Is $ f(x,y) = frac{|x|^{5/2} y}{x^4 + y^2} $ differentiable at $(0,0)$?












4












$begingroup$



Let $ f: mathbb{R}^2 rightarrow mathbb{R} $ be a function defined as :



$$
begin{cases}
f(x,y) = frac{|x|^{5/2} y}{x^4 + y^2} , & (x,y) not= (0,0) \
f(0,0) = 0
end{cases}
$$




  1. Compute $ frac{df}{dx}(0,0) $ and $frac{df}{dy}(0,0)$.

  2. Is $f$ differentiable at $(0,0)$?





I have problem with question 2.



For $f$ to be differentiable, the partial derivatives must exist and be continuous.



They exist because we have:



$ frac{df}{dx}(0,0) = lim_{x to 0} frac{f(x,0) - f(0,0)}{x} = 0$



$ frac{df}{dy}(0,0) = lim_{y to 0} frac{f(0,y) - f(0,0)}{y} = 0$



but I cannot compute the partial derivatives because of the absolute value.



How to know if the partial derivatives are continuous at $(0,0)$ ?






real-analysis multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Can you find the derivative of $xmapsto |x|$ where it exists?
    $endgroup$
    – Git Gud
    Dec 29 '18 at 21:05






  • 2




    $begingroup$
    I need to split the interval to have two differents cases. In this case I have the power $5/2$ which is the problem for me. Why my question is voted to be closed?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 21:08


















4












$begingroup$



Let $ f: mathbb{R}^2 rightarrow mathbb{R} $ be a function defined as :



$$
begin{cases}
f(x,y) = frac{|x|^{5/2} y}{x^4 + y^2} , & (x,y) not= (0,0) \
f(0,0) = 0
end{cases}
$$




  1. Compute $ frac{df}{dx}(0,0) $ and $frac{df}{dy}(0,0)$.

  2. Is $f$ differentiable at $(0,0)$?





I have problem with question 2.



For $f$ to be differentiable, the partial derivatives must exist and be continuous.



They exist because we have:



$ frac{df}{dx}(0,0) = lim_{x to 0} frac{f(x,0) - f(0,0)}{x} = 0$



$ frac{df}{dy}(0,0) = lim_{y to 0} frac{f(0,y) - f(0,0)}{y} = 0$



but I cannot compute the partial derivatives because of the absolute value.



How to know if the partial derivatives are continuous at $(0,0)$ ?






real-analysis multivariable-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Can you find the derivative of $xmapsto |x|$ where it exists?
    $endgroup$
    – Git Gud
    Dec 29 '18 at 21:05






  • 2




    $begingroup$
    I need to split the interval to have two differents cases. In this case I have the power $5/2$ which is the problem for me. Why my question is voted to be closed?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 21:08
















4












4








4


0



$begingroup$



Let $ f: mathbb{R}^2 rightarrow mathbb{R} $ be a function defined as :



$$
begin{cases}
f(x,y) = frac{|x|^{5/2} y}{x^4 + y^2} , & (x,y) not= (0,0) \
f(0,0) = 0
end{cases}
$$




  1. Compute $ frac{df}{dx}(0,0) $ and $frac{df}{dy}(0,0)$.

  2. Is $f$ differentiable at $(0,0)$?





I have problem with question 2.



For $f$ to be differentiable, the partial derivatives must exist and be continuous.



They exist because we have:



$ frac{df}{dx}(0,0) = lim_{x to 0} frac{f(x,0) - f(0,0)}{x} = 0$



$ frac{df}{dy}(0,0) = lim_{y to 0} frac{f(0,y) - f(0,0)}{y} = 0$



but I cannot compute the partial derivatives because of the absolute value.



How to know if the partial derivatives are continuous at $(0,0)$ ?






real-analysis multivariable-calculus







share|cite|improve this question











$endgroup$





Let $ f: mathbb{R}^2 rightarrow mathbb{R} $ be a function defined as :



$$
begin{cases}
f(x,y) = frac{|x|^{5/2} y}{x^4 + y^2} , & (x,y) not= (0,0) \
f(0,0) = 0
end{cases}
$$




  1. Compute $ frac{df}{dx}(0,0) $ and $frac{df}{dy}(0,0)$.

  2. Is $f$ differentiable at $(0,0)$?





I have problem with question 2.



