Still confused about the dx notation

[jaydnul]

So dx means an infinitesimally small change in x. Why is the slope written \frac{dy}{dx} instead of \frac{f(x)}{dx} since you are only making the x component infinitely small?

When you take the integral you do ∫f(x)dx not ∫dy*dx

 

[SteamKing]

f(x) is the value of function f at a particular value x. The slope of a line is defined as the rise, which is the change in f, divided by the run, which is the change in x, or dx in the limit.

Remember, teh derivative is defined as limit as h --> 0 of [f(x+h) - f(x)]/h. As h approaches 0 it becomes dxin differential notation, while [f(x+h)-f(x)] becomes dy, the change in the value of the function.

Conversely, when you integrate a function f, you are adding up the area of a series of strips which are f(x) tall by dx wide. The integral of dy*dx makes no sense and it appears you are confusing the concept of the derivative with the concept of the integral, and vice versa.

 

[johnqwertyful]

In real analysis (calculus) there is no such thing as an infinitesimal, period. dx is NOT an "infinitesimal", as a matter of fact dx by itself doesn't make sense. Anything you see in calculus where you treat dx as something to be moved around, you are seeing an abuse of notation.

 

[WannabeNewton]

In real analysis (calculus) there is no such thing as an infinitesimal, period. dx is NOT an "infinitesimal", as a matter of fact dx by itself doesn't make sense.

This is quite incorrect on several accounts:
http://en.wikipedia.org/wiki/Non-standard_analysis
http://en.wikipedia.org/wiki/Differential_form

 

[Mandelbroth]

In real analysis (calculus) there is no such thing as an infinitesimal, period. dx is NOT an "infinitesimal", as a matter of fact dx by itself doesn't make sense. Anything you see in calculus where you treat dx as something to be moved around, you are seeing an abuse of notation.

OP, please disregard this post.

There are two arguably "best" rigorous notions of the differential. The first, as an infinitesimal, is more intuitive (to some), but less useful. The second is, in my personal opinion, far superior and intuitive, but much more mathematically complicated.

Additionally, I find the infinitesimal approach to be less motivated than worth using. It's really a matter of preference for you at this point.

 

[johnqwertyful]

This is quite incorrect on several accounts:
http://en.wikipedia.org/wiki/Non-standard_analysis
http://en.wikipedia.org/wiki/Differential_form

That's why I said in real analysis/calculus. This is not true always, but within a standard analysis, what he is doing now, there are no infinitesimals.

Edit, "abuse of notation" is not necessarily a bad thing. In my physics classes, I'm not going to try to do things 100% rigorously and I abuse notation all the time.

 

[jaydnul]

OP, please disregard this post.

There are two arguably "best" rigorous notions of the differential. The first, as an infinitesimal, is more intuitive (to some), but less useful. The second is, in my personal opinion, far superior and intuitive, but much more mathematically complicated.

Additionally, I find the infinitesimal approach to be less motivated than worth using. It's really a matter of preference for you at this point.

Well don't keep me in suspense! :)

What's the second notion that is far superior and intuitive?

 

[Mandelbroth]

Well don't keep me in suspense! :)

What's the second notion that is far superior and intuitive?

...is too complicated for us to really go into too much depth with you now.

For differentiable manifolds $M$ and $N$ and a differentiable map $f:M\to N$, the differential $df$ defines at each point $p\in M$ a linear mapping $df_p:T_pM\to T_{f(p)}N$.

Differential forms generalize the concept of a differential of a scalar function.

It only gets more complicated from there. One of the best motivations for it is that we can put most integration theorems from analysis into the form $$\int\limits_{\Omega}d\alpha=\int\limits_{\partial\Omega}\alpha.$$

To really understand these things, it's best to leave them alone until college. If you want an example, suppose we have $f:\mathbb{R}\to\mathbb{R}$. Then, we have, for $x,v\in\mathbb{R}$, $df_x(v)=f'(x)v$.