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
edited Dec 14 '17 at 16:35
Martin Sleziak
44.7k9117272
asked Oct 1 '14 at 22:00
D-Kuang-RQD-Kuang-RQ
1536
$endgroup$
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.
add a comment |
$begingroup$
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
edited Dec 14 '17 at 16:35
Martin Sleziak
44.7k9117272
asked Oct 1 '14 at 22:00
D-Kuang-RQD-Kuang-RQ
1536
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 14 '17 at 16:35
Martin Sleziak
44.7k9117272
asked Oct 1 '14 at 22:00
D-Kuang-RQD-Kuang-RQ
1536
$endgroup$
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
edited Dec 14 '17 at 16:35
Martin Sleziak
44.7k9117272
asked Oct 1 '14 at 22:00
D-Kuang-RQD-Kuang-RQ
1536
edited Dec 14 '17 at 16:35
Martin Sleziak
44.7k9117272
asked Oct 1 '14 at 22:00
D-Kuang-RQD-Kuang-RQ
1536
edited Dec 14 '17 at 16:35
Martin Sleziak
44.7k9117272
edited Dec 14 '17 at 16:35
Martin Sleziak
44.7k9117272
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
asked Oct 1 '14 at 22:00
D-Kuang-RQD-Kuang-RQ
1536
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
add a comment |
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
add a comment |
1 Answer
1
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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.
answered Oct 1 '14 at 23:47
jm324354jm324354
1,75211125
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Oct 1 '14 at 23:47
jm324354jm324354
1,75211125
$endgroup$
add a comment |
$begingroup$
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.
answered Oct 1 '14 at 23:47
jm324354jm324354
1,75211125
$endgroup$
add a comment |
$begingroup$
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.
answered Oct 1 '14 at 23:47
jm324354jm324354
1,75211125
$endgroup$
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.
answered Oct 1 '14 at 23:47
jm324354jm324354
1,75211125
answered Oct 1 '14 at 23:47
jm324354jm324354
1,75211125
answered Oct 1 '14 at 23:47
jm324354jm324354
1,75211125
answered Oct 1 '14 at 23:47
jm324354jm324354
1,75211125
1,75211125
add a comment |
add a comment |
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