Understanding operator under a subtitution












3














in my notes, I have the following phrase:



With $x = e^t$ and $x frac{d}{dx} = frac{d}{dt}$



How, how come we get $x frac{d}{dx} = frac{d}{dt} $? I know if I diferentiate with respect to $t$ I obtain



$$ frac{d}{dt} x = x $$



I do not understand how this operators are defined? maybe I am misunderstading the notation?






calculus







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    3














    in my notes, I have the following phrase:



    With $x = e^t$ and $x frac{d}{dx} = frac{d}{dt}$



    How, how come we get $x frac{d}{dx} = frac{d}{dt} $? I know if I diferentiate with respect to $t$ I obtain



    $$ frac{d}{dt} x = x $$



    I do not understand how this operators are defined? maybe I am misunderstading the notation?






    calculus







    share|cite|improve this question

























      3












      3








      3


      1





      in my notes, I have the following phrase:



      With $x = e^t$ and $x frac{d}{dx} = frac{d}{dt}$



      How, how come we get $x frac{d}{dx} = frac{d}{dt} $? I know if I diferentiate with respect to $t$ I obtain



      $$ frac{d}{dt} x = x $$



      I do not understand how this operators are defined? maybe I am misunderstading the notation?






      calculus







      share|cite|improve this question













      in my notes, I have the following phrase:



      With $x = e^t$ and $x frac{d}{dx} = frac{d}{dt}$



      How, how come we get $x frac{d}{dx} = frac{d}{dt} $? I know if I diferentiate with respect to $t$ I obtain



      $$ frac{d}{dt} x = x $$



      I do not understand how this operators are defined? maybe I am misunderstading the notation?






      calculus




      calculus






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      share|cite|improve this question











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      asked 2 hours ago









      Jimmy SabaterJimmy Sabater

      1,997219




      1,997219






















          2 Answers
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          With



          $x = e^t tag 1$



          we have



          $dfrac{dx}{dt} = e^t = x; tag 2$



          thus for any function $f$



          $xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$



          by the chain rule. Thus,



          $xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$






          share|cite|improve this answer





























            2














            $frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              2














              With



              $x = e^t tag 1$



              we have



              $dfrac{dx}{dt} = e^t = x; tag 2$



              thus for any function $f$



              $xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$



              by the chain rule. Thus,



              $xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$






              share|cite|improve this answer


























                2














                With



                $x = e^t tag 1$



                we have



                $dfrac{dx}{dt} = e^t = x; tag 2$



                thus for any function $f$



                $xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$



                by the chain rule. Thus,



                $xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$






                share|cite|improve this answer
























                  2












                  2








                  2






                  With



                  $x = e^t tag 1$



                  we have



                  $dfrac{dx}{dt} = e^t = x; tag 2$



                  thus for any function $f$



                  $xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$



                  by the chain rule. Thus,



                  $xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$






                  share|cite|improve this answer












                  With



                  $x = e^t tag 1$



                  we have



                  $dfrac{dx}{dt} = e^t = x; tag 2$



                  thus for any function $f$



                  $xdfrac{df}{dx} = dfrac{dx}{dt} dfrac{df}{dx} = dfrac{df}{dt} tag 3$



                  by the chain rule. Thus,



                  $xdfrac{d}{dx} = dfrac{d}{dt}. tag 4$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Robert LewisRobert Lewis

                  44k22963




                  44k22963























                      2














                      $frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.






                      share|cite|improve this answer


























                        2














                        $frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.






                        share|cite|improve this answer
























                          2












                          2








                          2






                          $frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.






                          share|cite|improve this answer












                          $frac {dy} {dt} =frac {dy} {dx}frac {dx} {dt} =frac {dy} {dx} e^{t}=frac {dy} {dx} x$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 2 hours ago









                          Kavi Rama MurthyKavi Rama Murthy

                          51.8k32055




                          51.8k32055






























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