微积分_例题解-4

例4.ysin(2x+1),求d 解:把2x+1看成中间变量u,则 dy=d(sin u)=cos udu=coS(2x+1)d(2x+1) =coS(2x+1)2dx =2cos(2x+1)dx

结束 解 例4 y=sin(2x+1) 求dy 把2x+1看成中间变量u则 dy=d(sin u) =cos(2x+1)2dx =cos udu=cos(2x+1)d(2x+1) =2cos(2x+1)dx