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Solving for y prime

  1. Sep 30, 2007 #1
    1. The problem statement, all variables and given/known data
    find the derivative of...
    2 sinxcosy = 1


    3. The attempt at a solution
    (2 cosxcosy) * (cosy')
    cos y' = -2 cosxcosy
    y' = (-2 cosxcosy)/(cos)

    I know thats not right but I am not sure where I am making the mistake.
     
  2. jcsd
  3. Sep 30, 2007 #2

    EnumaElish

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    How do you find the derivative of f(x)g(y)?
     
  4. Sep 30, 2007 #3
    use the product rule, f(x)g(y') + g(y)f(x')
     
  5. Sep 30, 2007 #4
    you didn't do the product rule right
     
  6. Sep 30, 2007 #5
    Can you use the chain rule? i think thats what I was trying to do. I was taking the derivative of the outside then the derivative of the inside.
     
    Last edited: Sep 30, 2007
  7. Sep 30, 2007 #6

    EnumaElish

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    You can apply the chain rule to g(y) (for example, g'(y) dy/dx), but you still have a product to sort through.
     
  8. Sep 30, 2007 #7
    so if you use the product rule you would get...
    (2sinx -siny') + (2cosxcosy) = 0
     
  9. Sep 30, 2007 #8

    EnumaElish

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    Write out the formula for [f(x)g(y)]'. Then substitute in f(x) = 2 sin x, g(y) = cos y, f '(x) = ... and g'(y) = ...
     
    Last edited: Sep 30, 2007
  10. Sep 30, 2007 #9
    f(x)g(y') + f(x')g(y)
    sinxy' + 2cosxcosy
     
  11. Sep 30, 2007 #10
    okay thats not right... it should be...
    2cosxcosy - 2sinxsinyy' = 0
    2cosxcosy - 2y'sinxsiny = 0
    y' = -2cosxcosy/-2sinxsiny
     
  12. Sep 30, 2007 #11

    EnumaElish

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    It should be f(x)g'(y)y' + f '(x)g(y).
     
  13. Sep 30, 2007 #12
    oh ok...so is that answer right now?
    2cosxcosy - 2sinxsinyy' = 0
    2cosxcosy - 2y'sinxsiny = 0
    y' = -2cosxcosy/-2sinxsiny
     
  14. Sep 30, 2007 #13

    EnumaElish

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    Do you think it is right?

    Write out f(x)g'(y)y' + f '(x)g(y).

    State f(x) and g(x).

    State f '(x) and g'(y).

    Make the substitutions.
     
  15. Sep 30, 2007 #14
    yes I do think its right...

    2sinxcosy
    f(x) = 2sinx
    g(x) = cosy
    f'(x) = 2cosx
    g'(y) = -sinyy'

    2cosxcosy - 2sinxsinyy' = 0
    2cosxcosy - 2y'sinxsiny = 0
    y' = -2cosxcosy/-2sinxsiny
     
  16. Sep 30, 2007 #15

    EnumaElish

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    I think you're right. You can cancel out the minuses.
     
  17. Sep 30, 2007 #16
    ok thank you!
     
  18. Sep 30, 2007 #17
    you still have 1 more step

    what is cosine/sine?
     
  19. Sep 30, 2007 #18

    HallsofIvy

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    [2 sinxcosx]'= 2 [(sin(x)' cos(x)+ sin(x) cos(x)']
     
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