What does it mean to move a differential?

[helixkirby]

Not a homework question, I'm just curious, I know this has been asked a few times, but what exactly is happening when you move dx over to the other side of dy/dx=f'(x), is it like the point slope form ∆y=m∆x, or is it applying the differential to both sides, I've always been told a rigorous proof was necessary to show you could "multiply by dx", could someone possibly show me a proof of it, or even just explain why it works?

 

[Simon Bridge]

What's happening is that you are abusing the notation as a kind of shorthand.
Read "dy" as "a small change in y" and "dx" as "a small change in x" and you are just about there ...

Also see:
https://www.physicsforums.com/threads/abuse-of-liebniz-notation.318484/

 

[Delta2]

The way i ve been taught Leibniz's notation is by the definition of the differential of a function.

The differential df of a function f, is the function $df(\tau,x)=f'(x)\tau$ where $\tau$ is a variable independent of the variable x. However where things "go messy" with the notation is on the next step:

Consider the differential di of the identity function i(x)=x. It will be $di(\tau,x)=i'(x)\tau=\tau$ so if we allow the notation $di(\tau,x)=dx(\tau,x)$ (because i(x)=x) we end up with $dx(\tau,x)=\tau$ or abbreviated $dx=\tau$. It is important to notice that $dx$ simply stands for the differential of the identity function.

So going back to $df(\tau,x)=f'(x)\tau$ and using the fact that $dx=\tau$ we end up with $df(\tau,x)=f'(x)dx(\tau,x)$ or in compact notation $df=f'dx$. Dont forget that on this last equation on the RHS we have an ordinary multiplication of two functions (the function f' and the function dx) so we can write it as $\frac{df}{dx}=f'$.

Just to clear a thing, dx is independent of x, it is essentially the independent variable $\tau$.

 

[helixkirby]

Could you possibly show it in single variable, I just took the calc BC test and don't really know much about multi-variable calculus yet.

 

[Delta2]

Well it is essentially one variable calculus, $\tau$ is some sort of "artificial" variable to do the trick, we differentiate only with respect to x.

 

[Simon Bridge]

What you are used to thinking of as single variable calculus is "really" a shorthand way of doing the thing Delta² did in post #3.
You are not normally taught that stuff at your level because you are not used to thinking like that yet, so you get taught the shortcuts and rules without rigor.

 

[helixkirby]

Well, hopefully I get a 4 or 5 on my ap test, I actually did it instead of the ab test I was supposed to take, I really struggled with series though. Hopefully I can skip calc 2 in college. Multivariable sounds fun.