Differentiability of function of two variables at $(1,0)$Differentiablility of a function of two...
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Differentiability of function of two variables at $(1,0)$
Differentiablility of a function of two variablesdifferentiation-chain rule- function of two variablesContinuity and differentiability of the function $x|x|$Is $f(x,y)=frac{y^3-sin^3x}{x^2+y^2}$ differentiable at $(0,0)$?Differentiability of function definitionDifferentiability of piecewise functionsDifferentiability of a function $mathbb{R}^2tomathbb{R}$ at $(0,0)$Differentiability of logistic functionContinuity and differentiability of $f(x,y)$ at $(0,0)$Differentiability of $f$ at the origin
$begingroup$
Define $F:mathbb{R}^2tomathbb{R}$ by $F(x,y)=y^2+sqrt{y^2x^4}+x$.
Determine if F is differentiable at (1,0) or not.
To solve this problem, I have tried to applied $lim_{x to a} frac{||F(x)-F(a)-T(x-a)||}{||x-a||}$ to figure out whether the limit is equal to zero or not.
Then I got $lim_{(x,y) to (1,0)}frac{y^2+yx^2-y}{sqrt{(x-1)^2+y^2}}$ , the problem is I do not know how to solve such a limit or I used a wrong method to determine the differentiability.
In the other sub-part of the question, I have already solved the function is differentiable at (1,1) and (0,0).
real-analysis
edited yesterday
zhw.
74k43175
asked 2 days ago
AntonyAntony
61
New contributor
Antony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 3 more comments
$begingroup$
Define $F:mathbb{R}^2tomathbb{R}$ by $F(x,y)=y^2+sqrt{y^2x^4}+x$.
Determine if F is differentiable at (1,0) or not.
To solve this problem, I have tried to applied $lim_{x to a} frac{||F(x)-F(a)-T(x-a)||}{||x-a||}$ to figure out whether the limit is equal to zero or not.
Then I got $lim_{(x,y) to (1,0)}frac{y^2+yx^2-y}{sqrt{(x-1)^2+y^2}}$ , the problem is I do not know how to solve such a limit or I used a wrong method to determine the differentiability.
In the other sub-part of the question, I have already solved the function is differentiable at (1,1) and (0,0).
real-analysis
edited yesterday
zhw.
74k43175
asked 2 days ago
AntonyAntony
61
New contributor
Antony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
$F$ is not a vector valued function. What is $T$?
$endgroup$
– uniquesolution
2 days ago
$begingroup$
I thought the existence of partial derivatives at a point were a necessary condition for differentiability at that point. At $(0,0)$ they don't exist do they? so how did you prove it is differentiable at $(0,0)$? They don't seem to exist at $(1,0)$ either. Don't take my word, I'm also a student so hopefully someone will answer this soon (and as mentioned above, F is not a vector valued function since it takes values in reals - it is real valued.)
$endgroup$
– Displayname
2 days ago
$begingroup$
@Displayname The partials do exist at $(0,0).$
$endgroup$
– zhw.
2 days ago
$begingroup$
@zhw im a bit confused now...so what would the partial of $F$ w.r.t y be at (0,0)?
$endgroup$
– Displayname
2 days ago
1
$begingroup$
@Displayname $F(0,y)=y^2,$ so the partial of $F$ wrt $y$ is $0$ at the origin.
$endgroup$
– zhw.
yesterday
|
show 3 more comments
$begingroup$
Define $F:mathbb{R}^2tomathbb{R}$ by $F(x,y)=y^2+sqrt{y^2x^4}+x$.
Determine if F is differentiable at (1,0) or not.
To solve this problem, I have tried to applied $lim_{x to a} frac{||F(x)-F(a)-T(x-a)||}{||x-a||}$ to figure out whether the limit is equal to zero or not.
Then I got $lim_{(x,y) to (1,0)}frac{y^2+yx^2-y}{sqrt{(x-1)^2+y^2}}$ , the problem is I do not know how to solve such a limit or I used a wrong method to determine the differentiability.
In the other sub-part of the question, I have already solved the function is differentiable at (1,1) and (0,0).
real-analysis
edited yesterday
zhw.
74k43175
asked 2 days ago
AntonyAntony
61
New contributor
Antony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Define $F:mathbb{R}^2tomathbb{R}$ by $F(x,y)=y^2+sqrt{y^2x^4}+x$.
Determine if F is differentiable at (1,0) or not.
