The partial derivative












0












$begingroup$


Suppose the variables $x$ and $u$ are related by



$$x=u$$



Then I have a function $f=f(x)$ which does not explicitly depend on $u$.



Then is it true that $$frac{partial f}{partial u}=0$$?






partial-derivative







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    0












    $begingroup$


    Suppose the variables $x$ and $u$ are related by



    $$x=u$$



    Then I have a function $f=f(x)$ which does not explicitly depend on $u$.



    Then is it true that $$frac{partial f}{partial u}=0$$?






    partial-derivative







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose the variables $x$ and $u$ are related by



      $$x=u$$



      Then I have a function $f=f(x)$ which does not explicitly depend on $u$.



      Then is it true that $$frac{partial f}{partial u}=0$$?






      partial-derivative







      share|cite|improve this question









      $endgroup$




      Suppose the variables $x$ and $u$ are related by



      $$x=u$$



      Then I have a function $f=f(x)$ which does not explicitly depend on $u$.



      Then is it true that $$frac{partial f}{partial u}=0$$?






      partial-derivative




      partial-derivative






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      share|cite|improve this question











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      asked Dec 4 '18 at 2:19









      glowstonetreesglowstonetrees

      2,336418




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          $begingroup$

          $frac{partial f}{partial u} = frac{partial f}{partial x} frac{partial x}{partial u} = f'(x)$, since $frac{partial x}{partial u} = 1$.






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            1












            $begingroup$

            No, since you have only one variable, the symbol you should be using is



            $$
            frac{{rm d}}{{rm d}u}
            $$



            and



            $$
            require{cancel}
            frac{{rm d}f}{{rm d}u} = frac{{rm d}f}{{rm d}x}cancelto{1}{frac{{rm d}x}{{rm d}u}} = frac{{rm d}f}{{rm d}x}
            $$






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              $begingroup$

              $frac{partial f}{partial u} = frac{partial f}{partial x} frac{partial x}{partial u} = f'(x)$, since $frac{partial x}{partial u} = 1$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $frac{partial f}{partial u} = frac{partial f}{partial x} frac{partial x}{partial u} = f'(x)$, since $frac{partial x}{partial u} = 1$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $frac{partial f}{partial u} = frac{partial f}{partial x} frac{partial x}{partial u} = f'(x)$, since $frac{partial x}{partial u} = 1$.






                  share|cite|improve this answer









                  $endgroup$



                  $frac{partial f}{partial u} = frac{partial f}{partial x} frac{partial x}{partial u} = f'(x)$, since $frac{partial x}{partial u} = 1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 2:24









                  Aditya DuaAditya Dua

                  1,11418




                  1,11418























                      1












                      $begingroup$

                      No, since you have only one variable, the symbol you should be using is



                      $$
                      frac{{rm d}}{{rm d}u}
                      $$



                      and



                      $$
                      require{cancel}
                      frac{{rm d}f}{{rm d}u} = frac{{rm d}f}{{rm d}x}cancelto{1}{frac{{rm d}x}{{rm d}u}} = frac{{rm d}f}{{rm d}x}
                      $$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        No, since you have only one variable, the symbol you should be using is



                        $$
                        frac{{rm d}}{{rm d}u}
                        $$



                        and



                        $$
                        require{cancel}
                        frac{{rm d}f}{{rm d}u} = frac{{rm d}f}{{rm d}x}cancelto{1}{frac{{rm d}x}{{rm d}u}} = frac{{rm d}f}{{rm d}x}
                        $$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          No, since you have only one variable, the symbol you should be using is



                          $$
                          frac{{rm d}}{{rm d}u}
                          $$



                          and



                          $$
                          require{cancel}
                          frac{{rm d}f}{{rm d}u} = frac{{rm d}f}{{rm d}x}cancelto{1}{frac{{rm d}x}{{rm d}u}} = frac{{rm d}f}{{rm d}x}
                          $$






                          share|cite|improve this answer









                          $endgroup$



                          No, since you have only one variable, the symbol you should be using is



                          $$
                          frac{{rm d}}{{rm d}u}
                          $$



                          and



                          $$
                          require{cancel}
                          frac{{rm d}f}{{rm d}u} = frac{{rm d}f}{{rm d}x}cancelto{1}{frac{{rm d}x}{{rm d}u}} = frac{{rm d}f}{{rm d}x}
                          $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 4 '18 at 2:25









                          caveraccaverac

                          14.5k31130




                          14.5k31130






























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