Is the following function differentiable at (0,0)? [on hold] The Next CEO of Stack OverflowVerify if the function $f(x,y)$ is differentiable in $(x,y) = (0,0)$ and $(x,y) ne (0,0)$Prove the following function is differentiableIs this function differentiable in $(0,0)$Check whether the given function is differentiable at $(0,0)$Is the function differentiable at $(0,0)$Calculate the partial derivatives at $(0,0)$Function totally differentiable in $(0,0)$Differentiability of a multivariable function at (0,0)Is the function $(xy)^1/3$ differentiable at (0,0)?Is the following function continuously differentiable?
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Is the following function differentiable at (0,0)? [on hold]
The Next CEO of Stack OverflowVerify if the function $f(x,y)$ is differentiable in $(x,y) = (0,0)$ and $(x,y) ne (0,0)$Prove the following function is differentiableIs this function differentiable in $(0,0)$Check whether the given function is differentiable at $(0,0)$Is the function differentiable at $(0,0)$Calculate the partial derivatives at $(0,0)$Function totally differentiable in $(0,0)$Differentiability of a multivariable function at (0,0)Is the function $(xy)^1/3$ differentiable at (0,0)?Is the following function continuously differentiable?
$begingroup$
Is the following function differentiable at (0,0)?
I found that it is differentiable (using l`hopitals rule to calculate the limits ), am I correct?
calculus multivariable-calculus derivatives
asked Mar 28 at 3:10
hopefullyhopefully
301214
$endgroup$
put on hold as off-topic by Eevee Trainer, RRL, Delta-u, YiFan, Adrian Keister Mar 28 at 16:57
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." – Eevee Trainer, RRL, Delta-u, YiFan, Adrian Keister
add a comment |
$begingroup$
Is the following function differentiable at (0,0)?
I found that it is differentiable (using l`hopitals rule to calculate the limits ), am I correct?
calculus multivariable-calculus derivatives
asked Mar 28 at 3:10
hopefullyhopefully
301214
$endgroup$
put on hold as off-topic by Eevee Trainer, RRL, Delta-u, YiFan, Adrian Keister Mar 28 at 16:57
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." – Eevee Trainer, RRL, Delta-u, YiFan, Adrian Keister
add a comment |
$begingroup$
Is the following function differentiable at (0,0)?
I found that it is differentiable (using l`hopitals rule to calculate the limits ), am I correct?
calculus multivariable-calculus derivatives
asked Mar 28 at 3:10
hopefullyhopefully
301214
$endgroup$
Is the following function differentiable at (0,0)?
I found that it is differentiable (using l`hopitals rule to calculate the limits ), am I correct?
calculus multivariable-calculus derivatives
calculus multivariable-calculus derivatives
asked Mar 28 at 3:10
hopefullyhopefully
301214
asked Mar 28 at 3:10
hopefullyhopefully
301214
asked Mar 28 at 3:10
hopefullyhopefully
301214
asked Mar 28 at 3:10
hopefullyhopefully
301214
asked Mar 28 at 3:10
hopefullyhopefully
301214
301214
put on hold as off-topic by Eevee Trainer, RRL, Delta-u, YiFan, Adrian Keister Mar 28 at 16:57
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." – Eevee Trainer, RRL, Delta-u, YiFan, Adrian Keister
put on hold as off-topic by Eevee Trainer, RRL, Delta-u, YiFan, Adrian Keister Mar 28 at 16:57
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." – Eevee Trainer, RRL, Delta-u, YiFan, Adrian Keister
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The derivative is in fact $0$ (the zero linear transformation). To show that $frac f(x,y)-0sqrt x^2+y^2 to 0$ just put $u=frac 1 sqrt x^2+y^2$ and apply L'Hopital's Rule.
answered Mar 28 at 6:00
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
add a comment |
$begingroup$
I don't know what you mean to calculate the limit, but you need to apply the definition of differentiability to determine whether this function is differentiable. Namely, compute the limit: $$lim_(h,k)to(0,0)fracf(h,k)-f(0,0)-nabla f(0,0)cdot(h,k)$$
If it goes to zero, then this function is differentiable at (0,0).
