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

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          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

          1,1712417




          1,1712417












          • $begingroup$
            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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          Plaza Victoria

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          Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
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          In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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          How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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          1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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