A calculus problem by A Former Brilliant Member

Put, $x = y = 0$ then

$f(0 +0 ) = f(0)f'(0) + f(0)f'(0)$ $\Rightarrow f'(0) = \dfrac {1}{2}$ Since, $f(0) = 1.$

Now, put, $x = x$ and $y = 0$ then,

$f(x + 0) = f(x)f'(0) + f(0)f'(x)$ $\Rightarrow f(x) = \dfrac {f(x)}{2} + f'(x) \Rightarrow$ $f(x)=2f'(x).$

Now, $\dfrac {f'(x)}{f(x)} = \dfrac {1}{2}.$ Integrating on both sides,

$\ln(f(x)) = \dfrac {x}{2} + C.$ Substitute $f(0) = 1$ we get, $C = 0$

Therefore, $\ln(f(4)) = \dfrac {4}{2}$

$ANSWER = \boxed{2}$