已知函数f(x)可导,且对任何实数x,y满足：f(x+y)=e^xf(y)+e^yf(x)和f'(0)=e证明：f'(x)=f(x)+e^(x+1)e^x*f(x)

Question

已知函数f(x)可导,且对任何实数x,y满足：f(x+y)=e^xf(y)+e^yf(x)和f'(0)=e 证明：f'(x)=f(x)+e^(x+1)
e^x*f(x)

Answer 1

put x=y =0
f(0) = f(0) + f(0)
=>f(0) = 0
f'(x)= lim(y->0)[f(x+y) - f(x)]/y
= lim(y->0)[e^xf(y)+e^yf(x) - f(x)]/y
= e^x lim(y->0)[f(0+y)-f(0)]/y + f(x) lim(y->0)( e^y - 1)/y
= e^xf'(0) + f(x)
= e^(x+1) + f(x)