## Calculus, 10th Edition (Anton)

$=4sinxcosx$
Take out the $2$ from the derivative, it can be multiplied to the answer at the end. Splitting $sin^{2}(x)$ into $sinx\times\sin(x)$ allows one to do the product rule, stating $f=sinx$ and $g=sinx$. Using $(fg)'=f'g+g'f$, the answer becomes $2(sinxcosx+sinxcosx)=2(2sinxcosx)=4sinxcosx.$