Mixing
In the derivative of the product of two functions , why (dx)² is ignored?
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I was digging deeply in the fundamentals of calculus, I found that in the famous '3Blue1Brown' channel, when demonstrating the process of finding the derivative of the product of two functions , (dx) ² is ignored because it becomes so tiny when dx is going closer and closer to zero.
I think that our calculus can be more precise if we don't ignore it, actually I think we shouldn't ignore it as this is math and not physics or such, the question is how our calculus is still valid and reliable if we give up even a tiny bit of precision.
The following image shows the details of the demonstration:
calculus derivatives products
asked Jan 24 at 22:45
Hassen DhiaHassen Dhia
84
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add a comment |
$begingroup$
I was digging deeply in the fundamentals of calculus, I found that in the famous '3Blue1Brown' channel, when demonstrating the process of finding the derivative of the product of two functions , (dx) ² is ignored because it becomes so tiny when dx is going closer and closer to zero.
I think that our calculus can be more precise if we don't ignore it, actually I think we shouldn't ignore it as this is math and not physics or such, the question is how our calculus is still valid and reliable if we give up even a tiny bit of precision.
The following image shows the details of the demonstration:
calculus derivatives products
asked Jan 24 at 22:45
Hassen DhiaHassen Dhia
84
$endgroup$
1
$begingroup$
Because if you're going after that approach, then $dx$ is infinitesimal and the square of an infinitesimal is zero. Alternatively, write everything with $Delta x$ instead; at some point, you'll end up with $(Delta x)^2 / Delta x$, which tends to zero as $Delta x$ does.
$endgroup$
– T. Bongers
Jan 24 at 22:48
2
$begingroup$
We don't give up on the precision, it's just that it truly does vanish in the limit, as @T.Bongers explained. 3Blue1Brown is there just giving intuition, he's not explaining the actual formalization.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 22:50
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Clear , i would accept your answer if you add some reference to it , i don't want to delete this question , as someone else would find it useful , thanks ^^
$endgroup$
– Hassen Dhia
Jan 24 at 22:56
$begingroup$
The quadratic variation of x, y etc is zero.
$endgroup$
– Vim
Jan 25 at 0:25
add a comment |
$begingroup$
I was digging deeply in the fundamentals of calculus, I found that in the famous '3Blue1Brown' channel, when demonstrating the process of finding the derivative of the product of two functions , (dx) ² is ignored because it becomes so tiny when dx is going closer and closer to zero.
I think that our calculus can be more precise if we don't ignore it, actually I think we shouldn't ignore it as this is math and not physics or such, the question is how our calculus is still valid and reliable if we give up even a tiny bit of precision.
The following image shows the details of the demonstration:
calculus derivatives products
asked Jan 24 at 22:45
Hassen DhiaHassen Dhia
84
$endgroup$
I was digging deeply in the fundamentals of calculus, I found that in the famous '3Blue1Brown' channel, when demonstrating the process of finding the derivative of the product of two functions , (dx) ² is ignored because it becomes so tiny when dx is going closer and closer to zero.
I think that our calculus can be more precise if we don't ignore it, actually I think we shouldn't ignore it as this is math and not physics or such, the question is how our calculus is still valid and reliable if we give up even a tiny bit of precision.
The following image shows the details of the demonstration:
calculus derivatives products
calculus derivatives products
asked Jan 24 at 22:45
Hassen DhiaHassen Dhia
84
asked Jan 24 at 22:45
Hassen DhiaHassen Dhia
84
edited Jan 24 at 23:19
Hassen Dhia
asked Jan 24 at 22:45
Hassen DhiaHassen Dhia
84
asked Jan 24 at 22:45
Hassen DhiaHassen Dhia
84
asked Jan 24 at 22:45
Hassen DhiaHassen Dhia
84
84
1
$begingroup$
Because if you're going after that approach, then $dx$ is infinitesimal and the square of an infinitesimal is zero. Alternatively, write everything with $Delta x$ instead; at some point, you'll end up with $(Delta x)^2 / Delta x$, which tends to zero as $Delta x$ does.
$endgroup$
– T. Bongers
Jan 24 at 22:48
2
$begingroup$
We don't give up on the precision, it's just that it truly does vanish in the limit, as @T.Bongers explained. 3Blue1Brown is there just giving intuition, he's not explaining the actual formalization.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 22:50
$begingroup$
Clear , i would accept your answer if you add some reference to it , i don't want to delete this question , as someone else would find it useful , thanks ^^
$endgroup$
– Hassen Dhia
Jan 24 at 22:56
$begingroup$
The quadratic variation of x, y etc is zero.
