两边同时除以e^(x+y)得
f(x+y)/e^(x+y)=f(x)/e^x+f(y)/e^y
所以令f(x)/e^x=g(x),上式变成g(x+y)=g(x)+g(y).容易知道g(0)=0
题目已知f'(0)=2.
又f'(x)=(g(x)+g'(x))e^x,故得g(0)+g'(0)=2,g'(0)=2
g'(x)=lim(g(t+x)-g(x))/t=limg(t)/t=lim(g(t)-g(0))/(t-0)=g'(0)=2
所以g(x)=2x+g(0)=2x
f(x)=2xe^x