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Differentiability of $fracxy^2x^2+y^2$…
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1
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I want to prove that $$
f(x,y)=
begincases fracxy^2x^2+y^2 &(x,y)neq (0,0)\
0, &(x,y)=(0,0)
endcases
$$
is not differentiable at $(0,0)$.
I thought that I can prove that it is not continuous around $(0,0)$ but it certainly is!
So how can I prove that it is not differentiable?
real-analysis multivariable-calculus derivatives
edited Aug 2 at 10:04
José Carlos Santos
112k1696172
asked Aug 2 at 9:56
Marc
676421
add a comment |Â
up vote
1
down vote
favorite
I want to prove that $$
f(x,y)=
begincases fracxy^2x^2+y^2 &(x,y)neq (0,0)\
0, &(x,y)=(0,0)
endcases
$$
is not differentiable at $(0,0)$.
I thought that I can prove that it is not continuous around $(0,0)$ but it certainly is!
So how can I prove that it is not differentiable?
real-analysis multivariable-calculus derivatives
edited Aug 2 at 10:04
José Carlos Santos
112k1696172
asked Aug 2 at 9:56
Marc
676421
1
It could be helpful to calculate the limit along different directions approaching the origin.
– Math-fun
Aug 2 at 9:59
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to prove that $$
f(x,y)=
begincases fracxy^2x^2+y^2 &(x,y)neq (0,0)\
0, &(x,y)=(0,0)
endcases
$$
is not differentiable at $(0,0)$.
I thought that I can prove that it is not continuous around $(0,0)$ but it certainly is!
So how can I prove that it is not differentiable?
real-analysis multivariable-calculus derivatives
edited Aug 2 at 10:04
José Carlos Santos
112k1696172
asked Aug 2 at 9:56
Marc
676421
I want to prove that $$
f(x,y)=
begincases fracxy^2x^2+y^2 &(x,y)neq (0,0)\
0, &(x,y)=(0,0)
endcases
$$
is not differentiable at $(0,0)$.
I thought that I can prove that it is not continuous around $(0,0)$ but it certainly is!
So how can I prove that it is not differentiable?
real-analysis multivariable-calculus derivatives
edited Aug 2 at 10:04
José Carlos Santos
112k1696172
asked Aug 2 at 9:56
Marc
676421
edited Aug 2 at 10:04
José Carlos Santos
112k1696172
edited Aug 2 at 10:04
José Carlos Santos
112k1696172
edited Aug 2 at 10:04
José Carlos Santos
112k1696172
112k1696172
asked Aug 2 at 9:56
Marc
676421
asked Aug 2 at 9:56
Marc
676421
asked Aug 2 at 9:56
Marc
676421
676421
1
It could be helpful to calculate the limit along different directions approaching the origin.
– Math-fun
Aug 2 at 9:59
add a comment |Â
1
It could be helpful to calculate the limit along different directions approaching the origin.
– Math-fun
Aug 2 at 9:59
1
1
It could be helpful to calculate the limit along different directions approaching the origin.
– Math-fun
Aug 2 at 9:59
It could be helpful to calculate the limit along different directions approaching the origin.
– Math-fun
Aug 2 at 9:59
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
6
down vote
accepted
Since $f_x(0,0)=f_y(0,0)=0$, if $f$ was differentiable at $(0,0)$, $f'(0,0)$ would be the null function. Therefore$$lim_(x,y)to(0,0)fracsqrtx^2+y^2=0,$$which means that$$lim_(x,y)to(0,0)fracxy^2(x^2+y^2)^frac32=0.$$However, this is false. See what happens when $x=y$.
answered Aug 2 at 10:02
José Carlos Santos
112k1696172
add a comment |Â
up vote
2
down vote
At the origin the directional derivative in the direction of $(1,1)$ is $frac 1 2$ whereas the derivative in the direction of $x-$ axis and $y-$axis are $0$. This implies that the derivative does not exist.
answered Aug 2 at 10:02
Kavi Rama Murthy
19.2k2829
You wrote: " f does not a have a partial derivative w.r.t. $x$ at the origin.... " This is false ! We have , for $x ne 0$ that $frac f(x,0)-f(0,0) x=0$. Hence $f_x(0,0)$ exists and $=0$.
– Fred
Aug 2 at 10:22
@Fred Thanks for the correction.
