Partial Derivatives and Their Limits: $f(x,y)=frac{xy(x^2-y^2)}{(x^2+y^2)}$ [closed]












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So I have this function: $f(x,y)=frac{xy(x^2-y^2)}{(x^2+y^2)}$ for $(x,y)neq(0,0)$ and $f(x)=0$ for $(x,y)=(0,0)$


I am asked to show:

1) $frac{partial{f}}{partial{x}}(0,0)=frac{partial{f}}{partial{y}}(0,0)=0$

2) $frac{partial^2{f}}{partial{y^2}}(0,0) = 1$ and $frac{partial^2{f}}{partial{y^2}}(0,0) = -1$

3) Explain why the mixed partial derivatives are not equal



I have computed the first partial derivatives for $f(x)$, and I know that I need to use limit to show these partial derivatives' values at $(0,0)$, but I don't know how exactly I should approach them.






multivariable-calculus partial-derivative







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closed as off-topic by Did, metamorphy, Lord_Farin, Matthew Towers, José Carlos Santos Jan 2 at 17:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, metamorphy, Lord_Farin, Matthew Towers, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    $begingroup$
    $fleft(xright)$ needs to be $fleft(x,yright)$ since this is a function of two variables, $x:text{and}:y$...
    $endgroup$
    – jm324354
    Oct 1 '14 at 23:48
















-1












$begingroup$


So I have this function: $f(x,y)=frac{xy(x^2-y^2)}{(x^2+y^2)}$ for $(x,y)neq(0,0)$ and $f(x)=0$ for $(x,y)=(0,0)$


I am asked to show:

1) $frac{partial{f}}{partial{x}}(0,0)=frac{partial{f}}{partial{y}}(0,0)=0$

2) $frac{partial^2{f}}{partial{y^2}}(0,0) = 1$ and $frac{partial^2{f}}{partial{y^2}}(0,0) = -1$

3) Explain why the mixed partial derivatives are not equal



I have computed the first partial derivatives for $f(x)$, and I know that I need to use limit to show these partial derivatives' values at $(0,0)$, but I don't know how exactly I should approach them.






multivariable-calculus partial-derivative







share|cite|improve this question











$endgroup$



closed as off-topic by Did, metamorphy, Lord_Farin, Matthew Towers, José Carlos Santos Jan 2 at 17:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, metamorphy, Lord_Farin, Matthew Towers, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    $begingroup$
    $fleft(xright)$ needs to be $fleft(x,yright)$ since this is a function of two variables, $x:text{and}:y$...
    $endgroup$
    – jm324354
    Oct 1 '14 at 23:48














-1












-1








-1





$begingroup$


So I have this function: $f(x,y)=frac{xy(x^2-y^2)}{(x^2+y^2)}$ for $(x,y)neq(0,0)$ and $f(x)=0$ for $(x,y)=(0,0)$


I am asked to show:

1) $frac{partial{f}}{partial{x}}(0,0)=frac{partial{f}}{partial{y}}(0,0)=0$

2) $frac{partial^2{f}}{partial{y^2}}(0,0) = 1$ and $frac{partial^2{f}}{partial{y^2}}(0,0) = -1$

3) Explain why the mixed partial derivatives are not equal



I have computed the first partial derivatives for $f(x)$, and I know that I need to use limit to show these partial derivatives' values at $(0,0)$, but I don't know how exactly I should approach them.






multivariable-calculus partial-derivative







share|cite|improve this question











$endgroup$




So I have this function: $f(x,y)=frac{xy(x^2-y^2)}{(x^2+y^2)}$ for $(x,y)neq(0,0)$ and $f(x)=0$ for $(x,y)=(0,0)$


I am asked to show:

1) $frac{partial{f}}{partial{x}}(0,0)=frac{partial{f}}{partial{y}}(0,0)=0$

2) $frac{partial^2{f}}{partial{y^2}}(0,0) = 1$ and $frac{partial^2{f}}{partial{y^2}}(0,0) = -1$

3) Explain why the mixed partial derivatives are not equal



I have computed the first partial derivatives for $f(x)$, and I know that I need to use limit to show these partial derivatives' values at $(0,0)$, but I don't know how exactly I should approach them.






