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

      2,15841125




      2,15841125










      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











            $endgroup$













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              2 Answers
              2






              active

              oldest

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              2 Answers
              2






              active

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              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

                  33.5k12261




                  33.5k12261























                      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

                          52k43474




                          52k43474






























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