For $f$ to be differentiable, the partial derivatives must exist and be continuous.



They exist because we have:



$ frac{df}{dx}(0,0) = lim_{x to 0} frac{f(x,0) - f(0,0)}{x} = 0$



$ frac{df}{dy}(0,0) = lim_{y to 0} frac{f(0,y) - f(0,0)}{y} = 0$



but I cannot compute the partial derivatives because of the absolute value.



How to know if the partial derivatives are continuous at $(0,0)$ ?






real-analysis multivariable-calculus




real-analysis multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 22:23









A.Γ.

22.6k32656




22.6k32656










asked Dec 29 '18 at 20:47









Zouhair El YaagoubiZouhair El Yaagoubi

506411




506411








  • 1




    $begingroup$
    Can you find the derivative of $xmapsto |x|$ where it exists?
    $endgroup$
    – Git Gud
    Dec 29 '18 at 21:05






  • 2




    $begingroup$
    I need to split the interval to have two differents cases. In this case I have the power $5/2$ which is the problem for me. Why my question is voted to be closed?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 21:08
















  • 1




    $begingroup$
    Can you find the derivative of $xmapsto |x|$ where it exists?
    $endgroup$
    – Git Gud
    Dec 29 '18 at 21:05






  • 2




    $begingroup$
    I need to split the interval to have two differents cases. In this case I have the power $5/2$ which is the problem for me. Why my question is voted to be closed?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 21:08










1




1




$begingroup$
Can you find the derivative of $xmapsto |x|$ where it exists?
$endgroup$
– Git Gud
Dec 29 '18 at 21:05




$begingroup$
Can you find the derivative of $xmapsto |x|$ where it exists?
$endgroup$
– Git Gud
Dec 29 '18 at 21:05




2




2




$begingroup$
I need to split the interval to have two differents cases. In this case I have the power $5/2$ which is the problem for me. Why my question is voted to be closed?
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 21:08






$begingroup$
I need to split the interval to have two differents cases. In this case I have the power $5/2$ which is the problem for me. Why my question is voted to be closed?
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 21:08












3 Answers
3






active

oldest

votes


















1












$begingroup$

With your last edit the answer to 2) is negative. For differentiability we need by definition that
$$
rho(x,y)=frac{f(x,y)-f(0,0)-f'_x(0,0)x-f'_y(0,0)y}{sqrt{x^2+y^2}}to 0
$$
as $(x,y)to(0,0)$, however, with $y=x^2$ we get for $xto 0$
$$
rho(x,x^2)=frac{|x|^{1/2}x^4}{2x^4cdot |x|sqrt{1+x^2}}=frac{1}{2|x|^{1/2}sqrt{1+x^2}}notto 0.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why did divide by $sqrt{x^2 + y^2}$ ? I am not familiar with the function you considered.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:40










  • $begingroup$
    @ZouhairElYaagoubi What is your definition of a differentiable function? I am using the standard one. Here $|h|=|(x,y)|=sqrt{x^2+y^2}$.
    $endgroup$
    – A.Γ.
    Dec 29 '18 at 22:45












  • $begingroup$
    Ah! I see it now. Thank you so much for your help. Accepted answer.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 23:04



















4












$begingroup$

$f$ is differentiable at $(0,0)$ and the derivative is the always vanishing map as for $(x,y)neq(0,0)$ you have



$$0 le leftvert frac {f(x,y)}{sqrt{x^2+y^2}} rightvert = frac{vert x vert}{sqrt{x^2+y^2}} frac{vert xy vert}{x^2+y^2}sqrt{vert x vert}le frac{sqrt{vert x vert}}{2} to 0$$



As $(x,y) to (0,0)$.



