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Is the function $(xy)^1/3$ differentiable at (0,0)?

proving that a certain function is not differentiable at $(0,0)$Is this function differentiable?Check whether the given function is differentiable at $(0,0)$Function totally differentiable in $(0,0)$is the function $f$ differentiable at $(0,0)$?Two variables limit / differentiable functionWhat is the correct method to show that $f(x,y)$ is not-differentiable at $(0,0)$?Differentiability of a multivariable function at (0,0)Is $f(x,y)= fracx^3y^2-2y^4sqrt2x^2+y^4$ for $(x,y)neq (0,0)$, $f(0,0)=0$, defining a differentiable function?Is the function $f(x,y) = fracx^2y^2x^2y^2 + (y-x)^2$ if $(x,y) neq (0,0)$, $f(0,0) = 0$ differentiable? Continuous?

0

$begingroup$

Is the function $(xy)^1/3$ differentiable at $(0,0)$?

My trial:

No, as I calculated $r(x,y)$ and it turned out to be $(xy)^1/3$, then I calculated the limit $frac (xy)^2/3x^2 + y^2$ at (0,0) and it turned out that it does not exist hence the function is not differentiable at $(0,0)$ . Am I correct?

calculus limits multivariable-calculus derivatives

share|cite|improve this question

edited Mar 12 at 9:33

dmtri

1,6332521

asked Mar 12 at 9:05

HappyHappy

777

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I will attach the formula I am using.
  $endgroup$
  – Happy
  Mar 12 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Where did that second function come from?
  $endgroup$
  – Displayname
  Mar 12 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur I have edited my question
  $endgroup$
  – Happy
  Mar 12 at 9:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Displayname I have edited my question
  $endgroup$
  – Happy
  Mar 12 at 9:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes they are @dmtri
  $endgroup$
  – Happy
  Mar 12 at 9:25
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Is the function $(xy)^1/3$ differentiable at $(0,0)$?

My trial:

No, as I calculated $r(x,y)$ and it turned out to be $(xy)^1/3$, then I calculated the limit $frac (xy)^2/3x^2 + y^2$ at (0,0) and it turned out that it does not exist hence the function is not differentiable at $(0,0)$ . Am I correct?

calculus limits multivariable-calculus derivatives

share|cite|improve this question

edited Mar 12 at 9:33

dmtri

1,6332521

asked Mar 12 at 9:05

HappyHappy

777

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I will attach the formula I am using.
  $endgroup$
  – Happy
  Mar 12 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Where did that second function come from?
  $endgroup$
  – Displayname
  Mar 12 at 9:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Arthur I have edited my question
  $endgroup$
  – Happy
  Mar 12 at 9:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Displayname I have edited my question
  $endgroup$
  – Happy
  Mar 12 at 9:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes they are @dmtri
  $endgroup$
  – Happy
  Mar 12 at 9:25
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Is the function $(xy)^1/3$ differentiable at $(0,0)$?

My trial:

No, as I calculated $r(x,y)$ and it turned out to be $(xy)^1/3$, then I calculated the limit $frac (xy)^2/3x^2 + y^2$ at (0,0) and it turned out that it does not exist hence the function is not differentiable at $(0,0)$ . Am I correct?

calculus limits multivariable-calculus derivatives

share|cite|improve this question

edited Mar 12 at 9:33

dmtri

1,6332521

asked Mar 12 at 9:05

HappyHappy

777

$endgroup$

share|cite|improve this question

edited Mar 12 at 9:33

dmtri

1,6332521

asked Mar 12 at 9:05

HappyHappy

777

share|cite|improve this question

edited Mar 12 at 9:33

dmtri

1,6332521

asked Mar 12 at 9:05

HappyHappy

777

edited Mar 12 at 9:33

dmtri

1,6332521

edited Mar 12 at 9:33

dmtri

1,6332521

asked Mar 12 at 9:05

HappyHappy

777

asked Mar 12 at 9:05

HappyHappy

777

1 Answer
1

active

oldest

votes

4

$begingroup$

If a function is differentiable, then all its directional derivatives must exist (and they must also play along with one another in a particularily nice way, which is what that formula of yours is putting in rigorous terms).

Let's look at the line $x = y$. Does your function have a directional derivative along this line?

The (or rather, a) directional derivative along that line is given by
$$
lim_tto 0fracf(t, t) - f(0,0)t = lim_tto 0fract^2/3 - 0t = lim_tto 0left(t^-1/3right)
$$
which doesn't exist.