multivariable-calculus partial-derivative




multivariable-calculus partial-derivative






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share|cite|improve this question













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edited Dec 14 '17 at 16:35









Martin Sleziak

44.7k9117272




44.7k9117272










asked Oct 1 '14 at 22:00









D-Kuang-RQD-Kuang-RQ

1536




1536




closed as off-topic by Did, metamorphy, Lord_Farin, Matthew Towers, José Carlos Santos Jan 2 at 17:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, metamorphy, Lord_Farin, Matthew Towers, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Did, metamorphy, Lord_Farin, Matthew Towers, José Carlos Santos Jan 2 at 17:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, metamorphy, Lord_Farin, Matthew Towers, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    $fleft(xright)$ needs to be $fleft(x,yright)$ since this is a function of two variables, $x:text{and}:y$...
    $endgroup$
    – jm324354
    Oct 1 '14 at 23:48














  • 1




    $begingroup$
    $fleft(xright)$ needs to be $fleft(x,yright)$ since this is a function of two variables, $x:text{and}:y$...
    $endgroup$
    – jm324354
    Oct 1 '14 at 23:48








1




1




$begingroup$
$fleft(xright)$ needs to be $fleft(x,yright)$ since this is a function of two variables, $x:text{and}:y$...
$endgroup$
– jm324354
Oct 1 '14 at 23:48




$begingroup$
$fleft(xright)$ needs to be $fleft(x,yright)$ since this is a function of two variables, $x:text{and}:y$...
$endgroup$
– jm324354
Oct 1 '14 at 23:48










