Derivative Simplification












0












$begingroup$


Need help proving the following:



$frac{dp}{d ln(t)} = t frac {dp}{dt} $



This is what I have so far:



Applying chain rule.



$frac{dp}{d ln(t)} = frac {dp}{dt}frac {dt}{d ln(t)} $



Simplifying:



$frac {dp}{dt}frac {dt}{d ln(t)} = frac {dp}{dt}frac {1}{frac{d}{dt}} ln(t)$



Need assistance with the rest.
Thanks!






calculus







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  • $begingroup$
    $frac {dt}{d ln(t)}=frac 1{frac {d ln(t)}{dt}}$ could help
    $endgroup$
    – Claude Leibovici
    Dec 10 '18 at 8:35
















0












$begingroup$


Need help proving the following:



$frac{dp}{d ln(t)} = t frac {dp}{dt} $



This is what I have so far:



Applying chain rule.



$frac{dp}{d ln(t)} = frac {dp}{dt}frac {dt}{d ln(t)} $



Simplifying:



$frac {dp}{dt}frac {dt}{d ln(t)} = frac {dp}{dt}frac {1}{frac{d}{dt}} ln(t)$



Need assistance with the rest.
Thanks!






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    $frac {dt}{d ln(t)}=frac 1{frac {d ln(t)}{dt}}$ could help
    $endgroup$
    – Claude Leibovici
    Dec 10 '18 at 8:35














0












0








0





$begingroup$


Need help proving the following:



$frac{dp}{d ln(t)} = t frac {dp}{dt} $



This is what I have so far:



Applying chain rule.



$frac{dp}{d ln(t)} = frac {dp}{dt}frac {dt}{d ln(t)} $



Simplifying:



$frac {dp}{dt}frac {dt}{d ln(t)} = frac {dp}{dt}frac {1}{frac{d}{dt}} ln(t)$



Need assistance with the rest.
Thanks!






calculus







share|cite|improve this question









$endgroup$




Need help proving the following:



$frac{dp}{d ln(t)} = t frac {dp}{dt} $



This is what I have so far:



Applying chain rule.



$frac{dp}{d ln(t)} = frac {dp}{dt}frac {dt}{d ln(t)} $



Simplifying:



$frac {dp}{dt}frac {dt}{d ln(t)} = frac {dp}{dt}frac {1}{frac{d}{dt}} ln(t)$



Need assistance with the rest.
Thanks!






calculus




calculus






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asked Dec 10 '18 at 6:21









MrCleanMrClean

84




84












  • $begingroup$
    $frac {dt}{d ln(t)}=frac 1{frac {d ln(t)}{dt}}$ could help
    $endgroup$
    – Claude Leibovici
    Dec 10 '18 at 8:35


















  • $begingroup$
    $frac {dt}{d ln(t)}=frac 1{frac {d ln(t)}{dt}}$ could help
    $endgroup$
    – Claude Leibovici
    Dec 10 '18 at 8:35
















$begingroup$
$frac {dt}{d ln(t)}=frac 1{frac {d ln(t)}{dt}}$ could help
$endgroup$
– Claude Leibovici
Dec 10 '18 at 8:35




$begingroup$
$frac {dt}{d ln(t)}=frac 1{frac {d ln(t)}{dt}}$ could help
$endgroup$
– Claude Leibovici
Dec 10 '18 at 8:35










1 Answer
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$begingroup$

Leibniz's notations can sometimes be confusing. Writing something like $frac{1}{frac{mathrm{d}}{mathrm{d}t}}$ is terrible.



Let $p(t) = q(ln(t))$. Then



$$frac{mathrm{d}p(t)}{mathrm{d}ln(t)} = frac{mathrm{d} q(ln(t))}{mathrm{d} ln(t)} = q'(ln(t)).$$ Here we treat $ln(t)$ as a single variable.



Also,



$$frac{mathrm{d}p(t)}{mathrm{d}t} = frac{mathrm{d}q(ln(t))}{mathrm{d}t}= q'(ln(t))(ln(t))'=q'(ln(t))frac{1}{t}.$$






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    $begingroup$

    Leibniz's notations can sometimes be confusing. Writing something like $frac{1}{frac{mathrm{d}}{mathrm{d}t}}$ is terrible.



    Let $p(t) = q(ln(t))$. Then



    $$frac{mathrm{d}p(t)}{mathrm{d}ln(t)} = frac{mathrm{d} q(ln(t))}{mathrm{d} ln(t)} = q'(ln(t)).$$ Here we treat $ln(t)$ as a single variable.



    Also,



    $$frac{mathrm{d}p(t)}{mathrm{d}t} = frac{mathrm{d}q(ln(t))}{mathrm{d}t}= q'(ln(t))(ln(t))'=q'(ln(t))frac{1}{t}.$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Leibniz's notations can sometimes be confusing. Writing something like $frac{1}{frac{mathrm{d}}{mathrm{d}t}}$ is terrible.



      Let $p(t) = q(ln(t))$. Then



      $$frac{mathrm{d}p(t)}{mathrm{d}ln(t)} = frac{mathrm{d} q(ln(t))}{mathrm{d} ln(t)} = q'(ln(t)).$$ Here we treat $ln(t)$ as a single variable.



      Also,



      $$frac{mathrm{d}p(t)}{mathrm{d}t} = frac{mathrm{d}q(ln(t))}{mathrm{d}t}= q'(ln(t))(ln(t))'=q'(ln(t))frac{1}{t}.$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Leibniz's notations can sometimes be confusing. Writing something like $frac{1}{frac{mathrm{d}}{mathrm{d}t}}$ is terrible.



        Let $p(t) = q(ln(t))$. Then



        $$frac{mathrm{d}p(t)}{mathrm{d}ln(t)} = frac{mathrm{d} q(ln(t))}{mathrm{d} ln(t)} = q'(ln(t)).$$ Here we treat $ln(t)$ as a single variable.



        Also,



        $$frac{mathrm{d}p(t)}{mathrm{d}t} = frac{mathrm{d}q(ln(t))}{mathrm{d}t}= q'(ln(t))(ln(t))'=q'(ln(t))frac{1}{t}.$$






        share|cite|improve this answer









        $endgroup$



        Leibniz's notations can sometimes be confusing. Writing something like $frac{1}{frac{mathrm{d}}{mathrm{d}t}}$ is terrible.



        Let $p(t) = q(ln(t))$. Then



        $$frac{mathrm{d}p(t)}{mathrm{d}ln(t)} = frac{mathrm{d} q(ln(t))}{mathrm{d} ln(t)} = q'(ln(t)).$$ Here we treat $ln(t)$ as a single variable.



        Also,



        $$frac{mathrm{d}p(t)}{mathrm{d}t} = frac{mathrm{d}q(ln(t))}{mathrm{d}t}= q'(ln(t))(ln(t))'=q'(ln(t))frac{1}{t}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 6:35









        tonychow0929tonychow0929

        29825




        29825






























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