Question

求导！！！

求导！！！

若以下回答无法解决问题，邀请你更新回答

Answer 1

y= √x.[sin(2x+1)]^2
y'
= √x. d/dx [sin(2x+1)]^2 + [sin(2x+1)]^2 .d/dx {√x}
=√x. [2sin(2x+1)] .d/dx {sin(2x+1)} + [sin(2x+1)]^2/[ 2√x]
=√x. [2sin(2x+1)] .[2cos(2x+1)] + [sin(2x+1)]^2/[ 2√x]
=4√x.sin(2x+1) .cos(2x+1) + { [sin(2x+1)]^2/[ 2√x] }