f（x）为偶函数,在x=0处导数存在,证明x=0处导数为0

f'(0-)=lim(x→0-)(f(x)-f(0))/x
=lim(t→0+)(f(-t)-f(0))/(-t) (t=-x)
=-lim(t→0+)(f(t)-f(0))/t
=-f'(0+)
因为可导,所以f'(0-)=f'(0+),所以f'(0-)=f'(0+)=f'(0)=0