1 Answer
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For 1) I get $$frac{partial{f}}{partial x}= frac{3yx^2-y^3}{x^2+y^2}-frac{2x^4y-2x^2y^3}{left(x^2+y^2right)^2}, \
frac{partial f}{partial y}=frac{-3xy^2+x^3}{x^2+y^2}-frac{2x^3y-2xy^3}{left(x^2+y^2right)^2},$$
so we will need to evaluate following limits:
$$lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}left[frac{3yx^2-y^3}{x^2+y^2}-frac{2x^4y-2x^2y^3}{left(x^2+y^2right)^2}right], \
lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}left[frac{-3xy^2+x^3}{x^2+y^2}-frac{2x^3y-2xy^3}{left(x^2+y^2right)^2}right],$$
and according to the question show that they both $=0$. The first limit simplifies a bit to
$$ lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{-4y^3}{x^2+y^2}+lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{6x^2y^3+2y^5}{y^4+2x^2y+x^4}, $$
and the second goes to
$$ lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{-4y^2}{y^2+x^2}+lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}frac{2xy^3-2x^3y}{left(x^2+y^2right)^2}, $$
and now let just $y$ go to $0$ and observe that all of the limits will then be $0$, regardless of what $x$ is.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    For 1) I get $$frac{partial{f}}{partial x}= frac{3yx^2-y^3}{x^2+y^2}-frac{2x^4y-2x^2y^3}{left(x^2+y^2right)^2}, \
    frac{partial f}{partial y}=frac{-3xy^2+x^3}{x^2+y^2}-frac{2x^3y-2xy^3}{left(x^2+y^2right)^2},$$
    so we will need to evaluate following limits:
    $$lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}left[frac{3yx^2-y^3}{x^2+y^2}-frac{2x^4y-2x^2y^3}{left(x^2+y^2right)^2}right], \
    lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}left[frac{-3xy^2+x^3}{x^2+y^2}-frac{2x^3y-2xy^3}{left(x^2+y^2right)^2}right],$$
    and according to the question show that they both $=0$. The first limit simplifies a bit to
    $$ lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{-4y^3}{x^2+y^2}+lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{6x^2y^3+2y^5}{y^4+2x^2y+x^4}, $$
    and the second goes to
    $$ lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{-4y^2}{y^2+x^2}+lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}frac{2xy^3-2x^3y}{left(x^2+y^2right)^2}, $$
    and now let just $y$ go to $0$ and observe that all of the limits will then be $0$, regardless of what $x$ is.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      For 1) I get $$frac{partial{f}}{partial x}= frac{3yx^2-y^3}{x^2+y^2}-frac{2x^4y-2x^2y^3}{left(x^2+y^2right)^2}, \
      frac{partial f}{partial y}=frac{-3xy^2+x^3}{x^2+y^2}-frac{2x^3y-2xy^3}{left(x^2+y^2right)^2},$$
      so we will need to evaluate following limits:
      $$lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}left[frac{3yx^2-y^3}{x^2+y^2}-frac{2x^4y-2x^2y^3}{left(x^2+y^2right)^2}right], \
      lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}left[frac{-3xy^2+x^3}{x^2+y^2}-frac{2x^3y-2xy^3}{left(x^2+y^2right)^2}right],$$
      and according to the question show that they both $=0$. The first limit simplifies a bit to
      $$ lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{-4y^3}{x^2+y^2}+lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{6x^2y^3+2y^5}{y^4+2x^2y+x^4}, $$
      and the second goes to
      $$ lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{-4y^2}{y^2+x^2}+lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}frac{2xy^3-2x^3y}{left(x^2+y^2right)^2}, $$
      and now let just $y$ go to $0$ and observe that all of the limits will then be $0$, regardless of what $x$ is.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        For 1) I get $$frac{partial{f}}{partial x}= frac{3yx^2-y^3}{x^2+y^2}-frac{2x^4y-2x^2y^3}{left(x^2+y^2right)^2}, \
        frac{partial f}{partial y}=frac{-3xy^2+x^3}{x^2+y^2}-frac{2x^3y-2xy^3}{left(x^2+y^2right)^2},$$
        so we will need to evaluate following limits:
        $$lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}left[frac{3yx^2-y^3}{x^2+y^2}-frac{2x^4y-2x^2y^3}{left(x^2+y^2right)^2}right], \
        lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}left[frac{-3xy^2+x^3}{x^2+y^2}-frac{2x^3y-2xy^3}{left(x^2+y^2right)^2}right],$$
        and according to the question show that they both $=0$. The first limit simplifies a bit to
        $$ lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{-4y^3}{x^2+y^2}+lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{6x^2y^3+2y^5}{y^4+2x^2y+x^4}, $$
        and the second goes to
        $$ lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{-4y^2}{y^2+x^2}+lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}frac{2xy^3-2x^3y}{left(x^2+y^2right)^2}, $$
        and now let just $y$ go to $0$ and observe that all of the limits will then be $0$, regardless of what $x$ is.






        share|cite|improve this answer









        $endgroup$



        For 1) I get $$frac{partial{f}}{partial x}= frac{3yx^2-y^3}{x^2+y^2}-frac{2x^4y-2x^2y^3}{left(x^2+y^2right)^2}, \
        frac{partial f}{partial y}=frac{-3xy^2+x^3}{x^2+y^2}-frac{2x^3y-2xy^3}{left(x^2+y^2right)^2},$$
        so we will need to evaluate following limits:
        $$lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}left[frac{3yx^2-y^3}{x^2+y^2}-frac{2x^4y-2x^2y^3}{left(x^2+y^2right)^2}right], \
        lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}left[frac{-3xy^2+x^3}{x^2+y^2}-frac{2x^3y-2xy^3}{left(x^2+y^2right)^2}right],$$
        and according to the question show that they both $=0$. The first limit simplifies a bit to
        $$ lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{-4y^3}{x^2+y^2}+lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{6x^2y^3+2y^5}{y^4+2x^2y+x^4}, $$
        and the second goes to
        $$ lim_{substack{left(x,yright)rightarrowleft(0,0right) \ x,yneq 0}}frac{-4y^2}{y^2+x^2}+lim_{substack{left(x,yright)rightarrow left(0,0right) \ x,yneq 0}}frac{2xy^3-2x^3y}{left(x^2+y^2right)^2}, $$
        and now let just $y$ go to $0$ and observe that all of the limits will then be $0$, regardless of what $x$ is.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 1 '14 at 23:47









        jm324354jm324354

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        1,75211125















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