When is it legal to use $dx$?












3












$begingroup$


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










share|cite|improve this question









$endgroup$












  • $begingroup$
    $dx$ is the difference between 2 values of $x$, your case is obviously unfit for such use
    $endgroup$
    – Makina
    Dec 16 '18 at 15:58












  • $begingroup$
    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


















3












$begingroup$


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










share|cite|improve this question









$endgroup$












  • $begingroup$
    $dx$ is the difference between 2 values of $x$, your case is obviously unfit for such use
    $endgroup$
    – Makina
    Dec 16 '18 at 15:58












  • $begingroup$
    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
















3












3








3


2



$begingroup$


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










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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




















  • $begingroup$
    $dx$ 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












2 Answers
2






active

oldest

votes


















2












$begingroup$

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






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    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.



    enter image description here






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      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






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        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






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          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






          share|cite|improve this answer











          $endgroup$



          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







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 12:13

























          answered Dec 16 '18 at 17:06









          HenryHenry

          101k481168




          101k481168























              1












              $begingroup$

              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.



              enter image description here






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                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.



                enter image description here






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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.



                  enter image description here






                  share|cite|improve this answer











                  $endgroup$



                  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.



                  enter image description here







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 16 '18 at 17:25

























                  answered Dec 16 '18 at 17:13









                  Yves DaoustYves Daoust

                  130k676227




                  130k676227






























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