What exactly is the meaning of/why exactly do we write $dy = y'(x) dx$?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite
1












I have only taken basic real analysis, so to me $dfrac dydx$ is just a formal symbol which stands for the limit of the difference quotient. Now, I know that $dy = y'(x) dx$ has some intuitive meaning historically, but if nowdays it's just another representation for the equation $dfrac dydx = y'(x)$ (which does have a formal meaning), why does it even exist in modern mathematics?




real-analysis differential-equations analysis




share|cite|improve this question



















  • I think it's partly because "infinitesimal intuition" seems very clear sometimes. Thinking about tiny changes in $x$ and the corresponding tiny changes in $y$ is a clear way to understand intuitively many ideas in calculus.
    – littleO
    Jul 28 at 3:36











  • $dy=y'dx$ is not an "intuitive meaning". It is the precise relationship between the differential of $y$ and the differential of $x$. The one that you are calling an equation, $fracdydx=y'$, is the one that is saying nothing. It is just two names, from different notations, for the same thing put on both sides of an equal sign.
    – user578878
    Jul 28 at 3:45







  • 1




    This is a question that has been asked many many times. Search in the site. Better instead, pick up Calculus on Manifolds by Spivak. It is a short book. You will get there what is a differential, why $dy=y'dx$, or $dy=fracdydxdx$, or $dy=D_xy dx$, or $dy=dotydx$ (whichever notation you want for the derivative of $y$ with respect to $x$), etc.
    – user578878
    Jul 28 at 3:55







  • 1




    @nextpuzzle I am aware that differentials have to do with something called "differential forms", and I know that the symbol $dfrac dydx$ has some meaning in those terms as well. However, in the context of basic real analysis the definition of $dfrac dydx(t)$ is $lim_x to t dfrac f(x) - (t)x-t$. Is there a corresponding definition in basic real analysis of $dx$?
    – Ovi
    Jul 28 at 4:00










  • The definition of $fracdydx$ is the same that you are saying everywhere.
    – user578878
    Jul 28 at 4:03














up vote
0
down vote

favorite
1












I have only taken basic real analysis, so to me $dfrac dydx$ is just a formal symbol which stands for the limit of the difference quotient. Now, I know that $dy = y'(x) dx$ has some intuitive meaning historically, but if nowdays it's just another representation for the equation $dfrac dydx = y'(x)$ (which does have a formal meaning), why does it even exist in modern mathematics?




real-analysis differential-equations analysis




share|cite|improve this question



















  • I think it's partly because "infinitesimal intuition" seems very clear sometimes. Thinking about tiny changes in $x$ and the corresponding tiny changes in $y$ is a clear way to understand intuitively many ideas in calculus.
    – littleO
    Jul 28 at 3:36











  • $dy=y'dx$ is not an "intuitive meaning". It is the precise relationship between the differential of $y$ and the differential of $x$. The one that you are calling an equation, $fracdydx=y'$, is the one that is saying nothing. It is just two names, from different notations, for the same thing put on both sides of an equal sign.
    – user578878
    Jul 28 at 3:45







  • 1




    This is a question that has been asked many many times. Search in the site. Better instead, pick up Calculus on Manifolds by Spivak. It is a short book. You will get there what is a differential, why $dy=y'dx$, or $dy=fracdydxdx$, or $dy=D_xy dx$, or $dy=dotydx$ (whichever notation you want for the derivative of $y$ with respect to $x$), etc.
    – user578878
    Jul 28 at 3:55







  • 1




    @nextpuzzle I am aware that differentials have to do with something called "differential forms", and I know that the symbol $dfrac dydx$ has some meaning in those terms as well. However, in the context of basic real analysis the definition of $dfrac dydx(t)$ is $lim_x to t dfrac f(x) - (t)x-t$. Is there a corresponding definition in basic real analysis of $dx$?
    – Ovi
    Jul 28 at 4:00










  • The definition of $fracdydx$ is the same that you are saying everywhere.
    – user578878
    Jul 28 at 4:03












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I have only taken basic real analysis, so to me $dfrac dydx$ is just a formal symbol which stands for the limit of the difference quotient. Now, I know that $dy = y'(x) dx$ has some intuitive meaning historically, but if nowdays it's just another representation for the equation $dfrac dydx = y'(x)$ (which does have a formal meaning), why does it even exist in modern mathematics?




real-analysis differential-equations analysis




share|cite|improve this question











I have only taken basic real analysis, so to me $dfrac dydx$ is just a formal symbol which stands for the limit of the difference quotient. Now, I know that $dy = y'(x) dx$ has some intuitive meaning historically, but if nowdays it's just another representation for the equation $dfrac dydx = y'(x)$ (which does have a formal meaning), why does it even exist in modern mathematics?





real-analysis differential-equations analysis





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 28 at 3:32









Ovi

11.3k935105




11.3k935105











  • I think it's partly because "infinitesimal intuition" seems very clear sometimes. Thinking about tiny changes in $x$ and the corresponding tiny changes in $y$ is a clear way to understand intuitively many ideas in calculus.
    – littleO
    Jul 28 at 3:36











  • $dy=y'dx$ is not an "intuitive meaning". It is the precise relationship between the differential of $y$ and the differential of $x$. The one that you are calling an equation, $fracdydx=y'$, is the one that is saying nothing. It is just two names, from different notations, for the same thing put on both sides of an equal sign.
    – user578878
    Jul 28 at 3:45







