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

      15610




      15610






















          1 Answer
          1






          active

          oldest

          votes


















          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

          1,2042417




          1,2042417












          • $begingroup$
            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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          3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
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          Can I use Tabulator js library in my java Spring + Thymeleaf project?

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          0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
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