$endgroup$
– Vim
Jan 25 at 0:25
add a comment |
1
$begingroup$
Because if you're going after that approach, then $dx$ is infinitesimal and the square of an infinitesimal is zero. Alternatively, write everything with $Delta x$ instead; at some point, you'll end up with $(Delta x)^2 / Delta x$, which tends to zero as $Delta x$ does.
$endgroup$
– T. Bongers
Jan 24 at 22:48
2
$begingroup$
We don't give up on the precision, it's just that it truly does vanish in the limit, as @T.Bongers explained. 3Blue1Brown is there just giving intuition, he's not explaining the actual formalization.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 22:50
$begingroup$
Clear , i would accept your answer if you add some reference to it , i don't want to delete this question , as someone else would find it useful , thanks ^^
$endgroup$
– Hassen Dhia
Jan 24 at 22:56
$begingroup$
The quadratic variation of x, y etc is zero.
$endgroup$
– Vim
Jan 25 at 0:25
1
1
$begingroup$
Because if you're going after that approach, then $dx$ is infinitesimal and the square of an infinitesimal is zero. Alternatively, write everything with $Delta x$ instead; at some point, you'll end up with $(Delta x)^2 / Delta x$, which tends to zero as $Delta x$ does.
$endgroup$
– T. Bongers
Jan 24 at 22:48
$begingroup$
Because if you're going after that approach, then $dx$ is infinitesimal and the square of an infinitesimal is zero. Alternatively, write everything with $Delta x$ instead; at some point, you'll end up with $(Delta x)^2 / Delta x$, which tends to zero as $Delta x$ does.
$endgroup$
– T. Bongers
Jan 24 at 22:48
2
2
$begingroup$
We don't give up on the precision, it's just that it truly does vanish in the limit, as @T.Bongers explained. 3Blue1Brown is there just giving intuition, he's not explaining the actual formalization.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 22:50
$begingroup$
We don't give up on the precision, it's just that it truly does vanish in the limit, as @T.Bongers explained. 3Blue1Brown is there just giving intuition, he's not explaining the actual formalization.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 22:50
$begingroup$
Clear , i would accept your answer if you add some reference to it , i don't want to delete this question , as someone else would find it useful , thanks ^^
$endgroup$
– Hassen Dhia
Jan 24 at 22:56
$begingroup$
Clear , i would accept your answer if you add some reference to it , i don't want to delete this question , as someone else would find it useful , thanks ^^
$endgroup$
– Hassen Dhia
Jan 24 at 22:56
$begingroup$
The quadratic variation of x, y etc is zero.
$endgroup$
– Vim
Jan 25 at 0:25
$begingroup$
The quadratic variation of x, y etc is zero.
$endgroup$
– Vim
Jan 25 at 0:25
add a comment |
1 Answer
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$begingroup$
With all of the terms in $$df=sin(x)d(x^2)+x^2dsin(x)+d(x^2)dsin(x)$$ to move forward, you will divide by $dx$: $$frac{df}{dx}=sin(x)frac{d(x^2)}{dx}+x^2frac{dsin(x)}{dx}+d(x^2)frac{dsin(x)}{dx}$$ (I'm using $d$ as a positive quantity prior to taking the limit.)
Now take the limit as $dxto0$ (which compels $d(x^2)to0$).
$$frac{df}{dx}=sin(x)2x+x^2cos(x)+0cdotcos(x)$$
So after taking the limit, this additional term certainly contributes $0$. (Note that I could have paired the last denominator with $d(x^2)$ instead, but then $dsin(x)to0$ and the end result is the same.)
answered Jan 24 at 22:57
alex.jordanalex.jordan
39.5k560122
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1 Answer
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$begingroup$
With all of the terms in $$df=sin(x)d(x^2)+x^2dsin(x)+d(x^2)dsin(x)$$ to move forward, you will divide by $dx$: $$frac{df}{dx}=sin(x)frac{d(x^2)}{dx}+x^2frac{dsin(x)}{dx}+d(x^2)frac{dsin(x)}{dx}$$ (I'm using $d$ as a positive quantity prior to taking the limit.)
