## 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 Group Title

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

2. swin2013 Group Title

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 Group Title

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

4. swin2013 Group Title

yea

5. swin2013 Group Title

it's not really difficult lol

6. Chlorophyll Group Title

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

7. Chlorophyll Group Title

f'(x) = ( 2x³ -1 ) / 2x²

8. swin2013 Group Title

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

9. swin2013 Group Title

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 Group Title

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

11. swin2013 Group Title

oh wow. oooooooooook

12. swin2013 Group Title

well either way, the derivative you have is wrong

13. swin2013 Group Title

14. Chlorophyll Group Title

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

15. swin2013 Group Title

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

16. Chlorophyll Group Title

Of course not!

17. swin2013 Group Title

Ok then

18. Chlorophyll Group Title

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

19. swin2013 Group Title

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

20. swin2013 Group Title

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

21. Chlorophyll Group Title

It's a FRACTION!

22. Chlorophyll Group Title

You're poor soul !!!

23. swin2013 Group Title

ok lmao

24. zepdrix Group Title

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

25. swin2013 Group Title

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

26. swin2013 Group Title

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

27. zepdrix Group Title

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

28. zepdrix Group Title

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

29. zepdrix Group Title

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

30. zepdrix Group Title

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

31. swin2013 Group Title

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 Group Title

*1/2x^-1

33. zepdrix Group Title

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 Group Title

Yeeeee Yeah I got it :)) Thankss!

35. swin2013 Group Title

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

36. zepdrix Group Title

Yes. Understand how to proceed from here or no?

37. swin2013 Group Title

Yes I do, thank youss!!!