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Absolute value of first derivative interpretation, why is this true?
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in my book I have this:
Let $f: mathbbR to mathbbR$ be a function such that $f in C^1(mathbbR)$, and $x_0$ a point in which $f'(x_0) ne 0$
Then
$lim_deltato0$ $fracmu( f([x_0 - delta, x_0 + delta ] )mu( [x_0 - delta, x_0 + delta ]) = |f'(x_0)| $
Where $mu$ is the Lebesgue measure.
I feel pretty OK with that considering that the first derivative should measure the rate of change of the function in relation with the rate of change of the independent variable but why that's the exact expression formalizing this concept?
real-analysis measure-theory derivatives
asked Jul 31 at 14:18
Baffo rasta
14210
 |Â
show 2 more comments
up vote
0
down vote
favorite
in my book I have this:
Let $f: mathbbR to mathbbR$ be a function such that $f in C^1(mathbbR)$, and $x_0$ a point in which $f'(x_0) ne 0$
Then
$lim_deltato0$ $fracmu( f([x_0 - delta, x_0 + delta ] )mu( [x_0 - delta, x_0 + delta ]) = |f'(x_0)| $
Where $mu$ is the Lebesgue measure.
I feel pretty OK with that considering that the first derivative should measure the rate of change of the function in relation with the rate of change of the independent variable but why that's the exact expression formalizing this concept?
real-analysis measure-theory derivatives
asked Jul 31 at 14:18
Baffo rasta
14210
Is $mu$ supposed to more specifically be Lebesgue measure? This certainly isn't true for an arbitrary measure.
– Eric Wofsey
Jul 31 at 14:22
@EricWofsey I had the strong suspect that $mu$ was the Lebesgue measure but I wasn't completely sure, so I avoided to specify. I edit the text-
– Baffo rasta
Jul 31 at 14:23
do you know what $mu( [x_0 - delta, x_0 + delta ])$ means and how to calculate it?
– user190080
Jul 31 at 14:27
@user190080 It is a real interval so (correct me if I'm wrong) its Lebesgue measure should be $2delta$...
– Baffo rasta
Jul 31 at 14:28
No need for a correction, that's it - Eric's answer shows pretty nicely how straight forward some proofs are...sometimes it is really just plug-in
– user190080
Jul 31 at 14:37
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
in my book I have this:
Let $f: mathbbR to mathbbR$ be a function such that $f in C^1(mathbbR)$, and $x_0$ a point in which $f'(x_0) ne 0$
Then
$lim_deltato0$ $fracmu( f([x_0 - delta, x_0 + delta ] )mu( [x_0 - delta, x_0 + delta ]) = |f'(x_0)| $
Where $mu$ is the Lebesgue measure.
I feel pretty OK with that considering that the first derivative should measure the rate of change of the function in relation with the rate of change of the independent variable but why that's the exact expression formalizing this concept?
real-analysis measure-theory derivatives
asked Jul 31 at 14:18
Baffo rasta
14210
in my book I have this:
Let $f: mathbbR to mathbbR$ be a function such that $f in C^1(mathbbR)$, and $x_0$ a point in which $f'(x_0) ne 0$
Then
$lim_deltato0$ $fracmu( f([x_0 - delta, x_0 + delta ] )mu( [x_0 - delta, x_0 + delta ]) = |f'(x_0)| $
Where $mu$ is the Lebesgue measure.
I feel pretty OK with that considering that the first derivative should measure the rate of change of the function in relation with the rate of change of the independent variable but why that's the exact expression formalizing this concept?
real-analysis measure-theory derivatives
asked Jul 31 at 14:18
Baffo rasta
14210
edited Jul 31 at 14:24
asked Jul 31 at 14:18
Baffo rasta
14210
asked Jul 31 at 14:18
Baffo rasta
14210
asked Jul 31 at 14:18
Baffo rasta
14210
14210
Is $mu$ supposed to more specifically be Lebesgue measure? This certainly isn't true for an arbitrary measure.