  • 1




    This is a question that has been asked many many times. Search in the site. Better instead, pick up Calculus on Manifolds by Spivak. It is a short book. You will get there what is a differential, why $dy=y'dx$, or $dy=fracdydxdx$, or $dy=D_xy dx$, or $dy=dotydx$ (whichever notation you want for the derivative of $y$ with respect to $x$), etc.
    – user578878
    Jul 28 at 3:55







  • 1




    @nextpuzzle I am aware that differentials have to do with something called "differential forms", and I know that the symbol $dfrac dydx$ has some meaning in those terms as well. However, in the context of basic real analysis the definition of $dfrac dydx(t)$ is $lim_x to t dfrac f(x) - (t)x-t$. Is there a corresponding definition in basic real analysis of $dx$?
    – Ovi
    Jul 28 at 4:00










  • The definition of $fracdydx$ is the same that you are saying everywhere.
    – user578878
    Jul 28 at 4:03
















  • I think it's partly because "infinitesimal intuition" seems very clear sometimes. Thinking about tiny changes in $x$ and the corresponding tiny changes in $y$ is a clear way to understand intuitively many ideas in calculus.
    – littleO
    Jul 28 at 3:36











  • $dy=y'dx$ is not an "intuitive meaning". It is the precise relationship between the differential of $y$ and the differential of $x$. The one that you are calling an equation, $fracdydx=y'$, is the one that is saying nothing. It is just two names, from different notations, for the same thing put on both sides of an equal sign.
    – user578878
    Jul 28 at 3:45







  • 1




    This is a question that has been asked many many times. Search in the site. Better instead, pick up Calculus on Manifolds by Spivak. It is a short book. You will get there what is a differential, why $dy=y'dx$, or $dy=fracdydxdx$, or $dy=D_xy dx$, or $dy=dotydx$ (whichever notation you want for the derivative of $y$ with respect to $x$), etc.
    – user578878
    Jul 28 at 3:55







  • 1




    @nextpuzzle I am aware that differentials have to do with something called "differential forms", and I know that the symbol $dfrac dydx$ has some meaning in those terms as well. However, in the context of basic real analysis the definition of $dfrac dydx(t)$ is $lim_x to t dfrac f(x) - (t)x-t$. Is there a corresponding definition in basic real analysis of $dx$?
    – Ovi
    Jul 28 at 4:00










  • The definition of $fracdydx$ is the same that you are saying everywhere.
    – user578878
    Jul 28 at 4:03















I think it's partly because "infinitesimal intuition" seems very clear sometimes. Thinking about tiny changes in $x$ and the corresponding tiny changes in $y$ is a clear way to understand intuitively many ideas in calculus.
– littleO
Jul 28 at 3:36





I think it's partly because "infinitesimal intuition" seems very clear sometimes. Thinking about tiny changes in $x$ and the corresponding tiny changes in $y$ is a clear way to understand intuitively many ideas in calculus.
– littleO
Jul 28 at 3:36













$dy=y'dx$ is not an "intuitive meaning". It is the precise relationship between the differential of $y$ and the differential of $x$. The one that you are calling an equation, $fracdydx=y'$, is the one that is saying nothing. It is just two names, from different notations, for the same thing put on both sides of an equal sign.
– user578878
Jul 28 at 3:45





$dy=y'dx$ is not an "intuitive meaning". It is the precise relationship between the differential of $y$ and the differential of $x$. The one that you are calling an equation, $fracdydx=y'$, is the one that is saying nothing. It is just two names, from different notations, for the same thing put on both sides of an equal sign.
– user578878
Jul 28 at 3:45





1




1




This is a question that has been asked many many times. Search in the site. Better instead, pick up Calculus on Manifolds by Spivak. It is a short book. You will get there what is a differential, why $dy=y'dx$, or $dy=fracdydxdx$, or $dy=D_xy dx$, or $dy=dotydx$ (whichever notation you want for the derivative of $y$ with respect to $x$), etc.
– user578878
Jul 28 at 3:55





This is a question that has been asked many many times. Search in the site. Better instead, pick up Calculus on Manifolds by Spivak. It is a short book. You will get there what is a differential, why $dy=y'dx$, or $dy=fracdydxdx$, or $dy=D_xy dx$, or $dy=dotydx$ (whichever notation you want for the derivative of $y$ with respect to $x$), etc.
– user578878
Jul 28 at 3:55





1




1




@nextpuzzle I am aware that differentials have to do with something called "differential forms", and I know that the symbol $dfrac dydx$ has some meaning in those terms as well. However, in the context of basic real analysis the definition of $dfrac dydx(t)$ is $lim_x to t dfrac f(x) - (t)x-t$. Is there a corresponding definition in basic real analysis of $dx$?
– Ovi
Jul 28 at 4:00




@nextpuzzle I am aware that differentials have to do with something called "differential forms", and I know that the symbol $dfrac dydx$ has some meaning in those terms as well. However, in the context of basic real analysis the definition of $dfrac dydx(t)$ is $lim_x to t dfrac f(x) - (t)x-t$. Is there a corresponding definition in basic real analysis of $dx$?
– Ovi
Jul 28 at 4:00












The definition of $fracdydx$ is the same that you are saying everywhere.
– user578878
Jul 28 at 4:03




The definition of $fracdydx$ is the same that you are saying everywhere.
– user578878
Jul 28 at 4:03















active

oldest

votes











Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864963%2fwhat-exactly-is-the-meaning-of-why-exactly-do-we-write-dy-yx-dx%23new-answer', 'question_page');

);

Post as a guest






















active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes










 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864963%2fwhat-exactly-is-the-meaning-of-why-exactly-do-we-write-dy-yx-dx%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
Read more