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How do you prove that $x|x|$ is differentiable at all points?
Clash Royale CLAN TAG#URR8PPP
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In general how do you prove that if two functions are not differentiable at a point, then it is not necessary that their product is not differentiable at that point. (Which in $x|x|$ is $0$ )
calculus limits derivatives continuity
edited Jul 28 at 4:36
Asaf Karagila
291k31402732
asked Jul 28 at 0:20
Seylin
445
 |Â
show 1 more comment
up vote
3
down vote
favorite
In general how do you prove that if two functions are not differentiable at a point, then it is not necessary that their product is not differentiable at that point. (Which in $x|x|$ is $0$ )
calculus limits derivatives continuity
edited Jul 28 at 4:36
Asaf Karagila
291k31402732
asked Jul 28 at 0:20
Seylin
445
2
To prove your general statement, just use the example you just gave! Proof by counter example is legitimate.
– user496634
Jul 28 at 0:26
2
You can study the cases $x<0$, $x=0$, and $x>0$ separately. For $x<0$, then $x|x|=-x^2$, which is differentiable. For $x>0$, then $x|x|=x^2$ which is differentiable. For $x=0$ you prove it directly from the definition $(x|x|)'|_x=0=lim_hto0frac-0x=lim_hto0|x|=0$. Since the limit exists, the derivative exits.
– user578878
Jul 28 at 0:32
Is x=0 the only point of probable discontinuity? I mean to ask, are the probable points of discontinuity of a product function only those points of discontinuity of the individual functions or is it possible to have other points of discontinuity as well?
– Seylin
Jul 28 at 0:39
If you have more to ask about this topic, it is best to start up a new question, rather than to ask in a comment where no-one will notice it.
– Lee Mosher
Jul 28 at 0:48
1
@Seylin Yes, since the product of two continuous functions at a point is continuous at that same point, then the failure of continuity can only occur at points where at least one of the factors is not continuous. Likewise, for differentiability.
– user578878
Jul 28 at 1:12
 |Â
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
In general how do you prove that if two functions are not differentiable at a point, then it is not necessary that their product is not differentiable at that point. (Which in $x|x|$ is $0$ )
calculus limits derivatives continuity
edited Jul 28 at 4:36
Asaf Karagila
291k31402732
asked Jul 28 at 0:20
Seylin
445
In general how do you prove that if two functions are not differentiable at a point, then it is not necessary that their product is not differentiable at that point. (Which in $x|x|$ is $0$ )
calculus limits derivatives continuity
edited Jul 28 at 4:36
Asaf Karagila
291k31402732
asked Jul 28 at 0:20
Seylin
445
edited Jul 28 at 4:36
Asaf Karagila
291k31402732
edited Jul 28 at 4:36
Asaf Karagila
291k31402732
edited Jul 28 at 4:36
Asaf Karagila
291k31402732
291k31402732
asked Jul 28 at 0:20
Seylin
445
asked Jul 28 at 0:20
Seylin
445
asked Jul 28 at 0:20
Seylin
445
445
2
To prove your general statement, just use the example you just gave! Proof by counter example is legitimate.
– user496634
Jul 28 at 0:26
2
You can study the cases $x<0$, $x=0$, and $x>0$ separately. For $x<0$, then $x|x|=-x^2$, which is differentiable. For $x>0$, then $x|x|=x^2$ which is differentiable. For $x=0$ you prove it directly from the definition $(x|x|)'|_x=0=lim_hto0frac-0x=lim_hto0|x|=0$. Since the limit exists, the derivative exits.
– user578878
Jul 28 at 0:32
Is x=0 the only point of probable discontinuity? I mean to ask, are the probable points of discontinuity of a product function only those points of discontinuity of the individual functions or is it possible to have other points of discontinuity as well?
– Seylin
Jul 28 at 0:39
If you have more to ask about this topic, it is best to start up a new question, rather than to ask in a comment where no-one will notice it.
