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What is the difference between differentiability of a function and continuity of its derivative?
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I am sort of confused regarding differentiable functions, continuous derivatives, and continuous functions. And I just want to make sure I'm thinking about this correctly.
(1) If you have a function that's continuous everywhere, then this doesn't necessarily mean its derivative exists everywhere, correct? e.g., $$f(x) = |x|$$ has an undefined derivative at $x=0$
(2) So this above function, even though its continuous, does not have a continuous derivative?
(3) Now say you have a derivative that's continuous everywhere, then this doesn't necessarily mean the underlying function is continuous everywhere, correct? For example, consider
$$
f(x) = begincases
1 - x x<0 \
2 - x x geq 0
endcases
$$
So its derivative is -1 everywhere, hence continuous, but the function itself is not continuous?
So what does a function with a continuous derivative say about the underlying function?
calculus derivatives
asked 2 hours ago
Iamanon
194
add a comment |Â
up vote
2
down vote
favorite
I am sort of confused regarding differentiable functions, continuous derivatives, and continuous functions. And I just want to make sure I'm thinking about this correctly.
(1) If you have a function that's continuous everywhere, then this doesn't necessarily mean its derivative exists everywhere, correct? e.g., $$f(x) = |x|$$ has an undefined derivative at $x=0$
(2) So this above function, even though its continuous, does not have a continuous derivative?
(3) Now say you have a derivative that's continuous everywhere, then this doesn't necessarily mean the underlying function is continuous everywhere, correct? For example, consider
$$
f(x) = begincases
1 - x x<0 \
2 - x x geq 0
endcases
$$
So its derivative is -1 everywhere, hence continuous, but the function itself is not continuous?
So what does a function with a continuous derivative say about the underlying function?
calculus derivatives
asked 2 hours ago
Iamanon
194
Note that the derivative of $|x|$ isn't discontinuous. It's actually continuous. It just doesn't exist everywhere. There is a difference. It allows us to say things like "derivatives have the intermediate value property", meaning if they exist they cannot have jump discontinuities.
– Arthur
1 hour ago
In high school, I was taught that a function is continuous if you can draw the function without lifting up your pencil. So here, isn't the derivative of |x| discontinuous using the same logic, i.e., I can't draw it's derivative because I have to lift up my pencil at x=0?
– Iamanon
1 hour ago
That's not really the definition of "continuous". There is a formal definition (you may have seen it; it begins "for any $epsilon >0$, there is a $delta > 0$ such that ..."), and according to that definition, a function must be defined at a point before you can even begin to make sense of wether it's continuous or discontinuous there.
– Arthur
1 hour ago
Ah. So the derivative is undefined at x=0. But you said the derivative is continuous. So how can you make that assessment if the derivative is not defined there? or do you mean it's continuous only to the left and right of x=0?
– Iamanon
1 hour ago
1
It's continuous everywhere it's defined, which is everywhere continuity makes sense to ask about. Thus one can just say that it's continuous. It's the same for, say, $f(x) = frac1x$. That function is also continuous. It just isn't a function on all of the real numbers.
– Arthur
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am sort of confused regarding differentiable functions, continuous derivatives, and continuous functions. And I just want to make sure I'm thinking about this correctly.
(1) If you have a function that's continuous everywhere, then this doesn't necessarily mean its derivative exists everywhere, correct? e.g., $$f(x) = |x|$$ has an undefined derivative at $x=0$
(2) So this above function, even though its continuous, does not have a continuous derivative?
(3) Now say you have a derivative that's continuous everywhere, then this doesn't necessarily mean the underlying function is continuous everywhere, correct? For example, consider
$$
f(x) = begincases
1 - x x<0 \
2 - x x geq 0
endcases
$$
So its derivative is -1 everywhere, hence continuous, but the function itself is not continuous?
So what does a function with a continuous derivative say about the underlying function?
calculus derivatives
asked 2 hours ago
Iamanon
194
I am sort of confused regarding differentiable functions, continuous derivatives, and continuous functions. And I just want to make sure I'm thinking about this correctly.
(1) If you have a function that's continuous everywhere, then this doesn't necessarily mean its derivative exists everywhere, correct? e.g., $$f(x) = |x|$$ has an undefined derivative at $x=0$
(2) So this above function, even though its continuous, does not have a continuous derivative?
