求证：可导的奇函数其导数函数是偶函数

if f(x) is odd
then f(x) = -f(-x)
f'(x) = lim(y->0) [f(x+y) - f(x)]/ y
= lim(y->0) [-f(-x-y) + f(-x)]/ y ( f is odd)
= -lim(y->0)[ f(-x-y) - f(-x) ] /y
= lim(-y->0)[ f(-x-y) - f(-x)] / (-y)
= f'(-x)
=> 可导的奇函数其导数函数是偶函数