Understanding operator under a subtitution












5












$begingroup$


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?










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$endgroup$

















    5












    $begingroup$


    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?










    share|cite|improve this question









    $endgroup$















      5












      5








      5


      1



      $begingroup$


      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?










      share|cite|improve this question









      $endgroup$




      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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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 9 at 5:14









      Jimmy SabaterJimmy Sabater

      2,389319




      2,389319






















          2 Answers
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          10












          $begingroup$

          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









          $endgroup$





















            3












            $begingroup$

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






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              2 Answers
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              2 Answers
              2






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              10












              $begingroup$

              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









              $endgroup$


















                10












                $begingroup$

                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









                $endgroup$
















                  10












                  10








                  10





                  $begingroup$

                  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









                  $endgroup$



                  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 Jan 9 at 5:33









                  Robert LewisRobert Lewis

                  45.1k23065




                  45.1k23065























                      3












                      $begingroup$

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






                      share|cite|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

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






                        share|cite|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

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






                          share|cite|improve this answer









                          $endgroup$



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







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 9 at 5:22









                          Kavi Rama MurthyKavi Rama Murthy

                          55.7k42158




                          55.7k42158






























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