– Eric Wofsey
Jul 31 at 14:22
@EricWofsey I had the strong suspect that $mu$ was the Lebesgue measure but I wasn't completely sure, so I avoided to specify. I edit the text-
– Baffo rasta
Jul 31 at 14:23
do you know what $mu( [x_0 - delta, x_0 + delta ])$ means and how to calculate it?
– user190080
Jul 31 at 14:27
@user190080 It is a real interval so (correct me if I'm wrong) its Lebesgue measure should be $2delta$...
– Baffo rasta
Jul 31 at 14:28
No need for a correction, that's it - Eric's answer shows pretty nicely how straight forward some proofs are...sometimes it is really just plug-in
– user190080
Jul 31 at 14:37
 |Â
show 2 more comments
Is $mu$ supposed to more specifically be Lebesgue measure? This certainly isn't true for an arbitrary measure.
– Eric Wofsey
Jul 31 at 14:22
@EricWofsey I had the strong suspect that $mu$ was the Lebesgue measure but I wasn't completely sure, so I avoided to specify. I edit the text-
– Baffo rasta
Jul 31 at 14:23
do you know what $mu( [x_0 - delta, x_0 + delta ])$ means and how to calculate it?
– user190080
Jul 31 at 14:27
@user190080 It is a real interval so (correct me if I'm wrong) its Lebesgue measure should be $2delta$...
– Baffo rasta
Jul 31 at 14:28
No need for a correction, that's it - Eric's answer shows pretty nicely how straight forward some proofs are...sometimes it is really just plug-in
– user190080
Jul 31 at 14:37
Is $mu$ supposed to more specifically be Lebesgue measure? This certainly isn't true for an arbitrary measure.
– Eric Wofsey
Jul 31 at 14:22
Is $mu$ supposed to more specifically be Lebesgue measure? This certainly isn't true for an arbitrary measure.
– Eric Wofsey
Jul 31 at 14:22
@EricWofsey I had the strong suspect that $mu$ was the Lebesgue measure but I wasn't completely sure, so I avoided to specify. I edit the text-
– Baffo rasta
Jul 31 at 14:23
@EricWofsey I had the strong suspect that $mu$ was the Lebesgue measure but I wasn't completely sure, so I avoided to specify. I edit the text-
– Baffo rasta
Jul 31 at 14:23
do you know what $mu( [x_0 - delta, x_0 + delta ])$ means and how to calculate it?
– user190080
Jul 31 at 14:27
do you know what $mu( [x_0 - delta, x_0 + delta ])$ means and how to calculate it?
– user190080
Jul 31 at 14:27
@user190080 It is a real interval so (correct me if I'm wrong) its Lebesgue measure should be $2delta$...
– Baffo rasta
Jul 31 at 14:28
@user190080 It is a real interval so (correct me if I'm wrong) its Lebesgue measure should be $2delta$...
– Baffo rasta
Jul 31 at 14:28
No need for a correction, that's it - Eric's answer shows pretty nicely how straight forward some proofs are...sometimes it is really just plug-in
– user190080
Jul 31 at 14:37
No need for a correction, that's it - Eric's answer shows pretty nicely how straight forward some proofs are...sometimes it is really just plug-in
– user190080
Jul 31 at 14:37
 |Â
show 2 more comments
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
This is almost literally just a rephrasing of the usual definition of the derivative. If $f'(x_0)neq 0$, then $f$ is monotonic in a neighborhood of $x_0$; let's assume it's increasing. Then $f([x_0-delta,x_0+delta])$ is just the interval $[f(x_0-delta),f(x_0+delta)]$, so we are taking the limit of $$fracmu([f(x_0-delta),f(x_0+delta)])mu([x_0-delta,x_0+delta])=fracf(x_0+delta)-f(x_0-delta)2delta.$$
This is just like the difference quotient defining $f'(x_0)$, except that we are taking a "two-sided" difference where we compare $f(x_0+delta)$ and $f(x_0-delta)$ instead of comparing each of them to $f(x_0)$. Indeed, we can break this fraction up as $$fracf(x_0+delta)-f(x_0-delta)2delta=frac12left(fracf(x_0+delta)-f(x_0)delta+fracf(x_0-delta)-f(x_0)-deltaright)$$ which is just an average of two difference quotients (one for $h=delta$ and one for $h=-delta$). So, since each difference quotient is converging to $f'(x_0)$, so is our fraction.
