How to use the chain rule for change of variable












1












$begingroup$


I have asked this questions: Change of variables in differential equation?



...but after thinking about it, I am still a little confused of how to rigorously use the chain rule to calculate the derivative(s) of a function for a change of variable.



I have the following derivative:



$f(x) = frac{dw(x)}{dx}$



Now I introduce the change of variable: $hat{x}=frac{x}{L}$
and I apply the chain rule:




  1. I write: $g(hat{x}) = L hat{x} = x$

  2. I substitute: $f(g(hat{x})) = frac{dw(g(hat{x}))}{d(g(hat{x}))}$


...but this does not help me... I am confusing something.



I would be glad, if someone could show me in detail and step by step how to do this rigorously.



Thanks a lot.






derivatives chain-rule







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I have asked this questions: Change of variables in differential equation?



    ...but after thinking about it, I am still a little confused of how to rigorously use the chain rule to calculate the derivative(s) of a function for a change of variable.



    I have the following derivative:



    $f(x) = frac{dw(x)}{dx}$



    Now I introduce the change of variable: $hat{x}=frac{x}{L}$
    and I apply the chain rule:




    1. I write: $g(hat{x}) = L hat{x} = x$

    2. I substitute: $f(g(hat{x})) = frac{dw(g(hat{x}))}{d(g(hat{x}))}$


    ...but this does not help me... I am confusing something.



    I would be glad, if someone could show me in detail and step by step how to do this rigorously.



    Thanks a lot.






    derivatives chain-rule







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I have asked this questions: Change of variables in differential equation?



      ...but after thinking about it, I am still a little confused of how to rigorously use the chain rule to calculate the derivative(s) of a function for a change of variable.



      I have the following derivative:



      $f(x) = frac{dw(x)}{dx}$



      Now I introduce the change of variable: $hat{x}=frac{x}{L}$
      and I apply the chain rule:




      1. I write: $g(hat{x}) = L hat{x} = x$

      2. I substitute: $f(g(hat{x})) = frac{dw(g(hat{x}))}{d(g(hat{x}))}$


      ...but this does not help me... I am confusing something.



      I would be glad, if someone could show me in detail and step by step how to do this rigorously.



      Thanks a lot.






      derivatives chain-rule







      share|cite|improve this question









      $endgroup$




      I have asked this questions: Change of variables in differential equation?



      ...but after thinking about it, I am still a little confused of how to rigorously use the chain rule to calculate the derivative(s) of a function for a change of variable.



      I have the following derivative:



      $f(x) = frac{dw(x)}{dx}$



      Now I introduce the change of variable: $hat{x}=frac{x}{L}$
      and I apply the chain rule:




      1. I write: $g(hat{x}) = L hat{x} = x$

      2. I substitute: $f(g(hat{x})) = frac{dw(g(hat{x}))}{d(g(hat{x}))}$


      ...but this does not help me... I am confusing something.



      I would be glad, if someone could show me in detail and step by step how to do this rigorously.



      Thanks a lot.






      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










      asked Dec 15 '18 at 9:02









      jamesjames

      15110




      15110






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          Let's say you want to change $x$ for $y$, where $y$ is a function of $x$, i.e., $y=g(x)$. I will use $y$ for the nondimensional variable because it's way easier to type in my phone. The chain rule tells us that if we want to calculate the derivative
          $$frac{df}{dx}$$
          in terms of $y$, we need to use the formula
          $$frac{df}{dx}=frac{dy}{dx}frac{df}{dy}.$$
          The $dy/dx$ part is equal to $dg(x)/dx$, therefore,
          $$frac{df}{dx}=frac{dg(x)}{dx}frac{df}{dy}.$$
          See that it works for any possible change of variables. In your particular case, $g(x)=x/L$ and $dg/dx=1/L$. Therefore, we have
          $$frac{df}{dx}=frac{1}{L}frac{df}{dy}.$$
          It leads to
          $$frac{df}{dy}=Lw(g^{-1}(y)),$$
          in which $g^{-1}(y)=Ly$. Finally:
          $$frac{df}{dy}=Lw(Ly).$$
          It's up to you to show that this equation is dimensionally consistent. Remember: $L$ has the same dimension of $x$ and $y$ is nondimensional. Let me know if you have any question.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Many thnaks for your answer. There is one thing that I don't understand, if I take $frac{df}{dx}=frac{1}{L}frac{df}{dy}.$ and I know that $frac{df}{dy} = Lw(Ly)$, then the final result is: $frac{df}{dx} = w(Ly)$... ?
            $endgroup$
            – james
            Dec 15 '18 at 21:03










