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Derivative with Constant

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the max/min of the function for [0,pi/2]

    f(x)= kcos2xsin(x)

    2. Relevant equations

    Product Rule and chain rule

    3. The attempt at a solution

    f'(x)= k(2cosx(-sinx)(sinx)+cos2xcosx)

    f'(x)= kcosx(-2sin2x+cos2x)

    Set eqaution to 0

    (2sin2x+cos(x)2) = 0

    2(1-cos2x)+cos(x)2 = 0

    cos2x = 2
    Now I'm stuck. There should be a max or min.
  2. jcsd
  3. Apr 7, 2010 #2
    You appear to have made two errors here. You neglected to account for the other factor, that perhaps [tex]\cos x = 0[/tex]. You also seem to have dropped a minus sign; if the second factor is zero, you should have [tex]-2 \sin^2 x + \cos^2 x = 0[/tex].

    You may have an easier time with this second factor if you rewrite it so that only one trig function of [tex]x[/tex] appears.
  4. Apr 7, 2010 #3
    Man I hate when I forget a negative sign. I also scrwed up the interval is supposed to be (0,pi/2) instead of [0,pi/2]. Thank you for pointing that out.
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