Question

设函数f(x)在（-∞,+∞)内有定义，且对任意的实数x和y，满足f(x+y)=f(y)e^x+f(x)e^y,又已知f'(0)=e.

求f'(x)的表达式。

Answer 1

令x=y=0,得f(0)=0.由倒数的定义得
f'(x)=lim(y趋向于0)[f(x+y)-f(x)]/y
=lim(y趋向于0)[f(y)e^x+f(x)e^y-f(x)]/y
=lim(y趋向于0)[f(y)e^x]/y+lim(y趋向于0)[(e^y-1)f(x)]/y
=lim(y趋向于0){e^x[f(y)-f(0)]}/y+f(x)
=f'(x)e^x+f(x)
=e^(x+1)+f(x)
即f'(x)-f(x)=e^(x+1),由通解公式得
f(x)=xe^(x+1)+Ce^x.又f(0)=0,可得C=0.
故f(x)=xe^(x+1).所以f'(x)=(x+1)e^(x+1).

Answer 2

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)=e.
又f'(x)=(g(x)+g'(x))e^x，故得g(0)+g'(0)=e,g'(0)=e
g'(x)=lim(g(t+x)-g(x))/t=limg(t)/t=lim(g(t)-g(0))/(t-0)=g'(0)=e
所以g(x)=ex+g(0)=ex
f(x)=exe^x

f'(x)=e^x(ex+e)