          • $begingroup$
            @james You are right, but I can't see how this is useful. In general, one has the dimensional form of an equation and then obtains the nondimensional form of it. It's useful then to have "consistency" the variables, in the sense that you use only $x$ or only $y$; since they mean the same thing (say, the dimensional and nondimensional distance, respectively), you won't want to mix them. Therefore, while $df/dx=w(Ly)$ is true, I think that $df/dy=Lw(Ly)$ is far more useful.
            $endgroup$
            – rafa11111
            Dec 15 '18 at 23:55










          • $begingroup$
            I see. Okay, so if I apply the change of variable to the following integral $int _ { - frac { L } { 2 } } ^ { frac { L} { 2 } } w ^ { prime } ( x ) ^ { 2 } d x$, I would get: $int _ { - frac { 1 } { 2 } } ^ { frac { 1} { 2 } } frac{1}{L^2} w ^ { prime } ( Lhat{x} ) ^ { 2 } L d hat{x}$, right ?
            $endgroup$
            – james
            Dec 16 '18 at 9:12












          • $begingroup$
            @james Precisely!
            $endgroup$
            – rafa11111
            Dec 16 '18 at 10:46










          • $begingroup$
            Thank you so much for your help !
            $endgroup$
            – james
            Dec 16 '18 at 13:28











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          1












          $begingroup$

          Let's say you want to change $x$ for $y$, where $y$ is a function of $x$, i.e., $y=g(x)$. I will use $y$ for the nondimensional variable because it's way easier to type in my phone. The chain rule tells us that if we want to calculate the derivative
          $$frac{df}{dx}$$
          in terms of $y$, we need to use the formula
          $$frac{df}{dx}=frac{dy}{dx}frac{df}{dy}.$$
          The $dy/dx$ part is equal to $dg(x)/dx$, therefore,
          $$frac{df}{dx}=frac{dg(x)}{dx}frac{df}{dy}.$$
          See that it works for any possible change of variables. In your particular case, $g(x)=x/L$ and $dg/dx=1/L$. Therefore, we have
          $$frac{df}{dx}=frac{1}{L}frac{df}{dy}.$$
          It leads to
          $$frac{df}{dy}=Lw(g^{-1}(y)),$$
          in which $g^{-1}(y)=Ly$. Finally:
          $$frac{df}{dy}=Lw(Ly).$$
          It's up to you to show that this equation is dimensionally consistent. Remember: $L$ has the same dimension of $x$ and $y$ is nondimensional. Let me know if you have any question.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Many thnaks for your answer. There is one thing that I don't understand, if I take $frac{df}{dx}=frac{1}{L}frac{df}{dy}.$ and I know that $frac{df}{dy} = Lw(Ly)$, then the final result is: $frac{df}{dx} = w(Ly)$... ?
            $endgroup$
            – james
            Dec 15 '18 at 21:03










          • $begingroup$
            @james You are right, but I can't see how this is useful. In general, one has the dimensional form of an equation and then obtains the nondimensional form of it. It's useful then to have "consistency" the variables, in the sense that you use only $x$ or only $y$; since they mean the same thing (say, the dimensional and nondimensional distance, respectively), you won't want to mix them. Therefore, while $df/dx=w(Ly)$ is true, I think that $df/dy=Lw(Ly)$ is far more useful.
            $endgroup$
            – rafa11111
            Dec 15 '18 at 23:55










          • $begingroup$
            I see. Okay, so if I apply the change of variable to the following integral $int _ { - frac { L } { 2 } } ^ { frac { L} { 2 } } w ^ { prime } ( x ) ^ { 2 } d x$, I would get: $int _ { - frac { 1 } { 2 } } ^ { frac { 1} { 2 } } frac{1}{L^2} w ^ { prime } ( Lhat{x} ) ^ { 2 } L d hat{x}$, right ?
            $endgroup$
            – james
            Dec 16 '18 at 9:12