Therefore your function doesn't have directional derivatives in all directions, which means it cannot be differentiable.

share|cite|improve this answer

edited Mar 12 at 9:33

answered Mar 12 at 9:27

ArthurArthur

118k7118201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  but the theorem mentioned in the other answer is written in my book ..... may be I am mistaking the direction of the implication?
  $endgroup$
  – Happy
  Mar 12 at 9:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Happy Take a look at that theorem. What is $r(t, t)$? Does that look like something in my answer? I've basically done the same thing here as that theorem of yours does, only used different words to explain why it's right.
  $endgroup$
  – Arthur
  Mar 12 at 9:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  also the function $xy/sqrt x^2 + y^2$ if $x^2 + y^2 neq 1$ and 0 otherwise turned out to be not differentiable at (0,0) by the same method , am I correct?
  $endgroup$
  – Happy
  Mar 12 at 9:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Happy I think you mean "if $x^2 + y^2neq 0$". But that function is differentiable at $(0,0)$. Note that, say, along the line $x = y$, the same directional derivative as above becomes $lim_tto0fract^2$, which certainly exists.
  $endgroup$
  – Arthur
  Mar 12 at 9:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes you are correct .... you mean it is differentiable/
  $endgroup$
  – Happy
  Mar 12 at 10:08
  

  
  
  
  

 | 
show 2 more comments

Your Answer

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4

$begingroup$

If a function is differentiable, then all its directional derivatives must exist (and they must also play along with one another in a particularily nice way, which is what that formula of yours is putting in rigorous terms).

Let's look at the line $x = y$. Does your function have a directional derivative along this line?

The (or rather, a) directional derivative along that line is given by
$$
lim_tto 0fracf(t, t) - f(0,0)t = lim_tto 0fract^2/3 - 0t = lim_tto 0left(t^-1/3right)
$$
which doesn't exist.

Therefore your function doesn't have directional derivatives in all directions, which means it cannot be differentiable.

share|cite|improve this answer

edited Mar 12 at 9:33

answered Mar 12 at 9:27

ArthurArthur

118k7118201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  but the theorem mentioned in the other answer is written in my book ..... may be I am mistaking the direction of the implication?
  $endgroup$
  – Happy
  Mar 12 at 9:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Happy Take a look at that theorem. What is $r(t, t)$? Does that look like something in my answer? I've basically done the same thing here as that theorem of yours does, only used different words to explain why it's right.
  $endgroup$
  – Arthur
  Mar 12 at 9:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  also the function $xy/sqrt x^2 + y^2$ if $x^2 + y^2 neq 1$ and 0 otherwise turned out to be not differentiable at (0,0) by the same method , am I correct?
  $endgroup$
  – Happy
  Mar 12 at 9:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Happy I think you mean "if $x^2 + y^2neq 0$". But that function is differentiable at $(0,0)$. Note that, say, along the line $x = y$, the same directional derivative as above becomes $lim_tto0fract^2$, which certainly exists.
  $endgroup$
  – Arthur
  Mar 12 at 9:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes you are correct .... you mean it is differentiable/
  $endgroup$
  – Happy
  Mar 12 at 10:08
  

  
  
  
  

 | 
show 2 more comments

4

$begingroup$

If a function is differentiable, then all its directional derivatives must exist (and they must also play along with one another in a particularily nice way, which is what that formula of yours is putting in rigorous terms).

Let's look at the line $x = y$. Does your function have a directional derivative along this line?

The (or rather, a) directional derivative along that line is given by
$$
lim_tto 0fracf(t, t) - f(0,0)t = lim_tto 0fract^2/3 - 0t = lim_tto 0left(t^-1/3right)
$$
which doesn't exist.

Therefore your function doesn't have directional derivatives in all directions, which means it cannot be differentiable.

share|cite|improve this answer

edited Mar 12 at 9:33

answered Mar 12 at 9:27

ArthurArthur

118k7118201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  but the theorem mentioned in the other answer is written in my book ..... may be I am mistaking the direction of the implication?
  $endgroup$
  – Happy
  Mar 12 at 9:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Happy Take a look at that theorem. What is $r(t, t)$? Does that look like something in my answer? I've basically done the same thing here as that theorem of yours does, only used different words to explain why it's right.
  $endgroup$
  – Arthur
  Mar 12 at 9:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  also the function $xy/sqrt x^2 + y^2$ if $x^2 + y^2 neq 1$ and 0 otherwise turned out to be not differentiable at (0,0) by the same method , am I correct?
  $endgroup$
  – Happy
  Mar 12 at 9:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Happy I think you mean "if $x^2 + y^2neq 0$". But that function is differentiable at $(0,0)$. Note that, say, along the line $x = y$, the same directional derivative as above becomes $lim_tto0fract^2$, which certainly exists.
  $endgroup$
  – Arthur
  Mar 12 at 9:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes you are correct .... you mean it is differentiable/
  $endgroup$
  – Happy
  Mar 12 at 10:08
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

If a function is differentiable, then all its directional derivatives must exist (and they must also play along with one another in a particularily nice way, which is what that formula of yours is putting in rigorous terms).

Let's look at the line $x = y$. Does your function have a directional derivative along this line?

The (or rather, a) directional derivative along that line is given by
$$
lim_tto 0fracf(t, t) - f(0,0)t = lim_tto 0fract^2/3 - 0t = lim_tto 0left(t^-1/3right)
$$
which doesn't exist.

Therefore your function doesn't have directional derivatives in all directions, which means it cannot be differentiable.

share|cite|improve this answer

edited Mar 12 at 9:33

answered Mar 12 at 9:27

ArthurArthur

118k7118201

$endgroup$

share|cite|improve this answer

edited Mar 12 at 9:33

answered Mar 12 at 9:27

ArthurArthur

118k7118201

answered Mar 12 at 9:27

ArthurArthur

118k7118201

answered Mar 12 at 9:27

ArthurArthur

118k7118201