Partial Derivative and fraction












0












$begingroup$


In the George F Simmons Calculus with Analytical Geometry, textbook, it is mentioned that the Partial derivatives of a function cannot be treated as fractions as in case of single variable function.



In the example provided it has been proved that,



from
$pV = nRT$

where,
- p - pressure

- T - Temperature

- V - Volume

- n - number of moles

- R - Universal gas constant



$dfrac{partial{p}}{partial{T}} dfrac{partial{T}}{partial{V}} dfrac{partial{V}}{partial{p}} = -1$



which is fine, but in the explanation, it is given as,
as the right hand side is -1 and not 1 we cannot treat the LHS of the above statement as fractions



If I am not wrong the 1st statement would mean, we cannot expand a partial differential as say,
$dfrac{d y}{d t} = dfrac{dy}{dx}dfrac{dx}{dt}$, where, $y=y(x)$ and $x = x(t)$,for a single variable function and we cannot represent the same for partial differentials the same manner if $y = y(x,t)$



What what is the significance of +1 or -1 for a differential to represented as a fraction?










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    In the George F Simmons Calculus with Analytical Geometry, textbook, it is mentioned that the Partial derivatives of a function cannot be treated as fractions as in case of single variable function.



    In the example provided it has been proved that,



    from
    $pV = nRT$

    where,
    - p - pressure

    - T - Temperature

    - V - Volume

    - n - number of moles

    - R - Universal gas constant



    $dfrac{partial{p}}{partial{T}} dfrac{partial{T}}{partial{V}} dfrac{partial{V}}{partial{p}} = -1$



    which is fine, but in the explanation, it is given as,
    as the right hand side is -1 and not 1 we cannot treat the LHS of the above statement as fractions



    If I am not wrong the 1st statement would mean, we cannot expand a partial differential as say,
    $dfrac{d y}{d t} = dfrac{dy}{dx}dfrac{dx}{dt}$, where, $y=y(x)$ and $x = x(t)$,for a single variable function and we cannot represent the same for partial differentials the same manner if $y = y(x,t)$



    What what is the significance of +1 or -1 for a differential to represented as a fraction?










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      In the George F Simmons Calculus with Analytical Geometry, textbook, it is mentioned that the Partial derivatives of a function cannot be treated as fractions as in case of single variable function.



      In the example provided it has been proved that,



      from
      $pV = nRT$

      where,
      - p - pressure

      - T - Temperature

      - V - Volume

      - n - number of moles

      - R - Universal gas constant



      $dfrac{partial{p}}{partial{T}} dfrac{partial{T}}{partial{V}} dfrac{partial{V}}{partial{p}} = -1$



      which is fine, but in the explanation, it is given as,
      as the right hand side is -1 and not 1 we cannot treat the LHS of the above statement as fractions



      If I am not wrong the 1st statement would mean, we cannot expand a partial differential as say,
      $dfrac{d y}{d t} = dfrac{dy}{dx}dfrac{dx}{dt}$, where, $y=y(x)$ and $x = x(t)$,for a single variable function and we cannot represent the same for partial differentials the same manner if $y = y(x,t)$



      What what is the significance of +1 or -1 for a differential to represented as a fraction?










      share|cite|improve this question











      $endgroup$




      In the George F Simmons Calculus with Analytical Geometry, textbook, it is mentioned that the Partial derivatives of a function cannot be treated as fractions as in case of single variable function.



      In the example provided it has been proved that,



      from
      $pV = nRT$

      where,
      - p - pressure

      - T - Temperature

      - V - Volume

      - n - number of moles

      - R - Universal gas constant



      $dfrac{partial{p}}{partial{T}} dfrac{partial{T}}{partial{V}} dfrac{partial{V}}{partial{p}} = -1$



      which is fine, but in the explanation, it is given as,
      as the right hand side is -1 and not 1 we cannot treat the LHS of the above statement as fractions