– Lee Mosher
Jul 28 at 0:48
1
@Seylin Yes, since the product of two continuous functions at a point is continuous at that same point, then the failure of continuity can only occur at points where at least one of the factors is not continuous. Likewise, for differentiability.
– user578878
Jul 28 at 1:12
 |Â
show 1 more comment
2
To prove your general statement, just use the example you just gave! Proof by counter example is legitimate.
– user496634
Jul 28 at 0:26
2
You can study the cases $x<0$, $x=0$, and $x>0$ separately. For $x<0$, then $x|x|=-x^2$, which is differentiable. For $x>0$, then $x|x|=x^2$ which is differentiable. For $x=0$ you prove it directly from the definition $(x|x|)'|_x=0=lim_hto0frac-0x=lim_hto0|x|=0$. Since the limit exists, the derivative exits.
– user578878
Jul 28 at 0:32
Is x=0 the only point of probable discontinuity? I mean to ask, are the probable points of discontinuity of a product function only those points of discontinuity of the individual functions or is it possible to have other points of discontinuity as well?
– Seylin
Jul 28 at 0:39
If you have more to ask about this topic, it is best to start up a new question, rather than to ask in a comment where no-one will notice it.
– Lee Mosher
Jul 28 at 0:48
1
@Seylin Yes, since the product of two continuous functions at a point is continuous at that same point, then the failure of continuity can only occur at points where at least one of the factors is not continuous. Likewise, for differentiability.
– user578878
Jul 28 at 1:12
2
2
To prove your general statement, just use the example you just gave! Proof by counter example is legitimate.
– user496634
Jul 28 at 0:26
To prove your general statement, just use the example you just gave! Proof by counter example is legitimate.
– user496634
Jul 28 at 0:26
2
2
You can study the cases $x<0$, $x=0$, and $x>0$ separately. For $x<0$, then $x|x|=-x^2$, which is differentiable. For $x>0$, then $x|x|=x^2$ which is differentiable. For $x=0$ you prove it directly from the definition $(x|x|)'|_x=0=lim_hto0frac-0x=lim_hto0|x|=0$. Since the limit exists, the derivative exits.
– user578878
Jul 28 at 0:32
You can study the cases $x<0$, $x=0$, and $x>0$ separately. For $x<0$, then $x|x|=-x^2$, which is differentiable. For $x>0$, then $x|x|=x^2$ which is differentiable. For $x=0$ you prove it directly from the definition $(x|x|)'|_x=0=lim_hto0frac-0x=lim_hto0|x|=0$. Since the limit exists, the derivative exits.
– user578878
Jul 28 at 0:32
Is x=0 the only point of probable discontinuity? I mean to ask, are the probable points of discontinuity of a product function only those points of discontinuity of the individual functions or is it possible to have other points of discontinuity as well?
– Seylin
Jul 28 at 0:39
Is x=0 the only point of probable discontinuity? I mean to ask, are the probable points of discontinuity of a product function only those points of discontinuity of the individual functions or is it possible to have other points of discontinuity as well?
– Seylin
Jul 28 at 0:39
If you have more to ask about this topic, it is best to start up a new question, rather than to ask in a comment where no-one will notice it.
– Lee Mosher
Jul 28 at 0:48
If you have more to ask about this topic, it is best to start up a new question, rather than to ask in a comment where no-one will notice it.
– Lee Mosher
Jul 28 at 0:48
1
1
@Seylin Yes, since the product of two continuous functions at a point is continuous at that same point, then the failure of continuity can only occur at points where at least one of the factors is not continuous. Likewise, for differentiability.
– user578878
Jul 28 at 1:12
@Seylin Yes, since the product of two continuous functions at a point is continuous at that same point, then the failure of continuity can only occur at points where at least one of the factors is not continuous. Likewise, for differentiability.
