When is it legal to use $dx$?
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I still have some trouble grasping the true mathematical meaning of $dx$. I have the intuitive understanding that it is an "infinitely tiny amount of $x$" but I don't know in what cases I can use it.
Here is an example :
is $lim_{xto 0} frac{2x + 32x^2}{e^x}$ equivalent to $frac{2.dx + 32.dx^2}{e^{dx}}$ ?
Thanks a lot
calculus
$endgroup$
|
show 5 more comments
$begingroup$
I still have some trouble grasping the true mathematical meaning of $dx$. I have the intuitive understanding that it is an "infinitely tiny amount of $x$" but I don't know in what cases I can use it.
Here is an example :
is $lim_{xto 0} frac{2x + 32x^2}{e^x}$ equivalent to $frac{2.dx + 32.dx^2}{e^{dx}}$ ?
Thanks a lot
calculus
$endgroup$
$begingroup$
$dx$ is the difference between 2 values of $x$, your case is obviously unfit for such use
$endgroup$
– Makina
Dec 16 '18 at 15:58
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Then is $lim_{Delta xto 0} frac{2Delta x + 32(Delta x)^2}{e^{Delta x}}$ equivalent to $frac{2.dx + 32.dx^2}{e^{dx}}$ ?
$endgroup$
– s89ne
Dec 16 '18 at 16:00
$begingroup$
No, because the limit you gave either doesn’t exist or evaluates to a real number. What you wrote is an expression that’s neither. This feels similar to how people treat $frac{df}{dx}$. Luckily notation and theory work out to where you can treat it like a fraction in same cases without running into problems, but it’s still not correct.
$endgroup$
– Will Fisher
Dec 16 '18 at 16:01
2
$begingroup$
Depending on the context, $textrm{d}x$ on its own has different meanings. It can be a measure or a differential form (off the top of my head). I know no context where it would stand for an infinitely tiny amount of $x$. The usual notation for a very small amount is just $varepsilon$, but it is usually not infinitesimal. But even if $textrm{d}x$ was an actual infinitesimal, the two expressions would not be the same (not before applying standard part to the right hand side).
$endgroup$
– tomasz
Dec 16 '18 at 16:02
1
$begingroup$
What $dx$ precisely is seems confusing. Is there a rigorous mathematical definition of it, or should it always be understood intuitively depending on the context?
$endgroup$
– s89ne
Dec 16 '18 at 16:11
|
show 5 more comments
$begingroup$
I still have some trouble grasping the true mathematical meaning of $dx$. I have the intuitive understanding that it is an "infinitely tiny amount of $x$" but I don't know in what cases I can use it.
Here is an example :
is $lim_{xto 0} frac{2x + 32x^2}{e^x}$ equivalent to $frac{2.dx + 32.dx^2}{e^{dx}}$ ?
Thanks a lot
calculus
$endgroup$
I still have some trouble grasping the true mathematical meaning of $dx$. I have the intuitive understanding that it is an "infinitely tiny amount of $x$" but I don't know in what cases I can use it.
Here is an example :
is $lim_{xto 0} frac{2x + 32x^2}{e^x}$ equivalent to $frac{2.dx + 32.dx^2}{e^{dx}}$ ?
Thanks a lot
calculus
calculus
asked Dec 16 '18 at 15:56
s89nes89ne
556
556
$begingroup$
$dx$ is the difference between 2 values of $x$, your case is obviously unfit for such use
$endgroup$
– Makina
Dec 16 '18 at 15:58
$begingroup$
Then is $lim_{Delta xto 0} frac{2Delta x + 32(Delta x)^2}{e^{Delta x}}$ equivalent to $frac{2.dx + 32.dx^2}{e^{dx}}$ ?
$endgroup$
– s89ne
Dec 16 '18 at 16:00
$begingroup$
No, because the limit you gave either doesn’t exist or evaluates to a real number. What you wrote is an expression that’s neither. This feels similar to how people treat $frac{df}{dx}$. Luckily notation and theory work out to where you can treat it like a fraction in same cases without running into problems, but it’s still not correct.
