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

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).