To solve this problem, I have tried to applied $lim_{x to a} frac{||F(x)-F(a)-T(x-a)||}{||x-a||}$ to figure out whether the limit is equal to zero or not.
Then I got $lim_{(x,y) to (1,0)}frac{y^2+yx^2-y}{sqrt{(x-1)^2+y^2}}$ , the problem is I do not know how to solve such a limit or I used a wrong method to determine the differentiability.
In the other sub-part of the question, I have already solved the function is differentiable at (1,1) and (0,0).
real-analysis
real-analysis
edited yesterday
zhw.
74k43175
asked 2 days ago
AntonyAntony
61
New contributor
Antony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
zhw.
74k43175
asked 2 days ago
AntonyAntony
61
New contributor
Antony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
zhw.
74k43175
edited yesterday
zhw.
74k43175
edited yesterday
zhw.
74k43175
74k43175
asked 2 days ago
AntonyAntony
61
New contributor
Antony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
AntonyAntony
61
asked 2 days ago
AntonyAntony
61
61
New contributor
Antony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Antony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Antony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
$F$ is not a vector valued function. What is $T$?
$endgroup$
– uniquesolution
2 days ago
$begingroup$
I thought the existence of partial derivatives at a point were a necessary condition for differentiability at that point. At $(0,0)$ they don't exist do they? so how did you prove it is differentiable at $(0,0)$? They don't seem to exist at $(1,0)$ either. Don't take my word, I'm also a student so hopefully someone will answer this soon (and as mentioned above, F is not a vector valued function since it takes values in reals - it is real valued.)
$endgroup$
– Displayname
2 days ago
$begingroup$
@Displayname The partials do exist at $(0,0).$
$endgroup$
– zhw.
2 days ago
$begingroup$
@zhw im a bit confused now...so what would the partial of $F$ w.r.t y be at (0,0)?
$endgroup$
– Displayname
2 days ago
1
$begingroup$
@Displayname $F(0,y)=y^2,$ so the partial of $F$ wrt $y$ is $0$ at the origin.
$endgroup$
– zhw.
yesterday
|
show 3 more comments
1
$begingroup$
$F$ is not a vector valued function. What is $T$?
$endgroup$
– uniquesolution
2 days ago
$begingroup$
I thought the existence of partial derivatives at a point were a necessary condition for differentiability at that point. At $(0,0)$ they don't exist do they? so how did you prove it is differentiable at $(0,0)$? They don't seem to exist at $(1,0)$ either. Don't take my word, I'm also a student so hopefully someone will answer this soon (and as mentioned above, F is not a vector valued function since it takes values in reals - it is real valued.)
$endgroup$
– Displayname
2 days ago
$begingroup$
@Displayname The partials do exist at $(0,0).$
$endgroup$
– zhw.
2 days ago
$begingroup$
@zhw im a bit confused now...so what would the partial of $F$ w.r.t y be at (0,0)?
$endgroup$
– Displayname
2 days ago
1
$begingroup$
@Displayname $F(0,y)=y^2,$ so the partial of $F$ wrt $y$ is $0$ at the origin.
$endgroup$
– zhw.
yesterday
1
1
$begingroup$
$F$ is not a vector valued function. What is $T$?
$endgroup$
– uniquesolution
2 days ago
$begingroup$
$F$ is not a vector valued function. What is $T$?
$endgroup$
– uniquesolution
2 days ago
$begingroup$
I thought the existence of partial derivatives at a point were a necessary condition for differentiability at that point. At $(0,0)$ they don't exist do they? so how did you prove it is differentiable at $(0,0)$? They don't seem to exist at $(1,0)$ either. Don't take my word, I'm also a student so hopefully someone will answer this soon (and as mentioned above, F is not a vector valued function since it takes values in reals - it is real valued.)
$endgroup$
– Displayname
2 days ago
$begingroup$
I thought the existence of partial derivatives at a point were a necessary condition for differentiability at that point. At $(0,0)$ they don't exist do they? so how did you prove it is differentiable at $(0,0)$? They don't seem to exist at $(1,0)$ either. Don't take my word, I'm also a student so hopefully someone will answer this soon (and as mentioned above, F is not a vector valued function since it takes values in reals - it is real valued.)
$endgroup$
– Displayname
2 days ago
$begingroup$
@Displayname The partials do exist at $(0,0).$
$endgroup$
– zhw.
2 days ago
$begingroup$
@Displayname The partials do exist at $(0,0).$
$endgroup$
– zhw.
2 days ago
$begingroup$
@zhw im a bit confused now...so what would the partial of $F$ w.r.t y be at (0,0)?