(If $f$ is decreasing instead of increasing, we get essentially the same thing except with a minus sign since $f$ reverses the order of the endpoints of our interval, so we get $-f'(x_0)=|f'(x_0)|$.)
answered Jul 31 at 14:30
Eric Wofsey
161k12188297
I think this one is pretty clean, thanks for the answer.
– Baffo rasta
Jul 31 at 14:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This is almost literally just a rephrasing of the usual definition of the derivative. If $f'(x_0)neq 0$, then $f$ is monotonic in a neighborhood of $x_0$; let's assume it's increasing. Then $f([x_0-delta,x_0+delta])$ is just the interval $[f(x_0-delta),f(x_0+delta)]$, so we are taking the limit of $$fracmu([f(x_0-delta),f(x_0+delta)])mu([x_0-delta,x_0+delta])=fracf(x_0+delta)-f(x_0-delta)2delta.$$
This is just like the difference quotient defining $f'(x_0)$, except that we are taking a "two-sided" difference where we compare $f(x_0+delta)$ and $f(x_0-delta)$ instead of comparing each of them to $f(x_0)$. Indeed, we can break this fraction up as $$fracf(x_0+delta)-f(x_0-delta)2delta=frac12left(fracf(x_0+delta)-f(x_0)delta+fracf(x_0-delta)-f(x_0)-deltaright)$$ which is just an average of two difference quotients (one for $h=delta$ and one for $h=-delta$). So, since each difference quotient is converging to $f'(x_0)$, so is our fraction.
(If $f$ is decreasing instead of increasing, we get essentially the same thing except with a minus sign since $f$ reverses the order of the endpoints of our interval, so we get $-f'(x_0)=|f'(x_0)|$.)
answered Jul 31 at 14:30
Eric Wofsey
161k12188297
I think this one is pretty clean, thanks for the answer.
– Baffo rasta
Jul 31 at 14:35
add a comment |Â
up vote
2
down vote
accepted
This is almost literally just a rephrasing of the usual definition of the derivative. If $f'(x_0)neq 0$, then $f$ is monotonic in a neighborhood of $x_0$; let's assume it's increasing. Then $f([x_0-delta,x_0+delta])$ is just the interval $[f(x_0-delta),f(x_0+delta)]$, so we are taking the limit of $$fracmu([f(x_0-delta),f(x_0+delta)])mu([x_0-delta,x_0+delta])=fracf(x_0+delta)-f(x_0-delta)2delta.$$
This is just like the difference quotient defining $f'(x_0)$, except that we are taking a "two-sided" difference where we compare $f(x_0+delta)$ and $f(x_0-delta)$ instead of comparing each of them to $f(x_0)$. Indeed, we can break this fraction up as $$fracf(x_0+delta)-f(x_0-delta)2delta=frac12left(fracf(x_0+delta)-f(x_0)delta+fracf(x_0-delta)-f(x_0)-deltaright)$$ which is just an average of two difference quotients (one for $h=delta$ and one for $h=-delta$). So, since each difference quotient is converging to $f'(x_0)$, so is our fraction.
(If $f$ is decreasing instead of increasing, we get essentially the same thing except with a minus sign since $f$ reverses the order of the endpoints of our interval, so we get $-f'(x_0)=|f'(x_0)|$.)
answered Jul 31 at 14:30
Eric Wofsey
161k12188297
I think this one is pretty clean, thanks for the answer.