– user578878
Jul 28 at 1:12
 |Â
show 1 more comment
2 Answers
2
active
oldest
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up vote
-1
down vote
accepted
In this particular case, of course, we can find the derivative.
begineqnarray
fracddxx|x|&=&|x|fracdxdx+xfracddx\
&=&|x|+xcdotfracx\
&=&|x|left(1+fracxxright)
endeqnarray
which is undefined at $x=0$ but which has limit $0$ at $x=0$. Since the function is continuous at $x=0$ and has a horizontal tangent at $x=0$ it is the case that
$$ fracddxx|x|=2|x| $$
for all $x$.
ADDENDUM: Here is a complete proof that $fracddx(x|x|)=2|x|$ using the definition
$$ f^prime(x)=lim_hto0fracf(x+h)-f(x)h $$
begineqnarray
f^prime(x)&=&lim_hto0frac(x+h)hcdotcolorbluefrac\
&=&texta few omitted algebra steps\
&=&lim_hto0frac4(x+h)^3+6(x+h)^2h+4(x+h)h+2+h^3\
&=&frac4x^32x\
&=&2|x|
endeqnarray
answered Jul 28 at 0:33
John Wayland Bales
12.8k21135
3
"which has limit $2lvert xrvert$ at $x=0$" doesn't mean anything. Once you have taken the limit as $xto 0$, there is no meaning for the symbol $x$.
– Clement C.
Jul 28 at 0:46
1
Why do you not just write "which has limit $0$ at $x=0$" instead of going (now) to "limiting functions" (which has no sandard meaning, unless you are talking about asymptotic equivalence, in which case this is (1) overkill and (2) requiring to be specified)?
– Clement C.
Jul 28 at 1:01
@ClementC. Well, the point is that the function $y=2|x|$ is, in fact, the derivative of the function $y=x|x|$, including at $x=0$. I will amend my answer to make that explicit.
– John Wayland Bales
Jul 28 at 1:25
1
@JohnWaylandBales The issue with the limit is clear what you mean. I would say that it is practically an irrelevant concern. However, there is an omitted argument in the proof, which is significantly more advanced that any of the parts said explicitly. This is, the argument of why the existence of the limit of the derivative, plus continuity of the function, implies the existence of the derivative at the point. None of what is currently written ensures that the derivative exists at $x=0$. Hint: You will need to use, L'Hospital, or Mean Value Theorem, or prove it from scratch.
– user578878
Jul 28 at 1:43
This is mostly unnecessary. It's clear that $f(x)= x^2$ for $x>0,$ $ f(x)= -x^2$ for $x<0.$ Hence $f$ is differentiable everywhere in $mathbb Rsetminus 0.$ The only thing left to check is whether $f'(0)$ exists, which is covered in the answer of @ncmathsadist.
– zhw.
Jul 28 at 5:41
 |Â
show 1 more comment
up vote
11
down vote
You only need to check at $x = 0$ in which case
$$f'(0) = lim_hto 0h = lim_hto 0 |h| = 0.$$
The rest is clear.
answered Jul 28 at 0:51
ncmathsadist
41.3k25699
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
accepted
In this particular case, of course, we can find the derivative.
begineqnarray
fracddxx|x|&=&|x|fracdxdx+xfracddx\
&=&|x|+xcdotfracx\
&=&|x|left(1+fracxxright)
endeqnarray
which is undefined at $x=0$ but which has limit $0$ at $x=0$. Since the function is continuous at $x=0$ and has a horizontal tangent at $x=0$ it is the case that
$$ fracddxx|x|=2|x| $$
for all $x$.
ADDENDUM: Here is a complete proof that $fracddx(x|x|)=2|x|$ using the definition
$$ f^prime(x)=lim_hto0fracf(x+h)-f(x)h $$
begineqnarray
f^prime(x)&=&lim_hto0frac(x+h)hcdotcolorbluefrac\
&=&texta few omitted algebra steps\
&=&lim_hto0frac4(x+h)^3+6(x+h)^2h+4(x+h)h+2+h^3\
&=&frac4x^32x\
&=&2|x|
endeqnarray
answered Jul 28 at 0:33
John Wayland Bales
12.8k21135
3
"which has limit $2lvert xrvert$ at $x=0$" doesn't mean anything. Once you have taken the limit as $xto 0$, there is no meaning for the symbol $x$.
