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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    \[2\sqrt{x}+\sqrt{y}=5\]

  2. anonymous
    • 5 years ago
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    I am thinking it is \[1/4x^-1/2+1/2y^-1/2\]

  3. anonymous
    • 5 years ago
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    those should be raised to the -1/2 but it doesn't look like it

  4. amistre64
    • 5 years ago
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    implicit is the same process as explicity; you just keep the dy and dx in until the end

  5. amistre64
    • 5 years ago
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    2x^(1/2) + y^(1/2) = 5 ; derive everything with respect to x

  6. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    now dx = 1 and dy=y' solve for y' 1/sqrt(x) + y' (1/2sqrt(y)) = 0

  8. amistre64
    • 5 years ago
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    y' = [-1/sqrt(x)]/[1/2sqrt(y)] y ' = 2sqrt(y)/sqrt(x)

  9. amistre64
    • 5 years ago
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    -2sqrt(x)/......

  10. amistre64
    • 5 years ago
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    i lost the negative lol

  11. anonymous
    • 5 years ago
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    OK trying to process this really quick

  12. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    y = 2x dy = 2 dx right? dx=1 becasue it always changes the same with respect to itself

  14. amistre64
    • 5 years ago
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    dy is just another term for y'

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

  16. amistre64
    • 5 years ago
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    sorta, cant say im familiar with your notation there

  17. amistre64
    • 5 years ago
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    lets take itby bits smaller pieces so you can see it work

  18. anonymous
    • 5 years ago
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    well just the dirvitive of y/x times the starting function?

  19. amistre64
    • 5 years ago
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    3y^2 what is the derivative of this with respect to x?

  20. anonymous
    • 5 years ago
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    6y^1

  21. amistre64
    • 5 years ago
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    you threw out the dy...how come?

  22. anonymous
    • 5 years ago
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    well you don't need the raised to the power of one but

  23. amistre64
    • 5 years ago
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    3y^2 has no "x" in it; does it?

  24. anonymous
    • 5 years ago
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    no it doesn't

  25. amistre64
    • 5 years ago
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    so the dy/dx that you derive doesnt go away does it?

  26. amistre64
    • 5 years ago
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    the y with respect to x doesnt disappear does it?

  27. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    its the change in y with respect to x: dy/dx

  29. amistre64
    • 5 years ago
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    d (3y^2) dy (6y) ------- = --- dx dx

  30. anonymous
    • 5 years ago
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    ok i'm with you

  31. amistre64
    • 5 years ago
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    watch this one and Im sure your gonna slap yourself: d (5x^3) --------- = ? dx

  32. amistre64
    • 5 years ago
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    d (5x^3) dx 15x^2 --------- = ---- dx dx

  33. anonymous
    • 5 years ago
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    d (5x^3) dy (15x oh man! is that it?

  34. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    dx/dx = 1

  36. amistre64
    • 5 years ago
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    we tend to through out the "dx/dx" because anything times 1 is itself... but its still there :)

  37. amistre64
    • 5 years ago
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    ya with me?

  38. anonymous
    • 5 years ago
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    yea i think i am getting it now. I don't know why this is giving me such a hard time.

  39. anonymous
    • 5 years ago
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    so it's kinda like doing the point slope but with the change from f(x) to f'(x)

  40. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    not point slope but the slope finder

  42. amistre64
    • 5 years ago
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    the product rule for xy: Dx(xy) = x'y + xy'

  43. amistre64
    • 5 years ago
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    x' = 1 so its usually discarded

  44. amistre64
    • 5 years ago
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    the power rule for y^5: y' 5y^4

  45. anonymous
    • 5 years ago
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    so my dy/dx on that would be 5? because it's changed for a total of 5 in the coeficent?

  46. amistre64
    • 5 years ago
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    your dy/dx inthat one is undertermined until you kow what to plug in for a value of y

  47. amistre64
    • 5 years ago
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    but when you do, then you can determine dy/dx = y'

  48. amistre64
    • 5 years ago
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    5xy^2 = z derive with respect to t?

  49. amistre64
    • 5 years ago
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    5x2y y' + 5 x'y^2 = z'

  50. amistre64
    • 5 years ago
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    there are no "t" values so all your derivative bits stay put

  51. amistre64
    • 5 years ago
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    y' = dy/dt in this case x' = dx/dt z' = dz/dt

  52. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    the: 2sqrt(x) + sqrt(y) =5 one?

  54. anonymous
    • 5 years ago
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    or\[2\sqrt{x}*1/2+1/2y^1/2\]

  55. amistre64
    • 5 years ago
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    Dx(2sqrt(x)) + Dx(sqrt(y)) = Dx(5)

  56. amistre64
    • 5 years ago
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    x' (1/sqrt(x)) + y' (1/2sqrt(y)) = 0

  57. amistre64
    • 5 years ago
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    x' = dx/dx = 1 so we can ignore it.... or leave it in. solve for y'

  58. amistre64
    • 5 years ago
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    there is no f'(x)*f(x) + dy/dx...... your mixing notations and confusing yoursoef :)

  59. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    \[\frac{dx}{dx}\frac{1}{\sqrt{x}} + \frac{dy}{dx}\frac{1}{2 \sqrt{y}} = 0\]

  61. amistre64
    • 5 years ago
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    solve for dy/dx which is another notation for y'

  62. anonymous
    • 5 years ago
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    Alright i think i am getting it now. Fantasitic answer as always thanks a ton

  63. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    Yea i think i was reading into it more than it was really asking me to

  65. amistre64
    • 5 years ago
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    y^2 = 5x^2 + 6x +3

  66. amistre64
    • 5 years ago
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    \[\frac{dy}{dx}2y = \frac{dx}{dx}10x + \frac{dx}{dx}6 +\frac{dx}{dx}0\]

  67. amistre64
    • 5 years ago
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    y' 2y = 10x +6

  68. amistre64
    • 5 years ago
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    10x + 6 y' = --------- 2y

  69. amistre64
    • 5 years ago
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    its just deriving like normal; just keep the bit in that you were taught to throw out lol

  70. anonymous
    • 5 years ago
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    OH!!! so it's the change in y over the change in x

  71. amistre64
    • 5 years ago
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    yes :)

  72. anonymous
    • 5 years ago
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    i see what you did there

  73. amistre64
    • 5 years ago
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    no you see what youve always done there ;)

  74. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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
    • 5 years ago
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    Well if they did things the easy way there would be a lot of math teachers out of a job.

  77. amistre64
    • 5 years ago
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    lol maybe :)

  78. anonymous
    • 5 years ago
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    Thanks for your time great explianition.

  79. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    youre welcome :)

  81. anonymous
    • 5 years ago
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    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.

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