$endgroup$
– Will Fisher
Dec 16 '18 at 16:01
2
$begingroup$
Depending on the context, $textrm{d}x$ on its own has different meanings. It can be a measure or a differential form (off the top of my head). I know no context where it would stand for an infinitely tiny amount of $x$. The usual notation for a very small amount is just $varepsilon$, but it is usually not infinitesimal. But even if $textrm{d}x$ was an actual infinitesimal, the two expressions would not be the same (not before applying standard part to the right hand side).
$endgroup$
– tomasz
Dec 16 '18 at 16:02
1
$begingroup$
What $dx$ precisely is seems confusing. Is there a rigorous mathematical definition of it, or should it always be understood intuitively depending on the context?
$endgroup$
– s89ne
Dec 16 '18 at 16:11
|
show 5 more comments
$begingroup$
$dx$ is the difference between 2 values of $x$, your case is obviously unfit for such use
$endgroup$
– Makina
Dec 16 '18 at 15:58
$begingroup$
Then is $lim_{Delta xto 0} frac{2Delta x + 32(Delta x)^2}{e^{Delta x}}$ equivalent to $frac{2.dx + 32.dx^2}{e^{dx}}$ ?
$endgroup$
– s89ne
Dec 16 '18 at 16:00
$begingroup$
No, because the limit you gave either doesn’t exist or evaluates to a real number. What you wrote is an expression that’s neither. This feels similar to how people treat $frac{df}{dx}$. Luckily notation and theory work out to where you can treat it like a fraction in same cases without running into problems, but it’s still not correct.
$endgroup$
– Will Fisher
Dec 16 '18 at 16:01
2
$begingroup$
Depending on the context, $textrm{d}x$ on its own has different meanings. It can be a measure or a differential form (off the top of my head). I know no context where it would stand for an infinitely tiny amount of $x$. The usual notation for a very small amount is just $varepsilon$, but it is usually not infinitesimal. But even if $textrm{d}x$ was an actual infinitesimal, the two expressions would not be the same (not before applying standard part to the right hand side).
$endgroup$
– tomasz
Dec 16 '18 at 16:02
1
$begingroup$
What $dx$ precisely is seems confusing. Is there a rigorous mathematical definition of it, or should it always be understood intuitively depending on the context?
$endgroup$
– s89ne
Dec 16 '18 at 16:11
$begingroup$
$dx$ is the difference between 2 values of $x$, your case is obviously unfit for such use
$endgroup$
– Makina
Dec 16 '18 at 15:58
$begingroup$
$dx$ is the difference between 2 values of $x$, your case is obviously unfit for such use
$endgroup$
– Makina
Dec 16 '18 at 15:58
$begingroup$
Then is $lim_{Delta xto 0} frac{2Delta x + 32(Delta x)^2}{e^{Delta x}}$ equivalent to $frac{2.dx + 32.dx^2}{e^{dx}}$ ?
$endgroup$
– s89ne
Dec 16 '18 at 16:00
$begingroup$
Then is $lim_{Delta xto 0} frac{2Delta x + 32(Delta x)^2}{e^{Delta x}}$ equivalent to $frac{2.dx + 32.dx^2}{e^{dx}}$ ?
$endgroup$
– s89ne
Dec 16 '18 at 16:00
$begingroup$
No, because the limit you gave either doesn’t exist or evaluates to a real number. What you wrote is an expression that’s neither. This feels similar to how people treat $frac{df}{dx}$. Luckily notation and theory work out to where you can treat it like a fraction in same cases without running into problems, but it’s still not correct.
$endgroup$
– Will Fisher
Dec 16 '18 at 16:01
$begingroup$
No, because the limit you gave either doesn’t exist or evaluates to a real number. What you wrote is an expression that’s neither. This feels similar to how people treat $frac{df}{dx}$. Luckily notation and theory work out to where you can treat it like a fraction in same cases without running into problems, but it’s still not correct.
$endgroup$
– Will Fisher
Dec 16 '18 at 16:01
2
2
$begingroup$
Depending on the context, $textrm{d}x$ on its own has different meanings. It can be a measure or a differential form (off the top of my head). I know no context where it would stand for an infinitely tiny amount of $x$. The usual notation for a very small amount is just $varepsilon$, but it is usually not infinitesimal. But even if $textrm{d}x$ was an actual infinitesimal, the two expressions would not be the same (not before applying standard part to the right hand side).