(3) Now say you have a derivative that's continuous everywhere, then this doesn't necessarily mean the underlying function is continuous everywhere, correct? For example, consider
$$
f(x) = begincases
1 - x x<0 \
2 - x x geq 0
endcases
$$
So its derivative is -1 everywhere, hence continuous, but the function itself is not continuous?
So what does a function with a continuous derivative say about the underlying function?
calculus derivatives
calculus derivatives
asked 2 hours ago
Iamanon
194
asked 2 hours ago
Iamanon
194
asked 2 hours ago
Iamanon
194
asked 2 hours ago
Iamanon
194
asked 2 hours ago
Iamanon
194
194
Note that the derivative of $|x|$ isn't discontinuous. It's actually continuous. It just doesn't exist everywhere. There is a difference. It allows us to say things like "derivatives have the intermediate value property", meaning if they exist they cannot have jump discontinuities.
– Arthur
1 hour ago
In high school, I was taught that a function is continuous if you can draw the function without lifting up your pencil. So here, isn't the derivative of |x| discontinuous using the same logic, i.e., I can't draw it's derivative because I have to lift up my pencil at x=0?
– Iamanon
1 hour ago
That's not really the definition of "continuous". There is a formal definition (you may have seen it; it begins "for any $epsilon >0$, there is a $delta > 0$ such that ..."), and according to that definition, a function must be defined at a point before you can even begin to make sense of wether it's continuous or discontinuous there.
– Arthur
1 hour ago
Ah. So the derivative is undefined at x=0. But you said the derivative is continuous. So how can you make that assessment if the derivative is not defined there? or do you mean it's continuous only to the left and right of x=0?
– Iamanon
1 hour ago
1
It's continuous everywhere it's defined, which is everywhere continuity makes sense to ask about. Thus one can just say that it's continuous. It's the same for, say, $f(x) = frac1x$. That function is also continuous. It just isn't a function on all of the real numbers.
– Arthur
1 hour ago
add a comment |Â
Note that the derivative of $|x|$ isn't discontinuous. It's actually continuous. It just doesn't exist everywhere. There is a difference. It allows us to say things like "derivatives have the intermediate value property", meaning if they exist they cannot have jump discontinuities.
– Arthur
1 hour ago
In high school, I was taught that a function is continuous if you can draw the function without lifting up your pencil. So here, isn't the derivative of |x| discontinuous using the same logic, i.e., I can't draw it's derivative because I have to lift up my pencil at x=0?
– Iamanon
1 hour ago
That's not really the definition of "continuous". There is a formal definition (you may have seen it; it begins "for any $epsilon >0$, there is a $delta > 0$ such that ..."), and according to that definition, a function must be defined at a point before you can even begin to make sense of wether it's continuous or discontinuous there.
– Arthur
1 hour ago
Ah. So the derivative is undefined at x=0. But you said the derivative is continuous. So how can you make that assessment if the derivative is not defined there? or do you mean it's continuous only to the left and right of x=0?
– Iamanon
1 hour ago
1
It's continuous everywhere it's defined, which is everywhere continuity makes sense to ask about. Thus one can just say that it's continuous. It's the same for, say, $f(x) = frac1x$. That function is also continuous. It just isn't a function on all of the real numbers.
– Arthur
1 hour ago
Note that the derivative of $|x|$ isn't discontinuous. It's actually continuous. It just doesn't exist everywhere. There is a difference. It allows us to say things like "derivatives have the intermediate value property", meaning if they exist they cannot have jump discontinuities.
– Arthur
1 hour ago
Note that the derivative of $|x|$ isn't discontinuous. It's actually continuous. It just doesn't exist everywhere. There is a difference. It allows us to say things like "derivatives have the intermediate value property", meaning if they exist they cannot have jump discontinuities.
– Arthur
1 hour ago
In high school, I was taught that a function is continuous if you can draw the function without lifting up your pencil. So here, isn't the derivative of |x| discontinuous using the same logic, i.e., I can't draw it's derivative because I have to lift up my pencil at x=0?
– Iamanon
1 hour ago
In high school, I was taught that a function is continuous if you can draw the function without lifting up your pencil. So here, isn't the derivative of |x| discontinuous using the same logic, i.e., I can't draw it's derivative because I have to lift up my pencil at x=0?