– Clement C.
Jul 28 at 0:46
1
Why do you not just write "which has limit $0$ at $x=0$" instead of going (now) to "limiting functions" (which has no sandard meaning, unless you are talking about asymptotic equivalence, in which case this is (1) overkill and (2) requiring to be specified)?
– Clement C.
Jul 28 at 1:01
@ClementC. Well, the point is that the function $y=2|x|$ is, in fact, the derivative of the function $y=x|x|$, including at $x=0$. I will amend my answer to make that explicit.
– John Wayland Bales
Jul 28 at 1:25
1
@JohnWaylandBales The issue with the limit is clear what you mean. I would say that it is practically an irrelevant concern. However, there is an omitted argument in the proof, which is significantly more advanced that any of the parts said explicitly. This is, the argument of why the existence of the limit of the derivative, plus continuity of the function, implies the existence of the derivative at the point. None of what is currently written ensures that the derivative exists at $x=0$. Hint: You will need to use, L'Hospital, or Mean Value Theorem, or prove it from scratch.
– user578878
Jul 28 at 1:43
This is mostly unnecessary. It's clear that $f(x)= x^2$ for $x>0,$ $ f(x)= -x^2$ for $x<0.$ Hence $f$ is differentiable everywhere in $mathbb Rsetminus 0.$ The only thing left to check is whether $f'(0)$ exists, which is covered in the answer of @ncmathsadist.
– zhw.
Jul 28 at 5:41
 |Â
show 1 more comment
up vote
-1
down vote
accepted
In this particular case, of course, we can find the derivative.
begineqnarray
fracddxx|x|&=&|x|fracdxdx+xfracddx\
&=&|x|+xcdotfracx\
&=&|x|left(1+fracxxright)
endeqnarray
which is undefined at $x=0$ but which has limit $0$ at $x=0$. Since the function is continuous at $x=0$ and has a horizontal tangent at $x=0$ it is the case that
$$ fracddxx|x|=2|x| $$
for all $x$.
ADDENDUM: Here is a complete proof that $fracddx(x|x|)=2|x|$ using the definition
$$ f^prime(x)=lim_hto0fracf(x+h)-f(x)h $$
begineqnarray
f^prime(x)&=&lim_hto0frac(x+h)hcdotcolorbluefrac\
&=&texta few omitted algebra steps\
&=&lim_hto0frac4(x+h)^3+6(x+h)^2h+4(x+h)h+2+h^3\
&=&frac4x^32x\
&=&2|x|
endeqnarray
answered Jul 28 at 0:33
John Wayland Bales
12.8k21135
3
"which has limit $2lvert xrvert$ at $x=0$" doesn't mean anything. Once you have taken the limit as $xto 0$, there is no meaning for the symbol $x$.
– Clement C.
Jul 28 at 0:46
1
Why do you not just write "which has limit $0$ at $x=0$" instead of going (now) to "limiting functions" (which has no sandard meaning, unless you are talking about asymptotic equivalence, in which case this is (1) overkill and (2) requiring to be specified)?
– Clement C.
Jul 28 at 1:01
@ClementC. Well, the point is that the function $y=2|x|$ is, in fact, the derivative of the function $y=x|x|$, including at $x=0$. I will amend my answer to make that explicit.
– John Wayland Bales
Jul 28 at 1:25
1
@JohnWaylandBales The issue with the limit is clear what you mean. I would say that it is practically an irrelevant concern. However, there is an omitted argument in the proof, which is significantly more advanced that any of the parts said explicitly. This is, the argument of why the existence of the limit of the derivative, plus continuity of the function, implies the existence of the derivative at the point. None of what is currently written ensures that the derivative exists at $x=0$. Hint: You will need to use, L'Hospital, or Mean Value Theorem, or prove it from scratch.