$endgroup$
– Displayname
2 days ago
$begingroup$
@zhw im a bit confused now...so what would the partial of $F$ w.r.t y be at (0,0)?
$endgroup$
– Displayname
2 days ago
1
1
$begingroup$
@Displayname $F(0,y)=y^2,$ so the partial of $F$ wrt $y$ is $0$ at the origin.
$endgroup$
– zhw.
yesterday
$begingroup$
@Displayname $F(0,y)=y^2,$ so the partial of $F$ wrt $y$ is $0$ at the origin.
$endgroup$
– zhw.
yesterday
|
show 3 more comments
1 Answer
1
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votes
$begingroup$
Hint: $F(x,y)=y^2+ |y|x^2 +x.$ Does $partial F/partial y (1,0)$ exist?
answered 2 days ago
zhw.zhw.
74k43175
$endgroup$
$begingroup$
I see, as one of the partial derivative does not exist at (1,0), therefore F is not differentiable at (1,0). Thanks a lot for your help.
$endgroup$
– Antony
yesterday
add a comment |
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$begingroup$
Hint: $F(x,y)=y^2+ |y|x^2 +x.$ Does $partial F/partial y (1,0)$ exist?
answered 2 days ago
zhw.zhw.
74k43175
$endgroup$
$begingroup$
I see, as one of the partial derivative does not exist at (1,0), therefore F is not differentiable at (1,0). Thanks a lot for your help.
$endgroup$
– Antony
yesterday
add a comment |
$begingroup$
Hint: $F(x,y)=y^2+ |y|x^2 +x.$ Does $partial F/partial y (1,0)$ exist?
answered 2 days ago
zhw.zhw.
74k43175
$endgroup$
$begingroup$
I see, as one of the partial derivative does not exist at (1,0), therefore F is not differentiable at (1,0). Thanks a lot for your help.
$endgroup$
– Antony
yesterday
add a comment |
$begingroup$
Hint: $F(x,y)=y^2+ |y|x^2 +x.$ Does $partial F/partial y (1,0)$ exist?
answered 2 days ago
zhw.zhw.
74k43175
$endgroup$
Hint: $F(x,y)=y^2+ |y|x^2 +x.$ Does $partial F/partial y (1,0)$ exist?
answered 2 days ago
zhw.zhw.
74k43175
edited yesterday
answered 2 days ago
zhw.zhw.
74k43175
answered 2 days ago
zhw.zhw.
74k43175
answered 2 days ago
zhw.zhw.
74k43175
74k43175
$begingroup$
I see, as one of the partial derivative does not exist at (1,0), therefore F is not differentiable at (1,0). Thanks a lot for your help.
$endgroup$
– Antony
yesterday
add a comment |
$begingroup$
I see, as one of the partial derivative does not exist at (1,0), therefore F is not differentiable at (1,0). Thanks a lot for your help.
$endgroup$
– Antony
yesterday
$begingroup$
I see, as one of the partial derivative does not exist at (1,0), therefore F is not differentiable at (1,0). Thanks a lot for your help.
$endgroup$
– Antony
yesterday
$begingroup$
I see, as one of the partial derivative does not exist at (1,0), therefore F is not differentiable at (1,0). Thanks a lot for your help.
$endgroup$
– Antony
yesterday
add a comment |
Antony is a new contributor. Be nice, and check out our Code of Conduct.
Antony is a new contributor. Be nice, and check out our Code of Conduct.
Antony is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
$F$ is not a vector valued function. What is $T$?
$endgroup$
– uniquesolution
2 days ago
$begingroup$
I thought the existence of partial derivatives at a point were a necessary condition for differentiability at that point. At $(0,0)$ they don't exist do they? so how did you prove it is differentiable at $(0,0)$? They don't seem to exist at $(1,0)$ either. Don't take my word, I'm also a student so hopefully someone will answer this soon (and as mentioned above, F is not a vector valued function since it takes values in reals - it is real valued.)
$endgroup$
– Displayname
2 days ago
$begingroup$
@Displayname The partials do exist at $(0,0).$
$endgroup$
– zhw.
2 days ago
$begingroup$
@zhw im a bit confused now...so what would the partial of $F$ w.r.t y be at (0,0)?
$endgroup$
– Displayname
2 days ago
1
$begingroup$
@Displayname $F(0,y)=y^2,$ so the partial of $F$ wrt $y$ is $0$ at the origin.
$endgroup$
– zhw.
yesterday