$endgroup$
– tomasz
Dec 16 '18 at 16:02
$begingroup$
Depending on the context, $textrm{d}x$ on its own has different meanings. It can be a measure or a differential form (off the top of my head). I know no context where it would stand for an infinitely tiny amount of $x$. The usual notation for a very small amount is just $varepsilon$, but it is usually not infinitesimal. But even if $textrm{d}x$ was an actual infinitesimal, the two expressions would not be the same (not before applying standard part to the right hand side).
$endgroup$
– tomasz
Dec 16 '18 at 16:02
1
1
$begingroup$
What $dx$ precisely is seems confusing. Is there a rigorous mathematical definition of it, or should it always be understood intuitively depending on the context?
$endgroup$
– s89ne
Dec 16 '18 at 16:11
$begingroup$
What $dx$ precisely is seems confusing. Is there a rigorous mathematical definition of it, or should it always be understood intuitively depending on the context?
$endgroup$
– s89ne
Dec 16 '18 at 16:11
|
show 5 more comments
2 Answers
2
active
oldest
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$begingroup$
Let's start with your $f'(x)= limlimits_{Delta xto 0} frac{Delta (f(x))}{Delta x} = 2x$ when $f(x)=x^2$
Sometimes the derivative $f'(x)$ is written $frac{df}{dx}$ or $frac{d}{dx} f(x)$. All three of these are simply notation for $limlimits_{y to x} frac{f(y)-f(x)}{y-x}$ or $limlimits_{h to 0} frac{f(x+h)-f(x)}{h}$ and some people describe this as the limit of the ratio of the change of $f(x)$ to the change in $x$ leading to your $limlimits_{Delta xto 0} frac{Delta (f(x))}{Delta x}$ or $limlimits_{Delta xto 0} frac{Delta f}{Delta x}$: the notation $frac{df}{dx}$ is suggestive of this limit but the $dx$ does mean anything on its own here
Since the derivative operation can be applied to different functions, further notation lets $frac{d}{dx}$ represent a derivative operator. It can be applied more than once: for example a double derivative would involve a double operation and is sometimes written $f''(x)= frac{d^2}{dx^2}f(x)=frac{d^2f}{dx^2}$; with $f(x)=x^2$ you would get $frac{d^2f}{dx^2} = 2$. Again this is notation, and the $dx$ is part of the notation without having a stand-alone meaning; in particular it is not being squared
You will also see $dx$ as part of the notation for integrals: for example $int_6^9 x^2, dx = 171$. But while this might be suggestive of a sum (the long s in the integral sign) of lots of pieces for $f(x)$ with widths represented by $dx$, it is again notation and the $dx$ does not have an independent meaning here, just being part of a wider expression for the integral operation
Some people write expressions such as $d(fg)=f, dg + g, df$ for the product rule. Typically this is shorthand for $frac{d}{dx}(fg)=f frac{d}{dx}g + g frac{d}{dx}f$ and is another example of the use of notation
There is nothing to prevent people from giving a particular meaning to a bare $dx$ so long as they define it in a meaningful way. Then again it would still be notation, with that definition
$endgroup$
add a comment |
$begingroup$
The differential being an infinitely tiny amount is a misconception. In modern setting, the differential is a linear map, specifically the linear part of the variation of a function when you vary the variable. In other words, when a function is differentiable there is a constant $c$ such that
$$f(x+dx)=f(x)+ccdot dx+o(dx)$$ ($o(dx)$ is a sublinear term as a function of $dx$)
and by definition $$df(x)=ccdot dx.$$
The constant $c$ is well know to be the derivative of $f$, and $dx$ can be an arbitrarily large increment.
Actually, differentials have little to do with limits.
On the picture, the function is in blue. Its differential (at $x=1$) in magenta and the remainder in black.
