Mixing
differentiability of $(x^2-y)y^2/(x^4+y^2)$?
$begingroup$
$f(x,y)=(x^2-y)y^2/(x^4+y^2)$ for every $(x,y)neq (0,0)$ and $f(x,y)=(0,0)$ on $(0,0)$.
I am trying to find out whether this function is differentiable or not.
I proved that its continuous on the origin and I have calculated the definition of derivative (using limit) for different paths but I could not prove that the function is not differentiable so I am guessing that it is but I do not know how to prove it.
calculus multivariable-calculus derivatives
asked Jan 17 at 0:04
S.JafariS.Jafari
132
$endgroup$
add a comment |
$begingroup$
$f(x,y)=(x^2-y)y^2/(x^4+y^2)$ for every $(x,y)neq (0,0)$ and $f(x,y)=(0,0)$ on $(0,0)$.
I am trying to find out whether this function is differentiable or not.
I proved that its continuous on the origin and I have calculated the definition of derivative (using limit) for different paths but I could not prove that the function is not differentiable so I am guessing that it is but I do not know how to prove it.
calculus multivariable-calculus derivatives
asked Jan 17 at 0:04
S.JafariS.Jafari
132
$endgroup$
1
$begingroup$
What would be a strategy for proving the existence or nonexistence of a derivative? Did you compute partial derivatives and examine the behavior in a neighborhood of $(0,0)$? Why not show what you actually did compute. Otherwise this just looks like you want someone to solve the problem for you.
$endgroup$
– RRL
Jan 17 at 0:19
add a comment |
$begingroup$
$f(x,y)=(x^2-y)y^2/(x^4+y^2)$ for every $(x,y)neq (0,0)$ and $f(x,y)=(0,0)$ on $(0,0)$.
I am trying to find out whether this function is differentiable or not.
I proved that its continuous on the origin and I have calculated the definition of derivative (using limit) for different paths but I could not prove that the function is not differentiable so I am guessing that it is but I do not know how to prove it.
calculus multivariable-calculus derivatives
asked Jan 17 at 0:04
S.JafariS.Jafari
132
$endgroup$
$f(x,y)=(x^2-y)y^2/(x^4+y^2)$ for every $(x,y)neq (0,0)$ and $f(x,y)=(0,0)$ on $(0,0)$.
I am trying to find out whether this function is differentiable or not.
I proved that its continuous on the origin and I have calculated the definition of derivative (using limit) for different paths but I could not prove that the function is not differentiable so I am guessing that it is but I do not know how to prove it.
calculus multivariable-calculus derivatives
calculus multivariable-calculus derivatives
asked Jan 17 at 0:04
S.JafariS.Jafari
132
asked Jan 17 at 0:04
S.JafariS.Jafari
132
asked Jan 17 at 0:04
S.JafariS.Jafari
132
asked Jan 17 at 0:04
S.JafariS.Jafari
132
asked Jan 17 at 0:04
S.JafariS.Jafari
132
132
1
$begingroup$
What would be a strategy for proving the existence or nonexistence of a derivative? Did you compute partial derivatives and examine the behavior in a neighborhood of $(0,0)$? Why not show what you actually did compute. Otherwise this just looks like you want someone to solve the problem for you.
$endgroup$
– RRL
Jan 17 at 0:19
add a comment |
1
$begingroup$
What would be a strategy for proving the existence or nonexistence of a derivative? Did you compute partial derivatives and examine the behavior in a neighborhood of $(0,0)$? Why not show what you actually did compute. Otherwise this just looks like you want someone to solve the problem for you.
$endgroup$
– RRL
Jan 17 at 0:19
1
1
$begingroup$
What would be a strategy for proving the existence or nonexistence of a derivative? Did you compute partial derivatives and examine the behavior in a neighborhood of $(0,0)$? Why not show what you actually did compute. Otherwise this just looks like you want someone to solve the problem for you.
$endgroup$
– RRL
Jan 17 at 0:19
$begingroup$
What would be a strategy for proving the existence or nonexistence of a derivative? Did you compute partial derivatives and examine the behavior in a neighborhood of $(0,0)$? Why not show what you actually did compute. Otherwise this just looks like you want someone to solve the problem for you.
$endgroup$
– RRL
Jan 17 at 0:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I am going to introduce the definition of differentiability of a function. A function $f : mathbb{R}^n to mathbb{R}^m$ is differenctiable in $x_0in mathbb{R}^n$ if there exists a linear function $L:mathbb{R}^ntomathbb{R}^m$ such that:
$$
lim_{hto (0,0)}frac{left|left|,f(x_0+h)-f(x_0)-L(h)right|right|}{left|left|hright|right|}=0
$$
The matrix associated to the linear function $L$ is the Jacobian matrix.