Take care! There is indeed a theorem stating that the derivative exists and is continuous if and only if the partial derivatives exist and are continuous. But a map maybe be differentiable even if the partial derivatives exist but are not continuous.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @A.Γ. No, in the problem I have it is $x^4 + y^2$. Is it a mistake in the problem?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 21:49












  • $begingroup$
    I have made a mistake in the problem, the denominator is $ x^4 + y^2$.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:04



















1












$begingroup$

To calculate the partial derivatives, recall that the derivative of the absolute value is $|cdot|' = text{sign}$, on $mathbb{R}setminus {0}$.



Therefore for $(x,y) ne (0,0)$ we have



$$frac{partial f}{partial x}(x,y) = frac{frac{5}2(operatorname{sign} x)|x|^{3/2}y(x^4+y^2) - |x|^{5/2}ycdot 4x^3}{(x^4+y^2)^2}$$



$$frac{partial f}{partial y}(x,y) = frac{|x|^{5/2}(x^4+y^2) - |x|^{5/2}ycdot 2y}{(x^4+y^2)^2}$$



E.g. for $x > 0$ and $y = x^2$ we get



$$frac{partial f}{partial x}(x,x^2) = frac{5|x|^{7.5} - 4|x|^{7.5}}{4x^8} = frac1{4sqrt{|x|}} notto 0$$



as $(x,y) to (0,0)$ so $frac{partial f}{partial x}$ isn't continuous at $(0,0)$.



This says nothing about differentiability, but the other answers show that $f$ is not differentiable at $(0,0)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Using the same argment can we say that $f $ is not continuous on $(0,0)$ ?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:43






  • 1




    $begingroup$
    @ZouhairElYaagoubi $f$ is continuous at $(0,0)$: $$|f(x,y)| = frac{|x|^{5/2}|y|}{x^4+y^2} le frac{r^{5/2}r}{r^2} = r^{3/2} xrightarrow{rto 0} 0$$ where $r = sqrt{x^2+y^2}$ is the norm of $(x,y)$.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 22:45










  • $begingroup$
    Thank you so much for your help.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 23:03











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

With your last edit the answer to 2) is negative. For differentiability we need by definition that
$$
rho(x,y)=frac{f(x,y)-f(0,0)-f'_x(0,0)x-f'_y(0,0)y}{sqrt{x^2+y^2}}to 0
$$
as $(x,y)to(0,0)$, however, with $y=x^2$ we get for $xto 0$
$$
rho(x,x^2)=frac{|x|^{1/2}x^4}{2x^4cdot |x|sqrt{1+x^2}}=frac{1}{2|x|^{1/2}sqrt{1+x^2}}notto 0.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why did divide by $sqrt{x^2 + y^2}$ ? I am not familiar with the function you considered.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:40










  • $begingroup$
    @ZouhairElYaagoubi What is your definition of a differentiable function? I am using the standard one. Here $|h|=|(x,y)|=sqrt{x^2+y^2}$.
    $endgroup$
    – A.Γ.
    Dec 29 '18 at 22:45












  • $begingroup$
    Ah! I see it now. Thank you so much for your help. Accepted answer.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 23:04
















1












$begingroup$

With your last edit the answer to 2) is negative. For differentiability we need by definition that
$$
rho(x,y)=frac{f(x,y)-f(0,0)-f'_x(0,0)x-f'_y(0,0)y}{sqrt{x^2+y^2}}to 0
$$
as $(x,y)to(0,0)$, however, with $y=x^2$ we get for $xto 0$
$$
rho(x,x^2)=frac{|x|^{1/2}x^4}{2x^4cdot |x|sqrt{1+x^2}}=frac{1}{2|x|^{1/2}sqrt{1+x^2}}notto 0.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why did divide by $sqrt{x^2 + y^2}$ ? I am not familiar with the function you considered.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:40










  • $begingroup$
    @ZouhairElYaagoubi What is your definition of a differentiable function? I am using the standard one. Here $|h|=|(x,y)|=sqrt{x^2+y^2}$.
    $endgroup$
    – A.Γ.
    Dec 29 '18 at 22:45