– user578878
Jul 28 at 1:43
This is mostly unnecessary. It's clear that $f(x)= x^2$ for $x>0,$ $ f(x)= -x^2$ for $x<0.$ Hence $f$ is differentiable everywhere in $mathbb Rsetminus 0.$ The only thing left to check is whether $f'(0)$ exists, which is covered in the answer of @ncmathsadist.
– zhw.
Jul 28 at 5:41
 |Â
show 1 more comment
up vote
-1
down vote
accepted
up vote
-1
down vote
accepted
In this particular case, of course, we can find the derivative.
begineqnarray
fracddxx|x|&=&|x|fracdxdx+xfracddx\
&=&|x|+xcdotfracx\
&=&|x|left(1+fracxxright)
endeqnarray
which is undefined at $x=0$ but which has limit $0$ at $x=0$. Since the function is continuous at $x=0$ and has a horizontal tangent at $x=0$ it is the case that
$$ fracddxx|x|=2|x| $$
for all $x$.
ADDENDUM: Here is a complete proof that $fracddx(x|x|)=2|x|$ using the definition
$$ f^prime(x)=lim_hto0fracf(x+h)-f(x)h $$
begineqnarray
f^prime(x)&=&lim_hto0frac(x+h)hcdotcolorbluefrac\
&=&texta few omitted algebra steps\
&=&lim_hto0frac4(x+h)^3+6(x+h)^2h+4(x+h)h+2+h^3\
&=&frac4x^32x\
&=&2|x|
endeqnarray
answered Jul 28 at 0:33
John Wayland Bales
12.8k21135
In this particular case, of course, we can find the derivative.
begineqnarray
fracddxx|x|&=&|x|fracdxdx+xfracddx\
&=&|x|+xcdotfracx\
&=&|x|left(1+fracxxright)
endeqnarray
which is undefined at $x=0$ but which has limit $0$ at $x=0$. Since the function is continuous at $x=0$ and has a horizontal tangent at $x=0$ it is the case that
$$ fracddxx|x|=2|x| $$
for all $x$.
ADDENDUM: Here is a complete proof that $fracddx(x|x|)=2|x|$ using the definition
$$ f^prime(x)=lim_hto0fracf(x+h)-f(x)h $$
begineqnarray
f^prime(x)&=&lim_hto0frac(x+h)hcdotcolorbluefrac\
&=&texta few omitted algebra steps\
&=&lim_hto0frac4(x+h)^3+6(x+h)^2h+4(x+h)h+2+h^3\
&=&frac4x^32x\
&=&2|x|
endeqnarray
answered Jul 28 at 0:33
John Wayland Bales
12.8k21135
edited Jul 28 at 2:08
answered Jul 28 at 0:33
John Wayland Bales
12.8k21135
answered Jul 28 at 0:33
John Wayland Bales
12.8k21135
answered Jul 28 at 0:33
John Wayland Bales
12.8k21135
12.8k21135
3
"which has limit $2lvert xrvert$ at $x=0$" doesn't mean anything. Once you have taken the limit as $xto 0$, there is no meaning for the symbol $x$.
– Clement C.
Jul 28 at 0:46
1
Why do you not just write "which has limit $0$ at $x=0$" instead of going (now) to "limiting functions" (which has no sandard meaning, unless you are talking about asymptotic equivalence, in which case this is (1) overkill and (2) requiring to be specified)?
– Clement C.
Jul 28 at 1:01
@ClementC. Well, the point is that the function $y=2|x|$ is, in fact, the derivative of the function $y=x|x|$, including at $x=0$. I will amend my answer to make that explicit.
– John Wayland Bales
Jul 28 at 1:25
1
@JohnWaylandBales The issue with the limit is clear what you mean. I would say that it is practically an irrelevant concern. However, there is an omitted argument in the proof, which is significantly more advanced that any of the parts said explicitly. This is, the argument of why the existence of the limit of the derivative, plus continuity of the function, implies the existence of the derivative at the point. None of what is currently written ensures that the derivative exists at $x=0$. Hint: You will need to use, L'Hospital, or Mean Value Theorem, or prove it from scratch.