$endgroup$
add a comment |
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2 Answers
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$begingroup$
Let's start with your $f'(x)= limlimits_{Delta xto 0} frac{Delta (f(x))}{Delta x} = 2x$ when $f(x)=x^2$
Sometimes the derivative $f'(x)$ is written $frac{df}{dx}$ or $frac{d}{dx} f(x)$. All three of these are simply notation for $limlimits_{y to x} frac{f(y)-f(x)}{y-x}$ or $limlimits_{h to 0} frac{f(x+h)-f(x)}{h}$ and some people describe this as the limit of the ratio of the change of $f(x)$ to the change in $x$ leading to your $limlimits_{Delta xto 0} frac{Delta (f(x))}{Delta x}$ or $limlimits_{Delta xto 0} frac{Delta f}{Delta x}$: the notation $frac{df}{dx}$ is suggestive of this limit but the $dx$ does mean anything on its own here
Since the derivative operation can be applied to different functions, further notation lets $frac{d}{dx}$ represent a derivative operator. It can be applied more than once: for example a double derivative would involve a double operation and is sometimes written $f''(x)= frac{d^2}{dx^2}f(x)=frac{d^2f}{dx^2}$; with $f(x)=x^2$ you would get $frac{d^2f}{dx^2} = 2$. Again this is notation, and the $dx$ is part of the notation without having a stand-alone meaning; in particular it is not being squared
You will also see $dx$ as part of the notation for integrals: for example $int_6^9 x^2, dx = 171$. But while this might be suggestive of a sum (the long s in the integral sign) of lots of pieces for $f(x)$ with widths represented by $dx$, it is again notation and the $dx$ does not have an independent meaning here, just being part of a wider expression for the integral operation
Some people write expressions such as $d(fg)=f, dg + g, df$ for the product rule. Typically this is shorthand for $frac{d}{dx}(fg)=f frac{d}{dx}g + g frac{d}{dx}f$ and is another example of the use of notation
There is nothing to prevent people from giving a particular meaning to a bare $dx$ so long as they define it in a meaningful way. Then again it would still be notation, with that definition
$endgroup$
add a comment |
$begingroup$
Let's start with your $f'(x)= limlimits_{Delta xto 0} frac{Delta (f(x))}{Delta x} = 2x$ when $f(x)=x^2$
Sometimes the derivative $f'(x)$ is written $frac{df}{dx}$ or $frac{d}{dx} f(x)$. All three of these are simply notation for $limlimits_{y to x} frac{f(y)-f(x)}{y-x}$ or $limlimits_{h to 0} frac{f(x+h)-f(x)}{h}$ and some people describe this as the limit of the ratio of the change of $f(x)$ to the change in $x$ leading to your $limlimits_{Delta xto 0} frac{Delta (f(x))}{Delta x}$ or $limlimits_{Delta xto 0} frac{Delta f}{Delta x}$: the notation $frac{df}{dx}$ is suggestive of this limit but the $dx$ does mean anything on its own here
Since the derivative operation can be applied to different functions, further notation lets $frac{d}{dx}$ represent a derivative operator. It can be applied more than once: for example a double derivative would involve a double operation and is sometimes written $f''(x)= frac{d^2}{dx^2}f(x)=frac{d^2f}{dx^2}$; with $f(x)=x^2$ you would get $frac{d^2f}{dx^2} = 2$. Again this is notation, and the $dx$ is part of the notation without having a stand-alone meaning; in particular it is not being squared
You will also see $dx$ as part of the notation for integrals: for example $int_6^9 x^2, dx = 171$. But while this might be suggestive of a sum (the long s in the integral sign) of lots of pieces for $f(x)$ with widths represented by $dx$, it is again notation and the $dx$ does not have an independent meaning here, just being part of a wider expression for the integral operation
Some people write expressions such as $d(fg)=f, dg + g, df$ for the product rule. Typically this is shorthand for $frac{d}{dx}(fg)=f frac{d}{dx}g + g frac{d}{dx}f$ and is another example of the use of notation
There is nothing to prevent people from giving a particular meaning to a bare $dx$ so long as they define it in a meaningful way. Then again it would still be notation, with that definition
$endgroup$
add a comment |
$begingroup$
Let's start with your $f'(x)= limlimits_{Delta xto 0} frac{Delta (f(x))}{Delta x} = 2x$ when $f(x)=x^2$