  • $begingroup$
    Ah! I see it now. Thank you so much for your help. Accepted answer.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 23:04














1












1








1





$begingroup$

With your last edit the answer to 2) is negative. For differentiability we need by definition that
$$
rho(x,y)=frac{f(x,y)-f(0,0)-f'_x(0,0)x-f'_y(0,0)y}{sqrt{x^2+y^2}}to 0
$$
as $(x,y)to(0,0)$, however, with $y=x^2$ we get for $xto 0$
$$
rho(x,x^2)=frac{|x|^{1/2}x^4}{2x^4cdot |x|sqrt{1+x^2}}=frac{1}{2|x|^{1/2}sqrt{1+x^2}}notto 0.
$$






share|cite|improve this answer









$endgroup$



With your last edit the answer to 2) is negative. For differentiability we need by definition that
$$
rho(x,y)=frac{f(x,y)-f(0,0)-f'_x(0,0)x-f'_y(0,0)y}{sqrt{x^2+y^2}}to 0
$$
as $(x,y)to(0,0)$, however, with $y=x^2$ we get for $xto 0$
$$
rho(x,x^2)=frac{|x|^{1/2}x^4}{2x^4cdot |x|sqrt{1+x^2}}=frac{1}{2|x|^{1/2}sqrt{1+x^2}}notto 0.
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 22:32









A.Γ.A.Γ.

22.6k32656




22.6k32656












  • $begingroup$
    Why did divide by $sqrt{x^2 + y^2}$ ? I am not familiar with the function you considered.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:40










  • $begingroup$
    @ZouhairElYaagoubi What is your definition of a differentiable function? I am using the standard one. Here $|h|=|(x,y)|=sqrt{x^2+y^2}$.
    $endgroup$
    – A.Γ.
    Dec 29 '18 at 22:45












  • $begingroup$
    Ah! I see it now. Thank you so much for your help. Accepted answer.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 23:04


















  • $begingroup$
    Why did divide by $sqrt{x^2 + y^2}$ ? I am not familiar with the function you considered.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:40










  • $begingroup$
    @ZouhairElYaagoubi What is your definition of a differentiable function? I am using the standard one. Here $|h|=|(x,y)|=sqrt{x^2+y^2}$.
    $endgroup$
    – A.Γ.
    Dec 29 '18 at 22:45












  • $begingroup$
    Ah! I see it now. Thank you so much for your help. Accepted answer.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 23:04
















$begingroup$
Why did divide by $sqrt{x^2 + y^2}$ ? I am not familiar with the function you considered.
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 22:40




$begingroup$
Why did divide by $sqrt{x^2 + y^2}$ ? I am not familiar with the function you considered.
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 22:40












$begingroup$
@ZouhairElYaagoubi What is your definition of a differentiable function? I am using the standard one. Here $|h|=|(x,y)|=sqrt{x^2+y^2}$.
$endgroup$
– A.Γ.
Dec 29 '18 at 22:45






$begingroup$
@ZouhairElYaagoubi What is your definition of a differentiable function? I am using the standard one. Here $|h|=|(x,y)|=sqrt{x^2+y^2}$.
$endgroup$
– A.Γ.
Dec 29 '18 at 22:45














$begingroup$
Ah! I see it now. Thank you so much for your help. Accepted answer.
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 23:04




$begingroup$
Ah! I see it now. Thank you so much for your help. Accepted answer.
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 23:04











4












$begingroup$

$f$ is differentiable at $(0,0)$ and the derivative is the always vanishing map as for $(x,y)neq(0,0)$ you have



$$0 le leftvert frac {f(x,y)}{sqrt{x^2+y^2}} rightvert = frac{vert x vert}{sqrt{x^2+y^2}} frac{vert xy vert}{x^2+y^2}sqrt{vert x vert}le frac{sqrt{vert x vert}}{2} to 0$$



As $(x,y) to (0,0)$.