Now take the limit as $dxto0$ (which compels $d(x^2)to0$).
$$frac{df}{dx}=sin(x)2x+x^2cos(x)+0cdotcos(x)$$
So after taking the limit, this additional term certainly contributes $0$. (Note that I could have paired the last denominator with $d(x^2)$ instead, but then $dsin(x)to0$ and the end result is the same.)
answered Jan 24 at 22:57
alex.jordanalex.jordan
39.5k560122
$endgroup$
add a comment |
$begingroup$
With all of the terms in $$df=sin(x)d(x^2)+x^2dsin(x)+d(x^2)dsin(x)$$ to move forward, you will divide by $dx$: $$frac{df}{dx}=sin(x)frac{d(x^2)}{dx}+x^2frac{dsin(x)}{dx}+d(x^2)frac{dsin(x)}{dx}$$ (I'm using $d$ as a positive quantity prior to taking the limit.)
Now take the limit as $dxto0$ (which compels $d(x^2)to0$).
$$frac{df}{dx}=sin(x)2x+x^2cos(x)+0cdotcos(x)$$
So after taking the limit, this additional term certainly contributes $0$. (Note that I could have paired the last denominator with $d(x^2)$ instead, but then $dsin(x)to0$ and the end result is the same.)
answered Jan 24 at 22:57
alex.jordanalex.jordan
39.5k560122
$endgroup$
add a comment |
$begingroup$
With all of the terms in $$df=sin(x)d(x^2)+x^2dsin(x)+d(x^2)dsin(x)$$ to move forward, you will divide by $dx$: $$frac{df}{dx}=sin(x)frac{d(x^2)}{dx}+x^2frac{dsin(x)}{dx}+d(x^2)frac{dsin(x)}{dx}$$ (I'm using $d$ as a positive quantity prior to taking the limit.)
Now take the limit as $dxto0$ (which compels $d(x^2)to0$).
$$frac{df}{dx}=sin(x)2x+x^2cos(x)+0cdotcos(x)$$
So after taking the limit, this additional term certainly contributes $0$. (Note that I could have paired the last denominator with $d(x^2)$ instead, but then $dsin(x)to0$ and the end result is the same.)
answered Jan 24 at 22:57
alex.jordanalex.jordan
39.5k560122
$endgroup$
With all of the terms in $$df=sin(x)d(x^2)+x^2dsin(x)+d(x^2)dsin(x)$$ to move forward, you will divide by $dx$: $$frac{df}{dx}=sin(x)frac{d(x^2)}{dx}+x^2frac{dsin(x)}{dx}+d(x^2)frac{dsin(x)}{dx}$$ (I'm using $d$ as a positive quantity prior to taking the limit.)
Now take the limit as $dxto0$ (which compels $d(x^2)to0$).
$$frac{df}{dx}=sin(x)2x+x^2cos(x)+0cdotcos(x)$$
So after taking the limit, this additional term certainly contributes $0$. (Note that I could have paired the last denominator with $d(x^2)$ instead, but then $dsin(x)to0$ and the end result is the same.)
answered Jan 24 at 22:57
alex.jordanalex.jordan
39.5k560122
answered Jan 24 at 22:57
alex.jordanalex.jordan
39.5k560122
answered Jan 24 at 22:57
alex.jordanalex.jordan
39.5k560122
answered Jan 24 at 22:57
alex.jordanalex.jordan
39.5k560122
39.5k560122
add a comment |
add a comment |
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$begingroup$
Because if you're going after that approach, then $dx$ is infinitesimal and the square of an infinitesimal is zero. Alternatively, write everything with $Delta x$ instead; at some point, you'll end up with $(Delta x)^2 / Delta x$, which tends to zero as $Delta x$ does.
$endgroup$
– T. Bongers
Jan 24 at 22:48
2
$begingroup$
We don't give up on the precision, it's just that it truly does vanish in the limit, as @T.Bongers explained. 3Blue1Brown is there just giving intuition, he's not explaining the actual formalization.
$endgroup$
– nathan.j.mcdougall
Jan 24 at 22:50
$begingroup$
Clear , i would accept your answer if you add some reference to it , i don't want to delete this question , as someone else would find it useful , thanks ^^
$endgroup$
– Hassen Dhia
Jan 24 at 22:56
$begingroup$
The quadratic variation of x, y etc is zero.
$endgroup$
– Vim
Jan 25 at 0:25