– user578878
Jul 28 at 1:43
This is mostly unnecessary. It's clear that $f(x)= x^2$ for $x>0,$ $ f(x)= -x^2$ for $x<0.$ Hence $f$ is differentiable everywhere in $mathbb Rsetminus 0.$ The only thing left to check is whether $f'(0)$ exists, which is covered in the answer of @ncmathsadist.
– zhw.
Jul 28 at 5:41
 |Â
show 1 more comment
3
"which has limit $2lvert xrvert$ at $x=0$" doesn't mean anything. Once you have taken the limit as $xto 0$, there is no meaning for the symbol $x$.
– Clement C.
Jul 28 at 0:46
1
Why do you not just write "which has limit $0$ at $x=0$" instead of going (now) to "limiting functions" (which has no sandard meaning, unless you are talking about asymptotic equivalence, in which case this is (1) overkill and (2) requiring to be specified)?
– Clement C.
Jul 28 at 1:01
@ClementC. Well, the point is that the function $y=2|x|$ is, in fact, the derivative of the function $y=x|x|$, including at $x=0$. I will amend my answer to make that explicit.
– John Wayland Bales
Jul 28 at 1:25
1
@JohnWaylandBales The issue with the limit is clear what you mean. I would say that it is practically an irrelevant concern. However, there is an omitted argument in the proof, which is significantly more advanced that any of the parts said explicitly. This is, the argument of why the existence of the limit of the derivative, plus continuity of the function, implies the existence of the derivative at the point. None of what is currently written ensures that the derivative exists at $x=0$. Hint: You will need to use, L'Hospital, or Mean Value Theorem, or prove it from scratch.
– user578878
Jul 28 at 1:43
This is mostly unnecessary. It's clear that $f(x)= x^2$ for $x>0,$ $ f(x)= -x^2$ for $x<0.$ Hence $f$ is differentiable everywhere in $mathbb Rsetminus 0.$ The only thing left to check is whether $f'(0)$ exists, which is covered in the answer of @ncmathsadist.
– zhw.
Jul 28 at 5:41
3
3
"which has limit $2lvert xrvert$ at $x=0$" doesn't mean anything. Once you have taken the limit as $xto 0$, there is no meaning for the symbol $x$.
– Clement C.
Jul 28 at 0:46
"which has limit $2lvert xrvert$ at $x=0$" doesn't mean anything. Once you have taken the limit as $xto 0$, there is no meaning for the symbol $x$.
– Clement C.
Jul 28 at 0:46
1
1
Why do you not just write "which has limit $0$ at $x=0$" instead of going (now) to "limiting functions" (which has no sandard meaning, unless you are talking about asymptotic equivalence, in which case this is (1) overkill and (2) requiring to be specified)?
– Clement C.
Jul 28 at 1:01
Why do you not just write "which has limit $0$ at $x=0$" instead of going (now) to "limiting functions" (which has no sandard meaning, unless you are talking about asymptotic equivalence, in which case this is (1) overkill and (2) requiring to be specified)?
– Clement C.
Jul 28 at 1:01
@ClementC. Well, the point is that the function $y=2|x|$ is, in fact, the derivative of the function $y=x|x|$, including at $x=0$. I will amend my answer to make that explicit.
– John Wayland Bales
Jul 28 at 1:25
@ClementC. Well, the point is that the function $y=2|x|$ is, in fact, the derivative of the function $y=x|x|$, including at $x=0$. I will amend my answer to make that explicit.
– John Wayland Bales
Jul 28 at 1:25
1
1
@JohnWaylandBales The issue with the limit is clear what you mean. I would say that it is practically an irrelevant concern. However, there is an omitted argument in the proof, which is significantly more advanced that any of the parts said explicitly. This is, the argument of why the existence of the limit of the derivative, plus continuity of the function, implies the existence of the derivative at the point. None of what is currently written ensures that the derivative exists at $x=0$. Hint: You will need to use, L'Hospital, or Mean Value Theorem, or prove it from scratch.
