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Is this correct

  1. Oct 4, 2008 #1
    y=-1/2 x cosx



    y'=1/2 x sinx + cos x -1/2

    y"= 1/2 x cosx - sin x 1/2 + cos x + -1/2 - sin x

    y"= 1/2 x cos^2x


    I used the product rule (fg)'=fg'+gf'
     
  2. jcsd
  3. Oct 4, 2008 #2
    I am an amateur myself but I recall (hopefully) that cos(x) differentiates to -sin(x)

    EDIT: I didn't see the negative sorry
     
  4. Oct 4, 2008 #3

    statdad

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    Homework Helper

    If your first derivative is

    [tex]
    y' = \frac 1 2 x \sin x - \frac 1 2 \cos x
    [/tex]

    that is correct.

    I do not believe your second derivative is correct.
     
  5. Oct 4, 2008 #4
    Will the second derivative be
    y"=1/2 x cos x + sin x * 1/2 - 1/2 - sin x

    y"=1/2 x cos x + 1/2 sin x + 1/2 sin x
     
  6. Oct 4, 2008 #5
    Well I'm doing differential equations and I have to match it with Y" + y= sin x
     
  7. Oct 4, 2008 #6
    y''= (1/2)xcosx + sinx

    I didn't get the ^2x part.
     
  8. Oct 4, 2008 #7
    How'd you get that second derivative. I did it wrong how'd u do it, you did it right
     
  9. Oct 4, 2008 #8
    When you add Y"+y=sin x




    (1/2)xcosx + sinx + -1/2 x cosx

    which becomes sin x
     
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