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Is the following function differentiable at (0,0)? [on hold]

-3

$begingroup$

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

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asked Mar 28 at 3:10

hopefullyhopefully

301214

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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

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

add a comment | 

-3

$begingroup$

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

share|cite|improve this question

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

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

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

share|cite|improve this question

asked Mar 28 at 3:10

hopefullyhopefully

301214

$endgroup$

share|cite|improve this question

asked Mar 28 at 3:10

hopefullyhopefully

301214

share|cite|improve this question

asked Mar 28 at 3:10

hopefullyhopefully

301214

asked Mar 28 at 3:10

hopefullyhopefully

301214

asked Mar 28 at 3:10

hopefullyhopefully

301214

2 Answers
2

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oldest

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1

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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.

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answered Mar 28 at 6:00

Kavi Rama MurthyKavi Rama Murthy

71.4k53170

$endgroup$

add a comment | 

2

$begingroup$

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).

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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.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I already have done this
  $endgroup$
  – hopefully
  Mar 28 at 6:05
  

  
  
  
  

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered Mar 28 at 6:00

Kavi Rama MurthyKavi Rama Murthy

71.4k53170

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Mar 28 at 6:00

Kavi Rama MurthyKavi Rama Murthy

71.4k53170

$endgroup$

share|cite|improve this answer

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

2

$begingroup$

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).

share|cite|improve this answer

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 | 

2

$begingroup$

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).

share|cite|improve this answer

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).

share|cite|improve this answer

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.

$endgroup$

share|cite|improve this answer

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