## anonymous 5 years ago Can anyone break this down for me into steps? i need to find dy/dx by implicit differentiation?

1. anonymous

$2\sqrt{x}+\sqrt{y}=5$

2. anonymous

I am thinking it is $1/4x^-1/2+1/2y^-1/2$

3. anonymous

those should be raised to the -1/2 but it doesn't look like it

4. amistre64

implicit is the same process as explicity; you just keep the dy and dx in until the end

5. amistre64

2x^(1/2) + y^(1/2) = 5 ; derive everything with respect to x

6. amistre64

Dx(2x^(1/2)) = 1/sqrt(x) dx Dx(y^(1/2)) = 1/2sqrt(y) dy Dx(5) = 0 ....d5 is useless lol so it jsut 0

7. amistre64

now dx = 1 and dy=y' solve for y' 1/sqrt(x) + y' (1/2sqrt(y)) = 0

8. amistre64

y' = [-1/sqrt(x)]/[1/2sqrt(y)] y ' = 2sqrt(y)/sqrt(x)

9. amistre64

-2sqrt(x)/......

10. amistre64

i lost the negative lol

11. anonymous

OK trying to process this really quick

12. amistre64

its confusing at first becasue we are used to discarding that dx.... keep all the bits in and adjust for them in the end

13. amistre64

y = 2x dy = 2 dx right? dx=1 becasue it always changes the same with respect to itself

14. amistre64

dy is just another term for y'

15. anonymous

Alright, I think i almost with you. So when i break it down it's going dy/dx*f(x)

16. amistre64

sorta, cant say im familiar with your notation there

17. amistre64

lets take itby bits smaller pieces so you can see it work

18. anonymous

well just the dirvitive of y/x times the starting function?

19. amistre64

3y^2 what is the derivative of this with respect to x?

20. anonymous

6y^1

21. amistre64

you threw out the dy...how come?

22. anonymous

well you don't need the raised to the power of one but

23. amistre64

3y^2 has no "x" in it; does it?

24. anonymous

no it doesn't

25. amistre64

so the dy/dx that you derive doesnt go away does it?

26. amistre64

the y with respect to x doesnt disappear does it?

27. anonymous

Ok maybe this is where i am going wrong when i look at dy/dx i think of a fraction of the dirvitive of y over the dirvitive of x

28. amistre64

its the change in y with respect to x: dy/dx

29. amistre64

d (3y^2) dy (6y) ------- = --- dx dx

30. anonymous

ok i'm with you

31. amistre64

watch this one and Im sure your gonna slap yourself: d (5x^3) --------- = ? dx

32. amistre64

d (5x^3) dx 15x^2 --------- = ---- dx dx

33. anonymous

d (5x^3) dy (15x oh man! is that it?

34. amistre64

tell me; when x moves by any amount; x moves in respect to itself by the same amount right: lets say you are x and you move 5 feet to the right; how far have you moved?

35. amistre64

dx/dx = 1

36. amistre64

we tend to through out the "dx/dx" because anything times 1 is itself... but its still there :)

37. amistre64

ya with me?

38. anonymous

yea i think i am getting it now. I don't know why this is giving me such a hard time.

39. anonymous

so it's kinda like doing the point slope but with the change from f(x) to f'(x)

40. amistre64

its difficult becasue you are used to throwing away the "dx/dx" and keeping the dy/dx on the other side. Implicit is the same thing, but y aint by itself :)

41. anonymous

not point slope but the slope finder

42. amistre64

the product rule for xy: Dx(xy) = x'y + xy'

43. amistre64

x' = 1 so its usually discarded

44. amistre64

the power rule for y^5: y' 5y^4

45. anonymous

so my dy/dx on that would be 5? because it's changed for a total of 5 in the coeficent?

46. amistre64

your dy/dx inthat one is undertermined until you kow what to plug in for a value of y

47. amistre64

but when you do, then you can determine dy/dx = y'

48. amistre64

5xy^2 = z derive with respect to t?

49. amistre64

5x2y y' + 5 x'y^2 = z'

50. amistre64

there are no "t" values so all your derivative bits stay put

51. amistre64

y' = dy/dt in this case x' = dx/dt z' = dz/dt

52. anonymous

ok i think i am starting to get it so for my the equation i gave earlier it's going to work out to being f(x)+f'(x) +dy/dx+f'(y)

53. amistre64

the: 2sqrt(x) + sqrt(y) =5 one?

54. anonymous

or$2\sqrt{x}*1/2+1/2y^1/2$

55. amistre64

Dx(2sqrt(x)) + Dx(sqrt(y)) = Dx(5)

56. amistre64

x' (1/sqrt(x)) + y' (1/2sqrt(y)) = 0

57. amistre64

x' = dx/dx = 1 so we can ignore it.... or leave it in. solve for y'

58. amistre64

there is no f'(x)*f(x) + dy/dx...... your mixing notations and confusing yoursoef :)

59. anonymous

ok so it looks like you are just adding the dirivites inorder to solve this. am i hitting left field with that one?

60. amistre64

$\frac{dx}{dx}\frac{1}{\sqrt{x}} + \frac{dy}{dx}\frac{1}{2 \sqrt{y}} = 0$

61. amistre64

solve for dy/dx which is another notation for y'

62. anonymous

Alright i think i am getting it now. Fantasitic answer as always thanks a ton

63. amistre64

if this was all in the variables for x youd hav eno problem finding the derivative :) because you were taught to throw out that dx/dx

64. anonymous

Yea i think i was reading into it more than it was really asking me to

65. amistre64

y^2 = 5x^2 + 6x +3

66. amistre64

$\frac{dy}{dx}2y = \frac{dx}{dx}10x + \frac{dx}{dx}6 +\frac{dx}{dx}0$

67. amistre64

y' 2y = 10x +6

68. amistre64

10x + 6 y' = --------- 2y

69. amistre64

its just deriving like normal; just keep the bit in that you were taught to throw out lol

70. anonymous

OH!!! so it's the change in y over the change in x

71. amistre64

yes :)

72. anonymous

i see what you did there

73. amistre64

no you see what youve always done there ;)

74. anonymous

Yea it almost seems like it would have been eaiser to start off with this way instead of having to go back and relearn it

75. amistre64

exactly; I dont know wy most of math is taught backwards, giving you exceptions to the rules instead of letting you learn the rules as they are.

76. anonymous

Well if they did things the easy way there would be a lot of math teachers out of a job.

77. amistre64

lol maybe :)

78. anonymous

Thanks for your time great explianition.

79. amistre64

its like: heres a shortcut way of shortcut say thats only going to be helpful for about 1% of the math that is going to be done; learn it, memorize it and then forget it later on when you realize that its useless lol

80. amistre64

youre welcome :)

81. anonymous

yea thats pretty much how my math career seems to have gone so far. It seems to me that it would be better to learn a process that works 100% of the time then go back and learn all the shortcuts that will speed up your calculation time on equations. However, i have always been warned "becareful what you wish for" that might open open a can of worms that shouldn't be touched.