y=cosnxsin^nx求导谢谢PS答案是ncos(n+1).sin^n-1x为什么是这样?

Question

y=cosnxsin^nx 求导
谢谢
PS答案是ncos(n+1).sin^n-1 x 为什么是这样?

Answer 1

令u=cosnx v=sin^nx复合函数求导：[f(u(x))]'=f'(u)*u'(x)u'=(cosnx)'=(-sinnx)*n=-nsinnxv'=(sin^nx)'=[(sinx)^n]'=n(sinx)^(n-1)*cosxy=uvy'=(uv)'=u'v+uv'=(-nsinnx)*(sin^nx)+cosnx*[ncosx(sinx)^(n-1)]=-nsinn...