Then, let's calculate the Jacobian matrix of:
$$
f(x,y)=begin{cases}
displaystylefrac{(x^2-y)y^2}{x^4+y^2} & text{if } (x,y)neq (0,0) \
0 & text{if } (x,y)=(0,0)
end{cases}
$$
in $(x,y)=(0,0)$. First:
$$
frac{partial f}{partial x}(0,0) = lim_{hto 0} frac{0}{h^4}=0
$$
$$
frac{partial f}{partial y}(0,0) = lim_{hto 0} frac{-h^3}{h^2}=0
$$
Then the Jacobian matrix in $(0,0)$ of $f$ is $J_{(0,0)}(,f)=left(begin{array}{cc}
0 & 0
end{array}right)$, the if $L$ exists, it is $L(x,y)=left(begin{array}{cc}
0 & 0
end{array}right)left(begin{array}{c}
x \
y \
end{array}right)=0$. Then:
begin{eqnarray}
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)right|right|}{left|left|(x,y)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{(x,y)to (0,0)}left|frac{(x^2-y)y^2}{x^4+y^2},right|frac{1}{left|left|(x,y)right|right|} \
end{eqnarray}
If we use the path $y=x$:
begin{eqnarray}
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{xto 0}left|frac{(x^2-x)x^2}{x^4+x^2},right|frac{1}{left|left|(x,x)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{xto 0}left|frac{x^2-x}{x^2+1},right|frac{1}{|x|left|left|(1,1)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& frac{1}{left|left|(1,1)right|right|} lim_{xto 0}left|frac{x-1}{x^2+1},right| \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& frac{1}{left|left|(1,1)right|right|} neq 0 \
end{eqnarray}
Thus $f$ is not differentiable in $(0,0)$
answered Jan 17 at 1:02
El boritoEl borito
666216
$endgroup$
$begingroup$
It is fixed, thank you so much
$endgroup$
– El borito
Jan 17 at 4:35
1
$begingroup$
thank you. it really helped.
$endgroup$
– S.Jafari
Jan 17 at 19:36
add a comment |
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$begingroup$
I am going to introduce the definition of differentiability of a function. A function $f : mathbb{R}^n to mathbb{R}^m$ is differenctiable in $x_0in mathbb{R}^n$ if there exists a linear function $L:mathbb{R}^ntomathbb{R}^m$ such that:
$$
lim_{hto (0,0)}frac{left|left|,f(x_0+h)-f(x_0)-L(h)right|right|}{left|left|hright|right|}=0
$$
The matrix associated to the linear function $L$ is the Jacobian matrix.
Then, let's calculate the Jacobian matrix of:
$$
f(x,y)=begin{cases}
displaystylefrac{(x^2-y)y^2}{x^4+y^2} & text{if } (x,y)neq (0,0) \
0 & text{if } (x,y)=(0,0)
end{cases}
$$
in $(x,y)=(0,0)$. First:
$$
frac{partial f}{partial x}(0,0) = lim_{hto 0} frac{0}{h^4}=0
$$
$$
frac{partial f}{partial y}(0,0) = lim_{hto 0} frac{-h^3}{h^2}=0
$$
Then the Jacobian matrix in $(0,0)$ of $f$ is $J_{(0,0)}(,f)=left(begin{array}{cc}
0 & 0
end{array}right)$, the if $L$ exists, it is $L(x,y)=left(begin{array}{cc}
0 & 0
end{array}right)left(begin{array}{c}
x \
y \
end{array}right)=0$. Then:
begin{eqnarray}
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)right|right|}{left|left|(x,y)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{(x,y)to (0,0)}left|frac{(x^2-y)y^2}{x^4+y^2},right|frac{1}{left|left|(x,y)right|right|} \
end{eqnarray}
If we use the path $y=x$:
begin{eqnarray}
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{xto 0}left|frac{(x^2-x)x^2}{x^4+x^2},right|frac{1}{left|left|(x,x)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{xto 0}left|frac{x^2-x}{x^2+1},right|frac{1}{|x|left|left|(1,1)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& frac{1}{left|left|(1,1)right|right|} lim_{xto 0}left|frac{x-1}{x^2+1},right| \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& frac{1}{left|left|(1,1)right|right|} neq 0 \
end{eqnarray}
Thus $f$ is not differentiable in $(0,0)$
answered Jan 17 at 1:02
El boritoEl borito
666216
$endgroup$
$begingroup$
It is fixed, thank you so much
$endgroup$
– El borito
Jan 17 at 4:35
1
$begingroup$
thank you. it really helped.