      If I am not wrong the 1st statement would mean, we cannot expand a partial differential as say,
      $dfrac{d y}{d t} = dfrac{dy}{dx}dfrac{dx}{dt}$, where, $y=y(x)$ and $x = x(t)$,for a single variable function and we cannot represent the same for partial differentials the same manner if $y = y(x,t)$



      What what is the significance of +1 or -1 for a differential to represented as a fraction?







      pde partial-derivative






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 1 at 5:34







      Raptor

















      asked Jan 1 at 5:23









      RaptorRaptor

      3317




      3317






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          Yes. $$dfrac{partial{x}}{partial{y}} dfrac{partial{y}}{partial{x}} neq dfrac{partial{x}}{partial{z}}$$
          The equation above is an example to this phenomenon.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
            $endgroup$
            – Raptor
            Jan 1 at 5:47










          • $begingroup$
            Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
            $endgroup$
            – mouthetics
            Jan 1 at 5:50





















          0












          $begingroup$

          Although there's nothing wrong with the answer from mouthetics I would like to expand on something: ultimately what allows you to go from the lhs to rhs is the chain rule. For example, if $y = y(x, t)$ and $x = x(t)$ then



          $$
          frac{{rm d}y}{{rm d} t} = frac{partial y}{partial x}frac{{rm d}x}{{rm d}t} + frac{partial y}{partial t}
          $$



          However, consider this other case. Imagine a function $f(x, y, z) = 0$. Now suppose you can invert this relation to obtain $z = z(x, y)$, if you apply the chain rule you will get



          $$
          {rm d}z = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x{rm d}y tag{1}
          $$



          where I have used the script to indicate that variable remains fixed. Suppose now that you move along a path on the surface $f(x,y,z) = 0$, where $z$ is kept constant. In this case you can also invert the function $f$ to obtain $y = y(x)$ and



          $$
          {rm d}y = left(frac{partial y}{partial x}right)_z{rm d}x tag{2}
          $$



          Here it is more obvious why this notation is useful, it emphasizes the fact that we keep $z$ constant. Replace that in (1) with ${rm d}z = 0$ (remember, $z$ is constant) and you will get



          $$
          0 = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x left(frac{partial y}{partial x}right)_z{rm d}x tag{3}
          $$



          And from here is trivial to obtain



          $$
          left(frac{partial x}{partial y}right)_z left(frac{partial y}{partial z}right)_x left(frac{partial z}{partial x}right)_y = -1 tag{4}
          $$



          Note that this is a result of consistently using the chain rule, and keeping track what function depends on what variables.



          In your case just set $x = P, y = V, z = T$, the function $f(x,y,z)$ is called a equation of state






          share|cite|improve this answer









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058230%2fpartial-derivative-and-fraction%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            Yes. $$dfrac{partial{x}}{partial{y}} dfrac{partial{y}}{partial{x}} neq dfrac{partial{x}}{partial{z}}$$
            The equation above is an example to this phenomenon.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
              $endgroup$
              – Raptor
              Jan 1 at 5:47










            • $begingroup$
              Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
              $endgroup$
              – mouthetics
              Jan 1 at 5:50


















            0












            $begingroup$

            Yes. $$dfrac{partial{x}}{partial{y}} dfrac{partial{y}}{partial{x}} neq dfrac{partial{x}}{partial{z}}$$
            The equation above is an example to this phenomenon.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
              $endgroup$
              – Raptor
              Jan 1 at 5:47










            • $begingroup$
              Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
              $endgroup$
              – mouthetics
              Jan 1 at 5:50
















            0












            0








            0





            $begingroup$

            Yes. $$dfrac{partial{x}}{partial{y}} dfrac{partial{y}}{partial{x}} neq dfrac{partial{x}}{partial{z}}$$
            The equation above is an example to this phenomenon.






            share|cite|improve this answer









            $endgroup$



            Yes. $$dfrac{partial{x}}{partial{y}} dfrac{partial{y}}{partial{x}} neq dfrac{partial{x}}{partial{z}}$$
            The equation above is an example to this phenomenon.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 1 at 5:44









            moutheticsmouthetics

            50127




            50127












            • $begingroup$
              Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
              $endgroup$
              – Raptor
              Jan 1 at 5:47










