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# 能幫我解這題微分問題嗎

題組題:x及y為實數，則滿足f(xy)=f(x)+f(y)

1.求f(1)

2.求證f(x)=-f(1/x)

3.若f '(1)=a，求 f '(a)

想請問第三題之解及過程

### 1 Answer

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- 約翰教練Lv 48 years agoFavorite Answer
＠前言

考慮f在0的值,對於任一實數x皆有

f(0) = f(0*x) = f(0)+ f(x) ---> f(x) =0 ,因為對於任意實數x,f(x)皆為0故f(x)為零函數

我們考慮non-trivail case,假設該函數在0點未定義,則我們考慮非零點:

1.

f(1) = f(1*1) = f(1)+f(1) = 2*f(1) ---> f(1) = 0

2.

令x不等於0 ,由1.得知

0 = f(1) = f(x*(1/x)) = f(x)+f(1/x) ---> f(x) = -f(1/x)

3.

設f為C1實數函數, a不等於0

f(x) = f(x/a)+f(a)

f '(x) = f '(x/a)/a (連鎖律且f(a)為一常數)

令x=a

f '(a) = f '(1)/a = a/a = 1

2012-07-31 14:45:43 補充：

更正:第四行non-trivial

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