Find the equation for the line tangent to the curve at the given point?

- anonymous

Find the equation for the line tangent to the curve at the given point?

- chestercat

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- anonymous

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

- 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

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

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## More answers

- anonymous

yea

- anonymous

it's not really difficult lol

- anonymous

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

- anonymous

f'(x) = ( 2x³ -1 ) / 2x²

- anonymous

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

- 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

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

- anonymous

oh wow. oooooooooook

- anonymous

well either way, the derivative you have is wrong

- anonymous

@zepdrix can you please help me? :)

- anonymous

@swin2013 I'm not wrong at this trivial level :P

- anonymous

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

- anonymous

Of course not!

- anonymous

Ok then

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

It's a FRACTION!

- anonymous

You're poor soul !!!

- anonymous

ok lmao

- zepdrix

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

- anonymous

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

- anonymous

@zepdrix :) I'm not sure if you can still respond to this post since it's closed hahaha

- zepdrix

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

- 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

Remember, when you subtract one from a NEGATIVE number, it actually gets LARGER in the negative direction.

- zepdrix

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

- 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

*1/2x^-1

- 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

Yeeeee Yeah I got it :)) Thankss!

- anonymous

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

- zepdrix

Yes. Understand how to proceed from here or no?

- anonymous

Yes I do, thank youss!!!

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