y =sin²x， y' = xsinxcosx = sin(2x)， y" = 2cos(2x) = 2sin(2x+π/2)， y''' = (2^2)cos(2x+π/2) = (2^2)sin[2x+2(π/2)]，用归纳法证明，…… y^(n) = [2^(n-1)]sin[2x+(n-1)(π/2)]。
