f(0+0)=f(0)f(0) 即 f(0)=f(0)^2 则f(0)=0或1
若f(0)=0.
令x2=0 则f(x+0)=f(x)f(0)=0 即f(x)=0 与f`(0)=1矛盾
所以f(0)不为0.
f(0)=1
f(0)=f(x-x)=f(x)f(-x) =1 所以f(-x)=1/f(x)
f'(x-x)=f'(x)f(-x)+f(x)f'(-x)=f'(x)/f(x)+f(x)*[-f(x)^(-2)*f'(x)]=f'(x)/f(x)-f(x)/f'(x)=0
令f'(x)/f(x)=a
上述等式为a-1/a=0
解得 a=1 即f'(x)/f(x)=1
即f'(x)=f(x)