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y=x^3+1/2x @ x =1

* dy/dx = x-2/x^2

yea

it's not really difficult lol

No, it's not! It essentially need to be precisely though !

f'(x) = ( 2x³ -1 ) / 2x²

well the numerator is not in parenthesis, does that make a difference?

and even if you take a derivative for f(x) it wouldn't be 2x^3-1 right? it would be just 2x^3?

You need to use the equation editor if you don't know how to write properly!

oh wow. oooooooooook

well either way, the derivative you have is wrong

the derivative of x^3-1 is not 2x^3-1...

Of course not!

Ok then

It proves that you have no clue about the formula !!!

ooook. well i just quoted from what you said f'(x) was so yea

I came here to search for help @Chlorophyll, so i can just wait on the next person to help me

It's a FRACTION!

You're poor soul !!!

ok lmao

Hey were you able to figure this one out ok? c:

A little! sorry about the little mess earlier lolllll dy/dx = x - 1/2x^2

Yes you can, I still get pop-ups. It just doesn't show up on the main list.

\[\large \left(x^{-2}\right)\quad =\quad -2x^{-2-1}\quad = \quad -2x^{-3}\]Confused on this part?

*1/2x^-1

Yeeeee Yeah I got it :)) Thankss!

well idk does that look right? y' = x - 1/x^2?

Yes. Understand how to proceed from here or no?

Yes I do, thank youss!!!