## anonymous 4 years ago Use implicit differentiation to find an equation of the tangent line to the curve at the curve at the given point: x^2 + y^2 = (2x^2 + 2y^2 - x)^2

1. anonymous

... @ (0,1/2)

2. anonymous

not that i tmatters anyway i'm just a little confused regarding the differentiation here

3. anonymous

alright give me a second to remember lol

4. anonymous

alright i'll upload a picture on what to do

5. anonymous

and an example give me a sec

6. anonymous

cool thanks

7. anonymous

like i got as far as [4x+4y*y']/[4x+4y*y'-1] = 2x^2 + 2y^2 - x

8. anonymous

actualy i can't... i'll just type it out what my book says GUIDELINES FOR IMPLICIT DIFFERENTIATION 1. Differentiat both sides of the equatio with respects to x 2. collect al terms involve dy/dx on the left side of the equation and move all others to right 3.factor dy/dx out of the left side of the equation 4.solve for dy/dx by dividing both sides

9. anonymous

by factor

10. anonymous

right righ tok so and

11. anonymous

so what do i do from what i have am i good so far

12. anonymous

let me get a piece of paper lol

13. anonymous

ama differentiate for u

14. anonymous

thanks k cool coco

15. anonymous

love it

16. anonymous

so is coco checking if your differentiation is right?

17. anonymous

one question. just verifying... for the secon part, everyting is raised to teh 2nd power?

18. anonymous

i'm guessing

19. anonymous

oh, ok

20. anonymous

yes

21. anonymous

ouch

22. anonymous

umm. i suck @ algebra... and i think i am going to mess up if i do this. but, i am still going to go ahead and square all that. can someone then tell me if i was right?

23. anonymous

yeah i got completely something else than he did

24. anonymous

the second is a chain and the first is just regular

25. anonymous

i got 2x+2yy'=2(4x+4yy'-1)

26. anonymous

=2x+2yy'= 8x+8yy'-2

27. anonymous

if you do implicit differentiation, then don't you do (for example, the first part): 2xdx+2ydy... then get all teh dx's and dy's on one side and divide so that dy/dx

28. anonymous

yes

29. anonymous

$2yy'-8yy'=6x-2$

30. anonymous

ahh... i get it.. am so stupid! i was wondering where you got the chain rule from

31. anonymous

yep

32. anonymous

then you can pull out y'

33. anonymous

i think it's$\frac{-6x-2}{6y}$

34. anonymous

the - is outside parenthesis

35. anonymous

$-(6x-2)/6y$

36. anonymous

it's been a while though lol

37. anonymous

blah wait i messed up lol

38. anonymous

let me do this all over agani

39. anonymous

i forgot something

40. anonymous

it doesn't help that i'm also trying to help two people

41. anonymous

ok, sorry guys but i am lost lol. but i wanna know how it's done. so outkast, once you did the chain rule, did you multiply or what?

42. anonymous

my chain rule was messed up lol it should be $2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)$

43. anonymous

yes that's what i got

44. anonymous

then what did you do

45. anonymous

i'd say to combine the right side

46. anonymous

it's going to be ugly hahah

47. anonymous

dive 2xdx + 2ydy with 4xdx+4ydy...?

48. anonymous

divide*

49. anonymous

i think you'd have to combine the right side which is a pain

50. anonymous

Hey, do you know is there anyway i can like mark this question so that i can come back and look @ it 2morrow?

51. anonymous

no seriously?! ouch... calyne, your proff/teacher must be coldhearted to give you such a question.

52. anonymous

um i think it should still be in the top left forner under notifications

53. anonymous

54. anonymous

Oh, ok. thanks!

55. anonymous

$2x+2yy'=16x^3+16y^3y'+16x^2yy'+16y^2x-12x^2-4y^2-8xyy'+2x$ $2yy'=16x^3+16^3y'+16x^2yy'+16y^2x-12x^2-4y^2-8xyy'$

56. anonymous

...

57. anonymous

did you get what me and coco got about my last when you used the chain rule

58. anonymous

what

59. anonymous

you use the chain rule yeah du^2/du * du/dx first thing overall ik

60. anonymous

what you multiply that whole thing by dy/dx too

61. anonymous

so whta did you get after you take dy/dx of both sides

62. anonymous

what did you get damn

63. anonymous

using the chain rule i got 2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)

64. anonymous

sorry i got 2x+2yy' = 2(2x^2+2y^2-x) (4x + 4yy' - 1)

65. anonymous

yeahhhh

66. anonymous

alright so if you combine the terms on the right

67. anonymous

is what i did ... what did you do after

68. anonymous

idk that's gay

69. anonymous

i tried a bunch of different stuff what was your final answer i'll tell you if it's correct nah flutter that the answer is x+1/2. is that what you got. if not we're wasting time.

70. anonymous

not done yet haha

71. anonymous

yeah i have no idea.. even wolfram alpha has omething like my end answer

72. anonymous

unless your teacher is crazy and solved for like y lol

73. anonymous

wait the answer to the differentiation is that or the tangent line

74. anonymous

because if it's the tangent line my implicit differentiation may be correct