– Iamanon
1 hour ago
That's not really the definition of "continuous". There is a formal definition (you may have seen it; it begins "for any $epsilon >0$, there is a $delta > 0$ such that ..."), and according to that definition, a function must be defined at a point before you can even begin to make sense of wether it's continuous or discontinuous there.
– Arthur
1 hour ago
That's not really the definition of "continuous". There is a formal definition (you may have seen it; it begins "for any $epsilon >0$, there is a $delta > 0$ such that ..."), and according to that definition, a function must be defined at a point before you can even begin to make sense of wether it's continuous or discontinuous there.
– Arthur
1 hour ago
Ah. So the derivative is undefined at x=0. But you said the derivative is continuous. So how can you make that assessment if the derivative is not defined there? or do you mean it's continuous only to the left and right of x=0?
– Iamanon
1 hour ago
Ah. So the derivative is undefined at x=0. But you said the derivative is continuous. So how can you make that assessment if the derivative is not defined there? or do you mean it's continuous only to the left and right of x=0?
– Iamanon
1 hour ago
1
1
It's continuous everywhere it's defined, which is everywhere continuity makes sense to ask about. Thus one can just say that it's continuous. It's the same for, say, $f(x) = frac1x$. That function is also continuous. It just isn't a function on all of the real numbers.
– Arthur
1 hour ago
It's continuous everywhere it's defined, which is everywhere continuity makes sense to ask about. Thus one can just say that it's continuous. It's the same for, say, $f(x) = frac1x$. That function is also continuous. It just isn't a function on all of the real numbers.
– Arthur
1 hour ago
add a comment |Â
4 Answers
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A function may or may not be continuous.
If it is continuous, it may or may not be differentiable. $f(x) = |x|$ is a standard example of a function which is continuous, but not (everywhere) differentiable. However, any differentiable function is necessarily continuous.
If a function is differentiable, its derivative may or may not be continuous. This is a bit more subtle, and the standard example of a differentiable function with discontinuous derivative is a bit more complicated:
$$
f(x) = casesx^2sin(1/x) & if $xneq 0$\
0 & if $x = 0$
$$
It is differentiable everywhere, $f'(0) = 0$, but $f'(x)$ oscillates wildly between (a little less than) $-1$ and (a little more than) $1$ as $x$ comes closer and closer to $0$, so it isn't continuous.
answered 1 hour ago
Arthur
105k799182
I'd like to add that you can have a function $f$ that's continuous everywhere but differentiable nowhere - check out en.m.wikipedia.org/wiki/Weierstrass_function.
– J. C.
1 hour ago
For my example under (3), isn't that an example of a differentiable function where its derivative is defined everywhere and continuous everywhere, but the underlying function is not continuous?
– Iamanon
1 hour ago
@Iamanon No, your example at (3) doesn't have a derivative at $x = 0$. Note that derivative isn't quite the slope of a geometric tangent. It has a rigorous definition as a certain limit, and that limit doesn't exist at that point.
– Arthur
1 hour ago
Ahh. I thought that the derivative at x=0 was defined by whichever function was defined at x=0 (2-x in this case).
– Iamanon
1 hour ago
@Iamanon It does have a so-called right derivative at that point. That is the formal name of the idea you're talking about: if you're standing at that point, and only looking to the right (towards higher values of $x$), then that's what you get.
– Arthur
1 hour ago
 |Â
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1
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- Indeed.
- Indeed, the derivative does not even exist at $x=0$ as you argued in (1).
- Continuity is a necessary condition for differentiability (but it is not sufficient, as you have found in (1)). Your example is not differentiable at $0$, since it is discontinuous there. In particular, you can see this directly since
$$ lim_hto0^+ fracf(h)-f(0)h = lim_hto0^+ frac(2-h)-2h = -1, $$
whereas
$$lim_hto0^- fracf(h)-f(0)h = lim_hto0^- frac(1-h)-2h = lim_hto0^- frac-1-hh$$
does not even exist.
answered 1 hour ago
Sobi
3,0101517
add a comment |Â
up vote
1
down vote
As you have shown the continuity of the derivative function doesn't guarantee that the function is itself continuous since it can be piecewise continuous.
