Differentiability of $f$ at the origin












0












$begingroup$


I have $$
f(x,y) = cases{ dfrac{sin(x^2y^2)^{alpha}}{x^2y+y^3}& if $yne0$ \[6px]
x& if $y=0$ \
}
$$



The function is continuous at zero if $alpha>frac{3}{4}$ and $nabla f(0,0)=(1,0)$ but how can I prove differentiability at zero?



If I use the definition
$$lim_{(h,k)rightarrow(0,0)} frac{dfrac{sin(h^2k^2)^alpha}{k(h^2+k^2)}-h}{sqrt{h^2+k^2}}$$
is complicated.
Can I find a curve where $f$ isn't differentiable for $alpha >frac{3}{4}$?






real-analysis







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$endgroup$












  • $begingroup$
    I changed my answer; hope it helps.
    $endgroup$
    – zhw.
    Jan 10 at 18:03
















0












$begingroup$


I have $$
f(x,y) = cases{ dfrac{sin(x^2y^2)^{alpha}}{x^2y+y^3}& if $yne0$ \[6px]
x& if $y=0$ \
}
$$



The function is continuous at zero if $alpha>frac{3}{4}$ and $nabla f(0,0)=(1,0)$ but how can I prove differentiability at zero?



If I use the definition
$$lim_{(h,k)rightarrow(0,0)} frac{dfrac{sin(h^2k^2)^alpha}{k(h^2+k^2)}-h}{sqrt{h^2+k^2}}$$
is complicated.
Can I find a curve where $f$ isn't differentiable for $alpha >frac{3}{4}$?






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    I changed my answer; hope it helps.
    $endgroup$
    – zhw.
    Jan 10 at 18:03














0












0








0





$begingroup$


I have $$
f(x,y) = cases{ dfrac{sin(x^2y^2)^{alpha}}{x^2y+y^3}& if $yne0$ \[6px]
x& if $y=0$ \
}
$$



The function is continuous at zero if $alpha>frac{3}{4}$ and $nabla f(0,0)=(1,0)$ but how can I prove differentiability at zero?



If I use the definition
$$lim_{(h,k)rightarrow(0,0)} frac{dfrac{sin(h^2k^2)^alpha}{k(h^2+k^2)}-h}{sqrt{h^2+k^2}}$$
is complicated.
Can I find a curve where $f$ isn't differentiable for $alpha >frac{3}{4}$?






real-analysis







share|cite|improve this question











$endgroup$




I have $$
f(x,y) = cases{ dfrac{sin(x^2y^2)^{alpha}}{x^2y+y^3}& if $yne0$ \[6px]
x& if $y=0$ \
}
$$



The function is continuous at zero if $alpha>frac{3}{4}$ and $nabla f(0,0)=(1,0)$ but how can I prove differentiability at zero?



If I use the definition
$$lim_{(h,k)rightarrow(0,0)} frac{dfrac{sin(h^2k^2)^alpha}{k(h^2+k^2)}-h}{sqrt{h^2+k^2}}$$
is complicated.
Can I find a curve where $f$ isn't differentiable for $alpha >frac{3}{4}$?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













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edited Jan 10 at 14:07









egreg

181k1485203




181k1485203










asked Jan 9 at 21:03









Giulia B.Giulia B.

420211




420211












  • $begingroup$
    I changed my answer; hope it helps.
    $endgroup$
    – zhw.
    Jan 10 at 18:03


















  • $begingroup$
    I changed my answer; hope it helps.
    $endgroup$
    – zhw.
    Jan 10 at 18:03
















$begingroup$
I changed my answer; hope it helps.
$endgroup$
– zhw.
Jan 10 at 18:03




$begingroup$
I changed my answer; hope it helps.
$endgroup$
– zhw.
Jan 10 at 18:03










2 Answers
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$begingroup$

Differentiability is for $alpha> frac{3}{4} $ non only 1.Using the definition is complicate.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Suppose we look at $f$ along the curve $(x,x).$ Take your expression and let $h=k=x.$ We get



    $$frac{dfrac{sin[(x^4)^alpha]}{x^3}-x}{sqrt{2x^2}}= frac{sin [(x^4)^alpha]}{x^3sqrt 2|x|}-frac{x}{sqrt 2|x|}.$$



    This fails to have a limit as $xto 0 $ for $alpha>0,alpha ne 1.$ (Try $3/4<alpha < 1$ first; in this case the expression is unbounded near $0.$)






    share|cite|improve this answer











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      2 Answers
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      2 Answers
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      0












      $begingroup$

      Differentiability is for $alpha> frac{3}{4} $ non only 1.Using the definition is complicate.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Differentiability is for $alpha> frac{3}{4} $ non only 1.Using the definition is complicate.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Differentiability is for $alpha> frac{3}{4} $ non only 1.Using the definition is complicate.






          share|cite|improve this answer









          $endgroup$



          Differentiability is for $alpha> frac{3}{4} $ non only 1.Using the definition is complicate.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 6:29









          user495707user495707

          85




          85























              0












              $begingroup$

              Suppose we look at $f$ along the curve $(x,x).$ Take your expression and let $h=k=x.$ We get



              $$frac{dfrac{sin[(x^4)^alpha]}{x^3}-x}{sqrt{2x^2}}= frac{sin [(x^4)^alpha]}{x^3sqrt 2|x|}-frac{x}{sqrt 2|x|}.$$



              This fails to have a limit as $xto 0 $ for $alpha>0,alpha ne 1.$ (Try $3/4<alpha < 1$ first; in this case the expression is unbounded near $0.$)






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Suppose we look at $f$ along the curve $(x,x).$ Take your expression and let $h=k=x.$ We get



                $$frac{dfrac{sin[(x^4)^alpha]}{x^3}-x}{sqrt{2x^2}}= frac{sin [(x^4)^alpha]}{x^3sqrt 2|x|}-frac{x}{sqrt 2|x|}.$$



                This fails to have a limit as $xto 0 $ for $alpha>0,alpha ne 1.$ (Try $3/4<alpha < 1$ first; in this case the expression is unbounded near $0.$)






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Suppose we look at $f$ along the curve $(x,x).$ Take your expression and let $h=k=x.$ We get



                  $$frac{dfrac{sin[(x^4)^alpha]}{x^3}-x}{sqrt{2x^2}}= frac{sin [(x^4)^alpha]}{x^3sqrt 2|x|}-frac{x}{sqrt 2|x|}.$$



                  This fails to have a limit as $xto 0 $ for $alpha>0,alpha ne 1.$ (Try $3/4<alpha < 1$ first; in this case the expression is unbounded near $0.$)






                  share|cite|improve this answer











                  $endgroup$



                  Suppose we look at $f$ along the curve $(x,x).$ Take your expression and let $h=k=x.$ We get



                  $$frac{dfrac{sin[(x^4)^alpha]}{x^3}-x}{sqrt{2x^2}}= frac{sin [(x^4)^alpha]}{x^3sqrt 2|x|}-frac{x}{sqrt 2|x|}.$$



                  This fails to have a limit as $xto 0 $ for $alpha>0,alpha ne 1.$ (Try $3/4<alpha < 1$ first; in this case the expression is unbounded near $0.$)







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 10 at 17:59

























                  answered Jan 9 at 22:27









                  zhw.zhw.

                  72.6k43175




                  72.6k43175






























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