Sometimes the derivative $f'(x)$ is written $frac{df}{dx}$ or $frac{d}{dx} f(x)$. All three of these are simply notation for $limlimits_{y to x} frac{f(y)-f(x)}{y-x}$ or $limlimits_{h to 0} frac{f(x+h)-f(x)}{h}$ and some people describe this as the limit of the ratio of the change of $f(x)$ to the change in $x$ leading to your $limlimits_{Delta xto 0} frac{Delta (f(x))}{Delta x}$ or $limlimits_{Delta xto 0} frac{Delta f}{Delta x}$: the notation $frac{df}{dx}$ is suggestive of this limit but the $dx$ does mean anything on its own here
Since the derivative operation can be applied to different functions, further notation lets $frac{d}{dx}$ represent a derivative operator. It can be applied more than once: for example a double derivative would involve a double operation and is sometimes written $f''(x)= frac{d^2}{dx^2}f(x)=frac{d^2f}{dx^2}$; with $f(x)=x^2$ you would get $frac{d^2f}{dx^2} = 2$. Again this is notation, and the $dx$ is part of the notation without having a stand-alone meaning; in particular it is not being squared
You will also see $dx$ as part of the notation for integrals: for example $int_6^9 x^2, dx = 171$. But while this might be suggestive of a sum (the long s in the integral sign) of lots of pieces for $f(x)$ with widths represented by $dx$, it is again notation and the $dx$ does not have an independent meaning here, just being part of a wider expression for the integral operation
Some people write expressions such as $d(fg)=f, dg + g, df$ for the product rule. Typically this is shorthand for $frac{d}{dx}(fg)=f frac{d}{dx}g + g frac{d}{dx}f$ and is another example of the use of notation
There is nothing to prevent people from giving a particular meaning to a bare $dx$ so long as they define it in a meaningful way. Then again it would still be notation, with that definition
$endgroup$
Let's start with your $f'(x)= limlimits_{Delta xto 0} frac{Delta (f(x))}{Delta x} = 2x$ when $f(x)=x^2$
Sometimes the derivative $f'(x)$ is written $frac{df}{dx}$ or $frac{d}{dx} f(x)$. All three of these are simply notation for $limlimits_{y to x} frac{f(y)-f(x)}{y-x}$ or $limlimits_{h to 0} frac{f(x+h)-f(x)}{h}$ and some people describe this as the limit of the ratio of the change of $f(x)$ to the change in $x$ leading to your $limlimits_{Delta xto 0} frac{Delta (f(x))}{Delta x}$ or $limlimits_{Delta xto 0} frac{Delta f}{Delta x}$: the notation $frac{df}{dx}$ is suggestive of this limit but the $dx$ does mean anything on its own here
Since the derivative operation can be applied to different functions, further notation lets $frac{d}{dx}$ represent a derivative operator. It can be applied more than once: for example a double derivative would involve a double operation and is sometimes written $f''(x)= frac{d^2}{dx^2}f(x)=frac{d^2f}{dx^2}$; with $f(x)=x^2$ you would get $frac{d^2f}{dx^2} = 2$. Again this is notation, and the $dx$ is part of the notation without having a stand-alone meaning; in particular it is not being squared
You will also see $dx$ as part of the notation for integrals: for example $int_6^9 x^2, dx = 171$. But while this might be suggestive of a sum (the long s in the integral sign) of lots of pieces for $f(x)$ with widths represented by $dx$, it is again notation and the $dx$ does not have an independent meaning here, just being part of a wider expression for the integral operation
Some people write expressions such as $d(fg)=f, dg + g, df$ for the product rule. Typically this is shorthand for $frac{d}{dx}(fg)=f frac{d}{dx}g + g frac{d}{dx}f$ and is another example of the use of notation
There is nothing to prevent people from giving a particular meaning to a bare $dx$ so long as they define it in a meaningful way. Then again it would still be notation, with that definition
edited Dec 17 '18 at 12:13
answered Dec 16 '18 at 17:06
HenryHenry
101k481168
101k481168
add a comment |
add a comment |
$begingroup$
The differential being an infinitely tiny amount is a misconception. In modern setting, the differential is a linear map, specifically the linear part of the variation of a function when you vary the variable. In other words, when a function is differentiable there is a constant $c$ such that
$$f(x+dx)=f(x)+ccdot dx+o(dx)$$ ($o(dx)$ is a sublinear term as a function of $dx$)
and by definition $$df(x)=ccdot dx.$$
The constant $c$ is well know to be the derivative of $f$, and $dx$ can be an arbitrarily large increment.