– Baffo rasta
Jul 31 at 14:35
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This is almost literally just a rephrasing of the usual definition of the derivative. If $f'(x_0)neq 0$, then $f$ is monotonic in a neighborhood of $x_0$; let's assume it's increasing. Then $f([x_0-delta,x_0+delta])$ is just the interval $[f(x_0-delta),f(x_0+delta)]$, so we are taking the limit of $$fracmu([f(x_0-delta),f(x_0+delta)])mu([x_0-delta,x_0+delta])=fracf(x_0+delta)-f(x_0-delta)2delta.$$
This is just like the difference quotient defining $f'(x_0)$, except that we are taking a "two-sided" difference where we compare $f(x_0+delta)$ and $f(x_0-delta)$ instead of comparing each of them to $f(x_0)$. Indeed, we can break this fraction up as $$fracf(x_0+delta)-f(x_0-delta)2delta=frac12left(fracf(x_0+delta)-f(x_0)delta+fracf(x_0-delta)-f(x_0)-deltaright)$$ which is just an average of two difference quotients (one for $h=delta$ and one for $h=-delta$). So, since each difference quotient is converging to $f'(x_0)$, so is our fraction.
(If $f$ is decreasing instead of increasing, we get essentially the same thing except with a minus sign since $f$ reverses the order of the endpoints of our interval, so we get $-f'(x_0)=|f'(x_0)|$.)
answered Jul 31 at 14:30
Eric Wofsey
161k12188297
This is almost literally just a rephrasing of the usual definition of the derivative. If $f'(x_0)neq 0$, then $f$ is monotonic in a neighborhood of $x_0$; let's assume it's increasing. Then $f([x_0-delta,x_0+delta])$ is just the interval $[f(x_0-delta),f(x_0+delta)]$, so we are taking the limit of $$fracmu([f(x_0-delta),f(x_0+delta)])mu([x_0-delta,x_0+delta])=fracf(x_0+delta)-f(x_0-delta)2delta.$$
This is just like the difference quotient defining $f'(x_0)$, except that we are taking a "two-sided" difference where we compare $f(x_0+delta)$ and $f(x_0-delta)$ instead of comparing each of them to $f(x_0)$. Indeed, we can break this fraction up as $$fracf(x_0+delta)-f(x_0-delta)2delta=frac12left(fracf(x_0+delta)-f(x_0)delta+fracf(x_0-delta)-f(x_0)-deltaright)$$ which is just an average of two difference quotients (one for $h=delta$ and one for $h=-delta$). So, since each difference quotient is converging to $f'(x_0)$, so is our fraction.
(If $f$ is decreasing instead of increasing, we get essentially the same thing except with a minus sign since $f$ reverses the order of the endpoints of our interval, so we get $-f'(x_0)=|f'(x_0)|$.)
answered Jul 31 at 14:30
Eric Wofsey
161k12188297
answered Jul 31 at 14:30
Eric Wofsey
161k12188297
answered Jul 31 at 14:30
Eric Wofsey
161k12188297
answered Jul 31 at 14:30
Eric Wofsey
161k12188297
161k12188297
I think this one is pretty clean, thanks for the answer.
– Baffo rasta
Jul 31 at 14:35
add a comment |Â
I think this one is pretty clean, thanks for the answer.
– Baffo rasta
Jul 31 at 14:35
I think this one is pretty clean, thanks for the answer.
– Baffo rasta
Jul 31 at 14:35
I think this one is pretty clean, thanks for the answer.
– Baffo rasta
Jul 31 at 14:35
add a comment |Â
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Is $mu$ supposed to more specifically be Lebesgue measure? This certainly isn't true for an arbitrary measure.
– Eric Wofsey
Jul 31 at 14:22
@EricWofsey I had the strong suspect that $mu$ was the Lebesgue measure but I wasn't completely sure, so I avoided to specify. I edit the text-
– Baffo rasta
Jul 31 at 14:23
do you know what $mu( [x_0 - delta, x_0 + delta ])$ means and how to calculate it?
– user190080
Jul 31 at 14:27
@user190080 It is a real interval so (correct me if I'm wrong) its Lebesgue measure should be $2delta$...
– Baffo rasta
Jul 31 at 14:28
No need for a correction, that's it - Eric's answer shows pretty nicely how straight forward some proofs are...sometimes it is really just plug-in
– user190080
Jul 31 at 14:37