Take care! There is indeed a theorem stating that the derivative exists and is continuous if and only if the partial derivatives exist and are continuous. But a map maybe be differentiable even if the partial derivatives exist but are not continuous.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @A.Γ. No, in the problem I have it is $x^4 + y^2$. Is it a mistake in the problem?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 21:49












  • $begingroup$
    I have made a mistake in the problem, the denominator is $ x^4 + y^2$.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:04
















4












$begingroup$

$f$ is differentiable at $(0,0)$ and the derivative is the always vanishing map as for $(x,y)neq(0,0)$ you have



$$0 le leftvert frac {f(x,y)}{sqrt{x^2+y^2}} rightvert = frac{vert x vert}{sqrt{x^2+y^2}} frac{vert xy vert}{x^2+y^2}sqrt{vert x vert}le frac{sqrt{vert x vert}}{2} to 0$$



As $(x,y) to (0,0)$.



Take care! There is indeed a theorem stating that the derivative exists and is continuous if and only if the partial derivatives exist and are continuous. But a map maybe be differentiable even if the partial derivatives exist but are not continuous.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @A.Γ. No, in the problem I have it is $x^4 + y^2$. Is it a mistake in the problem?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 21:49












  • $begingroup$
    I have made a mistake in the problem, the denominator is $ x^4 + y^2$.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:04














4












4








4





$begingroup$

$f$ is differentiable at $(0,0)$ and the derivative is the always vanishing map as for $(x,y)neq(0,0)$ you have



$$0 le leftvert frac {f(x,y)}{sqrt{x^2+y^2}} rightvert = frac{vert x vert}{sqrt{x^2+y^2}} frac{vert xy vert}{x^2+y^2}sqrt{vert x vert}le frac{sqrt{vert x vert}}{2} to 0$$



As $(x,y) to (0,0)$.



Take care! There is indeed a theorem stating that the derivative exists and is continuous if and only if the partial derivatives exist and are continuous. But a map maybe be differentiable even if the partial derivatives exist but are not continuous.






share|cite|improve this answer









$endgroup$



$f$ is differentiable at $(0,0)$ and the derivative is the always vanishing map as for $(x,y)neq(0,0)$ you have



$$0 le leftvert frac {f(x,y)}{sqrt{x^2+y^2}} rightvert = frac{vert x vert}{sqrt{x^2+y^2}} frac{vert xy vert}{x^2+y^2}sqrt{vert x vert}le frac{sqrt{vert x vert}}{2} to 0$$



As $(x,y) to (0,0)$.



Take care! There is indeed a theorem stating that the derivative exists and is continuous if and only if the partial derivatives exist and are continuous. But a map maybe be differentiable even if the partial derivatives exist but are not continuous.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 21:08









mathcounterexamples.netmathcounterexamples.net

25.4k21953




25.4k21953












  • $begingroup$
    @A.Γ. No, in the problem I have it is $x^4 + y^2$. Is it a mistake in the problem?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 21:49












  • $begingroup$
    I have made a mistake in the problem, the denominator is $ x^4 + y^2$.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:04


















  • $begingroup$
    @A.Γ. No, in the problem I have it is $x^4 + y^2$. Is it a mistake in the problem?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 21:49












  • $begingroup$
    I have made a mistake in the problem, the denominator is $ x^4 + y^2$.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:04
















$begingroup$
@A.Γ. No, in the problem I have it is $x^4 + y^2$. Is it a mistake in the problem?
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 21:49






$begingroup$
@A.Γ. No, in the problem I have it is $x^4 + y^2$. Is it a mistake in the problem?
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 21:49














$begingroup$
I have made a mistake in the problem, the denominator is $ x^4 + y^2$.
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 22:04




$begingroup$
I have made a mistake in the problem, the denominator is $ x^4 + y^2$.
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 22:04











1












$begingroup$

To calculate the partial derivatives, recall that the derivative of the absolute value is $|cdot|' = text{sign}$, on $mathbb{R}setminus {0}$.