            • $begingroup$
              Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
              $endgroup$
              – mouthetics
              Jan 1 at 5:50




















            • $begingroup$
              Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
              $endgroup$
              – Raptor
              Jan 1 at 5:47










            • $begingroup$
              Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
              $endgroup$
              – mouthetics
              Jan 1 at 5:50


















            $begingroup$
            Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
            $endgroup$
            – Raptor
            Jan 1 at 5:47




            $begingroup$
            Does it just mean that the whole thing wont cancel out to give us 1 thus, cannot be represented as a fraction?
            $endgroup$
            – Raptor
            Jan 1 at 5:47












            $begingroup$
            Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
            $endgroup$
            – mouthetics
            Jan 1 at 5:50






            $begingroup$
            Yes true. You can't think of them as fraction, otherwise the product must be $1$ which is not the case.
            $endgroup$
            – mouthetics
            Jan 1 at 5:50













            0












            $begingroup$

            Although there's nothing wrong with the answer from mouthetics I would like to expand on something: ultimately what allows you to go from the lhs to rhs is the chain rule. For example, if $y = y(x, t)$ and $x = x(t)$ then



            $$
            frac{{rm d}y}{{rm d} t} = frac{partial y}{partial x}frac{{rm d}x}{{rm d}t} + frac{partial y}{partial t}
            $$



            However, consider this other case. Imagine a function $f(x, y, z) = 0$. Now suppose you can invert this relation to obtain $z = z(x, y)$, if you apply the chain rule you will get



            $$
            {rm d}z = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x{rm d}y tag{1}
            $$



            where I have used the script to indicate that variable remains fixed. Suppose now that you move along a path on the surface $f(x,y,z) = 0$, where $z$ is kept constant. In this case you can also invert the function $f$ to obtain $y = y(x)$ and



            $$
            {rm d}y = left(frac{partial y}{partial x}right)_z{rm d}x tag{2}
            $$



            Here it is more obvious why this notation is useful, it emphasizes the fact that we keep $z$ constant. Replace that in (1) with ${rm d}z = 0$ (remember, $z$ is constant) and you will get



            $$
            0 = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x left(frac{partial y}{partial x}right)_z{rm d}x tag{3}
            $$



            And from here is trivial to obtain



            $$
            left(frac{partial x}{partial y}right)_z left(frac{partial y}{partial z}right)_x left(frac{partial z}{partial x}right)_y = -1 tag{4}
            $$



            Note that this is a result of consistently using the chain rule, and keeping track what function depends on what variables.



            In your case just set $x = P, y = V, z = T$, the function $f(x,y,z)$ is called a equation of state






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Although there's nothing wrong with the answer from mouthetics I would like to expand on something: ultimately what allows you to go from the lhs to rhs is the chain rule. For example, if $y = y(x, t)$ and $x = x(t)$ then



              $$
              frac{{rm d}y}{{rm d} t} = frac{partial y}{partial x}frac{{rm d}x}{{rm d}t} + frac{partial y}{partial t}
              $$



              However, consider this other case. Imagine a function $f(x, y, z) = 0$. Now suppose you can invert this relation to obtain $z = z(x, y)$, if you apply the chain rule you will get



              $$
              {rm d}z = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x{rm d}y tag{1}
              $$



              where I have used the script to indicate that variable remains fixed. Suppose now that you move along a path on the surface $f(x,y,z) = 0$, where $z$ is kept constant. In this case you can also invert the function $f$ to obtain $y = y(x)$ and



              $$
              {rm d}y = left(frac{partial y}{partial x}right)_z{rm d}x tag{2}
              $$



              Here it is more obvious why this notation is useful, it emphasizes the fact that we keep $z$ constant. Replace that in (1) with ${rm d}z = 0$ (remember, $z$ is constant) and you will get



              $$
              0 = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x left(frac{partial y}{partial x}right)_z{rm d}x tag{3}
              $$



              And from here is trivial to obtain



              $$
              left(frac{partial x}{partial y}right)_z left(frac{partial y}{partial z}right)_x left(frac{partial z}{partial x}right)_y = -1 tag{4}
              $$



              Note that this is a result of consistently using the chain rule, and keeping track what function depends on what variables.



