Y=sin2x-cos2x Y=sin^2x-cos^2x +x Giúp em với ạ! 08/08/2021 Bởi Caroline Y=sin2x-cos2x Y=sin^2x-cos^2x +x Giúp em với ạ!
$y=\sin2x-\cos2x=\sqrt2\sin\Big(2x-\dfrac{\pi}{4}\Big)$ $\to y’=\sqrt2.\cos\Big(2x-\dfrac{\pi}{4}\Big).\Big(2x-\dfrac{\pi}{4}\Big)’=2\sqrt2\cos\Big(2x-\dfrac{\pi}{4}\Big)$ $y=\sin^2x-\cos^2x+x=x-(\cos^2x-\sin^2x)=x-\cos2x$ $\to y’=1+\sin2x.(2x)’=1+2\sin2x$ Bình luận
$y=\sin2x-\cos2x=\sqrt2\sin\Big(2x-\dfrac{\pi}{4}\Big)$
$\to y’=\sqrt2.\cos\Big(2x-\dfrac{\pi}{4}\Big).\Big(2x-\dfrac{\pi}{4}\Big)’=2\sqrt2\cos\Big(2x-\dfrac{\pi}{4}\Big)$
$y=\sin^2x-\cos^2x+x=x-(\cos^2x-\sin^2x)=x-\cos2x$
$\to y’=1+\sin2x.(2x)’=1+2\sin2x$