Simplify Derivative with Substitution











up vote
3
down vote

favorite












I try to evaluate:



$$ frac{partial}{partial x} log{u(x, y, z)}$$



Mathematica gives:



$$ frac{1}{x+y+z}$$



I want to simplify the expression with my function:



$$ frac{1}{u(x, y, z)}$$



How to do that?



Thanks.



u[x_, y_, z_] = x + y + z
Simplify[D[Log[u[x, y, z]], x]]









share|improve this question







New contributor




R zu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
























    up vote
    3
    down vote

    favorite












    I try to evaluate:



    $$ frac{partial}{partial x} log{u(x, y, z)}$$



    Mathematica gives:



    $$ frac{1}{x+y+z}$$



    I want to simplify the expression with my function:



    $$ frac{1}{u(x, y, z)}$$



    How to do that?



    Thanks.



    u[x_, y_, z_] = x + y + z
    Simplify[D[Log[u[x, y, z]], x]]









    share|improve this question







    New contributor




    R zu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      I try to evaluate:



      $$ frac{partial}{partial x} log{u(x, y, z)}$$



      Mathematica gives:



      $$ frac{1}{x+y+z}$$



      I want to simplify the expression with my function:



      $$ frac{1}{u(x, y, z)}$$



      How to do that?



      Thanks.



      u[x_, y_, z_] = x + y + z
      Simplify[D[Log[u[x, y, z]], x]]









      share|improve this question







      New contributor




      R zu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I try to evaluate:



      $$ frac{partial}{partial x} log{u(x, y, z)}$$



      Mathematica gives:



      $$ frac{1}{x+y+z}$$



      I want to simplify the expression with my function:



      $$ frac{1}{u(x, y, z)}$$



      How to do that?



      Thanks.



      u[x_, y_, z_] = x + y + z
      Simplify[D[Log[u[x, y, z]], x]]






      calculus-and-analysis simplifying-expressions






      share|improve this question







      New contributor




      R zu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question







      New contributor




      R zu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question






      New contributor




      R zu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 8 hours ago









      R zu

      1797




      1797




      New contributor




      R zu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      R zu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      R zu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          5
          down vote



          accepted










          D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]



          1/u[x, y, z]







          share|improve this answer























          • A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
            – R zu
            8 hours ago










          • @Rzu, good point.
            – kglr
            8 hours ago


















          up vote
          3
          down vote













          An alternative is to define UpValues instead of DownValues of u:



          Derivative[1, 0, 0][u] ^:= 1&
          Derivative[0, 1, 0][u] ^:= 1&
          Derivative[0, 0, 1][u] ^:= 1&

          D[Log[u[x, y, z]], x]



          1/u[x, y, z]







          share|improve this answer





















          • What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
            – R zu
            6 hours ago












          • @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
            – Carl Woll
            6 hours ago













          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: "387"
          };
          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',
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          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
          },
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });






          R zu is a new contributor. Be nice, and check out our Code of Conduct.










           

          draft saved


          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f186514%2fsimplify-derivative-with-substitution%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








          up vote
          5
          down vote



          accepted










          D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]



          1/u[x, y, z]







          share|improve this answer























          • A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
            – R zu
            8 hours ago










          • @Rzu, good point.
            – kglr
            8 hours ago















          up vote
          5
          down vote



          accepted










          D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]



          1/u[x, y, z]







          share|improve this answer























          • A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
            – R zu
            8 hours ago










          • @Rzu, good point.
            – kglr
            8 hours ago













          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]



          1/u[x, y, z]







          share|improve this answer














          D[Log[u[x, y, z]], x] /. u[x_, y_, z_] :> Defer[u[x, y, z]]



          1/u[x, y, z]








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 8 hours ago

























          answered 8 hours ago









          kglr

          172k8194399




          172k8194399












          • A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
            – R zu
            8 hours ago










          • @Rzu, good point.
            – kglr
            8 hours ago


















          • A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
            – R zu
            8 hours ago










          • @Rzu, good point.
            – kglr
            8 hours ago
















          A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
          – R zu
          8 hours ago




          A more general substitution: /. u[x_,y_,z_] -> Defer[u[x,y,z]]
          – R zu
          8 hours ago












          @Rzu, good point.
          – kglr
          8 hours ago




          @Rzu, good point.
          – kglr
          8 hours ago










          up vote
          3
          down vote













          An alternative is to define UpValues instead of DownValues of u:



          Derivative[1, 0, 0][u] ^:= 1&
          Derivative[0, 1, 0][u] ^:= 1&
          Derivative[0, 0, 1][u] ^:= 1&

          D[Log[u[x, y, z]], x]



          1/u[x, y, z]







          share|improve this answer





















          • What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
            – R zu
            6 hours ago












          • @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
            – Carl Woll
            6 hours ago

















          up vote
          3
          down vote













          An alternative is to define UpValues instead of DownValues of u:



          Derivative[1, 0, 0][u] ^:= 1&
          Derivative[0, 1, 0][u] ^:= 1&
          Derivative[0, 0, 1][u] ^:= 1&

          D[Log[u[x, y, z]], x]



          1/u[x, y, z]







          share|improve this answer





















          • What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
            – R zu
            6 hours ago












          • @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
            – Carl Woll
            6 hours ago















          up vote
          3
          down vote










          up vote
          3
          down vote









          An alternative is to define UpValues instead of DownValues of u:



          Derivative[1, 0, 0][u] ^:= 1&
          Derivative[0, 1, 0][u] ^:= 1&
          Derivative[0, 0, 1][u] ^:= 1&

          D[Log[u[x, y, z]], x]



          1/u[x, y, z]







          share|improve this answer












          An alternative is to define UpValues instead of DownValues of u:



          Derivative[1, 0, 0][u] ^:= 1&
          Derivative[0, 1, 0][u] ^:= 1&
          Derivative[0, 0, 1][u] ^:= 1&

          D[Log[u[x, y, z]], x]



          1/u[x, y, z]








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 6 hours ago









          Carl Woll

          65.6k285171




          65.6k285171












          • What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
            – R zu
            6 hours ago












          • @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
            – Carl Woll
            6 hours ago




















          • What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
            – R zu
            6 hours ago












          • @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
            – Carl Woll
            6 hours ago


















          What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
          – R zu
          6 hours ago






          What are UpValues and DownValues? The definition in the doc seems recursive: UpValue "gives a list of transformation rules corresponding to all upvalues defined for the symbol f. "
          – R zu
          6 hours ago














          @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
          – Carl Woll
          6 hours ago






          @Rzu Maybe you can check out the documentation for UpSetDelayed and TagSetDelayed.
          – Carl Woll
          6 hours ago












          R zu is a new contributor. Be nice, and check out our Code of Conduct.










           

          draft saved


          draft discarded


















          R zu is a new contributor. Be nice, and check out our Code of Conduct.













          R zu is a new contributor. Be nice, and check out our Code of Conduct.












          R zu is a new contributor. Be nice, and check out our Code of Conduct.















           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f186514%2fsimplify-derivative-with-substitution%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

          Lallio

          Futebolista

          Jornalista