Is $ f(x,y) = frac{|x|^{5/2} y}{x^4 + y^2} $ differentiable at $(0,0)$?
$begingroup$
Let $ f: mathbb{R}^2 rightarrow mathbb{R} $ be a function defined as :
$$
begin{cases}
f(x,y) = frac{|x|^{5/2} y}{x^4 + y^2} , & (x,y) not= (0,0) \
f(0,0) = 0
end{cases}
$$
- Compute $ frac{df}{dx}(0,0) $ and $frac{df}{dy}(0,0)$.
- 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
edited Dec 29 '18 at 22:23
A.Γ.
22.6k32656
asked Dec 29 '18 at 20:47
Zouhair El YaagoubiZouhair El Yaagoubi
506411
$endgroup$
add a comment |
$begingroup$
Let $ f: mathbb{R}^2 rightarrow mathbb{R} $ be a function defined as :
$$
begin{cases}
f(x,y) = frac{|x|^{5/2} y}{x^4 + y^2} , & (x,y) not= (0,0) \
f(0,0) = 0
end{cases}
$$
- Compute $ frac{df}{dx}(0,0) $ and $frac{df}{dy}(0,0)$.
- 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
edited Dec 29 '18 at 22:23
A.Γ.
22.6k32656
asked Dec 29 '18 at 20:47
Zouhair El YaagoubiZouhair El Yaagoubi
506411
$endgroup$
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
add a comment |
$begingroup$
Let $ f: mathbb{R}^2 rightarrow mathbb{R} $ be a function defined as :
$$
begin{cases}
f(x,y) = frac{|x|^{5/2} y}{x^4 + y^2} , & (x,y) not= (0,0) \
f(0,0) = 0
end{cases}
$$
- Compute $ frac{df}{dx}(0,0) $ and $frac{df}{dy}(0,0)$.
- 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
edited Dec 29 '18 at 22:23
A.Γ.
22.6k32656
asked Dec 29 '18 at 20:47
Zouhair El YaagoubiZouhair El Yaagoubi
506411
$endgroup$
Let $ f: mathbb{R}^2 rightarrow mathbb{R} $ be a function defined as :
$$
begin{cases}
f(x,y) = frac{|x|^{5/2} y}{x^4 + y^2} , & (x,y) not= (0,0) \
f(0,0) = 0
end{cases}
$$
- Compute $ frac{df}{dx}(0,0) $ and $frac{df}{dy}(0,0)$.
- 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
edited Dec 29 '18 at 22:23
A.Γ.
22.6k32656
asked Dec 29 '18 at 20:47
Zouhair El YaagoubiZouhair El Yaagoubi
506411
edited Dec 29 '18 at 22:23
A.Γ.
22.6k32656
asked Dec 29 '18 at 20:47
Zouhair El YaagoubiZouhair El Yaagoubi
506411
edited Dec 29 '18 at 22:23
A.Γ.
22.6k32656
edited Dec 29 '18 at 22:23
A.Γ.
22.6k32656
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
asked Dec 29 '18 at 20:47
Zouhair El YaagoubiZouhair El Yaagoubi
506411
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
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 $xto 0$
$$
rho(x,x^2)=frac{|x|^{1/2}x^4}{2x^4cdot |x|sqrt{1+x^2}}=frac{1}{2|x|^{1/2}sqrt{1+x^2}}notto 0.
$$
answered Dec 29 '18 at 22:32
A.Γ.A.Γ.
22.6k32656
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$f$ 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.
answered Dec 29 '18 at 21:08
mathcounterexamples.netmathcounterexamples.net
25.4k21953
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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 $(0,0)$.
answered Dec 29 '18 at 22:41
mechanodroidmechanodroid
27.1k62446
$endgroup$
$begingroup$
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
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $xto 0$
$$
rho(x,x^2)=frac{|x|^{1/2}x^4}{2x^4cdot |x|sqrt{1+x^2}}=frac{1}{2|x|^{1/2}sqrt{1+x^2}}notto 0.
$$
answered Dec 29 '18 at 22:32
A.Γ.A.Γ.
22.6k32656
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $xto 0$
$$
rho(x,x^2)=frac{|x|^{1/2}x^4}{2x^4cdot |x|sqrt{1+x^2}}=frac{1}{2|x|^{1/2}sqrt{1+x^2}}notto 0.
$$
answered Dec 29 '18 at 22:32
A.Γ.A.Γ.
22.6k32656
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $xto 0$
$$
rho(x,x^2)=frac{|x|^{1/2}x^4}{2x^4cdot |x|sqrt{1+x^2}}=frac{1}{2|x|^{1/2}sqrt{1+x^2}}notto 0.
$$
answered Dec 29 '18 at 22:32
A.Γ.A.Γ.
22.6k32656
$endgroup$
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 $xto 0$
$$
rho(x,x^2)=frac{|x|^{1/2}x^4}{2x^4cdot |x|sqrt{1+x^2}}=frac{1}{2|x|^{1/2}sqrt{1+x^2}}notto 0.
$$
answered Dec 29 '18 at 22:32
A.Γ.A.Γ.
22.6k32656
answered Dec 29 '18 at 22:32
A.Γ.A.Γ.
22.6k32656
answered Dec 29 '18 at 22:32
A.Γ.A.Γ.
22.6k32656
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
$f$ 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.
answered Dec 29 '18 at 21:08
mathcounterexamples.netmathcounterexamples.net
25.4k21953
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
$f$ 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.
answered Dec 29 '18 at 21:08
mathcounterexamples.netmathcounterexamples.net
25.4k21953
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
$f$ 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.
answered Dec 29 '18 at 21:08
mathcounterexamples.netmathcounterexamples.net
25.4k21953
$endgroup$
$f$ 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.
answered Dec 29 '18 at 21:08
mathcounterexamples.netmathcounterexamples.net
25.4k21953
answered Dec 29 '18 at 21:08
mathcounterexamples.netmathcounterexamples.net
25.4k21953
answered Dec 29 '18 at 21:08
mathcounterexamples.netmathcounterexamples.net
25.4k21953
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
add a comment |
$begingroup$
@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
add a comment |
$begingroup$
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 $(0,0)$.
answered Dec 29 '18 at 22:41
mechanodroidmechanodroid
27.1k62446
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $(0,0)$.
answered Dec 29 '18 at 22:41
mechanodroidmechanodroid
27.1k62446
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $(0,0)$.
answered Dec 29 '18 at 22:41
mechanodroidmechanodroid
27.1k62446
$endgroup$
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 $(0,0)$.
answered Dec 29 '18 at 22:41
mechanodroidmechanodroid
27.1k62446
edited Dec 29 '18 at 22:48
answered Dec 29 '18 at 22:41
mechanodroidmechanodroid
27.1k62446
answered Dec 29 '18 at 22:41
mechanodroidmechanodroid
27.1k62446
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
add a comment |
$begingroup$
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
add a comment |
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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