$endgroup$
– S.Jafari
Jan 17 at 19:36
add a comment |
$begingroup$
I am going to introduce the definition of differentiability of a function. A function $f : mathbb{R}^n to mathbb{R}^m$ is differenctiable in $x_0in mathbb{R}^n$ if there exists a linear function $L:mathbb{R}^ntomathbb{R}^m$ such that:
$$
lim_{hto (0,0)}frac{left|left|,f(x_0+h)-f(x_0)-L(h)right|right|}{left|left|hright|right|}=0
$$
The matrix associated to the linear function $L$ is the Jacobian matrix.
Then, let's calculate the Jacobian matrix of:
$$
f(x,y)=begin{cases}
displaystylefrac{(x^2-y)y^2}{x^4+y^2} & text{if } (x,y)neq (0,0) \
0 & text{if } (x,y)=(0,0)
end{cases}
$$
in $(x,y)=(0,0)$. First:
$$
frac{partial f}{partial x}(0,0) = lim_{hto 0} frac{0}{h^4}=0
$$
$$
frac{partial f}{partial y}(0,0) = lim_{hto 0} frac{-h^3}{h^2}=0
$$
Then the Jacobian matrix in $(0,0)$ of $f$ is $J_{(0,0)}(,f)=left(begin{array}{cc}
0 & 0
end{array}right)$, the if $L$ exists, it is $L(x,y)=left(begin{array}{cc}
0 & 0
end{array}right)left(begin{array}{c}
x \
y \
end{array}right)=0$. Then:
begin{eqnarray}
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)right|right|}{left|left|(x,y)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{(x,y)to (0,0)}left|frac{(x^2-y)y^2}{x^4+y^2},right|frac{1}{left|left|(x,y)right|right|} \
end{eqnarray}
If we use the path $y=x$:
begin{eqnarray}
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{xto 0}left|frac{(x^2-x)x^2}{x^4+x^2},right|frac{1}{left|left|(x,x)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{xto 0}left|frac{x^2-x}{x^2+1},right|frac{1}{|x|left|left|(1,1)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& frac{1}{left|left|(1,1)right|right|} lim_{xto 0}left|frac{x-1}{x^2+1},right| \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& frac{1}{left|left|(1,1)right|right|} neq 0 \
end{eqnarray}
Thus $f$ is not differentiable in $(0,0)$
answered Jan 17 at 1:02
El boritoEl borito
666216
$endgroup$
$begingroup$
It is fixed, thank you so much
$endgroup$
– El borito
Jan 17 at 4:35
1
$begingroup$
thank you. it really helped.
$endgroup$
– S.Jafari
Jan 17 at 19:36
add a comment |
$begingroup$
I am going to introduce the definition of differentiability of a function. A function $f : mathbb{R}^n to mathbb{R}^m$ is differenctiable in $x_0in mathbb{R}^n$ if there exists a linear function $L:mathbb{R}^ntomathbb{R}^m$ such that:
$$
lim_{hto (0,0)}frac{left|left|,f(x_0+h)-f(x_0)-L(h)right|right|}{left|left|hright|right|}=0
$$
The matrix associated to the linear function $L$ is the Jacobian matrix.
Then, let's calculate the Jacobian matrix of:
$$
f(x,y)=begin{cases}
displaystylefrac{(x^2-y)y^2}{x^4+y^2} & text{if } (x,y)neq (0,0) \
0 & text{if } (x,y)=(0,0)
end{cases}
$$
in $(x,y)=(0,0)$. First:
$$
frac{partial f}{partial x}(0,0) = lim_{hto 0} frac{0}{h^4}=0
$$
$$
frac{partial f}{partial y}(0,0) = lim_{hto 0} frac{-h^3}{h^2}=0
$$
Then the Jacobian matrix in $(0,0)$ of $f$ is $J_{(0,0)}(,f)=left(begin{array}{cc}
0 & 0
end{array}right)$, the if $L$ exists, it is $L(x,y)=left(begin{array}{cc}
0 & 0
end{array}right)left(begin{array}{c}
x \
y \
end{array}right)=0$. Then:
begin{eqnarray}
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)right|right|}{left|left|(x,y)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{(x,y)to (0,0)}left|frac{(x^2-y)y^2}{x^4+y^2},right|frac{1}{left|left|(x,y)right|right|} \
end{eqnarray}
If we use the path $y=x$:
begin{eqnarray}
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{xto 0}left|frac{(x^2-x)x^2}{x^4+x^2},right|frac{1}{left|left|(x,x)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{xto 0}left|frac{x^2-x}{x^2+1},right|frac{1}{|x|left|left|(1,1)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& frac{1}{left|left|(1,1)right|right|} lim_{xto 0}left|frac{x-1}{x^2+1},right| \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& frac{1}{left|left|(1,1)right|right|} neq 0 \
end{eqnarray}
Thus $f$ is not differentiable in $(0,0)$
answered Jan 17 at 1:02
El boritoEl borito
666216
$endgroup$
I am going to introduce the definition of differentiability of a function. A function $f : mathbb{R}^n to mathbb{R}^m$ is differenctiable in $x_0in mathbb{R}^n$ if there exists a linear function $L:mathbb{R}^ntomathbb{R}^m$ such that:
$$
lim_{hto (0,0)}frac{left|left|,f(x_0+h)-f(x_0)-L(h)right|right|}{left|left|hright|right|}=0
$$
The matrix associated to the linear function $L$ is the Jacobian matrix.
