## anonymous 3 years ago Find the equation for the line tangent to the curve at the given point?

1. anonymous

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

2. 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....

3. anonymous

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

4. anonymous

yea

5. anonymous

it's not really difficult lol

6. anonymous

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

7. anonymous

f'(x) = ( 2x³ -1 ) / 2x²

8. anonymous

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

9. 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?

10. anonymous

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

11. anonymous

oh wow. oooooooooook

12. anonymous

well either way, the derivative you have is wrong

13. anonymous

14. anonymous

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

15. anonymous

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

16. anonymous

Of course not!

17. anonymous

Ok then

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

It's a FRACTION!

22. anonymous

You're poor soul !!!

23. anonymous

ok lmao

24. zepdrix

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

25. anonymous

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

26. anonymous

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

27. zepdrix

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

28. 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.

29. zepdrix

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

30. zepdrix

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

31. 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

32. anonymous

*1/2x^-1

33. zepdrix

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

34. anonymous

Yeeeee Yeah I got it :)) Thankss!

35. anonymous

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

36. zepdrix

Yes. Understand how to proceed from here or no?

37. anonymous

Yes I do, thank youss!!!