WHat we can show is Continuous differentiability implies Lipschitz continuity.
answered 1 hour ago
gimusi
78.6k73990
add a comment |Â
up vote
1
down vote
The derivative is just another function, whether it's continuous doesn't affect the original function.
You have yourself given many examples on that.
Edit: in the further comments @gimusi has pointed we can show that Continuous differentiability implies Lipschitz continuity
answered 2 hours ago
Utsav Mangal
235
New contributor
Utsav Mangal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Can we say that the second derivative of |x| is continuous everywhere?
– Iamanon
2 hours ago
@Iamanon It is continuous (and equal to $0$) everywhere it's defined. It is, however, not defined at $x = 0$.
– Arthur
2 hours ago
So there an example of a function where its derivative is discontinuous but its second derivative is defined everywhere?
– Iamanon
1 hour ago
for a derivative to exist at a point it has to be continuous, so no @lamanon
– Utsav Mangal
1 hour ago
What about my example under (3)? We see that f(x) is discontinuous but the derivative is defined everywhere. never mind. Arthur, above, addressed this question.
– Iamanon
1 hour ago
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
A function may or may not be continuous.
If it is continuous, it may or may not be differentiable. $f(x) = |x|$ is a standard example of a function which is continuous, but not (everywhere) differentiable. However, any differentiable function is necessarily continuous.
If a function is differentiable, its derivative may or may not be continuous. This is a bit more subtle, and the standard example of a differentiable function with discontinuous derivative is a bit more complicated:
$$
f(x) = casesx^2sin(1/x) & if $xneq 0$\
0 & if $x = 0$
$$
It is differentiable everywhere, $f'(0) = 0$, but $f'(x)$ oscillates wildly between (a little less than) $-1$ and (a little more than) $1$ as $x$ comes closer and closer to $0$, so it isn't continuous.
answered 1 hour ago
Arthur
105k799182
I'd like to add that you can have a function $f$ that's continuous everywhere but differentiable nowhere - check out en.m.wikipedia.org/wiki/Weierstrass_function.
– J. C.
1 hour ago
For my example under (3), isn't that an example of a differentiable function where its derivative is defined everywhere and continuous everywhere, but the underlying function is not continuous?
– Iamanon
1 hour ago
@Iamanon No, your example at (3) doesn't have a derivative at $x = 0$. Note that derivative isn't quite the slope of a geometric tangent. It has a rigorous definition as a certain limit, and that limit doesn't exist at that point.
– Arthur
1 hour ago
Ahh. I thought that the derivative at x=0 was defined by whichever function was defined at x=0 (2-x in this case).
– Iamanon
1 hour ago
@Iamanon It does have a so-called right derivative at that point. That is the formal name of the idea you're talking about: if you're standing at that point, and only looking to the right (towards higher values of $x$), then that's what you get.
– Arthur
1 hour ago
 |Â
show 1 more comment
up vote
3
down vote
accepted
A function may or may not be continuous.
If it is continuous, it may or may not be differentiable. $f(x) = |x|$ is a standard example of a function which is continuous, but not (everywhere) differentiable. However, any differentiable function is necessarily continuous.
If a function is differentiable, its derivative may or may not be continuous. This is a bit more subtle, and the standard example of a differentiable function with discontinuous derivative is a bit more complicated:
$$
f(x) = casesx^2sin(1/x) & if $xneq 0$\
0 & if $x = 0$
$$
It is differentiable everywhere, $f'(0) = 0$, but $f'(x)$ oscillates wildly between (a little less than) $-1$ and (a little more than) $1$ as $x$ comes closer and closer to $0$, so it isn't continuous.
answered 1 hour ago
Arthur
105k799182
I'd like to add that you can have a function $f$ that's continuous everywhere but differentiable nowhere - check out en.m.wikipedia.org/wiki/Weierstrass_function.
– J. C.
1 hour ago
For my example under (3), isn't that an example of a differentiable function where its derivative is defined everywhere and continuous everywhere, but the underlying function is not continuous?
– Iamanon
1 hour ago
@Iamanon No, your example at (3) doesn't have a derivative at $x = 0$. Note that derivative isn't quite the slope of a geometric tangent. It has a rigorous definition as a certain limit, and that limit doesn't exist at that point.