Actually, differentials have little to do with limits.
On the picture, the function is in blue. Its differential (at $x=1$) in magenta and the remainder in black.
$endgroup$
add a comment |
$begingroup$
The differential being an infinitely tiny amount is a misconception. In modern setting, the differential is a linear map, specifically the linear part of the variation of a function when you vary the variable. In other words, when a function is differentiable there is a constant $c$ such that
$$f(x+dx)=f(x)+ccdot dx+o(dx)$$ ($o(dx)$ is a sublinear term as a function of $dx$)
and by definition $$df(x)=ccdot dx.$$
The constant $c$ is well know to be the derivative of $f$, and $dx$ can be an arbitrarily large increment.
Actually, differentials have little to do with limits.
On the picture, the function is in blue. Its differential (at $x=1$) in magenta and the remainder in black.
$endgroup$
add a comment |
$begingroup$
The differential being an infinitely tiny amount is a misconception. In modern setting, the differential is a linear map, specifically the linear part of the variation of a function when you vary the variable. In other words, when a function is differentiable there is a constant $c$ such that
$$f(x+dx)=f(x)+ccdot dx+o(dx)$$ ($o(dx)$ is a sublinear term as a function of $dx$)
and by definition $$df(x)=ccdot dx.$$
The constant $c$ is well know to be the derivative of $f$, and $dx$ can be an arbitrarily large increment.
Actually, differentials have little to do with limits.
On the picture, the function is in blue. Its differential (at $x=1$) in magenta and the remainder in black.
$endgroup$
The differential being an infinitely tiny amount is a misconception. In modern setting, the differential is a linear map, specifically the linear part of the variation of a function when you vary the variable. In other words, when a function is differentiable there is a constant $c$ such that
$$f(x+dx)=f(x)+ccdot dx+o(dx)$$ ($o(dx)$ is a sublinear term as a function of $dx$)
and by definition $$df(x)=ccdot dx.$$
The constant $c$ is well know to be the derivative of $f$, and $dx$ can be an arbitrarily large increment.
Actually, differentials have little to do with limits.
On the picture, the function is in blue. Its differential (at $x=1$) in magenta and the remainder in black.
edited Dec 16 '18 at 17:25
answered Dec 16 '18 at 17:13
Yves DaoustYves Daoust
130k676227
130k676227
add a comment |
add a comment |
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$dx$ is the difference between 2 values of $x$, your case is obviously unfit for such use
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– Makina
Dec 16 '18 at 15:58
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Then is $lim_{Delta xto 0} frac{2Delta x + 32(Delta x)^2}{e^{Delta x}}$ equivalent to $frac{2.dx + 32.dx^2}{e^{dx}}$ ?
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– s89ne
Dec 16 '18 at 16:00
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No, because the limit you gave either doesn’t exist or evaluates to a real number. What you wrote is an expression that’s neither. This feels similar to how people treat $frac{df}{dx}$. Luckily notation and theory work out to where you can treat it like a fraction in same cases without running into problems, but it’s still not correct.
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– Will Fisher
Dec 16 '18 at 16:01
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Depending on the context, $textrm{d}x$ on its own has different meanings. It can be a measure or a differential form (off the top of my head). I know no context where it would stand for an infinitely tiny amount of $x$. The usual notation for a very small amount is just $varepsilon$, but it is usually not infinitesimal. But even if $textrm{d}x$ was an actual infinitesimal, the two expressions would not be the same (not before applying standard part to the right hand side).
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– tomasz
Dec 16 '18 at 16:02
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What $dx$ precisely is seems confusing. Is there a rigorous mathematical definition of it, or should it always be understood intuitively depending on the context?
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– s89ne
Dec 16 '18 at 16:11