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Graph of a trigonometric function

  1. Nov 27, 2011 #1
    How do you graph f'(x) when f(x)= sin(x+sin2x), 0≤x≤∏ ?

    If I calculate f'(x) first by the chain rule, I get f'(x)=cos(x+sin2x)(1+cos2x), where to proceed ?

    Thanks for any help in advance.
     
  2. jcsd
  3. Nov 27, 2011 #2
    Do not forget the '2'.

    f'(x)=(1+2cos2x)cos(x+sin2x)
     
    Last edited: Nov 27, 2011
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