– Kavi Rama Murthy
Aug 2 at 12:14
add a comment |Â
up vote
1
down vote
Another approach: assume that $f$ is differentiable at $(0,0)$. For each $v=(u,w) in mathbb R^2$ with $u^2+w^2=1$ we then have
$$(*) quad fracpartial fpartial v(0,0)=v cdot gradf(0,0).$$
Since
$$ (**) quadfracpartial fpartial v(0,0)= lim_t to 0fracf(tv)-f(0,0)t=fracuw^2u^2+w^2=uw^2,$$
we see that $gradf(0,0)=(0,0)$ and therefor, by $(*)$
$$fracpartial fpartial v(0,0)=0.$$
But if $u=w= frac1sqrt2$, then by $(**)$
$$fracpartial fpartial v(0,0) ne 0.$$
A contradiction !
Remark: the solution of Jose Carlos Santos shorter. But my solution shows that a function $f$ can have directional derivatives at $x_0$ in all directions $v$, without being differentiable at $x_0$.
answered Aug 2 at 10:57
Fred
37k1237
add a comment |Â
up vote
1
down vote
Look at the directional derivative:
$$d_mathbfvf=limlimits_tto 0fracf(mathbfx+tmathbfv)-f(mathbfx)t$$
in your case you have for $mathbfx=(0,0)$
$$d_(a,b)f=limlimits_tto 0fracfract^3ab^2t^2(a^2+b^2)t=fracab^2a^2+b^2$$ Thus the directional derivative exists. However, it is not a linear function, and therefore $f$ is not differentiable.
answered Aug 2 at 12:14
Rene Schipperus
31.9k11958
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Since $f_x(0,0)=f_y(0,0)=0$, if $f$ was differentiable at $(0,0)$, $f'(0,0)$ would be the null function. Therefore$$lim_(x,y)to(0,0)fracsqrtx^2+y^2=0,$$which means that$$lim_(x,y)to(0,0)fracxy^2(x^2+y^2)^frac32=0.$$However, this is false. See what happens when $x=y$.
answered Aug 2 at 10:02
José Carlos Santos
112k1696172
add a comment |Â
up vote
6
down vote
accepted
Since $f_x(0,0)=f_y(0,0)=0$, if $f$ was differentiable at $(0,0)$, $f'(0,0)$ would be the null function. Therefore$$lim_(x,y)to(0,0)fracsqrtx^2+y^2=0,$$which means that$$lim_(x,y)to(0,0)fracxy^2(x^2+y^2)^frac32=0.$$However, this is false. See what happens when $x=y$.
answered Aug 2 at 10:02
José Carlos Santos
112k1696172
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Since $f_x(0,0)=f_y(0,0)=0$, if $f$ was differentiable at $(0,0)$, $f'(0,0)$ would be the null function. Therefore$$lim_(x,y)to(0,0)fracsqrtx^2+y^2=0,$$which means that$$lim_(x,y)to(0,0)fracxy^2(x^2+y^2)^frac32=0.$$However, this is false. See what happens when $x=y$.
answered Aug 2 at 10:02
José Carlos Santos
112k1696172
Since $f_x(0,0)=f_y(0,0)=0$, if $f$ was differentiable at $(0,0)$, $f'(0,0)$ would be the null function. Therefore$$lim_(x,y)to(0,0)fracsqrtx^2+y^2=0,$$which means that$$lim_(x,y)to(0,0)fracxy^2(x^2+y^2)^frac32=0.$$However, this is false. See what happens when $x=y$.
answered Aug 2 at 10:02
José Carlos Santos
112k1696172
answered Aug 2 at 10:02
José Carlos Santos
112k1696172
answered Aug 2 at 10:02
José Carlos Santos
112k1696172
answered Aug 2 at 10:02
José Carlos Santos
112k1696172
112k1696172
add a comment |Â
add a comment |Â
up vote
2
down vote
At the origin the directional derivative in the direction of $(1,1)$ is $frac 1 2$ whereas the derivative in the direction of $x-$ axis and $y-$axis are $0$. This implies that the derivative does not exist.
answered Aug 2 at 10:02
Kavi Rama Murthy
19.2k2829
You wrote: " f does not a have a partial derivative w.r.t. $x$ at the origin.... " This is false ! We have , for $x ne 0$ that $frac f(x,0)-f(0,0) x=0$. Hence $f_x(0,0)$ exists and $=0$.
– Fred
Aug 2 at 10:22
@Fred Thanks for the correction.
– Kavi Rama Murthy
Aug 2 at 12:14
add a comment |Â
up vote
2
down vote
At the origin the directional derivative in the direction of $(1,1)$ is $frac 1 2$ whereas the derivative in the direction of $x-$ axis and $y-$axis are $0$. This implies that the derivative does not exist.
answered Aug 2 at 10:02
Kavi Rama Murthy
19.2k2829
You wrote: " f does not a have a partial derivative w.r.t. $x$ at the origin.... " This is false ! We have , for $x ne 0$ that $frac f(x,0)-f(0,0) x=0$. Hence $f_x(0,0)$ exists and $=0$.
