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anonymous
 4 years ago
Find the equation for the line tangent to the curve at the given point?
anonymous
 4 years ago
Find the equation for the line tangent to the curve at the given point?

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anonymous
 4 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....

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

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

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

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

anonymous
 4 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?

anonymous
 4 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!

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

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

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

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

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

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

anonymous
 4 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

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

anonymous
 4 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

anonymous
 4 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
 4 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
 4 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
 4 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
 4 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?

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1Oh i wrote it wrong? :C woops...\[\large y=\frac{ 1 }{ 2 } x ^{2} + \frac{ 1 }{ 2 }x\] \[\large y'=xx^{2}\]

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes I do, thank youss!!!
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