What Is the Rigorous Definition of dy in Leibniz Notation?

[kotreny]

dy = lim \Deltax-->0 (f(x+\Deltax) - f(x))

dx = lim \Deltax-->0 (\Deltax)

Therefore dy/dx is f'(x) if f(x) = y

Is all of this true? I'm tired of integrating with variable substitution and not knowing what du by itself really means. People are always saying that Leibniz notation doesn't literally represent a fraction, and the seemingly algebraic manipulation is more complicated than it looks. But can't anyone say what exactly is going on? Maybe give me the rigorous definition of "dy" if I was wrong. Thanks.

Oh, and if the above is correct, then can we say that dy/dx is a fraction of limits?

 

[g_edgar]

dy = lim \Deltax-->0 (f(x+\Deltax) - f(x))

dx = lim \Deltax-->0 (\Deltax)

Both of those limits are zero.

Therefore dy/dx is f'(x) if f(x) = y

dy/dx is 0/0 in your guess, not useful by itself

Is all of this true?

No

I'm tired of integrating with variable substitution and not knowing what du by itself really means.

Some calculus textbooks invent meanings for it. But the meaning of du by itself used in modern mathematics has to wait for differential topology, a more advanced subject.

People are always saying that Leibniz notation doesn't literally represent a fraction, and the seemingly algebraic manipulation is more complicated than it looks. But can't anyone say what exactly is going on? Maybe give me the rigorous definition of "dy" if I was wrong. Thanks.

Oh, and if the above is correct, then can we say that dy/dx is a fraction of limits?

 

[lurflurf]

The meaning of dy depends on context. In most introductory calculus books dy would be defined by
dy=f'(x)dx
dy is a linearization of Δy
with this definition it is correct to regard dy/dx as a fraction with
dy/dx=f'(x)dx/dx=f'(x)

 

[kotreny]

Thanks for the help guys; I've been throwing my limit properties around too carelessly! Well, live and learn...

Differential Topology huh? I guess I'll stick with visualizing infinitesimals until then, like Leibniz, but always remembering that it's not the final word. Hopefully that won't lead me far astray.

The meaning of dy depends on context. In most introductory calculus books dy would be defined by
dy=f'(x)dx
dy is a linearization of Δy
with this definition it is correct to regard dy/dx as a fraction with
dy/dx=f'(x)dx/dx=f'(x)

What's dx then? Seems circular to me. Probably everything will be revealed when I have a bare-bones definition of differentials (maybe using differential topology.)


By the way, ever heard the Simpsons joke, where the derivative of r³/3 is r²*dr, or "r*dr*r" (hardy-har-har)? This must be wrong, because that's not the derivative, it's the differential of r³/3 or something, right? I used to think this joke was pretty funny, but now I hate it, because it confused me a little.

Or am I confused now?

 

[lurflurf]

It is not circular, but it does not allow one to do the wacky stuff like proving the chain rule by cancellation. To do the wacky stuff you need nonstandard analysis which requires a different foundation than standard analysis including logic and ultrafilters. You may be disappointed to know that the differential geometry definition of differential is not so different from the one given in elementaryy calculus. That is true infinitesimals are not used, but such definitions are closely connected to what the differential is, a linearization of the difference. There is also an algebraic definition using dual numbers.

The idea of a differential defined in elementary calculus as
dy=f'(x)dx
is we want the linear function closest to f near x
y=mx+b
we see that
m=f'(x)
b=f(x)-xf'(x)
we chose new coordinates with origin (x,f(x)) so the equation is
dy=f(x)dx
we now have a function that looks everywhere how f looks near (x,f(x))
we also sometimes use it to approximate the difference
dy~Δy
f'(x)dx~f(x+dx)-f(x)

 

[kotreny]

Having a strong, sturdy understanding of differentials must be key to mastering calculus, so I'm not going to stop til I get there!

The idea of a differential defined in elementary calculus as
dy=f'(x)dx
is we want the linear function closest to f near x
y=mx+b
we see that
m=f'(x)
b=f(x)-xf'(x)
we chose new coordinates with origin (x,f(x)) so the equation is
dy=f(x)dx
we now have a function that looks everywhere how f looks near (x,f(x))
we also sometimes use it to approximate the difference
dy~Δy
f'(x)dx~f(x+dx)-f(x)

Let me get this straight.

Take a function and a line tangent to it at x, and let a change in y be "Δy" and a change in y on the tangent line be "Δy_lin"=f'(x)Δx

and Δy~Δy_lin

which means you can approximate a change in y linearly with f'(x)Δx

so a differential means that dy=Δy=Δy_lin=f'(x)dx.
A differential is so infinitesimally small that it makes the linear approximation an "equality".

Is that it?

P.S. How do you write "Δ"? (I copy and pasted that)

 

[lurflurf]

A better way to think about it is that the differential picks oup an approximation. Once it has done the picking we need not and perhaps should not regard the differential as being small.
example
d(sin(x))=cos(x)dx
is saying
let us pick out the fuction that behaves everywhere how [sin(x+dx)-sin(x)] behaves when dx is "small"
ok here it is cos(x)dx
once this is done it is no longer required that dx is small
we might have x=1 dx=1 so
d(sin(x))|x,dx=1=cos(x)dx|x,dx=1=cos(1)~0.54030230586813971740093660744298
even though
Δ(sin(x))|x,dx=1=sin(x+dx)-sin(x)=sin(2)-sin(1)~0.067826442017785188743517544281446
thus the approximation would not be considered good in many application
or x=1 dx=10^100
d(sin(x))|x=1,dx=10^100=cos(x)dx|x=1,dx=10^100=cos(1)*10^100
which is even worse

 

[kotreny]

So basically, the differential starts out as a linear approximation, as I said earlier, only it doesn't end by turning the approximation into an equality?

Take a function and a line tangent to it at x, and let a change in y be "Δy" and a change in y on the tangent line be "Δy_lin"=f'(x)Δx

and Δy~Δy_lin

which means you can approximate a change in y linearly with f'(x)Δx

So far, so good.

so a differential means that dy=Δy=Δy_lin=f'(x)dx.
A differential is so infinitesimally small that it makes the linear approximation an "equality".

Instead, it may conclude with dy=Δy ~ Δy_lin=f'(x)dx

Meaning dy is the actual change in y, which can be estimated linearly with f'(x)dx.
Or is dy=Δy_lin, which then estimates the real change in y?

And is this truly the rigorous, mathematical definition? It sounds more like the physical interpretation, where a differential is simply smaller than can be measured by the instrument. While this interpretation will likely form the basis of my intuition, right now I'm curious about the abstract formality.

we might have x=1 dx=1 so
d(sin(x))|x,dx=1=cos(x)dx|x,dx=1=cos(1)~0.54030230 586813971740093660744298
even though
Δ(sin(x))|x,dx=1=sin(x+dx)-sin(x)=sin(2)-sin(1)~0.067826442017785188743517544281446
thus the approximation would not be considered good in many application

Your reference to applications and arbitrarily assigned values of dx suggested to me the physical sciences.