– Fred
Aug 2 at 10:22
@Fred Thanks for the correction.
– Kavi Rama Murthy
Aug 2 at 12:14
add a comment |Â
up vote
2
down vote
up vote
2
down vote
At the origin the directional derivative in the direction of $(1,1)$ is $frac 1 2$ whereas the derivative in the direction of $x-$ axis and $y-$axis are $0$. This implies that the derivative does not exist.
answered Aug 2 at 10:02
Kavi Rama Murthy
19.2k2829
At the origin the directional derivative in the direction of $(1,1)$ is $frac 1 2$ whereas the derivative in the direction of $x-$ axis and $y-$axis are $0$. This implies that the derivative does not exist.
answered Aug 2 at 10:02
Kavi Rama Murthy
19.2k2829
edited Aug 2 at 12:25
answered Aug 2 at 10:02
Kavi Rama Murthy
19.2k2829
answered Aug 2 at 10:02
Kavi Rama Murthy
19.2k2829
answered Aug 2 at 10:02
Kavi Rama Murthy
19.2k2829
19.2k2829
You wrote: " f does not a have a partial derivative w.r.t. $x$ at the origin.... " This is false ! We have , for $x ne 0$ that $frac f(x,0)-f(0,0) x=0$. Hence $f_x(0,0)$ exists and $=0$.
– Fred
Aug 2 at 10:22
@Fred Thanks for the correction.
– Kavi Rama Murthy
Aug 2 at 12:14
add a comment |Â
You wrote: " f does not a have a partial derivative w.r.t. $x$ at the origin.... " This is false ! We have , for $x ne 0$ that $frac f(x,0)-f(0,0) x=0$. Hence $f_x(0,0)$ exists and $=0$.
– Fred
Aug 2 at 10:22
@Fred Thanks for the correction.
– Kavi Rama Murthy
Aug 2 at 12:14
You wrote: " f does not a have a partial derivative w.r.t. $x$ at the origin.... " This is false ! We have , for $x ne 0$ that $frac f(x,0)-f(0,0) x=0$. Hence $f_x(0,0)$ exists and $=0$.
– Fred
Aug 2 at 10:22
You wrote: " f does not a have a partial derivative w.r.t. $x$ at the origin.... " This is false ! We have , for $x ne 0$ that $frac f(x,0)-f(0,0) x=0$. Hence $f_x(0,0)$ exists and $=0$.
– Fred
Aug 2 at 10:22
@Fred Thanks for the correction.
– Kavi Rama Murthy
Aug 2 at 12:14
@Fred Thanks for the correction.
– Kavi Rama Murthy
Aug 2 at 12:14
add a comment |Â
up vote
1
down vote
Another approach: assume that $f$ is differentiable at $(0,0)$. For each $v=(u,w) in mathbb R^2$ with $u^2+w^2=1$ we then have
$$(*) quad fracpartial fpartial v(0,0)=v cdot gradf(0,0).$$
Since
$$ (**) quadfracpartial fpartial v(0,0)= lim_t to 0fracf(tv)-f(0,0)t=fracuw^2u^2+w^2=uw^2,$$
we see that $gradf(0,0)=(0,0)$ and therefor, by $(*)$
$$fracpartial fpartial v(0,0)=0.$$
But if $u=w= frac1sqrt2$, then by $(**)$
$$fracpartial fpartial v(0,0) ne 0.$$
A contradiction !
Remark: the solution of Jose Carlos Santos shorter. But my solution shows that a function $f$ can have directional derivatives at $x_0$ in all directions $v$, without being differentiable at $x_0$.
answered Aug 2 at 10:57
Fred
37k1237
add a comment |Â
up vote
1
down vote
Another approach: assume that $f$ is differentiable at $(0,0)$. For each $v=(u,w) in mathbb R^2$ with $u^2+w^2=1$ we then have
$$(*) quad fracpartial fpartial v(0,0)=v cdot gradf(0,0).$$
Since
$$ (**) quadfracpartial fpartial v(0,0)= lim_t to 0fracf(tv)-f(0,0)t=fracuw^2u^2+w^2=uw^2,$$
we see that $gradf(0,0)=(0,0)$ and therefor, by $(*)$
$$fracpartial fpartial v(0,0)=0.$$
But if $u=w= frac1sqrt2$, then by $(**)$
$$fracpartial fpartial v(0,0) ne 0.$$
A contradiction !
