If $frac{x}{y^frac{n-1}{n}}$ is constant, how do I prove $frac{dx}{x}=frac{n-1}{n}frac{dy}{y}$?












1














From:



$frac{x}{y^frac{n-1}{n}}=constant$



To:



$frac{dx}{x}=frac{n-1}{n}frac{dy}{y}$



It's from a turbomachinery lecture. The text I'm studying says that binomial expansion is used, but I don't get it.






calculus derivatives







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    Take logarithms first and differentiate both sides.
    – Claude Leibovici
    Dec 1 at 18:06
















1














From:



$frac{x}{y^frac{n-1}{n}}=constant$



To:



$frac{dx}{x}=frac{n-1}{n}frac{dy}{y}$



It's from a turbomachinery lecture. The text I'm studying says that binomial expansion is used, but I don't get it.






calculus derivatives







share|cite|improve this question




















  • 3




    Take logarithms first and differentiate both sides.
    – Claude Leibovici
    Dec 1 at 18:06














1












1








1


1





From:



$frac{x}{y^frac{n-1}{n}}=constant$



To:



$frac{dx}{x}=frac{n-1}{n}frac{dy}{y}$



It's from a turbomachinery lecture. The text I'm studying says that binomial expansion is used, but I don't get it.






calculus derivatives







share|cite|improve this question















From:



$frac{x}{y^frac{n-1}{n}}=constant$



To:



$frac{dx}{x}=frac{n-1}{n}frac{dy}{y}$



It's from a turbomachinery lecture. The text I'm studying says that binomial expansion is used, but I don't get it.






calculus derivatives




calculus derivatives






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edited Dec 1 at 18:56









Jam

4,94811431




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asked Dec 1 at 17:59









user2235427

1083




1083








  • 3




    Take logarithms first and differentiate both sides.
    – Claude Leibovici
    Dec 1 at 18:06














  • 3




    Take logarithms first and differentiate both sides.
    – Claude Leibovici
    Dec 1 at 18:06








3




3




Take logarithms first and differentiate both sides.
– Claude Leibovici
Dec 1 at 18:06




Take logarithms first and differentiate both sides.
– Claude Leibovici
Dec 1 at 18:06










2 Answers
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Let $x=ccdot y^frac{n-1}{n}$. Then
begin{align}
frac{dx}{dy}&=frac{n-1}{n}ccdot y^frac{-1}{n}\
&=frac{n-1}{n}ccdot y^frac{-1}{n} cdot frac{y}{y}\
&=frac{n-1}{n}ccdot y^frac{n-1}{n} cdot frac{1}{y}\
&=frac{n-1}{n} frac{x}{y}\
end{align}






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    1














    We want



    $dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{y}{x}; tag 0$



    from



    $dfrac{x}{y^{frac{n - 1}{n}}} = alpha = text{constant}, tag 1$



    assuming of course $x$, $y$, and $n$ are all in range sufficient to legitimize our operations, we have:



    $x = alpha y^{frac{n - 1}{n}}; tag 2$



    $x^n = alpha^n y^{n - 1}; tag 3$



    $nx^{n - 1} = alpha^n (n - 1)y^{n - 2} dfrac{dy}{dx}; tag 4$



    $alpha^n dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{x^{n - 1}}{y^{n - 2}}; tag 5$



    from (3),



    $alpha^n = dfrac{x^n}{y^{n - 1}}; tag 6$



    thus, dividing (5) by (6),



    $dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{x^{n - 1}}{y^{n - 2}} dfrac{y^{n - 1}}{x^n} = dfrac{n}{n - 1} dfrac{y}{x}, tag 7$



    which we may write in differential form to obtain



    $dfrac{dx}{x} = dfrac{n - 1}{n} dfrac{dy}{y}; tag 8$



    as per request. $OEDelta$






    share|cite|improve this answer





















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      2 Answers
      2






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      oldest

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      2 Answers
      2






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      active

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      active

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      4














      Let $x=ccdot y^frac{n-1}{n}$. Then
      begin{align}
      frac{dx}{dy}&=frac{n-1}{n}ccdot y^frac{-1}{n}\
      &=frac{n-1}{n}ccdot y^frac{-1}{n} cdot frac{y}{y}\
      &=frac{n-1}{n}ccdot y^frac{n-1}{n} cdot frac{1}{y}\
      &=frac{n-1}{n} frac{x}{y}\
      end{align}






      share|cite|improve this answer


























        4














        Let $x=ccdot y^frac{n-1}{n}$. Then
        begin{align}
        frac{dx}{dy}&=frac{n-1}{n}ccdot y^frac{-1}{n}\
        &=frac{n-1}{n}ccdot y^frac{-1}{n} cdot frac{y}{y}\
        &=frac{n-1}{n}ccdot y^frac{n-1}{n} cdot frac{1}{y}\
        &=frac{n-1}{n} frac{x}{y}\
        end{align}






        share|cite|improve this answer
























          4












          4








          4






          Let $x=ccdot y^frac{n-1}{n}$. Then
          begin{align}
          frac{dx}{dy}&=frac{n-1}{n}ccdot y^frac{-1}{n}\
          &=frac{n-1}{n}ccdot y^frac{-1}{n} cdot frac{y}{y}\
          &=frac{n-1}{n}ccdot y^frac{n-1}{n} cdot frac{1}{y}\
          &=frac{n-1}{n} frac{x}{y}\
          end{align}






          share|cite|improve this answer












          Let $x=ccdot y^frac{n-1}{n}$. Then
          begin{align}
          frac{dx}{dy}&=frac{n-1}{n}ccdot y^frac{-1}{n}\
          &=frac{n-1}{n}ccdot y^frac{-1}{n} cdot frac{y}{y}\
          &=frac{n-1}{n}ccdot y^frac{n-1}{n} cdot frac{1}{y}\
          &=frac{n-1}{n} frac{x}{y}\
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 at 18:12









