Question

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

Answer 1

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

Answer 2

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....

Answer 3

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

Answer 4

yea

Answer 5

it's not really difficult lol

Answer 6

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

Answer 7

f'(x) = ( 2x³ -1 ) / 2x²

Answer 8

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

Answer 9

and even if you take a derivative for f(x) it wouldn't be 2x^3-1 right? it would be just 2x^3?

Answer 10

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

Answer 11

oh wow. oooooooooook

Answer 12

well either way, the derivative you have is wrong

Answer 13

@zepdrix can you please help me? :)

Answer 14

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

Answer 15

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

Answer 16

Of course not!

Answer 17

Ok then

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

It's a FRACTION!

Answer 22

You're poor soul !!!

Answer 23

ok lmao

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

\[\large y=\frac{x^3+1}{2x}\qquad \rightarrow \qquad y=\frac{1}{2}x^2+\frac{1}{2}x^{-2}\]
\[\large y'=x-x^{-3}\]I think this is what you should be getting.

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

*1/2x^-1

Answer 33

Oh i wrote it wrong? :C woops...\[\large y=\frac{ 1 }{ 2 } x ^{2} + \frac{ 1 }{ 2 }x\]
\[\large y'=x-x^{-2}\]

Answer 34

Yeeeee Yeah I got it :)) Thankss!

Answer 35

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

Answer 36

Yes. Understand how to proceed from here or no?

Answer 37

Yes I do, thank youss!!!