– Arthur
1 hour ago
Ahh. I thought that the derivative at x=0 was defined by whichever function was defined at x=0 (2-x in this case).
– Iamanon
1 hour ago
@Iamanon It does have a so-called right derivative at that point. That is the formal name of the idea you're talking about: if you're standing at that point, and only looking to the right (towards higher values of $x$), then that's what you get.
– Arthur
1 hour ago
 |Â
show 1 more comment
up vote
3
down vote
accepted
up vote
3
down vote
accepted
A function may or may not be continuous.
If it is continuous, it may or may not be differentiable. $f(x) = |x|$ is a standard example of a function which is continuous, but not (everywhere) differentiable. However, any differentiable function is necessarily continuous.
If a function is differentiable, its derivative may or may not be continuous. This is a bit more subtle, and the standard example of a differentiable function with discontinuous derivative is a bit more complicated:
$$
f(x) = casesx^2sin(1/x) & if $xneq 0$\
0 & if $x = 0$
$$
It is differentiable everywhere, $f'(0) = 0$, but $f'(x)$ oscillates wildly between (a little less than) $-1$ and (a little more than) $1$ as $x$ comes closer and closer to $0$, so it isn't continuous.
answered 1 hour ago
Arthur
105k799182
A function may or may not be continuous.
If it is continuous, it may or may not be differentiable. $f(x) = |x|$ is a standard example of a function which is continuous, but not (everywhere) differentiable. However, any differentiable function is necessarily continuous.
If a function is differentiable, its derivative may or may not be continuous. This is a bit more subtle, and the standard example of a differentiable function with discontinuous derivative is a bit more complicated:
$$
f(x) = casesx^2sin(1/x) & if $xneq 0$\
0 & if $x = 0$
$$
It is differentiable everywhere, $f'(0) = 0$, but $f'(x)$ oscillates wildly between (a little less than) $-1$ and (a little more than) $1$ as $x$ comes closer and closer to $0$, so it isn't continuous.
answered 1 hour ago
Arthur
105k799182
answered 1 hour ago
Arthur
105k799182
answered 1 hour ago
Arthur
105k799182
answered 1 hour ago
Arthur
105k799182
105k799182
I'd like to add that you can have a function $f$ that's continuous everywhere but differentiable nowhere - check out en.m.wikipedia.org/wiki/Weierstrass_function.
– J. C.
1 hour ago
For my example under (3), isn't that an example of a differentiable function where its derivative is defined everywhere and continuous everywhere, but the underlying function is not continuous?
– Iamanon
1 hour ago
@Iamanon No, your example at (3) doesn't have a derivative at $x = 0$. Note that derivative isn't quite the slope of a geometric tangent. It has a rigorous definition as a certain limit, and that limit doesn't exist at that point.
– Arthur
1 hour ago
Ahh. I thought that the derivative at x=0 was defined by whichever function was defined at x=0 (2-x in this case).
– Iamanon
1 hour ago
@Iamanon It does have a so-called right derivative at that point. That is the formal name of the idea you're talking about: if you're standing at that point, and only looking to the right (towards higher values of $x$), then that's what you get.
– Arthur
1 hour ago
 |Â
show 1 more comment
I'd like to add that you can have a function $f$ that's continuous everywhere but differentiable nowhere - check out en.m.wikipedia.org/wiki/Weierstrass_function.
– J. C.
1 hour ago
For my example under (3), isn't that an example of a differentiable function where its derivative is defined everywhere and continuous everywhere, but the underlying function is not continuous?
– Iamanon
1 hour ago
@Iamanon No, your example at (3) doesn't have a derivative at $x = 0$. Note that derivative isn't quite the slope of a geometric tangent. It has a rigorous definition as a certain limit, and that limit doesn't exist at that point.
– Arthur
1 hour ago
Ahh. I thought that the derivative at x=0 was defined by whichever function was defined at x=0 (2-x in this case).
– Iamanon
1 hour ago
@Iamanon It does have a so-called right derivative at that point. That is the formal name of the idea you're talking about: if you're standing at that point, and only looking to the right (towards higher values of $x$), then that's what you get.