          Thomas Shelby

          1,342216




          1,342216























              1














              We want



              $dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{y}{x}; tag 0$



              from



              $dfrac{x}{y^{frac{n - 1}{n}}} = alpha = text{constant}, tag 1$



              assuming of course $x$, $y$, and $n$ are all in range sufficient to legitimize our operations, we have:



              $x = alpha y^{frac{n - 1}{n}}; tag 2$



              $x^n = alpha^n y^{n - 1}; tag 3$



              $nx^{n - 1} = alpha^n (n - 1)y^{n - 2} dfrac{dy}{dx}; tag 4$



              $alpha^n dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{x^{n - 1}}{y^{n - 2}}; tag 5$



              from (3),



              $alpha^n = dfrac{x^n}{y^{n - 1}}; tag 6$



              thus, dividing (5) by (6),



              $dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{x^{n - 1}}{y^{n - 2}} dfrac{y^{n - 1}}{x^n} = dfrac{n}{n - 1} dfrac{y}{x}, tag 7$



              which we may write in differential form to obtain



              $dfrac{dx}{x} = dfrac{n - 1}{n} dfrac{dy}{y}; tag 8$



              as per request. $OEDelta$






              share|cite|improve this answer


























                1














                We want



                $dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{y}{x}; tag 0$



                from



                $dfrac{x}{y^{frac{n - 1}{n}}} = alpha = text{constant}, tag 1$



                assuming of course $x$, $y$, and $n$ are all in range sufficient to legitimize our operations, we have:



                $x = alpha y^{frac{n - 1}{n}}; tag 2$



                $x^n = alpha^n y^{n - 1}; tag 3$



                $nx^{n - 1} = alpha^n (n - 1)y^{n - 2} dfrac{dy}{dx}; tag 4$



                $alpha^n dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{x^{n - 1}}{y^{n - 2}}; tag 5$



                from (3),



                $alpha^n = dfrac{x^n}{y^{n - 1}}; tag 6$



                thus, dividing (5) by (6),



                $dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{x^{n - 1}}{y^{n - 2}} dfrac{y^{n - 1}}{x^n} = dfrac{n}{n - 1} dfrac{y}{x}, tag 7$



                which we may write in differential form to obtain



                $dfrac{dx}{x} = dfrac{n - 1}{n} dfrac{dy}{y}; tag 8$



                as per request. $OEDelta$






                share|cite|improve this answer
























                  1












                  1








                  1






                  We want



                  $dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{y}{x}; tag 0$



                  from



                  $dfrac{x}{y^{frac{n - 1}{n}}} = alpha = text{constant}, tag 1$



                  assuming of course $x$, $y$, and $n$ are all in range sufficient to legitimize our operations, we have:



                  $x = alpha y^{frac{n - 1}{n}}; tag 2$



                  $x^n = alpha^n y^{n - 1}; tag 3$



                  $nx^{n - 1} = alpha^n (n - 1)y^{n - 2} dfrac{dy}{dx}; tag 4$



                  $alpha^n dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{x^{n - 1}}{y^{n - 2}}; tag 5$



                  from (3),



                  $alpha^n = dfrac{x^n}{y^{n - 1}}; tag 6$



                  thus, dividing (5) by (6),



                  $dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{x^{n - 1}}{y^{n - 2}} dfrac{y^{n - 1}}{x^n} = dfrac{n}{n - 1} dfrac{y}{x}, tag 7$



                  which we may write in differential form to obtain



                  $dfrac{dx}{x} = dfrac{n - 1}{n} dfrac{dy}{y}; tag 8$



                  as per request. $OEDelta$






                  share|cite|improve this answer












                  We want



                  $dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{y}{x}; tag 0$



                  from



                  $dfrac{x}{y^{frac{n - 1}{n}}} = alpha = text{constant}, tag 1$



                  assuming of course $x$, $y$, and $n$ are all in range sufficient to legitimize our operations, we have:



                  $x = alpha y^{frac{n - 1}{n}}; tag 2$



                  $x^n = alpha^n y^{n - 1}; tag 3$



                  $nx^{n - 1} = alpha^n (n - 1)y^{n - 2} dfrac{dy}{dx}; tag 4$



                  $alpha^n dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{x^{n - 1}}{y^{n - 2}}; tag 5$



                  from (3),



                  $alpha^n = dfrac{x^n}{y^{n - 1}}; tag 6$



                  thus, dividing (5) by (6),



                  $dfrac{dy}{dx} = dfrac{n}{n - 1} dfrac{x^{n - 1}}{y^{n - 2}} dfrac{y^{n - 1}}{x^n} = dfrac{n}{n - 1} dfrac{y}{x}, tag 7$



                  which we may write in differential form to obtain



                  $dfrac{dx}{x} = dfrac{n - 1}{n} dfrac{dy}{y}; tag 8$



                  as per request. $OEDelta$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 1 at 19:16









                  Robert Lewis

                  43.5k22863




                  43.5k22863






























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