## swin2013 Group Title Find the equation for the line tangent to the curve at the given point? one year ago one year ago

1. swin2013

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

2. swin2013

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. swin2013

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

4. swin2013

yea

5. swin2013

it's not really difficult lol

6. Chlorophyll

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

7. Chlorophyll

f'(x) = ( 2x³ -1 ) / 2x²

8. swin2013

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

9. swin2013

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. Chlorophyll

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

11. swin2013

oh wow. oooooooooook

12. swin2013

well either way, the derivative you have is wrong

13. swin2013

14. Chlorophyll

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

15. swin2013

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

16. Chlorophyll

Of course not!

17. swin2013

Ok then

18. Chlorophyll

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

19. swin2013

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

20. swin2013

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

21. Chlorophyll

It's a FRACTION!

22. Chlorophyll

You're poor soul !!!

23. swin2013

ok lmao

24. zepdrix

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

25. swin2013

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

26. swin2013

@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. swin2013

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. swin2013

*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. swin2013

Yeeeee Yeah I got it :)) Thankss!

35. swin2013

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

36. zepdrix

Yes. Understand how to proceed from here or no?

37. swin2013

Yes I do, thank youss!!!