Partial derivative of x - is quotient rule necessary?












2












$begingroup$


Let
$$u(x,y)=frac{x}{x^2+y^2}$$



I'm trying to determine if the given function is harmonic. I know that the 2nd partial derivative with respect to $x$ should, when added to the 2nd partial derivative of $y$, equal $0$.



However, I'm kind of stuck. I'm using the quotient rule to solve for the partial derivative of $x$, but is this the right way to take a partial derivative of a quotient?






multivariable-calculus derivatives partial-derivative







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  • 1




    $begingroup$
    This is not a harmonic analysis question. Please read tag descriptions before using.
    $endgroup$
    – Matt Rosenzweig
    Feb 2 '16 at 2:11










  • $begingroup$
    @MattRosenzweig this is not possible on the mobile app, my apologies
    $endgroup$
    – whatwhatwhat
    Feb 2 '16 at 2:13
















2












$begingroup$


Let
$$u(x,y)=frac{x}{x^2+y^2}$$



I'm trying to determine if the given function is harmonic. I know that the 2nd partial derivative with respect to $x$ should, when added to the 2nd partial derivative of $y$, equal $0$.



However, I'm kind of stuck. I'm using the quotient rule to solve for the partial derivative of $x$, but is this the right way to take a partial derivative of a quotient?






multivariable-calculus derivatives partial-derivative







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is not a harmonic analysis question. Please read tag descriptions before using.
    $endgroup$
    – Matt Rosenzweig
    Feb 2 '16 at 2:11










  • $begingroup$
    @MattRosenzweig this is not possible on the mobile app, my apologies
    $endgroup$
    – whatwhatwhat
    Feb 2 '16 at 2:13














2












2








2





$begingroup$


Let
$$u(x,y)=frac{x}{x^2+y^2}$$



I'm trying to determine if the given function is harmonic. I know that the 2nd partial derivative with respect to $x$ should, when added to the 2nd partial derivative of $y$, equal $0$.



However, I'm kind of stuck. I'm using the quotient rule to solve for the partial derivative of $x$, but is this the right way to take a partial derivative of a quotient?






multivariable-calculus derivatives partial-derivative







share|cite|improve this question











$endgroup$




Let
$$u(x,y)=frac{x}{x^2+y^2}$$



I'm trying to determine if the given function is harmonic. I know that the 2nd partial derivative with respect to $x$ should, when added to the 2nd partial derivative of $y$, equal $0$.



However, I'm kind of stuck. I'm using the quotient rule to solve for the partial derivative of $x$, but is this the right way to take a partial derivative of a quotient?






multivariable-calculus derivatives partial-derivative




multivariable-calculus derivatives partial-derivative






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 15:15









Martin Sleziak

45k10122277




45k10122277










asked Feb 1 '16 at 19:30









whatwhatwhatwhatwhatwhat

782823




782823








  • 1




    $begingroup$
    This is not a harmonic analysis question. Please read tag descriptions before using.
    $endgroup$
    – Matt Rosenzweig
    Feb 2 '16 at 2:11










  • $begingroup$
    @MattRosenzweig this is not possible on the mobile app, my apologies
    $endgroup$
    – whatwhatwhat
    Feb 2 '16 at 2:13














  • 1




    $begingroup$
    This is not a harmonic analysis question. Please read tag descriptions before using.
    $endgroup$
    – Matt Rosenzweig
    Feb 2 '16 at 2:11










  • $begingroup$
    @MattRosenzweig this is not possible on the mobile app, my apologies
    $endgroup$
    – whatwhatwhat
    Feb 2 '16 at 2:13








1




1




$begingroup$
This is not a harmonic analysis question. Please read tag descriptions before using.
$endgroup$
– Matt Rosenzweig
Feb 2 '16 at 2:11




$begingroup$
This is not a harmonic analysis question. Please read tag descriptions before using.
$endgroup$
– Matt Rosenzweig
Feb 2 '16 at 2:11












$begingroup$
@MattRosenzweig this is not possible on the mobile app, my apologies
$endgroup$
– whatwhatwhat
Feb 2 '16 at 2:13




$begingroup$
@MattRosenzweig this is not possible on the mobile app, my apologies
$endgroup$
– whatwhatwhat
Feb 2 '16 at 2:13










1 Answer
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$begingroup$

When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.



Now, you can use the product rule if you choose; you just have to rewrite:



$$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$



Then
$$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$



which you can simplify.






share|cite|improve this answer











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    $begingroup$

    When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.



    Now, you can use the product rule if you choose; you just have to rewrite:



    $$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$



    Then
    $$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$



    which you can simplify.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.



      Now, you can use the product rule if you choose; you just have to rewrite:



      $$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$



      Then
      $$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$



      which you can simplify.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.



        Now, you can use the product rule if you choose; you just have to rewrite:



        $$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$



        Then
        $$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$



        which you can simplify.






        share|cite|improve this answer











        $endgroup$



        When you take a partial derivative of a multivariate function, you are simply "fixing" the variables you don't need and differentiating with respect to the variable you do. Thus since you have a rational function with respect to $x$, you simply fix $y$ and differentiate using the quotient rule.



        Now, you can use the product rule if you choose; you just have to rewrite:



        $$u(x,y)=frac{x}{x^2+y^2}=xleft(x^2+y^2right)^{-1}$$



        Then
        $$frac{partial{u}}{partial{x}}=xcdot(-2x)(x^2+y^2)^{-2}+(x^2+y^2)^{-1}$$



        which you can simplify.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 1 '16 at 20:38

























        answered Feb 1 '16 at 20:05









        IcemanIceman

        764821




        764821






























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