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