answered Mar 28 at 4:00
UnbelieveTableUnbelieveTable
965
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UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
I already have done this
$endgroup$
– hopefully
Mar 28 at 6:05
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The derivative is in fact $0$ (the zero linear transformation). To show that $frac f(x,y)-0sqrt x^2+y^2 to 0$ just put $u=frac 1 sqrt x^2+y^2$ and apply L'Hopital's Rule.
answered Mar 28 at 6:00
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
add a comment |
$begingroup$
The derivative is in fact $0$ (the zero linear transformation). To show that $frac f(x,y)-0sqrt x^2+y^2 to 0$ just put $u=frac 1 sqrt x^2+y^2$ and apply L'Hopital's Rule.
answered Mar 28 at 6:00
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
add a comment |
$begingroup$
The derivative is in fact $0$ (the zero linear transformation). To show that $frac f(x,y)-0sqrt x^2+y^2 to 0$ just put $u=frac 1 sqrt x^2+y^2$ and apply L'Hopital's Rule.
answered Mar 28 at 6:00
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
The derivative is in fact $0$ (the zero linear transformation). To show that $frac f(x,y)-0sqrt x^2+y^2 to 0$ just put $u=frac 1 sqrt x^2+y^2$ and apply L'Hopital's Rule.
answered Mar 28 at 6:00
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
answered Mar 28 at 6:00
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
answered Mar 28 at 6:00
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
answered Mar 28 at 6:00
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
71.4k53170
add a comment |
add a comment |
$begingroup$
I don't know what you mean to calculate the limit, but you need to apply the definition of differentiability to determine whether this function is differentiable. Namely, compute the limit: $$lim_(h,k)to(0,0)fracf(h,k)-f(0,0)-nabla f(0,0)cdot(h,k)$$
If it goes to zero, then this function is differentiable at (0,0).
answered Mar 28 at 4:00
UnbelieveTableUnbelieveTable
965
New contributor
UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I already have done this
$endgroup$
– hopefully
Mar 28 at 6:05
add a comment |
$begingroup$
I don't know what you mean to calculate the limit, but you need to apply the definition of differentiability to determine whether this function is differentiable. Namely, compute the limit: $$lim_(h,k)to(0,0)fracf(h,k)-f(0,0)-nabla f(0,0)cdot(h,k)$$
If it goes to zero, then this function is differentiable at (0,0).
answered Mar 28 at 4:00
UnbelieveTableUnbelieveTable
965
New contributor
UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I already have done this
$endgroup$
– hopefully
Mar 28 at 6:05
add a comment |
$begingroup$
I don't know what you mean to calculate the limit, but you need to apply the definition of differentiability to determine whether this function is differentiable. Namely, compute the limit: $$lim_(h,k)to(0,0)fracf(h,k)-f(0,0)-nabla f(0,0)cdot(h,k)$$
If it goes to zero, then this function is differentiable at (0,0).
answered Mar 28 at 4:00
UnbelieveTableUnbelieveTable
965
New contributor
UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I don't know what you mean to calculate the limit, but you need to apply the definition of differentiability to determine whether this function is differentiable. Namely, compute the limit: $$lim_(h,k)to(0,0)fracf(h,k)-f(0,0)-nabla f(0,0)cdot(h,k)$$
If it goes to zero, then this function is differentiable at (0,0).
answered Mar 28 at 4:00
UnbelieveTableUnbelieveTable
965
New contributor
UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 28 at 4:00
UnbelieveTableUnbelieveTable
965
New contributor
UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 28 at 4:00
UnbelieveTableUnbelieveTable
965
answered Mar 28 at 4:00
UnbelieveTableUnbelieveTable
965
965
New contributor
UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I already have done this
$endgroup$
– hopefully
Mar 28 at 6:05
add a comment |
$begingroup$
I already have done this
$endgroup$
– hopefully
Mar 28 at 6:05
$begingroup$
I already have done this
$endgroup$
– hopefully
Mar 28 at 6:05
$begingroup$
I already have done this
$endgroup$
– hopefully
Mar 28 at 6:05
add a comment |
