partial{F}}{partial{y}}}{frac{partial{F}}{partial{z}}} = frac{partial{F}}{partial{y}} cdot frac{partial{z}}{partial{F}} = frac{partial{z}}{partial{y}}$$



Therefore,



$$frac{partial{z}}{partial{y}} = frac{-x^2-2yz}{e^z-y^2}$$



However, the exercise cites the answer as



$$frac{x^2+2yz}{e^z-y^2}$$



For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.






multivariable-calculus chain-rule




multivariable-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 10 '18 at 5:09









ArtArt

32618




32618












  • $begingroup$
    I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
    $endgroup$
    – Dair
    Dec 10 '18 at 5:17










  • $begingroup$
    I think so too. But, is there some reasoning as to why the negative just cancels out?
    $endgroup$
    – Art
    Dec 10 '18 at 5:21


















  • $begingroup$
    I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
    $endgroup$
    – Dair
    Dec 10 '18 at 5:17










  • $begingroup$
    I think so too. But, is there some reasoning as to why the negative just cancels out?
    $endgroup$
    – Art
    Dec 10 '18 at 5:21
















$begingroup$
I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
$endgroup$
– Dair
Dec 10 '18 at 5:17




$begingroup$
I think the problem is that partial derivatives are not fractions... They don't obey the same cancellation laws as fractions.
$endgroup$
– Dair
Dec 10 '18 at 5:17












$begingroup$
I think so too. But, is there some reasoning as to why the negative just cancels out?
$endgroup$
– Art
Dec 10 '18 at 5:21




$begingroup$
I think so too. But, is there some reasoning as to why the negative just cancels out?
$endgroup$
– Art
Dec 10 '18 at 5:21










2 Answers
2






active

oldest

votes


















0












$begingroup$

Looking through my copy of Stewart's calculus we have



$$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$



and similarly



$$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$



So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.



    $e^z-x^2y-y^2z = 0$



    $z_ye^z-x^2-2yz-y^2z_y=0$



    $z_y(e^z-y^2)=x^2+2yz$



    Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$



    Added



    This proves the minus sign in the formula you used.






    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%2f3033477%2ffrac-partialz-partialy-if-z-is-given-implicitly%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









      0












      $begingroup$

      Looking through my copy of Stewart's calculus we have



      $$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$



      and similarly



      $$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$



      So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Looking through my copy of Stewart's calculus we have



        $$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$



        and similarly



        $$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$



        So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Looking through my copy of Stewart's calculus we have



          $$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$



          and similarly



          $$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$



          So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.






          share|cite|improve this answer









          $endgroup$



          Looking through my copy of Stewart's calculus we have



          $$frac{partial z}{partial x} = -frac{frac{partial F}{partial x}}{frac{partial F}{partial z}}$$



          and similarly



          $$frac{partial z}{partial y} = -frac{frac{partial F}{partial y}}{frac{partial F}{partial z}}$$



          So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 5:43









          carsandpulsarscarsandpulsars

          37817




          37817























              0












              $begingroup$

              Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.



              $e^z-x^2y-y^2z = 0$



              $z_ye^z-x^2-2yz-y^2z_y=0$



              $z_y(e^z-y^2)=x^2+2yz$



              Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$



              Added



              This proves the minus sign in the formula you used.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.



                $e^z-x^2y-y^2z = 0$



                $z_ye^z-x^2-2yz-y^2z_y=0$



                $z_y(e^z-y^2)=x^2+2yz$



                Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$



                Added



                This proves the minus sign in the formula you used.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.



                  $e^z-x^2y-y^2z = 0$



                  $z_ye^z-x^2-2yz-y^2z_y=0$



                  $z_y(e^z-y^2)=x^2+2yz$



                  Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$



                  Added



                  This proves the minus sign in the formula you used.






                  share|cite|improve this answer









                  $endgroup$



                  Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.



                  $e^z-x^2y-y^2z = 0$



                  $z_ye^z-x^2-2yz-y^2z_y=0$



                  $z_y(e^z-y^2)=x^2+2yz$



                  Leading to the given answer: $z_y=dfrac{x^2+2yz}{e^z-y^2}$



                  Added



                  This proves the minus sign in the formula you used.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 5:44









                  Rafa BudríaRafa Budría

                  5,9101825




                  5,9101825






























                      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%2f3033477%2ffrac-partialz-partialy-if-z-is-given-implicitly%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

                      Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

                      Image
                      2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
                      Read more

                      ComboBox Display Member on multiple fields

                      Image
                      2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
                      Read more

                      Is it possible to collect Nectar points via Trainline?

                      Image
                      2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
                      Read more