设y=cos(2x+1),求dy y=cos²2x，则dy=
y=cos(2x\uff0b1),\u6c42dy\uff1fy=cos(2x+1)
dy=-2sin(2x+1)
\u5e0c\u671b\u56de\u7b54\u5bf9\u4f60\u6709\u5e2e\u52a9\u3002
y=cos²2x\uff0c
dy/dx=(cos²2x)'=2cos2x(cos2x)'=2cos2x(-sin2x)(2x)'=-4sin2xcos2x=-2sin4x,
dy=-2sin4xdx
\u6216\u8005
如下
求导，加dx就行了，求导的时候先把2x+1看成u，也就是dy=（cosu）'（u）'dx，答案就是-2sin(2x+1)dx
解：dy=-sin(2x+1)*2dx=-2sin(2x+1)dx。
图
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绛旓細瑙ｏ細浠(x)=y=sin(2x+1)f'(x)=[sin(2x+1)]'=cos(2x+1)路(2x+1)'=cos(2x+1)路2 =2cos(2x+1)鍑芥暟鐨勫鍑芥暟涓簓=2cos(2x+1)
