Why can dy/dx be treated as a fraction only in certain contexts?

>being this much of a brainlet

how would you do it without dx=du part?

stupid person here, why is he assuming dx/du = 1? Am i missing something here?

kek

Do you know how to differentiate a linear function?

sort of, i know that you need to break down the equations into the basic forms of integrals to solve them. Haven't messed around with calculus for a while now (economics major lel)

[eqn]\mathcal{I}(x)=\int f(\varphi(x))\varphi'(x)\,\mathrm{d}x[/eqn] where [math]f(x)=(x-1)/\sqrt{x}[/math] and [math]\varphi(x)=x+1[/math]. By the chain rule
[eqn]\mathcal{I}(x)=\int \frac{u-1}{\sqrt{u}}\,\mathrm{d}u=\cdots[/eqn]

Derivate u - 1. You'll get du. Then just divide both sides by du and you'll get dx/du = 1

By doing the exact same thing and just not writing down that step. He's being a pedantic ass, and will get over it after a couple years in undergrad.

your first line is completely generic and would be written over and over again with different functions f and phi. It's not really necessary to write.

it just works. For example: take
[eqn]\frac{dy}{dx}=2[/eqn]
now cancel the "d"
[eqn]\frac{y}{x}=2[/eqn]
and solve for y:
[eqn]y=2x[/eqn]
so it's true. taking the derivative of y=2x gives you 2.

...

you may have done something

This is the most retarded thread i have seen here in a long time. So by the logic I see here d is just some variable. Pro-tip its not!

These silly physics students. I'm always dying a little inside if i see people calculating with differentials as with numbers, even through it works fine in some context if everything is nice enough.

lmfao

What's wrong with saying [math]\frac {\mathrm{d}x} {\mathrm{d}u} = 1 \implies \mathrm{d}x=\mathrm{d}u[/math]? Can you give a case where this isn't true?

Top kek

thanks breh

>pi cannot be written as a fraction
are you retarded?
pi/1? 2pi/2?

dx and du are just notation without meaning (unless you're working with differential forms, which you aren't)

pi cannot be written as a fraction of integers

This is definitely cancer

this

22/7
Did you pass 7th grade maths?

[eqn]\frac{22}{7} \gt \pi[/eqn]

sounds like you never went further than 7th grade

>just notation without meaning

this is an oxymoron, you oxymoron. dx means a tiny change in the x direction, du means a tiny change in the u direction. dx = du means that a tiny change in x leads to a proportional tiny change in u.

Is this place full of highschoolers or what?

You are a retard, it's the same as x'(u)

>[math]\frac{dy}{dx} = \frac{y}{x}[/math]

If only this was true, calculus would be a lot simpler.

/thread

And who said dy or dx are integers?

Actual mathematicians don't care, and they use the same kind of shortcuts when they know they'll work. It's just first year undergrads who are overly obsessed with rigor.

>t. physist

Good thread.

>implying mathematicians ever do integrals

Lol

it works bc of the chain rule

I mean, they do come up. Even in shit like number theory.

it literally is a fraction though

LMAFO

I literally write the exact same way as left side

with the let statement and therefores

I even do the dx/du=1 shit

shit ive realized i actually never understood this. How come its sometimes valid to separate it and other times not? What makes it rigorous?

suppose you have some integral [eqn] I=\int f(g(x))g'(x)\,dx\,. [/eqn] then by the chain rule
[eqn] I=\int f(u)\,du\quad (u=g(x))\,, [/eqn]
which is integration by substitution.
by the other way you have something like: let [math] u=g(x) [/math]. then [math] \frac{du}{dx}=g'(x) [/math] so [math] dx=\frac{du}{g'(x)} [/math]. therefore [eqn] I=\int f(u)g'(x)\,\frac{du}{g'(x)}=\int f(u)\,du\,, [/eqn]
which is the same result.

idk about differential equations as it varies in each case, but the "proof" will be similar.

i should add that that doesnt make the method rigorous, but it shows why it works