Question

这道题的导数怎么求

Answer 1

本题有个陷阱就是gx是连续但未必可导，所以对于fx不能直接用公式法来求，而应该用导数的定义来求。
f'(0)=lim(x→0)[f(x)-f(0)]/(x-0)
=lim(x→0)[g(x)sin2x-g(0)sin0]/x
=lim(x→0)[g(x)sin2x]/x
=2lim(x→0)[g(x)·(sin2x/2x)]
=2lim(x→0)g(x)·lim(x→0)(sin2x/2x)
=2g(0).

Answer 2

因为g(x)在x=0处连续，所以：lim(x→0)g(x)=g(0);
所以：f'(0)=lim(x→0)[f(x)-f(0)]/(x-0)
=lim(x→0)[g(x)sin2x-g(0)sin0]/x
=lim(x→0)[g(x)sin2x]/x
=2lim(x→0)[g(x)·(sin2x/2x)]
=2lim(x→0)g(x)·lim(x→0)(sin2x/2x)
=2g(0).