f(x1x2)=f(x1)+f(x2),f'(1)=1,求f(x)导数

Question

f(x1x2)=f(x1)+f(x2),f'(1)=1 ,求f(x)导数

Answer 1

f(1)=f(1)+f(1),得：f(1)=0
当x≠0时,f(1/x)+f(x)=f(1)=0
因此f(1/x)与f(x)互为相反数,即f(1/x)=-f(x)
f '(x)=lim[h→0] [f(x+h)-f(x)]/h
=lim[h→0] [f(x+h)+f(1/x)]/h
=lim[h→0] f[(x+h)/x]/h
=lim[h→0] f(1+h/x)/h
=lim[h→0] [f(1+h/x)-f(1)]/h
=(1/x)lim[h→0] [f(1+h/x)-f(1)] / (h/x)
=(1/x)f '(1)
=1/x
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