          • $begingroup$
            @james Precisely!
            $endgroup$
            – rafa11111
            Dec 16 '18 at 10:46










          • $begingroup$
            Thank you so much for your help !
            $endgroup$
            – james
            Dec 16 '18 at 13:28
















          1












          $begingroup$

          Let's say you want to change $x$ for $y$, where $y$ is a function of $x$, i.e., $y=g(x)$. I will use $y$ for the nondimensional variable because it's way easier to type in my phone. The chain rule tells us that if we want to calculate the derivative
          $$frac{df}{dx}$$
          in terms of $y$, we need to use the formula
          $$frac{df}{dx}=frac{dy}{dx}frac{df}{dy}.$$
          The $dy/dx$ part is equal to $dg(x)/dx$, therefore,
          $$frac{df}{dx}=frac{dg(x)}{dx}frac{df}{dy}.$$
          See that it works for any possible change of variables. In your particular case, $g(x)=x/L$ and $dg/dx=1/L$. Therefore, we have
          $$frac{df}{dx}=frac{1}{L}frac{df}{dy}.$$
          It leads to
          $$frac{df}{dy}=Lw(g^{-1}(y)),$$
          in which $g^{-1}(y)=Ly$. Finally:
          $$frac{df}{dy}=Lw(Ly).$$
          It's up to you to show that this equation is dimensionally consistent. Remember: $L$ has the same dimension of $x$ and $y$ is nondimensional. Let me know if you have any question.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Many thnaks for your answer. There is one thing that I don't understand, if I take $frac{df}{dx}=frac{1}{L}frac{df}{dy}.$ and I know that $frac{df}{dy} = Lw(Ly)$, then the final result is: $frac{df}{dx} = w(Ly)$... ?
            $endgroup$
            – james
            Dec 15 '18 at 21:03










          • $begingroup$
            @james You are right, but I can't see how this is useful. In general, one has the dimensional form of an equation and then obtains the nondimensional form of it. It's useful then to have "consistency" the variables, in the sense that you use only $x$ or only $y$; since they mean the same thing (say, the dimensional and nondimensional distance, respectively), you won't want to mix them. Therefore, while $df/dx=w(Ly)$ is true, I think that $df/dy=Lw(Ly)$ is far more useful.
            $endgroup$
            – rafa11111
            Dec 15 '18 at 23:55










          • $begingroup$
            I see. Okay, so if I apply the change of variable to the following integral $int _ { - frac { L } { 2 } } ^ { frac { L} { 2 } } w ^ { prime } ( x ) ^ { 2 } d x$, I would get: $int _ { - frac { 1 } { 2 } } ^ { frac { 1} { 2 } } frac{1}{L^2} w ^ { prime } ( Lhat{x} ) ^ { 2 } L d hat{x}$, right ?
            $endgroup$
            – james
            Dec 16 '18 at 9:12












          • $begingroup$
            @james Precisely!
            $endgroup$
            – rafa11111
            Dec 16 '18 at 10:46










          • $begingroup$
            Thank you so much for your help !
            $endgroup$
            – james
            Dec 16 '18 at 13:28














          1












          1








          1





          $begingroup$

          Let's say you want to change $x$ for $y$, where $y$ is a function of $x$, i.e., $y=g(x)$. I will use $y$ for the nondimensional variable because it's way easier to type in my phone. The chain rule tells us that if we want to calculate the derivative
          $$frac{df}{dx}$$
          in terms of $y$, we need to use the formula
          $$frac{df}{dx}=frac{dy}{dx}frac{df}{dy}.$$
          The $dy/dx$ part is equal to $dg(x)/dx$, therefore,
          $$frac{df}{dx}=frac{dg(x)}{dx}frac{df}{dy}.$$
          See that it works for any possible change of variables. In your particular case, $g(x)=x/L$ and $dg/dx=1/L$. Therefore, we have
          $$frac{df}{dx}=frac{1}{L}frac{df}{dy}.$$
          It leads to
          $$frac{df}{dy}=Lw(g^{-1}(y)),$$
          in which $g^{-1}(y)=Ly$. Finally:
          $$frac{df}{dy}=Lw(Ly).$$
          It's up to you to show that this equation is dimensionally consistent. Remember: $L$ has the same dimension of $x$ and $y$ is nondimensional. Let me know if you have any question.






