f(x)对任意实数x1,x2,有f(x1+x2)=f(x1)*f(x2),取x2=0,f(x1)=f(x1)f(0)所以f(0)=1所以f'(x)=lim(h->0)[f(x+h)-f(x)]/h=lim(h->0)[f(x))f(h)-f(x)]/h=f(x)lim(h->0)[f(h)-1]/h=f(x)lim(h->0)[f(h)-f(0)]/h=f(x)*f'(0)=...