A function fff is differentiable and f(x+y)=f(x)f(y)f(x+y) = f(x)f(y)f(x+y)=f(x)f(y). If f′(0)=1f'(0) = 1f′(0)=1, find f′(x)f'(x)f′(x).
f(x)f(x)f(x)
xf(x)x f(x)xf(x)
f′(0)f(x)f'(0)f(x)f′(0)f(x)
exe^xex