Re: [微分] 關於這題的証明
※ 引述《ghost17612 (就是要ROCK)》之銘言：
: 若f(x)定義於 x 屬於 R 且滿足
: f(x+y)=f(x)f(y)
: 當f(x)在X屬於R可微分且f(0)不等於0
: 求證 f'(a)=f(a)f'(0) a屬於R , 利用f'(a)的定義
: 感謝!!
1. f(0+0)=f(0)f(0)
=> f(0)=(f(0))^2
Since f(0) =/= 0 , so dividing f(0) , we have f(0)=1
2. Since f'(0) exists
f(0+h)-f(0)
so lim ───── exists
h→0 h
f(h)-1
i.e. lim ──── exists and equals to f'(0)
h→0 h
3. For a€Real number
f(a+h) - f(a) f(a)*f(h)-f(a) f(a)*(f(h)-1)
─────── = ─────── = ─────── ---(*)
h h h
f(h)-1
Since lim ──── = f'(0)
h→0 h
so (*) goes to f(a)f'(0) as h→0
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11/26 01:41, , 1F
11/26 01:41, 1F
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