Therefore for $(x,y) ne (0,0)$ we have



$$frac{partial f}{partial x}(x,y) = frac{frac{5}2(operatorname{sign} x)|x|^{3/2}y(x^4+y^2) - |x|^{5/2}ycdot 4x^3}{(x^4+y^2)^2}$$



$$frac{partial f}{partial y}(x,y) = frac{|x|^{5/2}(x^4+y^2) - |x|^{5/2}ycdot 2y}{(x^4+y^2)^2}$$



E.g. for $x > 0$ and $y = x^2$ we get



$$frac{partial f}{partial x}(x,x^2) = frac{5|x|^{7.5} - 4|x|^{7.5}}{4x^8} = frac1{4sqrt{|x|}} notto 0$$



as $(x,y) to (0,0)$ so $frac{partial f}{partial x}$ isn't continuous at $(0,0)$.



This says nothing about differentiability, but the other answers show that $f$ is not differentiable at $(0,0)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Using the same argment can we say that $f $ is not continuous on $(0,0)$ ?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:43






  • 1




    $begingroup$
    @ZouhairElYaagoubi $f$ is continuous at $(0,0)$: $$|f(x,y)| = frac{|x|^{5/2}|y|}{x^4+y^2} le frac{r^{5/2}r}{r^2} = r^{3/2} xrightarrow{rto 0} 0$$ where $r = sqrt{x^2+y^2}$ is the norm of $(x,y)$.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 22:45










  • $begingroup$
    Thank you so much for your help.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 23:03
















1












$begingroup$

To calculate the partial derivatives, recall that the derivative of the absolute value is $|cdot|' = text{sign}$, on $mathbb{R}setminus {0}$.



Therefore for $(x,y) ne (0,0)$ we have



$$frac{partial f}{partial x}(x,y) = frac{frac{5}2(operatorname{sign} x)|x|^{3/2}y(x^4+y^2) - |x|^{5/2}ycdot 4x^3}{(x^4+y^2)^2}$$



$$frac{partial f}{partial y}(x,y) = frac{|x|^{5/2}(x^4+y^2) - |x|^{5/2}ycdot 2y}{(x^4+y^2)^2}$$



E.g. for $x > 0$ and $y = x^2$ we get



$$frac{partial f}{partial x}(x,x^2) = frac{5|x|^{7.5} - 4|x|^{7.5}}{4x^8} = frac1{4sqrt{|x|}} notto 0$$



as $(x,y) to (0,0)$ so $frac{partial f}{partial x}$ isn't continuous at $(0,0)$.



This says nothing about differentiability, but the other answers show that $f$ is not differentiable at $(0,0)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Using the same argment can we say that $f $ is not continuous on $(0,0)$ ?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:43






  • 1




    $begingroup$
    @ZouhairElYaagoubi $f$ is continuous at $(0,0)$: $$|f(x,y)| = frac{|x|^{5/2}|y|}{x^4+y^2} le frac{r^{5/2}r}{r^2} = r^{3/2} xrightarrow{rto 0} 0$$ where $r = sqrt{x^2+y^2}$ is the norm of $(x,y)$.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 22:45










  • $begingroup$
    Thank you so much for your help.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 23:03














1












1








1





$begingroup$

To calculate the partial derivatives, recall that the derivative of the absolute value is $|cdot|' = text{sign}$, on $mathbb{R}setminus {0}$.



Therefore for $(x,y) ne (0,0)$ we have



$$frac{partial f}{partial x}(x,y) = frac{frac{5}2(operatorname{sign} x)|x|^{3/2}y(x^4+y^2) - |x|^{5/2}ycdot 4x^3}{(x^4+y^2)^2}$$



$$frac{partial f}{partial y}(x,y) = frac{|x|^{5/2}(x^4+y^2) - |x|^{5/2}ycdot 2y}{(x^4+y^2)^2}$$



E.g. for $x > 0$ and $y = x^2$ we get



$$frac{partial f}{partial x}(x,x^2) = frac{5|x|^{7.5} - 4|x|^{7.5}}{4x^8} = frac1{4sqrt{|x|}} notto 0$$



as $(x,y) to (0,0)$ so $frac{partial f}{partial x}$ isn't continuous at $(0,0)$.



