A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0My work: y = x^3/2x + 1/2x ====> i split the fractions up dy/dx = x = 2/x^2 ====> this is where I messed up I think....

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0it's not really difficult lol

Chlorophyll
 2 years ago
Best ResponseYou've already chosen the best response.0No, it's not! It essentially need to be precisely though !

Chlorophyll
 2 years ago
Best ResponseYou've already chosen the best response.0f'(x) = ( 2x³ 1 ) / 2x²

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0well the numerator is not in parenthesis, does that make a difference?

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0and even if you take a derivative for f(x) it wouldn't be 2x^31 right? it would be just 2x^3?

Chlorophyll
 2 years ago
Best ResponseYou've already chosen the best response.0You need to use the equation editor if you don't know how to write properly!

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0well either way, the derivative you have is wrong

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0@zepdrix can you please help me? :)

Chlorophyll
 2 years ago
Best ResponseYou've already chosen the best response.0@swin2013 I'm not wrong at this trivial level :P

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0the derivative of x^31 is not 2x^31...

Chlorophyll
 2 years ago
Best ResponseYou've already chosen the best response.0It proves that you have no clue about the formula !!!

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0ooook. well i just quoted from what you said f'(x) was so yea

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0I came here to search for help @Chlorophyll, so i can just wait on the next person to help me

Chlorophyll
 2 years ago
Best ResponseYou've already chosen the best response.0You're poor soul !!!

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Hey were you able to figure this one out ok? c:

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0A little! sorry about the little mess earlier lolllll dy/dx = x  1/2x^2

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0@zepdrix :) I'm not sure if you can still respond to this post since it's closed hahaha

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Yes you can, I still get popups. It just doesn't show up on the main list.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large y=\frac{x^3+1}{2x}\qquad \rightarrow \qquad y=\frac{1}{2}x^2+\frac{1}{2}x^{2}\] \[\large y'=xx^{3}\]I think this is what you should be getting.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Remember, when you subtract one from a NEGATIVE number, it actually gets LARGER in the negative direction.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large \left(x^{2}\right)\quad =\quad 2x^{21}\quad = \quad 2x^{3}\]Confused on this part?

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0I get how you got 1/2 x^2 but not 1/2 x^2 i split up the fractions to kinda help me see it :/ so I got \[\frac{ x ^{3} }{ 2x } + \frac{ 1 }{ 2x }\] which is \[\frac{ 1 }{ 2 } x ^{2} + \frac{ 1 }{ 2 }x? \] :( sorry lolll

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Oh i wrote it wrong? :C woops...\[\large y=\frac{ 1 }{ 2 } x ^{2} + \frac{ 1 }{ 2 }x\] \[\large y'=xx^{2}\]

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0Yeeeee Yeah I got it :)) Thankss!

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0well idk does that look right? y' = x  1/x^2?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Yes. Understand how to proceed from here or no?

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0Yes I do, thank youss!!!
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.