685. $y=\sin ^{2}(2 x-1)$, i.e. $y=u^{2}$, where $u=\sin (2 x-1)$.
Solution. $y^{\prime}=2 u u^{\prime}$ or $y^{\prime}=2 \sin (2 x-1) \cdot[\sin (2 x-1)]^{\prime}=$ $=2 \sin (2 x-1) \cdot \cos (2 x-1)(2 x-1)^{\prime}=4 \sin (2 x-1) \cdot \cos (2 x-1)=$ $=2 \sin 2(2 x-1)$.