– Arthur
1 hour ago
I'd like to add that you can have a function $f$ that's continuous everywhere but differentiable nowhere - check out en.m.wikipedia.org/wiki/Weierstrass_function.
– J. C.
1 hour ago
I'd like to add that you can have a function $f$ that's continuous everywhere but differentiable nowhere - check out en.m.wikipedia.org/wiki/Weierstrass_function.
– J. C.
1 hour ago
For my example under (3), isn't that an example of a differentiable function where its derivative is defined everywhere and continuous everywhere, but the underlying function is not continuous?
– Iamanon
1 hour ago
For my example under (3), isn't that an example of a differentiable function where its derivative is defined everywhere and continuous everywhere, but the underlying function is not continuous?
– Iamanon
1 hour ago
@Iamanon No, your example at (3) doesn't have a derivative at $x = 0$. Note that derivative isn't quite the slope of a geometric tangent. It has a rigorous definition as a certain limit, and that limit doesn't exist at that point.
– Arthur
1 hour ago
@Iamanon No, your example at (3) doesn't have a derivative at $x = 0$. Note that derivative isn't quite the slope of a geometric tangent. It has a rigorous definition as a certain limit, and that limit doesn't exist at that point.
– Arthur
1 hour ago
Ahh. I thought that the derivative at x=0 was defined by whichever function was defined at x=0 (2-x in this case).
– Iamanon
1 hour ago
Ahh. I thought that the derivative at x=0 was defined by whichever function was defined at x=0 (2-x in this case).
– Iamanon
1 hour ago
@Iamanon It does have a so-called right derivative at that point. That is the formal name of the idea you're talking about: if you're standing at that point, and only looking to the right (towards higher values of $x$), then that's what you get.
– Arthur
1 hour ago
@Iamanon It does have a so-called right derivative at that point. That is the formal name of the idea you're talking about: if you're standing at that point, and only looking to the right (towards higher values of $x$), then that's what you get.
– Arthur
1 hour ago
 |Â
show 1 more comment
up vote
1
down vote
- Indeed.
- Indeed, the derivative does not even exist at $x=0$ as you argued in (1).
- Continuity is a necessary condition for differentiability (but it is not sufficient, as you have found in (1)). Your example is not differentiable at $0$, since it is discontinuous there. In particular, you can see this directly since
$$ lim_hto0^+ fracf(h)-f(0)h = lim_hto0^+ frac(2-h)-2h = -1, $$
whereas
$$lim_hto0^- fracf(h)-f(0)h = lim_hto0^- frac(1-h)-2h = lim_hto0^- frac-1-hh$$
does not even exist.
answered 1 hour ago
Sobi
3,0101517
add a comment |Â
up vote
1
down vote
- Indeed.
- Indeed, the derivative does not even exist at $x=0$ as you argued in (1).
- Continuity is a necessary condition for differentiability (but it is not sufficient, as you have found in (1)). Your example is not differentiable at $0$, since it is discontinuous there. In particular, you can see this directly since
$$ lim_hto0^+ fracf(h)-f(0)h = lim_hto0^+ frac(2-h)-2h = -1, $$
whereas
$$lim_hto0^- fracf(h)-f(0)h = lim_hto0^- frac(1-h)-2h = lim_hto0^- frac-1-hh$$
does not even exist.
answered 1 hour ago
Sobi
3,0101517
add a comment |Â
up vote
1
down vote
up vote
1
down vote
- Indeed.
- Indeed, the derivative does not even exist at $x=0$ as you argued in (1).
- Continuity is a necessary condition for differentiability (but it is not sufficient, as you have found in (1)). Your example is not differentiable at $0$, since it is discontinuous there. In particular, you can see this directly since
$$ lim_hto0^+ fracf(h)-f(0)h = lim_hto0^+ frac(2-h)-2h = -1, $$
whereas
$$lim_hto0^- fracf(h)-f(0)h = lim_hto0^- frac(1-h)-2h = lim_hto0^- frac-1-hh$$
does not even exist.
answered 1 hour ago
Sobi
3,0101517
- Indeed.
- Indeed, the derivative does not even exist at $x=0$ as you argued in (1).
