## anonymous 4 years ago using implicit differentiation, find dy/dx of (x^2)y+x(y^2) = 6

1. hba

The easiest way to do this is to use implicit differentiation on it as it stands. You differentiate using the normal rules but you treat y as an unknown function of x. This means f(y) differentates to f '(y)*(dy/dx) by the chain rule. xy + x + y^2 = 6 First implicit differentiation gives [1*y + x*1*(dy/dx)] + 1 + 2y*(dy/dx) = 0 .....[from xy].....................[from y^2] You can rearrange this to give dy/dx = (function of x and y) Second implicit differentiation gives dy/dx + [1*(dy/dx) + x*(d2y/dx2)] + [2*(dy/dx)*(dy/dx) + 2y*(d2y/dx2)] = 0 ...................[from x*(dy/dx)]......................[from 2y*(dy/dx)] Now you replace each dy/dx by its function of x and y found above. Then rearrange it to give d2y/dx2 = (function of x and y).

2. anonymous

i don't understand how you split it up, i'm confused.... that honestly made no sense

3. anonymous

yay now my head hurts even worse

4. hba

How Much More Questions Do You Have ?

5. hba

Or Is This The Last Question ?

6. anonymous

this is my first question...

7. hba

8. anonymous

i've got about 16 more for this section, then 5 on a second section on it and a worksheet

9. anonymous

i'm a junior in highschool taking calculus.. this isn't making sense

10. hba

Oh Wow Genius :D

11. anonymous

obviously not if this isn't making sense. sorry it's just frustrating!

12. anonymous

and it's not making a bit of sense

13. hba

Well, (x^2)y + x(y^2) = 6 d/dx[(x^2)y + x(y^2)] = d/dx (6)

14. hba

2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0

15. anonymous

i'm lost stop

16. anonymous

how'd you get 2xy + dy/dx(x^2 + y^2 + 2y(dydx)x=0?

17. hba

You dont know how to differentiate ?

18. anonymous

I do, but it's confusing me

19. anonymous

the d/dx of x^2y?

20. anonymous

you use the multiplication rule right? udv + vdu?

21. hba

$x ^{2}y+xy^2=6$

22. anonymous

x^2(1) + y(2x) = 2xy + x^2

23. hba

Yes I am Using Product Rule

24. anonymous

then how did you get 2xy then dy/dx x^2?

25. anonymous

or do you get dy/dx and all that stuff in parenthasis because you're finding the d/dx of y

26. hba

$\frac{ dy }{ dx }[(x^2)y + x(y^2)] =\frac{ dy }{ dx }6$

27. hba

Now Use Product Rule

28. hba

$2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0$

29. hba

Now Take The Terms Including dy/dx On One Side

30. hba

$dy/dx(x^2)+2y(dy/dx)x = -y^2-2xy$

31. hba

Now Take Common dy/dx

32. hba

$dy/dx(x^2+2xy)=-y^2-2xy$

33. anonymous

then divide?

34. hba

So,$\frac{ dy }{ dx }=\frac{ -(y^2+2xy)}{x^2+2xy }$

35. hba

36. anonymous

question, when using the product rule

37. anonymous

how do you know which one to put dy/dx with

38. anonymous

like when we had x^2y and got dy/dx(x^2) + 2xy

39. anonymous

why'd we put dy/dx with x^2 and not 2xy?

40. hba

Well Its Your Own Decision Whether to Put U first or V First If We talk about the formula

41. hba

|dw:1350260697079:dw|

42. anonymous

okay but then which one do i put the dy/dx with?

43. anonymous

like the next problem is y2 = (x+1)/(x-1)

44. anonymous

so would i do 2y (dy/dx) = vdu - udv/v^2

45. hba

Depends Completely On You

46. anonymous

so dy/dx2y = (x+1)(1) - (x-1)(1) / (x-1)^2

47. hba

48. anonymous

Hey so wht do u need help with

49. anonymous

Hello?

50. anonymous

I'm here. I have a question on where to put the dy/dx when solving

51. anonymous

right now i'm working on y^2 = x-1/x+1

52. anonymous

so i take dy/dx y^2 = (x+1)(1) - (x-1)(1) / (x+1)^2 right?

53. anonymous

right

54. anonymous

and the left side will equal 2y when i take dy/dx of it

55. anonymous

now where do i keep dy/dx, when solving?

56. anonymous

57. anonymous

where do i put it? because i know i need to get it alone

58. anonymous

I sent it to u on ur messages

59. anonymous

60. anonymous

the new one is

61. anonymous

62. anonymous

y^2 = x-1/x+1

63. anonymous

i'm not trying to come off as rude, i'm sorry!

64. anonymous

just frustrated

65. hba

I am Back

66. hba

$y^2 = (x-1) /(x+1)$

67. hba

differentiate both sides with respect to x

68. anonymous

oh okay i was lost there

69. anonymous

i see where shes at now

70. hba

$2y dy/dx = [(x+1)-(x-1)] /(x+1)^2$

71. hba

$2y dy/dx = 2/(x+1)^2$

72. hba

$dy/dx = y / (x+1)^2$

73. hba

@jamroz Got It :)

74. anonymous

so you keep dy/dx with the y on the left

75. anonymous

but why? cause i had gotten that far, i just didn't know which side to leave dy/dx with?

76. hba

Now You Got Which Side You Had To Keep It :) Anyways I am Off To Sleep, Aww My Head Hurts Anyways Bye :)

77. anonymous

thank you so much!