$y=\sin2x-\cos2x=\sqrt2\sin\Big(2x-\dfrac{\pi}{4}\Big)$
$\to y'=\sqrt2.\cos\Big(2x-\dfrac{\pi}{4}\Big).\Big(2x-\dfrac{\pi}{4}\Big)'=2\sqrt2\cos\Big(2x-\dfrac{\pi}{4}\Big)$
$y=\sin^2x-\cos^2x+x=x-(\cos^2x-\sin^2x)=x-\cos2x$
$\to y'=1+\sin2x.(2x)'=1+2\sin2x$