How to compute $u_x(sqrt {(x^2+y^2)})=$?












1












$begingroup$


This is in reference to this question



$g(z) = f(|z|)$ is not holomorphic for a non constant function $f$



If $u$ is a function of single variable $x$ and if I want to differentiate $u(sqrt {(x^2+y^2)}$ with respect to $x$



then what will be $u_x(sqrt {(x^2+y^2)})=$?



It is given in the question that linked above
that $u_x(sqrt {(x^2+y^2)})=u_xtimes frac{x}{sqrt {(x^2+y^2)}}$ but If I take $u(x)=x^2+x$ then it becomes false



What is the correct way to solve it?



Any help.






complex-analysis derivatives pde







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    This is in reference to this question



    $g(z) = f(|z|)$ is not holomorphic for a non constant function $f$



    If $u$ is a function of single variable $x$ and if I want to differentiate $u(sqrt {(x^2+y^2)}$ with respect to $x$



    then what will be $u_x(sqrt {(x^2+y^2)})=$?



    It is given in the question that linked above
    that $u_x(sqrt {(x^2+y^2)})=u_xtimes frac{x}{sqrt {(x^2+y^2)}}$ but If I take $u(x)=x^2+x$ then it becomes false



    What is the correct way to solve it?



    Any help.






    complex-analysis derivatives pde







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      This is in reference to this question



      $g(z) = f(|z|)$ is not holomorphic for a non constant function $f$



      If $u$ is a function of single variable $x$ and if I want to differentiate $u(sqrt {(x^2+y^2)}$ with respect to $x$



      then what will be $u_x(sqrt {(x^2+y^2)})=$?



      It is given in the question that linked above
      that $u_x(sqrt {(x^2+y^2)})=u_xtimes frac{x}{sqrt {(x^2+y^2)}}$ but If I take $u(x)=x^2+x$ then it becomes false



      What is the correct way to solve it?



      Any help.






      complex-analysis derivatives pde







      share|cite|improve this question









      $endgroup$




      This is in reference to this question



      $g(z) = f(|z|)$ is not holomorphic for a non constant function $f$



      If $u$ is a function of single variable $x$ and if I want to differentiate $u(sqrt {(x^2+y^2)}$ with respect to $x$



      then what will be $u_x(sqrt {(x^2+y^2)})=$?



      It is given in the question that linked above
      that $u_x(sqrt {(x^2+y^2)})=u_xtimes frac{x}{sqrt {(x^2+y^2)}}$ but If I take $u(x)=x^2+x$ then it becomes false



      What is the correct way to solve it?



      Any help.






      complex-analysis derivatives pde




      complex-analysis derivatives pde






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 20 at 5:19







      user596656





























          1 Answer
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          $begingroup$

          What happens when I do this with $u(x)=x^2+x$. Write
          $$f(x,y)=uleft(sqrt{x^2+y^2}right)=x^2+y^2+sqrt{x^2+y^2}.$$
          Then
          $$fracpartial{partial x}f(x,y)=2x+frac{x}{sqrt{x^2+y^2}}.$$
          But $u'(x)=2x+1$, and
          $$u'left(sqrt{x^2+y^2}right)fracpartial{partial x}sqrt{x^2+y^2}
          =left(2sqrt{x^2+y^2}+1right)frac{x}{sqrt{x^2+y^2}}
          =2x+frac{x}{sqrt{x^2+y^2}}.$$



          I don't see any problem....






          share|cite|improve this answer









          $endgroup$













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            1 Answer
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            1 Answer
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            active

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            1












            $begingroup$

            What happens when I do this with $u(x)=x^2+x$. Write
            $$f(x,y)=uleft(sqrt{x^2+y^2}right)=x^2+y^2+sqrt{x^2+y^2}.$$
            Then
            $$fracpartial{partial x}f(x,y)=2x+frac{x}{sqrt{x^2+y^2}}.$$
            But $u'(x)=2x+1$, and
            $$u'left(sqrt{x^2+y^2}right)fracpartial{partial x}sqrt{x^2+y^2}
            =left(2sqrt{x^2+y^2}+1right)frac{x}{sqrt{x^2+y^2}}
            =2x+frac{x}{sqrt{x^2+y^2}}.$$



            I don't see any problem....






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              What happens when I do this with $u(x)=x^2+x$. Write
              $$f(x,y)=uleft(sqrt{x^2+y^2}right)=x^2+y^2+sqrt{x^2+y^2}.$$
              Then
              $$fracpartial{partial x}f(x,y)=2x+frac{x}{sqrt{x^2+y^2}}.$$
              But $u'(x)=2x+1$, and
              $$u'left(sqrt{x^2+y^2}right)fracpartial{partial x}sqrt{x^2+y^2}
              =left(2sqrt{x^2+y^2}+1right)frac{x}{sqrt{x^2+y^2}}
              =2x+frac{x}{sqrt{x^2+y^2}}.$$



              I don't see any problem....






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                What happens when I do this with $u(x)=x^2+x$. Write
                $$f(x,y)=uleft(sqrt{x^2+y^2}right)=x^2+y^2+sqrt{x^2+y^2}.$$
                Then
                $$fracpartial{partial x}f(x,y)=2x+frac{x}{sqrt{x^2+y^2}}.$$
                But $u'(x)=2x+1$, and
                $$u'left(sqrt{x^2+y^2}right)fracpartial{partial x}sqrt{x^2+y^2}
                =left(2sqrt{x^2+y^2}+1right)frac{x}{sqrt{x^2+y^2}}
                =2x+frac{x}{sqrt{x^2+y^2}}.$$



                I don't see any problem....






                share|cite|improve this answer









                $endgroup$



                What happens when I do this with $u(x)=x^2+x$. Write
                $$f(x,y)=uleft(sqrt{x^2+y^2}right)=x^2+y^2+sqrt{x^2+y^2}.$$
                Then
                $$fracpartial{partial x}f(x,y)=2x+frac{x}{sqrt{x^2+y^2}}.$$
                But $u'(x)=2x+1$, and
                $$u'left(sqrt{x^2+y^2}right)fracpartial{partial x}sqrt{x^2+y^2}
                =left(2sqrt{x^2+y^2}+1right)frac{x}{sqrt{x^2+y^2}}
                =2x+frac{x}{sqrt{x^2+y^2}}.$$



                I don't see any problem....







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 20 at 6:02









                Lord Shark the UnknownLord Shark the Unknown

                105k1160133




                105k1160133






























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