This says nothing about differentiability, but the other answers show that $f$ is not differentiable at $(0,0)$.






share|cite|improve this answer











$endgroup$



To calculate the partial derivatives, recall that the derivative of the absolute value is $|cdot|' = text{sign}$, on $mathbb{R}setminus {0}$.



Therefore for $(x,y) ne (0,0)$ we have



$$frac{partial f}{partial x}(x,y) = frac{frac{5}2(operatorname{sign} x)|x|^{3/2}y(x^4+y^2) - |x|^{5/2}ycdot 4x^3}{(x^4+y^2)^2}$$



$$frac{partial f}{partial y}(x,y) = frac{|x|^{5/2}(x^4+y^2) - |x|^{5/2}ycdot 2y}{(x^4+y^2)^2}$$



E.g. for $x > 0$ and $y = x^2$ we get



$$frac{partial f}{partial x}(x,x^2) = frac{5|x|^{7.5} - 4|x|^{7.5}}{4x^8} = frac1{4sqrt{|x|}} notto 0$$



as $(x,y) to (0,0)$ so $frac{partial f}{partial x}$ isn't continuous at $(0,0)$.



This says nothing about differentiability, but the other answers show that $f$ is not differentiable at $(0,0)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 22:48

























answered Dec 29 '18 at 22:41









mechanodroidmechanodroid

27.1k62446




27.1k62446












  • $begingroup$
    Using the same argment can we say that $f $ is not continuous on $(0,0)$ ?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:43






  • 1




    $begingroup$
    @ZouhairElYaagoubi $f$ is continuous at $(0,0)$: $$|f(x,y)| = frac{|x|^{5/2}|y|}{x^4+y^2} le frac{r^{5/2}r}{r^2} = r^{3/2} xrightarrow{rto 0} 0$$ where $r = sqrt{x^2+y^2}$ is the norm of $(x,y)$.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 22:45










  • $begingroup$
    Thank you so much for your help.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 23:03


















  • $begingroup$
    Using the same argment can we say that $f $ is not continuous on $(0,0)$ ?
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 22:43






  • 1




    $begingroup$
    @ZouhairElYaagoubi $f$ is continuous at $(0,0)$: $$|f(x,y)| = frac{|x|^{5/2}|y|}{x^4+y^2} le frac{r^{5/2}r}{r^2} = r^{3/2} xrightarrow{rto 0} 0$$ where $r = sqrt{x^2+y^2}$ is the norm of $(x,y)$.
    $endgroup$
    – mechanodroid
    Dec 29 '18 at 22:45










  • $begingroup$
    Thank you so much for your help.
    $endgroup$
    – Zouhair El Yaagoubi
    Dec 29 '18 at 23:03
















$begingroup$
Using the same argment can we say that $f $ is not continuous on $(0,0)$ ?
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 22:43




$begingroup$
Using the same argment can we say that $f $ is not continuous on $(0,0)$ ?
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 22:43




1




1




$begingroup$
@ZouhairElYaagoubi $f$ is continuous at $(0,0)$: $$|f(x,y)| = frac{|x|^{5/2}|y|}{x^4+y^2} le frac{r^{5/2}r}{r^2} = r^{3/2} xrightarrow{rto 0} 0$$ where $r = sqrt{x^2+y^2}$ is the norm of $(x,y)$.
$endgroup$
– mechanodroid
Dec 29 '18 at 22:45




$begingroup$
@ZouhairElYaagoubi $f$ is continuous at $(0,0)$: $$|f(x,y)| = frac{|x|^{5/2}|y|}{x^4+y^2} le frac{r^{5/2}r}{r^2} = r^{3/2} xrightarrow{rto 0} 0$$ where $r = sqrt{x^2+y^2}$ is the norm of $(x,y)$.
$endgroup$
– mechanodroid
Dec 29 '18 at 22:45












$begingroup$
Thank you so much for your help.
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 23:03




$begingroup$
Thank you so much for your help.
$endgroup$
– Zouhair El Yaagoubi
Dec 29 '18 at 23:03


















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