Remark: the solution of Jose Carlos Santos shorter. But my solution shows that a function $f$ can have directional derivatives at $x_0$ in all directions $v$, without being differentiable at $x_0$.
answered Aug 2 at 10:57
Fred
37k1237
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Another approach: assume that $f$ is differentiable at $(0,0)$. For each $v=(u,w) in mathbb R^2$ with $u^2+w^2=1$ we then have
$$(*) quad fracpartial fpartial v(0,0)=v cdot gradf(0,0).$$
Since
$$ (**) quadfracpartial fpartial v(0,0)= lim_t to 0fracf(tv)-f(0,0)t=fracuw^2u^2+w^2=uw^2,$$
we see that $gradf(0,0)=(0,0)$ and therefor, by $(*)$
$$fracpartial fpartial v(0,0)=0.$$
But if $u=w= frac1sqrt2$, then by $(**)$
$$fracpartial fpartial v(0,0) ne 0.$$
A contradiction !
Remark: the solution of Jose Carlos Santos shorter. But my solution shows that a function $f$ can have directional derivatives at $x_0$ in all directions $v$, without being differentiable at $x_0$.
answered Aug 2 at 10:57
Fred
37k1237
Another approach: assume that $f$ is differentiable at $(0,0)$. For each $v=(u,w) in mathbb R^2$ with $u^2+w^2=1$ we then have
$$(*) quad fracpartial fpartial v(0,0)=v cdot gradf(0,0).$$
Since
$$ (**) quadfracpartial fpartial v(0,0)= lim_t to 0fracf(tv)-f(0,0)t=fracuw^2u^2+w^2=uw^2,$$
we see that $gradf(0,0)=(0,0)$ and therefor, by $(*)$
$$fracpartial fpartial v(0,0)=0.$$
But if $u=w= frac1sqrt2$, then by $(**)$
$$fracpartial fpartial v(0,0) ne 0.$$
A contradiction !
Remark: the solution of Jose Carlos Santos shorter. But my solution shows that a function $f$ can have directional derivatives at $x_0$ in all directions $v$, without being differentiable at $x_0$.
answered Aug 2 at 10:57
Fred
37k1237
answered Aug 2 at 10:57
Fred
37k1237
answered Aug 2 at 10:57
Fred
37k1237
answered Aug 2 at 10:57
Fred
37k1237
37k1237
add a comment |Â
add a comment |Â
up vote
1
down vote
Look at the directional derivative:
$$d_mathbfvf=limlimits_tto 0fracf(mathbfx+tmathbfv)-f(mathbfx)t$$
in your case you have for $mathbfx=(0,0)$
$$d_(a,b)f=limlimits_tto 0fracfract^3ab^2t^2(a^2+b^2)t=fracab^2a^2+b^2$$ Thus the directional derivative exists. However, it is not a linear function, and therefore $f$ is not differentiable.
answered Aug 2 at 12:14
Rene Schipperus
31.9k11958
add a comment |Â
up vote
1
down vote
Look at the directional derivative:
$$d_mathbfvf=limlimits_tto 0fracf(mathbfx+tmathbfv)-f(mathbfx)t$$
in your case you have for $mathbfx=(0,0)$
$$d_(a,b)f=limlimits_tto 0fracfract^3ab^2t^2(a^2+b^2)t=fracab^2a^2+b^2$$ Thus the directional derivative exists. However, it is not a linear function, and therefore $f$ is not differentiable.
answered Aug 2 at 12:14
Rene Schipperus
31.9k11958
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Look at the directional derivative:
$$d_mathbfvf=limlimits_tto 0fracf(mathbfx+tmathbfv)-f(mathbfx)t$$
in your case you have for $mathbfx=(0,0)$
$$d_(a,b)f=limlimits_tto 0fracfract^3ab^2t^2(a^2+b^2)t=fracab^2a^2+b^2$$ Thus the directional derivative exists. However, it is not a linear function, and therefore $f$ is not differentiable.
answered Aug 2 at 12:14
Rene Schipperus
31.9k11958
Look at the directional derivative:
$$d_mathbfvf=limlimits_tto 0fracf(mathbfx+tmathbfv)-f(mathbfx)t$$
in your case you have for $mathbfx=(0,0)$
$$d_(a,b)f=limlimits_tto 0fracfract^3ab^2t^2(a^2+b^2)t=fracab^2a^2+b^2$$ Thus the directional derivative exists. However, it is not a linear function, and therefore $f$ is not differentiable.
answered Aug 2 at 12:14
Rene Schipperus
31.9k11958
answered Aug 2 at 12:14
Rene Schipperus
31.9k11958
answered Aug 2 at 12:14
Rene Schipperus
31.9k11958
answered Aug 2 at 12:14
Rene Schipperus
31.9k11958
31.9k11958
add a comment |Â
add a comment |Â
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1
It could be helpful to calculate the limit along different directions approaching the origin.
– Math-fun
Aug 2 at 9:59