          share|cite|improve this answer









          $endgroup$



          Let's say you want to change $x$ for $y$, where $y$ is a function of $x$, i.e., $y=g(x)$. I will use $y$ for the nondimensional variable because it's way easier to type in my phone. The chain rule tells us that if we want to calculate the derivative
          $$frac{df}{dx}$$
          in terms of $y$, we need to use the formula
          $$frac{df}{dx}=frac{dy}{dx}frac{df}{dy}.$$
          The $dy/dx$ part is equal to $dg(x)/dx$, therefore,
          $$frac{df}{dx}=frac{dg(x)}{dx}frac{df}{dy}.$$
          See that it works for any possible change of variables. In your particular case, $g(x)=x/L$ and $dg/dx=1/L$. Therefore, we have
          $$frac{df}{dx}=frac{1}{L}frac{df}{dy}.$$
          It leads to
          $$frac{df}{dy}=Lw(g^{-1}(y)),$$
          in which $g^{-1}(y)=Ly$. Finally:
          $$frac{df}{dy}=Lw(Ly).$$
          It's up to you to show that this equation is dimensionally consistent. Remember: $L$ has the same dimension of $x$ and $y$ is nondimensional. Let me know if you have any question.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 19:46









          rafa11111rafa11111

          1,1612417




          1,1612417












          • $begingroup$
            Many thnaks for your answer. There is one thing that I don't understand, if I take $frac{df}{dx}=frac{1}{L}frac{df}{dy}.$ and I know that $frac{df}{dy} = Lw(Ly)$, then the final result is: $frac{df}{dx} = w(Ly)$... ?
            $endgroup$
            – james
            Dec 15 '18 at 21:03










          • $begingroup$
            @james You are right, but I can't see how this is useful. In general, one has the dimensional form of an equation and then obtains the nondimensional form of it. It's useful then to have "consistency" the variables, in the sense that you use only $x$ or only $y$; since they mean the same thing (say, the dimensional and nondimensional distance, respectively), you won't want to mix them. Therefore, while $df/dx=w(Ly)$ is true, I think that $df/dy=Lw(Ly)$ is far more useful.
            $endgroup$
            – rafa11111
            Dec 15 '18 at 23:55










          • $begingroup$
            I see. Okay, so if I apply the change of variable to the following integral $int _ { - frac { L } { 2 } } ^ { frac { L} { 2 } } w ^ { prime } ( x ) ^ { 2 } d x$, I would get: $int _ { - frac { 1 } { 2 } } ^ { frac { 1} { 2 } } frac{1}{L^2} w ^ { prime } ( Lhat{x} ) ^ { 2 } L d hat{x}$, right ?
            $endgroup$
            – james
            Dec 16 '18 at 9:12












          • $begingroup$
            @james Precisely!
            $endgroup$
            – rafa11111
            Dec 16 '18 at 10:46










          • $begingroup$
            Thank you so much for your help !
            $endgroup$
            – james
            Dec 16 '18 at 13:28


















          • $begingroup$
            Many thnaks for your answer. There is one thing that I don't understand, if I take $frac{df}{dx}=frac{1}{L}frac{df}{dy}.$ and I know that $frac{df}{dy} = Lw(Ly)$, then the final result is: $frac{df}{dx} = w(Ly)$... ?
            $endgroup$
            – james
            Dec 15 '18 at 21:03










          • $begingroup$
            @james You are right, but I can't see how this is useful. In general, one has the dimensional form of an equation and then obtains the nondimensional form of it. It's useful then to have "consistency" the variables, in the sense that you use only $x$ or only $y$; since they mean the same thing (say, the dimensional and nondimensional distance, respectively), you won't want to mix them. Therefore, while $df/dx=w(Ly)$ is true, I think that $df/dy=Lw(Ly)$ is far more useful.
            $endgroup$
            – rafa11111
            Dec 15 '18 at 23:55










