# Graph of a trigonometric function

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.

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.
Do not forget the '2'.

f'(x)=(1+2cos2x)cos(x+sin2x)

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