- Continuity is a necessary condition for differentiability (but it is not sufficient, as you have found in (1)). Your example is not differentiable at $0$, since it is discontinuous there. In particular, you can see this directly since
$$ lim_hto0^+ fracf(h)-f(0)h = lim_hto0^+ frac(2-h)-2h = -1, $$
whereas
$$lim_hto0^- fracf(h)-f(0)h = lim_hto0^- frac(1-h)-2h = lim_hto0^- frac-1-hh$$
does not even exist.
answered 1 hour ago
Sobi
3,0101517
answered 1 hour ago
Sobi
3,0101517
answered 1 hour ago
Sobi
3,0101517
answered 1 hour ago
Sobi
3,0101517
3,0101517
add a comment |Â
add a comment |Â
up vote
1
down vote
As you have shown the continuity of the derivative function doesn't guarantee that the function is itself continuous since it can be piecewise continuous.
WHat we can show is Continuous differentiability implies Lipschitz continuity.
answered 1 hour ago
gimusi
78.6k73990
add a comment |Â
up vote
1
down vote
As you have shown the continuity of the derivative function doesn't guarantee that the function is itself continuous since it can be piecewise continuous.
WHat we can show is Continuous differentiability implies Lipschitz continuity.
answered 1 hour ago
gimusi
78.6k73990
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As you have shown the continuity of the derivative function doesn't guarantee that the function is itself continuous since it can be piecewise continuous.
WHat we can show is Continuous differentiability implies Lipschitz continuity.
answered 1 hour ago
gimusi
78.6k73990
As you have shown the continuity of the derivative function doesn't guarantee that the function is itself continuous since it can be piecewise continuous.
WHat we can show is Continuous differentiability implies Lipschitz continuity.
answered 1 hour ago
gimusi
78.6k73990
answered 1 hour ago
gimusi
78.6k73990
answered 1 hour ago
gimusi
78.6k73990
answered 1 hour ago
gimusi
78.6k73990
78.6k73990
add a comment |Â
add a comment |Â
up vote
1
down vote
The derivative is just another function, whether it's continuous doesn't affect the original function.
You have yourself given many examples on that.
Edit: in the further comments @gimusi has pointed we can show that Continuous differentiability implies Lipschitz continuity
answered 2 hours ago
Utsav Mangal
235
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Can we say that the second derivative of |x| is continuous everywhere?
– Iamanon
2 hours ago
@Iamanon It is continuous (and equal to $0$) everywhere it's defined. It is, however, not defined at $x = 0$.
– Arthur
2 hours ago
So there an example of a function where its derivative is discontinuous but its second derivative is defined everywhere?
– Iamanon
1 hour ago
for a derivative to exist at a point it has to be continuous, so no @lamanon
– Utsav Mangal
1 hour ago
What about my example under (3)? We see that f(x) is discontinuous but the derivative is defined everywhere. never mind. Arthur, above, addressed this question.
– Iamanon
1 hour ago
add a comment |Â
up vote
1
down vote
The derivative is just another function, whether it's continuous doesn't affect the original function.
You have yourself given many examples on that.
Edit: in the further comments @gimusi has pointed we can show that Continuous differentiability implies Lipschitz continuity
answered 2 hours ago
Utsav Mangal
235
New contributor
Utsav Mangal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Can we say that the second derivative of |x| is continuous everywhere?
– Iamanon
2 hours ago
@Iamanon It is continuous (and equal to $0$) everywhere it's defined. It is, however, not defined at $x = 0$.
– Arthur
2 hours ago
So there an example of a function where its derivative is discontinuous but its second derivative is defined everywhere?
– Iamanon
1 hour ago
for a derivative to exist at a point it has to be continuous, so no @lamanon
– Utsav Mangal
1 hour ago
What about my example under (3)? We see that f(x) is discontinuous but the derivative is defined everywhere. never mind. Arthur, above, addressed this question.
– Iamanon
1 hour ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The derivative is just another function, whether it's continuous doesn't affect the original function.
You have yourself given many examples on that.
Edit: in the further comments @gimusi has pointed we can show that Continuous differentiability implies Lipschitz continuity
answered 2 hours ago
Utsav Mangal
235
New contributor
Utsav Mangal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The derivative is just another function, whether it's continuous doesn't affect the original function.
You have yourself given many examples on that.