              In your case just set $x = P, y = V, z = T$, the function $f(x,y,z)$ is called a equation of state






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Although there's nothing wrong with the answer from mouthetics I would like to expand on something: ultimately what allows you to go from the lhs to rhs is the chain rule. For example, if $y = y(x, t)$ and $x = x(t)$ then



                $$
                frac{{rm d}y}{{rm d} t} = frac{partial y}{partial x}frac{{rm d}x}{{rm d}t} + frac{partial y}{partial t}
                $$



                However, consider this other case. Imagine a function $f(x, y, z) = 0$. Now suppose you can invert this relation to obtain $z = z(x, y)$, if you apply the chain rule you will get



                $$
                {rm d}z = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x{rm d}y tag{1}
                $$



                where I have used the script to indicate that variable remains fixed. Suppose now that you move along a path on the surface $f(x,y,z) = 0$, where $z$ is kept constant. In this case you can also invert the function $f$ to obtain $y = y(x)$ and



                $$
                {rm d}y = left(frac{partial y}{partial x}right)_z{rm d}x tag{2}
                $$



                Here it is more obvious why this notation is useful, it emphasizes the fact that we keep $z$ constant. Replace that in (1) with ${rm d}z = 0$ (remember, $z$ is constant) and you will get



                $$
                0 = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x left(frac{partial y}{partial x}right)_z{rm d}x tag{3}
                $$



                And from here is trivial to obtain



                $$
                left(frac{partial x}{partial y}right)_z left(frac{partial y}{partial z}right)_x left(frac{partial z}{partial x}right)_y = -1 tag{4}
                $$



                Note that this is a result of consistently using the chain rule, and keeping track what function depends on what variables.



                In your case just set $x = P, y = V, z = T$, the function $f(x,y,z)$ is called a equation of state






                share|cite|improve this answer









                $endgroup$



                Although there's nothing wrong with the answer from mouthetics I would like to expand on something: ultimately what allows you to go from the lhs to rhs is the chain rule. For example, if $y = y(x, t)$ and $x = x(t)$ then



                $$
                frac{{rm d}y}{{rm d} t} = frac{partial y}{partial x}frac{{rm d}x}{{rm d}t} + frac{partial y}{partial t}
                $$



                However, consider this other case. Imagine a function $f(x, y, z) = 0$. Now suppose you can invert this relation to obtain $z = z(x, y)$, if you apply the chain rule you will get



                $$
                {rm d}z = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x{rm d}y tag{1}
                $$



                where I have used the script to indicate that variable remains fixed. Suppose now that you move along a path on the surface $f(x,y,z) = 0$, where $z$ is kept constant. In this case you can also invert the function $f$ to obtain $y = y(x)$ and



                $$
                {rm d}y = left(frac{partial y}{partial x}right)_z{rm d}x tag{2}
                $$



                Here it is more obvious why this notation is useful, it emphasizes the fact that we keep $z$ constant. Replace that in (1) with ${rm d}z = 0$ (remember, $z$ is constant) and you will get



                $$
                0 = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x left(frac{partial y}{partial x}right)_z{rm d}x tag{3}
                $$



                And from here is trivial to obtain



                $$
                left(frac{partial x}{partial y}right)_z left(frac{partial y}{partial z}right)_x left(frac{partial z}{partial x}right)_y = -1 tag{4}
                $$



                Note that this is a result of consistently using the chain rule, and keeping track what function depends on what variables.



                In your case just set $x = P, y = V, z = T$, the function $f(x,y,z)$ is called a equation of state







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 1 at 10:49









                caveraccaverac

                14.4k31130




                14.4k31130






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058230%2fpartial-derivative-and-fraction%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Human spaceflight

                    Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                    張江高科駅