- #1
Duane
- 8
- 0
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.
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.