          • $begingroup$
            I see. Okay, so if I apply the change of variable to the following integral $int _ { - frac { L } { 2 } } ^ { frac { L} { 2 } } w ^ { prime } ( x ) ^ { 2 } d x$, I would get: $int _ { - frac { 1 } { 2 } } ^ { frac { 1} { 2 } } frac{1}{L^2} w ^ { prime } ( Lhat{x} ) ^ { 2 } L d hat{x}$, right ?
            $endgroup$
            – james
            Dec 16 '18 at 9:12












          • $begingroup$
            @james Precisely!
            $endgroup$
            – rafa11111
            Dec 16 '18 at 10:46










          • $begingroup$
            Thank you so much for your help !
            $endgroup$
            – james
            Dec 16 '18 at 13:28
















          $begingroup$
          Many thnaks for your answer. There is one thing that I don't understand, if I take $frac{df}{dx}=frac{1}{L}frac{df}{dy}.$ and I know that $frac{df}{dy} = Lw(Ly)$, then the final result is: $frac{df}{dx} = w(Ly)$... ?
          $endgroup$
          – james
          Dec 15 '18 at 21:03




          $begingroup$
          Many thnaks for your answer. There is one thing that I don't understand, if I take $frac{df}{dx}=frac{1}{L}frac{df}{dy}.$ and I know that $frac{df}{dy} = Lw(Ly)$, then the final result is: $frac{df}{dx} = w(Ly)$... ?
          $endgroup$
          – james
          Dec 15 '18 at 21:03












          $begingroup$
          @james You are right, but I can't see how this is useful. In general, one has the dimensional form of an equation and then obtains the nondimensional form of it. It's useful then to have "consistency" the variables, in the sense that you use only $x$ or only $y$; since they mean the same thing (say, the dimensional and nondimensional distance, respectively), you won't want to mix them. Therefore, while $df/dx=w(Ly)$ is true, I think that $df/dy=Lw(Ly)$ is far more useful.
          $endgroup$
          – rafa11111
          Dec 15 '18 at 23:55




          $begingroup$
          @james You are right, but I can't see how this is useful. In general, one has the dimensional form of an equation and then obtains the nondimensional form of it. It's useful then to have "consistency" the variables, in the sense that you use only $x$ or only $y$; since they mean the same thing (say, the dimensional and nondimensional distance, respectively), you won't want to mix them. Therefore, while $df/dx=w(Ly)$ is true, I think that $df/dy=Lw(Ly)$ is far more useful.
          $endgroup$
          – rafa11111
          Dec 15 '18 at 23:55












          $begingroup$
          I see. Okay, so if I apply the change of variable to the following integral $int _ { - frac { L } { 2 } } ^ { frac { L} { 2 } } w ^ { prime } ( x ) ^ { 2 } d x$, I would get: $int _ { - frac { 1 } { 2 } } ^ { frac { 1} { 2 } } frac{1}{L^2} w ^ { prime } ( Lhat{x} ) ^ { 2 } L d hat{x}$, right ?
          $endgroup$
          – james
          Dec 16 '18 at 9:12






          $begingroup$
          I see. Okay, so if I apply the change of variable to the following integral $int _ { - frac { L } { 2 } } ^ { frac { L} { 2 } } w ^ { prime } ( x ) ^ { 2 } d x$, I would get: $int _ { - frac { 1 } { 2 } } ^ { frac { 1} { 2 } } frac{1}{L^2} w ^ { prime } ( Lhat{x} ) ^ { 2 } L d hat{x}$, right ?
          $endgroup$
          – james
          Dec 16 '18 at 9:12














          $begingroup$
          @james Precisely!
          $endgroup$
          – rafa11111
          Dec 16 '18 at 10:46




          $begingroup$
          @james Precisely!
          $endgroup$
          – rafa11111
          Dec 16 '18 at 10:46












          $begingroup$
          Thank you so much for your help !
          $endgroup$
          – james
          Dec 16 '18 at 13:28




          $begingroup$
          Thank you so much for your help !
          $endgroup$
          – james
          Dec 16 '18 at 13:28


















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          Bundesstraße 106

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          Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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          Verónica Boquete

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          Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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          Ida-Boy-Ed-Garten

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          Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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