[ SECRET POST #2317 ]

[case]

⌈ Secret Post #2317 ⌋

Warning: Some secrets are NOT worksafe and may contain SPOILERS.

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Notes:

Secrets Left to Post: 02 pages, 048 secrets from Secret Submission Post #331.
Secrets Not Posted: [ 0 - broken links ], [ 1 - not!secrets ], [ 0 - not!fandom ], [ 0 - too big ], [ 0 - repeat ].
Current Secret Submissions Post: here.
Suggestions, comments, and concerns should go here.

Comments

Re: I need help

[(Anonymous)]

This might be a stupid question, but how did you go from du/dx=2 to dx=du/2?

Like, how did you decide that it was ^du/_dx?

Re: I need help

[(Anonymous)]

Integration by parts and u-substitution are the times you're allowed to treat du/dx the derivative as a fraction. Most of the time, you're not supposed to do that, but in this case you just go ahead and pretend it is. :)

Does that help?

Re: I need help

[(Anonymous)]

So there's no reason, it just is? And it's like that?

THAT MAKES SO MUCH MORE SENSE OMG.

I THOUGHT the teachers were just assuming everyone knew why and not understanding when I was asking (I can't words sometimes), but that's just one of the rules of integral calculus?

Re: I need help

[dreemyweird]

Not really; it's because they're rates of change. Hence here they behave as parts of a fraction. Say, if u changes twice as fast as x, we can safely assume the relationship above is true.

Re: I need help

[(Anonymous)]

Oh. Okay, that makes sense!

Re: I need help

[dreemyweird]

Because that's the way it works when one differentiates a function with two unknowns. You kind of need to acknowledge the fact that you're differentiating the function with respect to x, not to u. So, when you have u, you can't just write "du", you need to add this /dx. Technically speaking, this is not a fraction, but it does behave as one.
Nope, I'm glad if I'm of any help)

Re: I need help

[(Anonymous)]

I feel really stupid now, because I didn't get that part, but now I've gotten it explained to me, and I feel so much better omg.

Thank you!

Re: I need help

[dreemyweird]

You're welcome!