Edit: in the further comments @gimusi has pointed we can show that Continuous differentiability implies Lipschitz continuity
answered 2 hours ago
Utsav Mangal
235
New contributor
Utsav Mangal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
answered 2 hours ago
Utsav Mangal
235
New contributor
Utsav Mangal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Utsav Mangal
235
answered 2 hours ago
Utsav Mangal
235
235
New contributor
Utsav Mangal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Utsav Mangal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Utsav Mangal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Can we say that the second derivative of |x| is continuous everywhere?
– Iamanon
2 hours ago
@Iamanon It is continuous (and equal to $0$) everywhere it's defined. It is, however, not defined at $x = 0$.
– Arthur
2 hours ago
So there an example of a function where its derivative is discontinuous but its second derivative is defined everywhere?
– Iamanon
1 hour ago
for a derivative to exist at a point it has to be continuous, so no @lamanon
– Utsav Mangal
1 hour ago
What about my example under (3)? We see that f(x) is discontinuous but the derivative is defined everywhere. never mind. Arthur, above, addressed this question.
– Iamanon
1 hour ago
add a comment |Â
Can we say that the second derivative of |x| is continuous everywhere?
– Iamanon
2 hours ago
@Iamanon It is continuous (and equal to $0$) everywhere it's defined. It is, however, not defined at $x = 0$.
– Arthur
2 hours ago
So there an example of a function where its derivative is discontinuous but its second derivative is defined everywhere?
– Iamanon
1 hour ago
for a derivative to exist at a point it has to be continuous, so no @lamanon
– Utsav Mangal
1 hour ago
What about my example under (3)? We see that f(x) is discontinuous but the derivative is defined everywhere. never mind. Arthur, above, addressed this question.
– Iamanon
1 hour ago
Can we say that the second derivative of |x| is continuous everywhere?
– Iamanon
2 hours ago
Can we say that the second derivative of |x| is continuous everywhere?
– Iamanon
2 hours ago
@Iamanon It is continuous (and equal to $0$) everywhere it's defined. It is, however, not defined at $x = 0$.
– Arthur
2 hours ago
@Iamanon It is continuous (and equal to $0$) everywhere it's defined. It is, however, not defined at $x = 0$.
– Arthur
2 hours ago
So there an example of a function where its derivative is discontinuous but its second derivative is defined everywhere?
– Iamanon
1 hour ago
So there an example of a function where its derivative is discontinuous but its second derivative is defined everywhere?
– Iamanon
1 hour ago
for a derivative to exist at a point it has to be continuous, so no @lamanon
– Utsav Mangal
1 hour ago
for a derivative to exist at a point it has to be continuous, so no @lamanon
– Utsav Mangal
1 hour ago
What about my example under (3)? We see that f(x) is discontinuous but the derivative is defined everywhere. never mind. Arthur, above, addressed this question.
– Iamanon
1 hour ago
What about my example under (3)? We see that f(x) is discontinuous but the derivative is defined everywhere. never mind. Arthur, above, addressed this question.
– Iamanon
1 hour ago
add a comment |Â
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Note that the derivative of $|x|$ isn't discontinuous. It's actually continuous. It just doesn't exist everywhere. There is a difference. It allows us to say things like "derivatives have the intermediate value property", meaning if they exist they cannot have jump discontinuities.
– Arthur
1 hour ago
In high school, I was taught that a function is continuous if you can draw the function without lifting up your pencil. So here, isn't the derivative of |x| discontinuous using the same logic, i.e., I can't draw it's derivative because I have to lift up my pencil at x=0?
– Iamanon
1 hour ago
That's not really the definition of "continuous". There is a formal definition (you may have seen it; it begins "for any $epsilon >0$, there is a $delta > 0$ such that ..."), and according to that definition, a function must be defined at a point before you can even begin to make sense of wether it's continuous or discontinuous there.
– Arthur
1 hour ago
Ah. So the derivative is undefined at x=0. But you said the derivative is continuous. So how can you make that assessment if the derivative is not defined there? or do you mean it's continuous only to the left and right of x=0?
– Iamanon
1 hour ago
1
It's continuous everywhere it's defined, which is everywhere continuity makes sense to ask about. Thus one can just say that it's continuous. It's the same for, say, $f(x) = frac1x$. That function is also continuous. It just isn't a function on all of the real numbers.
– Arthur
1 hour ago