f(x)=2sinx+x^3|sinx|则在x=0处存在的最高阶导数n是多少?希望有详细解答过程,谢谢!

当x＞0时,f(x)=2sinx+x^3sinx,则由莱布尼茨法则
f(n)(x)=2sin(x+nπ/2)+x^3sin(x+nπ/2)+3nx^2sin[x+(n-1)π/2]+3n(n-1)xsin[x+(n-2)π/2]+n(n-1)(n-2)sin[x+(n-3)π/2],n≥3
显然有limf(n)(0+)=2sin(nπ/2)+n(n-1)(n-2)sin[(n-3)π/2]
同理,当x＜0,n≥3时,解得limf(n)(0-)=2sin(nπ/2)-n(n-1)(n-2)sin[(n-3)π/2]
令limf(n)(0+)=limf(n)(0-),解得sin[(n-3)π/2]=0,n≥3
显然n=4时limf(n)(0+)≠limf(n)(0-),即f(4)(0)不存在
而n≤3时恒有limf(n)(0+)=limf(n)(0-)=2sin(nπ/2)
故n最大为3