Question

一道高数导数证明题

设函数f(x)可导，对于任意x1,x2，恒有f(x1+x2)=f(x1)f(x2)且f'(0)=1
证明：f'(x)=f(x)

Answer 1

第一步：f'(0) = lim(f(0+t)-f(0))/t = lim(f(t)-f(0))/t =1
第二步：f'(x) = lim(f(x+t)-f(x))/t = lim(f(x)f(t)-f(t))/t=limf(x)(f(t)-1)/t
=f(x)lim(f(t)-f(0)+f(0)-1)/t=f(x)lim(f(t)-f(0))/t + f(x)lim(f(0)-1)/t
=f(x) + lim(f(x)f(0)-f(x))/t = f(x) + lim(f(x+0)-f(x))/t = f(x) + lim(f(x)-f(x))/t
= f(x)