Then, let's calculate the Jacobian matrix of:
$$
f(x,y)=begin{cases}
displaystylefrac{(x^2-y)y^2}{x^4+y^2} & text{if } (x,y)neq (0,0) \
0 & text{if } (x,y)=(0,0)
end{cases}
$$
in $(x,y)=(0,0)$. First:
$$
frac{partial f}{partial x}(0,0) = lim_{hto 0} frac{0}{h^4}=0
$$
$$
frac{partial f}{partial y}(0,0) = lim_{hto 0} frac{-h^3}{h^2}=0
$$
Then the Jacobian matrix in $(0,0)$ of $f$ is $J_{(0,0)}(,f)=left(begin{array}{cc}
0 & 0
end{array}right)$, the if $L$ exists, it is $L(x,y)=left(begin{array}{cc}
0 & 0
end{array}right)left(begin{array}{c}
x \
y \
end{array}right)=0$. Then:
begin{eqnarray}
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)right|right|}{left|left|(x,y)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{(x,y)to (0,0)}left|frac{(x^2-y)y^2}{x^4+y^2},right|frac{1}{left|left|(x,y)right|right|} \
end{eqnarray}
If we use the path $y=x$:
begin{eqnarray}
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{xto 0}left|frac{(x^2-x)x^2}{x^4+x^2},right|frac{1}{left|left|(x,x)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& lim_{xto 0}left|frac{x^2-x}{x^2+1},right|frac{1}{|x|left|left|(1,1)right|right|} \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& frac{1}{left|left|(1,1)right|right|} lim_{xto 0}left|frac{x-1}{x^2+1},right| \
lim_{(x,y)to (0,0)}frac{left|left|,f(x,y)-f(0,0)-L(x,y)right|right|}{left|left|(x,y)right|right|} &=& frac{1}{left|left|(1,1)right|right|} neq 0 \
end{eqnarray}
Thus $f$ is not differentiable in $(0,0)$
answered Jan 17 at 1:02
El boritoEl borito
666216
edited Jan 17 at 20:31
answered Jan 17 at 1:02
El boritoEl borito
666216
answered Jan 17 at 1:02
El boritoEl borito
666216
answered Jan 17 at 1:02
El boritoEl borito
666216
666216
$begingroup$
It is fixed, thank you so much
$endgroup$
– El borito
Jan 17 at 4:35
1
$begingroup$
thank you. it really helped.
$endgroup$
– S.Jafari
Jan 17 at 19:36
add a comment |
$begingroup$
It is fixed, thank you so much
$endgroup$
– El borito
Jan 17 at 4:35
1
$begingroup$
thank you. it really helped.
$endgroup$
– S.Jafari
Jan 17 at 19:36
$begingroup$
It is fixed, thank you so much
$endgroup$
– El borito
Jan 17 at 4:35
$begingroup$
It is fixed, thank you so much
$endgroup$
– El borito
Jan 17 at 4:35
1
1
$begingroup$
thank you. it really helped.
$endgroup$
– S.Jafari
Jan 17 at 19:36
$begingroup$
thank you. it really helped.
$endgroup$
– S.Jafari
Jan 17 at 19:36
add a comment |
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$begingroup$
What would be a strategy for proving the existence or nonexistence of a derivative? Did you compute partial derivatives and examine the behavior in a neighborhood of $(0,0)$? Why not show what you actually did compute. Otherwise this just looks like you want someone to solve the problem for you.
$endgroup$
– RRL
Jan 17 at 0:19