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Differentiating a trig function

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Find dy/dx, given
    y = sin( 3 x ) cos( 6 x )

    2. Relevant equations
    product rule= f'g+g'f

    3. The attempt at a solution
    I used the product rule and got this, but its supposedly wrong, what have i done wrong??????((3*cos(x))*(6*cos(x)))+((6*sin(-x))*(3*sin(x)))
     
  2. jcsd
  3. Apr 26, 2010 #2

    gabbagabbahey

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    For starters, [itex]\frac{d}{dx}\sin(3x)=3\cos(3x)\neq3\cos(x)[/itex]
     
  4. Apr 26, 2010 #3

    Char. Limit

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    Here's a general rule that may help you figure it out:

    [tex]\frac{d}{dx}(sin(ax)) = a \times cos(ax)[/tex]
     
  5. Apr 26, 2010 #4
  6. Apr 26, 2010 #5
    wtf, it still says its wrong,
    ((3*cos(3*x))*(6*cos(6*x)))+((6*sin(-6*x))*(3*sin(3*x)))
    what am i still doing wrong?
     
  7. Apr 26, 2010 #6

    gabbagabbahey

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    Where are the extra factors of 6 in your first term and 3 in your second term coming from?
     
  8. Apr 26, 2010 #7
    sorry i ment
    ((3*cos(3*x))*(cos(6*x)))+((6*sin(-6*x))*(sin(3*x)))
    but its still wrong, can you simplify this?
     
  9. Apr 26, 2010 #8

    gabbagabbahey

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    For starters, [itex]\sin(-6x)=-\sin(6x)[/itex]...can you think of a trig identity that involves something like [itex]\cos(a)\cos(b)-\sin(a)\sin(b)[/itex]?:wink:
     
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