anonymous
  • anonymous
Find the equation for the line tangent to the curve at the given point?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
y=x^3+1/2x @ x =1
anonymous
  • anonymous
My 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
  • anonymous
* dy/dx = x-2/x^2

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anonymous
  • anonymous
yea
anonymous
  • anonymous
it's not really difficult lol
anonymous
  • anonymous
No, it's not! It essentially need to be precisely though !
anonymous
  • anonymous
f'(x) = ( 2x³ -1 ) / 2x²
anonymous
  • anonymous
well the numerator is not in parenthesis, does that make a difference?
anonymous
  • anonymous
and even if you take a derivative for f(x) it wouldn't be 2x^3-1 right? it would be just 2x^3?
anonymous
  • anonymous
You need to use the equation editor if you don't know how to write properly!
anonymous
  • anonymous
oh wow. oooooooooook
anonymous
  • anonymous
well either way, the derivative you have is wrong
anonymous
  • anonymous
@zepdrix can you please help me? :)
anonymous
  • anonymous
@swin2013 I'm not wrong at this trivial level :P
anonymous
  • anonymous
the derivative of x^3-1 is not 2x^3-1...
anonymous
  • anonymous
Of course not!
anonymous
  • anonymous
Ok then
anonymous
  • anonymous
It proves that you have no clue about the formula !!!
anonymous
  • anonymous
ooook. well i just quoted from what you said f'(x) was so yea
anonymous
  • anonymous
I came here to search for help @Chlorophyll, so i can just wait on the next person to help me
anonymous
  • anonymous
It's a FRACTION!
anonymous
  • anonymous
You're poor soul !!!
anonymous
  • anonymous
ok lmao
zepdrix
  • zepdrix
Hey were you able to figure this one out ok? c:
anonymous
  • anonymous
A little! sorry about the little mess earlier lolllll dy/dx = x - 1/2x^2
anonymous
  • anonymous
@zepdrix :) I'm not sure if you can still respond to this post since it's closed hahaha
zepdrix
  • zepdrix
Yes you can, I still get pop-ups. It just doesn't show up on the main list.
zepdrix
  • zepdrix
\[\large y=\frac{x^3+1}{2x}\qquad \rightarrow \qquad y=\frac{1}{2}x^2+\frac{1}{2}x^{-2}\] \[\large y'=x-x^{-3}\]I think this is what you should be getting.
zepdrix
  • zepdrix
Remember, when you subtract one from a NEGATIVE number, it actually gets LARGER in the negative direction.
zepdrix
  • zepdrix
\[\large \left(x^{-2}\right)\quad =\quad -2x^{-2-1}\quad = \quad -2x^{-3}\]Confused on this part?
anonymous
  • anonymous
I 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
anonymous
  • anonymous
*1/2x^-1
zepdrix
  • zepdrix
Oh i wrote it wrong? :C woops...\[\large y=\frac{ 1 }{ 2 } x ^{2} + \frac{ 1 }{ 2 }x\] \[\large y'=x-x^{-2}\]
anonymous
  • anonymous
Yeeeee Yeah I got it :)) Thankss!
anonymous
  • anonymous
well idk does that look right? y' = x - 1/x^2?
zepdrix
  • zepdrix
Yes. Understand how to proceed from here or no?
anonymous
  • anonymous
Yes I do, thank youss!!!

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