– user578878
Jul 28 at 1:43
@JohnWaylandBales The issue with the limit is clear what you mean. I would say that it is practically an irrelevant concern. However, there is an omitted argument in the proof, which is significantly more advanced that any of the parts said explicitly. This is, the argument of why the existence of the limit of the derivative, plus continuity of the function, implies the existence of the derivative at the point. None of what is currently written ensures that the derivative exists at $x=0$. Hint: You will need to use, L'Hospital, or Mean Value Theorem, or prove it from scratch.
– user578878
Jul 28 at 1:43
This is mostly unnecessary. It's clear that $f(x)= x^2$ for $x>0,$ $ f(x)= -x^2$ for $x<0.$ Hence $f$ is differentiable everywhere in $mathbb Rsetminus 0.$ The only thing left to check is whether $f'(0)$ exists, which is covered in the answer of @ncmathsadist.
– zhw.
Jul 28 at 5:41
This is mostly unnecessary. It's clear that $f(x)= x^2$ for $x>0,$ $ f(x)= -x^2$ for $x<0.$ Hence $f$ is differentiable everywhere in $mathbb Rsetminus 0.$ The only thing left to check is whether $f'(0)$ exists, which is covered in the answer of @ncmathsadist.
– zhw.
Jul 28 at 5:41
 |Â
show 1 more comment
up vote
11
down vote
You only need to check at $x = 0$ in which case
$$f'(0) = lim_hto 0h = lim_hto 0 |h| = 0.$$
The rest is clear.
answered Jul 28 at 0:51
ncmathsadist
41.3k25699
add a comment |Â
up vote
11
down vote
You only need to check at $x = 0$ in which case
$$f'(0) = lim_hto 0h = lim_hto 0 |h| = 0.$$
The rest is clear.
answered Jul 28 at 0:51
ncmathsadist
41.3k25699
add a comment |Â
up vote
11
down vote
up vote
11
down vote
You only need to check at $x = 0$ in which case
$$f'(0) = lim_hto 0h = lim_hto 0 |h| = 0.$$
The rest is clear.
answered Jul 28 at 0:51
ncmathsadist
41.3k25699
You only need to check at $x = 0$ in which case
$$f'(0) = lim_hto 0h = lim_hto 0 |h| = 0.$$
The rest is clear.
answered Jul 28 at 0:51
ncmathsadist
41.3k25699
answered Jul 28 at 0:51
ncmathsadist
41.3k25699
answered Jul 28 at 0:51
ncmathsadist
41.3k25699
answered Jul 28 at 0:51
ncmathsadist
41.3k25699
41.3k25699
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2
To prove your general statement, just use the example you just gave! Proof by counter example is legitimate.
– user496634
Jul 28 at 0:26
2
You can study the cases $x<0$, $x=0$, and $x>0$ separately. For $x<0$, then $x|x|=-x^2$, which is differentiable. For $x>0$, then $x|x|=x^2$ which is differentiable. For $x=0$ you prove it directly from the definition $(x|x|)'|_x=0=lim_hto0frac-0x=lim_hto0|x|=0$. Since the limit exists, the derivative exits.
– user578878
Jul 28 at 0:32
Is x=0 the only point of probable discontinuity? I mean to ask, are the probable points of discontinuity of a product function only those points of discontinuity of the individual functions or is it possible to have other points of discontinuity as well?
– Seylin
Jul 28 at 0:39
If you have more to ask about this topic, it is best to start up a new question, rather than to ask in a comment where no-one will notice it.
– Lee Mosher
Jul 28 at 0:48
1
@Seylin Yes, since the product of two continuous functions at a point is continuous at that same point, then the failure of continuity can only occur at points where at least one of the factors is not continuous. Likewise, for differentiability.
– user578878
Jul 28 at 1:12