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求证:可导的奇函数其导数函数是偶函数
人气:474 ℃ 时间:2019-08-18 06:12:32
解答
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
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