Partial derivatives of the dependent variable.












0












$begingroup$


I have a function of the form $$f(x,y)=frac{1}{exp((y-mu)/tau)+1 }$$ where $mu,tau,m$ are constants. Also $y^2=x^2+m^2$. How do I calculate $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$. Can I just not substitute the value of y in the expression?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
    $endgroup$
    – Devashish Kaushik
    Dec 29 '18 at 10:02












  • $begingroup$
    I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
    $endgroup$
    – Icchyamoy
    Dec 29 '18 at 10:05










  • $begingroup$
    I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
    $endgroup$
    – Devashish Kaushik
    Dec 29 '18 at 10:06
















0












$begingroup$


I have a function of the form $$f(x,y)=frac{1}{exp((y-mu)/tau)+1 }$$ where $mu,tau,m$ are constants. Also $y^2=x^2+m^2$. How do I calculate $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$. Can I just not substitute the value of y in the expression?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
    $endgroup$
    – Devashish Kaushik
    Dec 29 '18 at 10:02












  • $begingroup$
    I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
    $endgroup$
    – Icchyamoy
    Dec 29 '18 at 10:05










  • $begingroup$
    I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
    $endgroup$
    – Devashish Kaushik
    Dec 29 '18 at 10:06














0












0








0





$begingroup$


I have a function of the form $$f(x,y)=frac{1}{exp((y-mu)/tau)+1 }$$ where $mu,tau,m$ are constants. Also $y^2=x^2+m^2$. How do I calculate $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$. Can I just not substitute the value of y in the expression?










share|cite|improve this question











$endgroup$




I have a function of the form $$f(x,y)=frac{1}{exp((y-mu)/tau)+1 }$$ where $mu,tau,m$ are constants. Also $y^2=x^2+m^2$. How do I calculate $frac{partial f}{partial x}$ and $frac{partial f}{partial y}$. Can I just not substitute the value of y in the expression?







multivariable-calculus partial-derivative






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 10:55









Atheesh Rathnaweera

31




31










asked Dec 29 '18 at 9:51









IcchyamoyIcchyamoy

11




11












  • $begingroup$
    Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
    $endgroup$
    – Devashish Kaushik
    Dec 29 '18 at 10:02












  • $begingroup$
    I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
    $endgroup$
    – Icchyamoy
    Dec 29 '18 at 10:05










  • $begingroup$
    I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
    $endgroup$
    – Devashish Kaushik
    Dec 29 '18 at 10:06


















  • $begingroup$
    Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
    $endgroup$
    – Devashish Kaushik
    Dec 29 '18 at 10:02












  • $begingroup$
    I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
    $endgroup$
    – Icchyamoy
    Dec 29 '18 at 10:05










  • $begingroup$
    I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
    $endgroup$
    – Devashish Kaushik
    Dec 29 '18 at 10:06
















$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02






$begingroup$
Are you sure that the question is complete? Since it seems that $f$ is a function of $y$ only and so we just have to replace $partial$ by $text{d}$.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:02














$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05




$begingroup$
I am sure that this is the question. btw, are you trying to tell me that $frac{partial f}{partial x}=0$ for the given expression? Why cannot I substitute the expression of y(x) here?
$endgroup$
– Icchyamoy
Dec 29 '18 at 10:05












$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06




$begingroup$
I mean that the partial derivative is the same as the regular one. And yes, you can definitely substitute.
$endgroup$
– Devashish Kaushik
Dec 29 '18 at 10:06










1 Answer
1






active

oldest

votes


















0












$begingroup$

If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.



$$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$



The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.






share|cite|improve this answer









$endgroup$














    Your Answer








    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%2f3055692%2fpartial-derivatives-of-the-dependent-variable%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.



    $$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$



    The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.



      $$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$



      The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.



        $$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$



        The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.






        share|cite|improve this answer









        $endgroup$



        If $mu$, $tau$,and $m$ are really constants, then $f$ is purely a function of $y$, i.e.



        $$f(x,y)=f(y)=frac{1}{exp((y-mu)/tau)+1 }= frac{1}{exp((sqrt{x^2 + m^2} - mu ) / tau ) -1}$$



        The partial derivatives are then just the regular derivatives, so you just differentiate as you would normally.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 10:11









        Devashish KaushikDevashish Kaushik

        638220




        638220






























            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%2f3055692%2fpartial-derivatives-of-the-dependent-variable%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

            Bundesstraße 106

            Verónica Boquete

            Ida-Boy-Ed-Garten